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Looking for 100 KHz - 10 MHz, 10W
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artie
Jul 24, 2013, 9:33:21 PM7/24/13
to
Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
watts into an inductive load for a physics package at a startup. We're
using function generators currently (HP 3314A for example) but we need
more power, as in up to around 10 watts (into nominally 50 Ohms).
The signal source is easy, we've got that -- it's getting from the mW
range to the 10 W range that's the difficult part, or difficult without
spending more than say $200.
Being a startup, cheap is important, or we'd go for a used ENI wideband
amp (I found an ENI 320L, 250 KHz - 110 MHz, 20W for $1450, out of our
price range).
Something off the shelf would be good, something easy to replicate
would be acceptable. Output doesn't have to be really clean, so AB2 is
acceptable.
Suggestions?
--
watts into an inductive load for a physics package at a startup. We're
using function generators currently (HP 3314A for example) but we need
more power, as in up to around 10 watts (into nominally 50 Ohms).
The signal source is easy, we've got that -- it's getting from the mW
range to the 10 W range that's the difficult part, or difficult without
spending more than say $200.
Being a startup, cheap is important, or we'd go for a used ENI wideband
amp (I found an ENI 320L, 250 KHz - 110 MHz, 20W for $1450, out of our
price range).
Something off the shelf would be good, something easy to replicate
would be acceptable. Output doesn't have to be really clean, so AB2 is
acceptable.
Suggestions?
--
Phil Hobbs
Jul 24, 2013, 10:11:16 PM7/24/13
to
moving fast than saving a small amount of money.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058
hobbs at electrooptical dot net
http://electrooptical.net
Joerg
Jul 24, 2013, 10:36:21 PM7/24/13
to
then look in the ham radio area. This is overkill but shows the idea,
broadband amplifiers:
http://www.ebay.com/itm/ws/eBayISAPI.dll?ViewItem&item=180525812983&item=180525812983
But you'll have to invest a lot of sweat equity because ham gear for
shortwave goes from 1.8MHz or 3.5MHz to 30MHz, typically. So you'll
have to redesign RF transformers and such. A 10:1 frequency range isn't
exactly trivial, you need to have someone on board who knows RF very well.
You can also do it without transformers, discrete transistor-level
design. But that's hard work.
--
Regards, Joerg
http://www.analogconsultants.com/
Jeff Liebermann
Jul 24, 2013, 11:42:55 PM7/24/13
to
On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
wrote:
>Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
>watts into an inductive load for a physics package at a startup. We're
>using function generators currently (HP 3314A for example) but we need
>more power, as in up to around 10 watts (into nominally 50 Ohms).
>
>The signal source is easy, we've got that -- it's getting from the mW
>range to the 10 W range that's the difficult part, or difficult without
>spending more than say $200.
I don't think you'll find anything with the required gain (about
30dB), bandwidth (about 7 octaves), and power (about +40dBm) for $200.
More like about $1,000 packaged and ready to play.
>Being a startup, cheap is important, or we'd go for a used ENI wideband
>amp (I found an ENI 320L, 250 KHz - 110 MHz, 20W for $1450, out of our
>price range).
Also try Amplifier Research:
<http://www.ar-amps.com/html/12100_rf_amplifier.asp>
or Mini-Circuits:
<http://www.minicircuits.com/products/Amplifiers.shtml>
They're not cheap.
>Something off the shelf would be good, something easy to replicate
>would be acceptable. Output doesn't have to be really clean, so AB2 is
>acceptable.
>
>Suggestions?
100 Khz to 10 Mhz is 7 octaves, which is not going to be easy to
deliver. The low end is the big problem, requiring big ferrites. Are
you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
would be much easier and cheaper.
You also didn't say anything about linearity. If you're just pumping
a sine wave into your fizzix package, almost anything will work.
However, if you're AM modulating the function generator, you'll need
some linearity specs. You may find yourself needing a 40-100 watt
amplifier, just to get 10 watts of "clean" and distortion free RF.
As Phil suggested, startups have better things to do than save a
pennies or reinvent the wheel. Find someone with a decent HF ham
radio transmitter, or linear amplifier, that puts out about 100 watts,
and you're done. There are some that might function down to 100 KHz
if you don't care about harmonic content and linearity.
Maybe something like this home built unit:
<http://www.nt7s.com/blog/2009/06/homebrew-20-watt-linear-amp/>
Well, maybe one of these:
<http://www.communication-concepts.com/index.php/amplifiers.html>
Again, watch out for the low end frequency range. Going from the
rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
wrote:
>Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
>watts into an inductive load for a physics package at a startup. We're
>using function generators currently (HP 3314A for example) but we need
>more power, as in up to around 10 watts (into nominally 50 Ohms).
>
>The signal source is easy, we've got that -- it's getting from the mW
>range to the 10 W range that's the difficult part, or difficult without
>spending more than say $200.
30dB), bandwidth (about 7 octaves), and power (about +40dBm) for $200.
More like about $1,000 packaged and ready to play.
>Being a startup, cheap is important, or we'd go for a used ENI wideband
>amp (I found an ENI 320L, 250 KHz - 110 MHz, 20W for $1450, out of our
>price range).
<http://www.ar-amps.com/html/12100_rf_amplifier.asp>
or Mini-Circuits:
<http://www.minicircuits.com/products/Amplifiers.shtml>
They're not cheap.
>Something off the shelf would be good, something easy to replicate
>would be acceptable. Output doesn't have to be really clean, so AB2 is
>acceptable.
>
>Suggestions?
deliver. The low end is the big problem, requiring big ferrites. Are
you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
would be much easier and cheaper.
You also didn't say anything about linearity. If you're just pumping
a sine wave into your fizzix package, almost anything will work.
However, if you're AM modulating the function generator, you'll need
some linearity specs. You may find yourself needing a 40-100 watt
amplifier, just to get 10 watts of "clean" and distortion free RF.
As Phil suggested, startups have better things to do than save a
pennies or reinvent the wheel. Find someone with a decent HF ham
radio transmitter, or linear amplifier, that puts out about 100 watts,
and you're done. There are some that might function down to 100 KHz
if you don't care about harmonic content and linearity.
Maybe something like this home built unit:
<http://www.nt7s.com/blog/2009/06/homebrew-20-watt-linear-amp/>
Well, maybe one of these:
<http://www.communication-concepts.com/index.php/amplifiers.html>
Again, watch out for the low end frequency range. Going from the
rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
John Larkin
Jul 25, 2013, 12:56:53 AM7/25/13
to
On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com> wrote:
If you do want to build something, take a look at the Apex opamps.
http://www.apexanalog.com/apex-products/pa107/
Note the "order free samples" tab.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
upsid...@downunder.com
Jul 25, 2013, 2:14:28 AM7/25/13
to
On Wed, 24 Jul 2013 20:42:55 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
>Again, watch out for the low end frequency range. Going from the
>rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
Use two amplifiers, one for the low end and the other for the high end
and a crossover filter.
The frequency drop at the extremes might be enough to have a gradual
crossover response, but of course, you should not feed large power
levels out of the design range, so some external crossover filtering
may be needed.
Some power measurement device at the output can be used to control the
amplifier input level to get a constant output, despite the filters.
wrote:
>Again, watch out for the low end frequency range. Going from the
>rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
and a crossover filter.
The frequency drop at the extremes might be enough to have a gradual
crossover response, but of course, you should not feed large power
levels out of the design range, so some external crossover filtering
may be needed.
Some power measurement device at the output can be used to control the
amplifier input level to get a constant output, despite the filters.
Phil Allison
Jul 25, 2013, 3:13:30 AM7/25/13
to
"John Larkin"
>
> Anything that you do yourself will take time.
>
> If you do want to build something, take a look at the Apex opamps.
>
> http://www.apexanalog.com/apex-products/pa107/
>
> Note the "order free samples" tab.
** 10W into 50 ohms ( = 64V p-p) at 10MHz requires a slew rate of 2000 V/uS.
> Anything that you do yourself will take time.
>
> If you do want to build something, take a look at the Apex opamps.
>
> http://www.apexanalog.com/apex-products/pa107/
>
> Note the "order free samples" tab.
So you are safe enough there.
But you are gonna HAVE to wangle a free sample to stay under your fanciful
budget of $200.
.... Phil
mike
Jul 25, 2013, 3:32:09 AM7/25/13
to
I've spent a lot of time with customers who wanted "just like that,
only a lot more of it!" Round numbers are a clue.
After some discussion, you find you that they really needed 4 watts
at 3MHz. +/-
Stated another way, it's easy to add "cushion" to the spec that
moves it from "simple" to nearly "impossible".
Does it have to be 50 ohms?
Could you drive both ends with a bridge? Cuts your voltage
requirement in half.
Brute force is not always the shortest distance to success.
upsid...@downunder.com
Jul 25, 2013, 4:04:11 AM7/25/13
to
On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
wrote:
wrote:
>Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
>watts into an inductive load for a physics package at a startup.
An ordinary solid state RF power transmitter is usually designed for
>watts into an inductive load for a physics package at a startup.
50 ohm resistive load.
Trying to use one with some inductive load at unknown impedance will
cause high SWR and a puff of smoke or the activation of any protection
circuit, which drops the power.
You might have to use an "antenna tuner" (transmatch) so that the
amplifier sees a 50 ohm resistive load. Depending of the impedance
response of the load, you may have to switch in various inductors to
cover that huge frequency range.
Phil Allison
Jul 25, 2013, 6:55:11 AM7/25/13
to
"mike"
>
>>
One of the more difficult aspects of any project is getting the spec right.
>
** Halleluiah !!!!
>>
One of the more difficult aspects of any project is getting the spec right.
>
> I've spent a lot of time with customers who wanted "just like that,
> only a lot more of it!" Round numbers are a clue.
> After some discussion, you find you that they really needed 4 watts
> at 3MHz. +/-
> Stated another way, it's easy to add "cushion" to the spec that
> moves it from "simple" to nearly "impossible".
Or... not think - as the case may be.
> Does it have to be 50 ohms?
> Could you drive both ends with a bridge? Cuts your voltage
> requirement in half.
> Brute force is not always the shortest distance to success.
Might be an engineering axiom.
... Phil
Baron
Jul 25, 2013, 6:59:06 AM7/25/13
to
upsid...@downunder.com Inscribed thus:
My guess is that he wants to do plasma generation !
--
Best Regards:
Baron.
--
Best Regards:
Baron.
Tim Wescott
Jul 25, 2013, 7:23:48 AM7/25/13
to
Cheap is important, but if it's a serious startup then your time is
important, too. If you're a degreed engineer in the US, $1450 will pay
for somewhere between one and three days of your time (remember to figure
that your overhead costs as least as much as your salary when you do your
computations), and there's no way that you'll put this together in three
days.
There might be something available on eBay for less. But remember that
if you spend three solid days getting something for free, you've just
spent more than that $1450 instrument.
--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
Adrian Tuddenham
Jul 25, 2013, 7:45:18 AM7/25/13
to
artie <art...@gNOSPAMmail.com> wrote:
> Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
> watts into an inductive load for a physics package at a startup.
Wasn't that what Sinclair used to claim for an OC28 driven by a couple
> Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
> watts into an inductive load for a physics package at a startup.
of 'red-spot' rejects ?
--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
Martin Brown
Jul 25, 2013, 8:04:05 AM7/25/13
to
the frequency is on the low side for sensible inductive coupling.
A typical ICP plasma is driven by at least a few hundred watts and a
peak output of 1kW or more to handle starting transients and solvents.
Also he is likely to become severely unpopular if it isn't very well
screened. Spurious broadcasting into the MW and ADSL frequency band.
27 & 40MHz are the preferred (cheap amp) frequencies for ICP.
Depending how long it is needed for hiring the kit @ ~10% pcm may well
be easier or looking for something suitable secondhand on eBay.
--
Regards,
Martin Brown
Klaus Kragelund
Jul 25, 2013, 8:24:19 AM7/25/13
to
A LT1210 55MHz single chip solution for 14USD:
http://www.digikey.com/product-detail/en/LT1210CT7%23PBF/LT1210CT7%23PBF-ND/891044
http://cds.linear.com/docs/en/datasheet/1210fa.pdf
Just add a power supply that can handle 10W (and a heat sink)
Cheers
Klaus
Klaus Kragelund
Jul 25, 2013, 8:29:06 AM7/25/13
to
http://www.linear.com/demo?demo_board=&part_number=&category_id=1154&companion_board=&software_id=
Phil Allison
Jul 25, 2013, 8:41:24 AM7/25/13
to
"Baron"
>
> My guess is that he wants to do plasma generation !
I reckon he is having another crack at "Cold Fusion ".
Even the great Amar Bose fell for that one ......
... Phil
Spehro Pefhany
Jul 25, 2013, 9:35:00 AM7/25/13
to
On Thu, 25 Jul 2013 05:24:19 -0700 (PDT), Klaus Kragelund
<klau...@hotmail.com> wrote:
>>
>
>Maybe I am missing somthing, but 10MHz is really not that big of a deal.
>
>A LT1210 55MHz single chip solution for 14USD:
>
>http://www.digikey.com/product-detail/en/LT1210CT7%23PBF/LT1210CT7%23PBF-ND/891044
>
>http://cds.linear.com/docs/en/datasheet/1210fa.pdf
>
>Just add a power supply that can handle 10W (and a heat sink)
>
>Cheers
>
>Klaus
Ha, the 'typical app' on page 16 is almost the exact stated
<klau...@hotmail.com> wrote:
>>
>
>Maybe I am missing somthing, but 10MHz is really not that big of a deal.
>
>A LT1210 55MHz single chip solution for 14USD:
>
>http://www.digikey.com/product-detail/en/LT1210CT7%23PBF/LT1210CT7%23PBF-ND/891044
>
>http://cds.linear.com/docs/en/datasheet/1210fa.pdf
>
>Just add a power supply that can handle 10W (and a heat sink)
>
>Cheers
>
>Klaus
requirement (9W into 50 ohms over 100kHz-10MHz) using a COTS
transformer and two chips in a bridged configuration.
http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf
Looks like Coiltronics have renamed their Versa Pac series at some
point in the last 16 years.. hopefully the CTX-01-13033-X2 isn't
discontinued. If it's off the shelf, that would be pretty fast to
throw together.
Martin Brown
Jul 25, 2013, 9:51:02 AM7/25/13
to
*everyone* fell for it at first. You could not buy heavy water or
palladium for months after their announcement for love nor money. Every
lab with the reagents had a go at it since it was so easy to do!
Sadly no-one else was able to make it work.
--
Regards,
Martin Brown
Baron
Jul 25, 2013, 11:55:42 AM7/25/13
to
Martin Brown Inscribed thus:
> On 25/07/2013 11:59, Baron wrote:
>> upsid...@downunder.com Inscribed thus:
>>
>>> On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
>>> wrote:
>>>
>>>> Need to provide a sine wave in the range 100 KHz - 10 MHz at around
>>>> 10 watts into an inductive load for a physics package at a startup.
>>>
>>> An ordinary solid state RF power transmitter is usually designed for
>>> 50 ohm resistive load.
>>>
>>> Trying to use one with some inductive load at unknown impedance will
>>> cause high SWR and a puff of smoke or the activation of any
>>> protection circuit, which drops the power.
>>>
>>> You might have to use an "antenna tuner" (transmatch) so that the
>>> amplifier sees a 50 ohm resistive load. Depending of the impedance
>>> response of the load, you may have to switch in various inductors to
>>> cover that huge frequency range.
>>
>> My guess is that he wants to do plasma generation !
>
> In which case he is underpowered by a couple of orders of magnitude
> and the frequency is on the low side for sensible inductive coupling.
>
> A typical ICP plasma is driven by at least a few hundred watts and a
> peak output of 1kW or more to handle starting transients and solvents.
I agree 100W is a bit low for plasma, and 13.5Mhz springs to mind.
> Also he is likely to become severely unpopular if it isn't very well
> screened. Spurious broadcasting into the MW and ADSL frequency band.
> 27 & 40MHz are the preferred (cheap amp) frequencies for ICP.
>
> Depending how long it is needed for hiring the kit @ ~10% pcm may well
> be easier or looking for something suitable secondhand on eBay.
>
--
Best Regards:
Baron.
> On 25/07/2013 11:59, Baron wrote:
>> upsid...@downunder.com Inscribed thus:
>>
>>> On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
>>> wrote:
>>>
>>>> Need to provide a sine wave in the range 100 KHz - 10 MHz at around
>>>> 10 watts into an inductive load for a physics package at a startup.
>>>
>>> An ordinary solid state RF power transmitter is usually designed for
>>> 50 ohm resistive load.
>>>
>>> Trying to use one with some inductive load at unknown impedance will
>>> cause high SWR and a puff of smoke or the activation of any
>>> protection circuit, which drops the power.
>>>
>>> You might have to use an "antenna tuner" (transmatch) so that the
>>> amplifier sees a 50 ohm resistive load. Depending of the impedance
>>> response of the load, you may have to switch in various inductors to
>>> cover that huge frequency range.
>>
>> My guess is that he wants to do plasma generation !
>
> In which case he is underpowered by a couple of orders of magnitude
> and the frequency is on the low side for sensible inductive coupling.
>
> A typical ICP plasma is driven by at least a few hundred watts and a
> peak output of 1kW or more to handle starting transients and solvents.
> Also he is likely to become severely unpopular if it isn't very well
> screened. Spurious broadcasting into the MW and ADSL frequency band.
> 27 & 40MHz are the preferred (cheap amp) frequencies for ICP.
>
> Depending how long it is needed for hiring the kit @ ~10% pcm may well
> be easier or looking for something suitable secondhand on eBay.
>
--
Baron.
Jeff Liebermann
Jul 25, 2013, 12:08:12 PM7/25/13
to
On Thu, 25 Jul 2013 09:14:28 +0300, upsid...@downunder.com wrote:
>On Wed, 24 Jul 2013 20:42:55 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>
>>Again, watch out for the low end frequency range. Going from the
>>rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
>
>Use two amplifiers, one for the low end and the other for the high end
>and a crossover filter.
>
>The frequency drop at the extremes might be enough to have a gradual
>crossover response, but of course, you should not feed large power
>levels out of the design range, so some external crossover filtering
>may be needed.
Yeah, that's one way. However, the OP was asking for a ready to use
>On Wed, 24 Jul 2013 20:42:55 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>
>>Again, watch out for the low end frequency range. Going from the
>>rated 1.5 Mhz down to your 0.1 Mhz is NOT easy.
>
>Use two amplifiers, one for the low end and the other for the high end
>and a crossover filter.
>
>The frequency drop at the extremes might be enough to have a gradual
>crossover response, but of course, you should not feed large power
>levels out of the design range, so some external crossover filtering
>may be needed.
package, that he could use for his startup, not a design or
construction effort. I couldn't find anything that meets all his
requirements, especially the $200 target price.[1]
If I were building a 0.1 to 10 MHz 10 watt amplifier with 30-40dB
gain, I would probably do it with a direct coupled amp that would work
down to DC. That would eliminate most of the ferrites and all the
transformers. 10 watts is about 64v peak-to-peak, so this should be
doable with a +/- 48v bipolar power source and some high voltage power
devices.
>Some power measurement device at the output can be used to control the
>amplifier input level to get a constant output, despite the filters.
are places where it causes more problems than it solves. I've found
it easier to gain level a direct coupled amplifier to 1dB over 7
octaves, than it is to build a 7 octave RF detector, loop amp, and
gain control circuit, to obtain the same effect. (A marginal analogy
is whether it's easier to control the speed of a car with the
accelerator pedal or with the brake).
[1] My rule of thumb for renting or leasing test equipment is that it
typically costs about 15% of the purchase price per month to rent.
That means if I need to use the item for more than 7-9 months, I might
as well borrow the money and buy it. The actual percentage varies
with the popularity of the equipment, calibration requirements, and
vendor. Do the math first.
Jeff Liebermann
Jul 25, 2013, 12:23:37 PM7/25/13
to
On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
wrote:
wrote:
>Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
>watts into an inductive load for a physics package at a startup. We're
>using function generators currently (HP 3314A for example) but we need
>more power, as in up to around 10 watts (into nominally 50 Ohms).
Looking at the data sheet for the HP3314A,
>watts into an inductive load for a physics package at a startup. We're
>using function generators currently (HP 3314A for example) but we need
>more power, as in up to around 10 watts (into nominally 50 Ohms).
<http://www.davmar.org/Spec/HP3314A.pdf>
I notice that the normal output level is 10v p-p (250mw). With option
001 it goes to 30v p-p (2 watts). If you happen to have Option 001,
you could probably get 10 watts out with just a broadband step up
transformer, a simple emitter follower using a high voltage power
xsister, and some ferrite beads.
John Larkin
Jul 25, 2013, 9:30:35 PM7/25/13
to
on a now-defunct (IBM?) high voltage cmos process. I think that they may be
regular surface-mount nowadays, under all that epoxy.
artie
Jul 25, 2013, 11:25:52 PM7/25/13
to
In article <9u72v8h1go3so37le...@4ax.com>, Spehro Pefhany
<spef...@interlogDOTyou.knowwhat> wrote:
> http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf
This is what I was looking for -- I've been using my HF transceiver
with the output power cranked down, but it won't go below 1.8 MHz for
transmit. The matching networks are easy.
We'll try the LT1210!
Many thanks--
--
<spef...@interlogDOTyou.knowwhat> wrote:
> http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf
This is what I was looking for -- I've been using my HF transceiver
with the output power cranked down, but it won't go below 1.8 MHz for
transmit. The matching networks are easy.
We'll try the LT1210!
Many thanks--
--
artie
Jul 25, 2013, 11:29:30 PM7/25/13
to
In article <8sj2v81qu3l32h53g...@4ax.com>, Jeff
follower and beads weren't satisfactory. Ideally, the source plus amp
fits in a small enclosure -- the DDS generator and the LT1210 should do
the trick.
Thanks--
--
Namaste--
Liebermann <je...@cruzio.com> wrote:
> On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
> wrote:
>
> >Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
> >watts into an inductive load for a physics package at a startup. We're
> >using function generators currently (HP 3314A for example) but we need
> >more power, as in up to around 10 watts (into nominally 50 Ohms).
>
> Looking at the data sheet for the HP3314A,
> <http://www.davmar.org/Spec/HP3314A.pdf>
> I notice that the normal output level is 10v p-p (250mw). With option
> 001 it goes to 30v p-p (2 watts). If you happen to have Option 001,
> you could probably get 10 watts out with just a broadband step up
> transformer, a simple emitter follower using a high voltage power
> xsister, and some ferrite beads.
Unfortunately it's not the opt 001 model. Results with an emitter
> On Wed, 24 Jul 2013 18:33:21 -0700, artie <art...@gNOSPAMmail.com>
> wrote:
>
> >Need to provide a sine wave in the range 100 KHz - 10 MHz at around 10
> >watts into an inductive load for a physics package at a startup. We're
> >using function generators currently (HP 3314A for example) but we need
> >more power, as in up to around 10 watts (into nominally 50 Ohms).
>
> Looking at the data sheet for the HP3314A,
> <http://www.davmar.org/Spec/HP3314A.pdf>
> I notice that the normal output level is 10v p-p (250mw). With option
> 001 it goes to 30v p-p (2 watts). If you happen to have Option 001,
> you could probably get 10 watts out with just a broadband step up
> transformer, a simple emitter follower using a high voltage power
> xsister, and some ferrite beads.
follower and beads weren't satisfactory. Ideally, the source plus amp
fits in a small enclosure -- the DDS generator and the LT1210 should do
the trick.
Thanks--
--
Namaste--
George Herold
Jul 26, 2013, 12:17:43 AM7/26/13
to
I'm a class A kinda guy, so just an opamp (or more for gain) into a transistor (with a heat sink)
To OP:
There's a nice trick for driving non-resistive loads, you put the same R and other equivalent Z (L or C) in series, and that combo in parallel with the load, and it makes the load look resistive to the amp. (you should do the math at least once.)
What's the physics package? (or can't you tell?)
I'm assuming the L is an air coil. What's L and R?
(or if not that, what's the R/L time?)
George H.
Tim Wescott
Jul 26, 2013, 6:51:35 AM7/26/13
to
when you set impedances (and when you do your power calculations), and
when you think of the necessary heat sinking, think "humongous".
I can't remember the calculations for power dissipation from a class B
stage (which is more or less what that thing is, or should be), but if
you're getting 10W out you're probably going to be burning up at least
10W in the device. That means that you'll be getting at least a 50
degree rise from tab to junction in a TO220 case. That, in turn, means
that (assuming a 125C junction temperature rating, which I'm too lazy to
check for) you need to hold the tab temperature below 75C.
Consider the temperature around your heat sink, and size appropriately.
And don't forget your heat sink goo.
Jeff Liebermann
Jul 26, 2013, 12:14:23 PM7/26/13
to
On Thu, 25 Jul 2013 21:17:43 -0700 (PDT), George Herold
<ghe...@teachspin.com> wrote:
>On Wednesday, July 24, 2013 11:42:55 PM UTC-4, Jeff Liebermann wrote:
>> 100 Khz to 10 Mhz is 7 octaves, which is not going to be easy to
>> deliver. The low end is the big problem, requiring big ferrites. Are
>> you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
>> would be much easier and cheaper.
>Oh, And I was thinking the low end was easier.
No. If he builds something in the style of a conventional 2-30MHz FET
<ghe...@teachspin.com> wrote:
>On Wednesday, July 24, 2013 11:42:55 PM UTC-4, Jeff Liebermann wrote:
>> 100 Khz to 10 Mhz is 7 octaves, which is not going to be easy to
>> deliver. The low end is the big problem, requiring big ferrites. Are
>> you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
>> would be much easier and cheaper.
>Oh, And I was thinking the low end was easier.
or bipolar RF amplifier, the matching transformers between stages and
at the output rely on the ferrites to deliver enough inductance and
coupling at the lower frequencies. It usually ends up looking like
this:
<http://www.radioaficion.com/HamNews/images/11-2012/TEN-TEC_418_Internal%20Photos%202-1.jpg>
<http://www.tentec.com/products/Model-418--160%252d6-Meter-Solid-State-Linear-Amplifier.html>
That's 1.6 to 54 MHz with 100 watts out.
At 17-30 MHz, there's probably enough wire in the transformers to
produce usable inductance without any ferrites. Below about 8 MHz,
it's either add much more wire, or add ferrites. The above amp
probably works from 3 to 29 MHz (ham bands). To make it work down to
0.1 MHz, my guess(tm) is that two different ferrite materials will be
needed and that the ferrites in the output section will be 4 times as
large. The inductance requirements are less using FET devices, but
still tend to be rather large.
There's also the problem of gain. If I double the bandwidth of the
amplifier, the gain goes down by about 6 dBm. (There are resonant
tricks to squeeze some more gain out at the high frequency end.) So,
if I start with a 4 octave (2-30MHz) amplifier design, and want to
extend it down to 0.1 Mhz or a total of 8.2 octaves, it will have
about 12.6 dB less gain. 12.6 dB gain is fairly cheap at low power
levels, but rather expensive at 10 watts and up. In this case,
dropping the upper end from 30 Mhz to 10 MHz for 0.1 to 10 Mhz range
results in about 7 octaves which only saves 3.6dB of gain.
> I'm a class A kinda guy, so just an opamp (or more for gain)
>into a transistor (with a heat sink)
except for efficiency. If OP wants 10 watts of distortion and
intermod free output, which is likely if he's feeding it with a
function generator, he's likely to get only about 15% amplifier
efficiency with class A. That means he has to burn 67 watts of heat
in order to get 10 watts of clean output.
67 watts is going to require some efforts in heat disposal. For one
time prototypes, I'm partial to a small aluminum (for heat transfer)
can filled with mineral oil. The power amp gets immersed in the
mineral oil. However, that's not enough for 67 watts, so the aluminum
can is dumped into a much larger bucket of water. No need for gallons
of mineral oil and the water can be cooled down with a garden hose.
George Herold
Jul 26, 2013, 1:37:02 PM7/26/13
to
On Friday, July 26, 2013 12:14:23 PM UTC-4, Jeff Liebermann wrote:
> On Thu, 25 Jul 2013 21:17:43 -0700 (PDT), George Herold
>
> <ghe...@teachspin.com> wrote:
>
>
>
> >On Wednesday, July 24, 2013 11:42:55 PM UTC-4, Jeff Liebermann wrote:
>
>
>
> >> 100 Khz to 10 Mhz is 7 octaves, which is not going to be easy to
>
> >> deliver. The low end is the big problem, requiring big ferrites. Are
>
> >> you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
>
> >> would be much easier and cheaper.
>
>
>
> >Oh, And I was thinking the low end was easier.
>
>
>
> No. If he builds something in the style of a conventional 2-30MHz FET
>
> or bipolar RF amplifier, the matching transformers between stages and
>
> at the output rely on the ferrites to deliver enough inductance and
>
> coupling at the lower frequencies. It usually ends up looking like
>
> this:
>
> <http://www.radioaficion.com/HamNews/images/11-2012/TEN-TEC_418_Internal%20Photos%202-1.jpg>
>
> <http://www.tentec.com/products/Model-418--160%252d6-Meter-Solid-State-Linear-Amplifier.html>
>
Hi Jeff,
> On Thu, 25 Jul 2013 21:17:43 -0700 (PDT), George Herold
>
> <ghe...@teachspin.com> wrote:
>
>
>
> >On Wednesday, July 24, 2013 11:42:55 PM UTC-4, Jeff Liebermann wrote:
>
>
>
> >> 100 Khz to 10 Mhz is 7 octaves, which is not going to be easy to
>
> >> deliver. The low end is the big problem, requiring big ferrites. Are
>
> >> you sure you need 100 KHz? If it was only 1 to 10 Mhz (4 octaves), it
>
> >> would be much easier and cheaper.
>
>
>
> >Oh, And I was thinking the low end was easier.
>
>
>
> No. If he builds something in the style of a conventional 2-30MHz FET
>
> or bipolar RF amplifier, the matching transformers between stages and
>
> at the output rely on the ferrites to deliver enough inductance and
>
> coupling at the lower frequencies. It usually ends up looking like
>
> this:
>
> <http://www.radioaficion.com/HamNews/images/11-2012/TEN-TEC_418_Internal%20Photos%202-1.jpg>
>
> <http://www.tentec.com/products/Model-418--160%252d6-Meter-Solid-State-Linear-Amplifier.html>
>
Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system. It looked something like those pictures.. not at all easy to do.
(Well at least it took me a while. I was learning RF at the same time.)
So ideally 50% efficiency.. but of course it will be a bit worse than that in practice. (I may be making some huge blunder here, but then (hopefully) I'll be corrected and learn something.)
Upon further reflection, (sorry for thinking out loud) I was picturing an opamp->transistor, voltage to current source for driving the inductor, since that's what I often use for driving coils.
Hey, if the OP's load is a coil then he may want a current source.
(no change in the B-field as the coil warms up.)
George H.
Jeff Liebermann
Jul 27, 2013, 12:37:26 AM7/27/13
to
On Fri, 26 Jul 2013 10:37:02 -0700 (PDT), George Herold
<ghe...@teachspin.com> wrote:
>Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>It looked something like those pictures.. not at all easy to do.
>(Well at least it took me a while. I was learning RF at the same time.)
We may have been competitors. I did an NMR driver and power amp
<ghe...@teachspin.com> wrote:
>Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>It looked something like those pictures.. not at all easy to do.
>(Well at least it took me a while. I was learning RF at the same time.)
system using then new VMOS devices. My guess(tm) was about 1979.
Getting ultra low distortion of -60dB 3rd order IM at rated power gave
me headaches and nightmares.
>> If OP wants 10 watts of distortion and
>> intermod free output, which is likely if he's feeding it with a
>> function generator, he's likely to get only about 15% amplifier
>> efficiency with class A. That means he has to burn 67 watts of heat
>> in order to get 10 watts of clean output.
>Hmm is it that bad?
>(the efficiency) I was thinking about just
>biasing the transistor at the mid current point. And then the
>sine wave turns it on or off.. but the average power loss still
>looks to be about 1/2.
>So ideally 50% efficiency.. but of course it will be a bit worse
>than that in practice. (I may be making some huge blunder here,
>but then (hopefully) I'll be corrected and learn something.)
efficiency, I end up driving the output from rail to rail. That's
fine for getting the maximum output from the device, but not so fine
for getting the best linearity, lowest intermod, and minimum
distortion. Generally, the output swing needs to stay away from the
rails. A safe guess(tm) would be about 50% voltage swing, which is
1/4th of the maximum output power. For example, for 10 watts of clean
output, I would need an amp capable of producing 40 watts. Of course,
if one doesn't care about distortion, intermod, and such, a 10 watt
amp would do just fine.
As a side note, the bandwidth should be at least 3 times the maximum
usable frequency. If the function generator is expected to be usable
to 10 MHz, the amp has to go up to at least 30 Mhz in order to amplify
the 3rd harmonic necessary for something resembling a square wave.
I dunno about using the LT1210. From the data sheet:
<http://cds.linear.com/docs/en/datasheet/1210fa.pdf>
the maximum output swing is +/- 11.5V. Into a 50 ohm load, that's:
(22.5v / 2 * 0.707)^2 / 50 = 0.4 watts rms
So much for direct drive. 0.4 watts seems a bit less than the
required 10 watts output power.
However, the LT1210 can drive a 10 ohm load with a 22.5v swing. Using
a 4:1 impedance ratio output transformer, the maximum output would be:
(22.5 / 2 * 0.707)^2 / 12.5 ohms = 1.6 watts rms
which is still shy of 10 watts by a large amount.
In my never humble opinion, getting rid the transformers and ferrites
is a necessity. To deliver 10 watts into 50 ohms, one needs:
(10 * 50)^0.5 = 22.4 v rms * 2.828 = 63.3 volts p-p
which means a +/- 50v power supply and output swing.
>Upon further reflection, (sorry for thinking out loud) I was
>picturing an opamp->transistor, voltage to current source for
>driving the inductor, since that's what I often use for driving coils.
>Hey, if the OP's load is a coil then he may want a current source.
>(no change in the B-field as the coil warms up.)
the common term for rubidium and cesium fountains used in frequency
standards. I don't recall the excitation frequency (and am too lazy
to Google for the numbers).
John S
Jul 27, 2013, 8:35:14 AM7/27/13
to
sheet you reference above. There is the schematic of the amplifier
requested.
Jeff Liebermann
Jul 27, 2013, 12:03:19 PM7/27/13
to
On Fri, 26 Jul 2013 21:37:26 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
>On Fri, 26 Jul 2013 10:37:02 -0700 (PDT), George Herold
><ghe...@teachspin.com> wrote:
>
>>Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>>It looked something like those pictures.. not at all easy to do.
>>(Well at least it took me a while. I was learning RF at the same time.)
>
>We may have been competitors. I did an NMR driver and power amp
>system using then new VMOS devices. My guess(tm) was about 1979.
>Getting ultra low distortion of -60dB 3rd order IM at rated power gave
>me headaches and nightmares.
Memory fault from 30 years ago. The 1979 version used bipolar
devices, which were difficult to work with and difficult to produce
low IM levels. The 1981 version used Siliconix VMOS-FETs, which was
much easier.
>... A safe guess(tm) would be about 50% voltage swing, which is
With a rail to rail output voltage swing, the output power of a
Class-A state will be the same its quiescent DC power dissipation
resulting in the theoretical maximum efficiency of 50%. If I cut the
output swing in half, the output power is reduced to 25% but the full
DC power dissipation remains the same. The overall efficiency is now
12.5%. So, in order to get 10 watts of squeaky clean output, one
needs to build an 80 watt amplifier.
To be fair, this is probably overkill for the OP's purpose. A
reduction to 70% of rail to rail output voltage swing will result in
50% of the output power, 25% efficiency, and requiring only a 40 watt
amp.
wrote:
>On Fri, 26 Jul 2013 10:37:02 -0700 (PDT), George Herold
><ghe...@teachspin.com> wrote:
>
>>Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>>It looked something like those pictures.. not at all easy to do.
>>(Well at least it took me a while. I was learning RF at the same time.)
>
>We may have been competitors. I did an NMR driver and power amp
>system using then new VMOS devices. My guess(tm) was about 1979.
>Getting ultra low distortion of -60dB 3rd order IM at rated power gave
>me headaches and nightmares.
devices, which were difficult to work with and difficult to produce
low IM levels. The 1981 version used Siliconix VMOS-FETs, which was
much easier.
>... A safe guess(tm) would be about 50% voltage swing, which is
>1/4th of the maximum output power. For example, for 10 watts of clean
>output, I would need an amp capable of producing 40 watts.
Oops, arithmetic error.
>output, I would need an amp capable of producing 40 watts.
With a rail to rail output voltage swing, the output power of a
Class-A state will be the same its quiescent DC power dissipation
resulting in the theoretical maximum efficiency of 50%. If I cut the
output swing in half, the output power is reduced to 25% but the full
DC power dissipation remains the same. The overall efficiency is now
12.5%. So, in order to get 10 watts of squeaky clean output, one
needs to build an 80 watt amplifier.
To be fair, this is probably overkill for the OP's purpose. A
reduction to 70% of rail to rail output voltage swing will result in
50% of the output power, 25% efficiency, and requiring only a 40 watt
amp.
Jeff Liebermann
Jul 27, 2013, 12:12:11 PM7/27/13
to
On Sat, 27 Jul 2013 07:35:14 -0500, John S <Sop...@invalid.org>
wrote:
>On 7/26/2013 11:37 PM, Jeff Liebermann wrote:
>> I dunno about using the LT1210. From the data sheet:
>> <http://cds.linear.com/docs/en/datasheet/1210fa.pdf>
>> the maximum output swing is +/- 11.5V. Into a 50 ohm load, that's:
>> (22.5v / 2 * 0.707)^2 / 50 = 0.4 watts rms
>> So much for direct drive. 0.4 watts seems a bit less than the
>> required 10 watts output power.
>>
>> However, the LT1210 can drive a 10 ohm load with a 22.5v swing. Using
>> a 4:1 impedance ratio output transformer, the maximum output would be:
>> (22.5 / 2 * 0.707)^2 / 12.5 ohms = 1.6 watts rms
>> which is still shy of 10 watts by a large amount.
>>
>> In my never humble opinion, getting rid the transformers and ferrites
>> is a necessity. To deliver 10 watts into 50 ohms, one needs:
>> (10 * 50)^0.5 = 22.4 v rms * 2.828 = 63.3 volts p-p
>> which means a +/- 50v power supply and output swing.
Oops. That should be +/- 35 volt power supply.
>As was pointed out elsewhere in this thread, see page 16 of the data
>sheet you reference above. There is the schematic of the amplifier
>requested.
<http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf>
Pg 6 thru Pg 8.
The schematic is almost exactly what the OP wants, except for one
little problem. There wasn't enough output voltage swing necessary to
produce the required 10 watts of RF output power. That's what I
calculated above. Otherwise, it's a fine circuit for producing 400
milliwatts of RF maximum.
There's also the not so minor problem of the bandwidth, which seems to
literally die at 10 MHz. If the OP is producing sine waves with his
function generator, that will work fine. However, if he's doing
ramps, square waves, pulses, and complex waveforms, he'll need at
least 30 MHz of bandwidth.
wrote:
>On 7/26/2013 11:37 PM, Jeff Liebermann wrote:
>> I dunno about using the LT1210. From the data sheet:
>> <http://cds.linear.com/docs/en/datasheet/1210fa.pdf>
>> the maximum output swing is +/- 11.5V. Into a 50 ohm load, that's:
>> (22.5v / 2 * 0.707)^2 / 50 = 0.4 watts rms
>> So much for direct drive. 0.4 watts seems a bit less than the
>> required 10 watts output power.
>>
>> However, the LT1210 can drive a 10 ohm load with a 22.5v swing. Using
>> a 4:1 impedance ratio output transformer, the maximum output would be:
>> (22.5 / 2 * 0.707)^2 / 12.5 ohms = 1.6 watts rms
>> which is still shy of 10 watts by a large amount.
>>
>> In my never humble opinion, getting rid the transformers and ferrites
>> is a necessity. To deliver 10 watts into 50 ohms, one needs:
>> (10 * 50)^0.5 = 22.4 v rms * 2.828 = 63.3 volts p-p
>> which means a +/- 50v power supply and output swing.
>As was pointed out elsewhere in this thread, see page 16 of the data
>sheet you reference above. There is the schematic of the amplifier
>requested.
Pg 6 thru Pg 8.
The schematic is almost exactly what the OP wants, except for one
little problem. There wasn't enough output voltage swing necessary to
produce the required 10 watts of RF output power. That's what I
calculated above. Otherwise, it's a fine circuit for producing 400
milliwatts of RF maximum.
There's also the not so minor problem of the bandwidth, which seems to
literally die at 10 MHz. If the OP is producing sine waves with his
function generator, that will work fine. However, if he's doing
ramps, square waves, pulses, and complex waveforms, he'll need at
least 30 MHz of bandwidth.
John S
Jul 27, 2013, 3:44:34 PM7/27/13
to
I don't know exactly what the problem is with your analysis and I really
don't want to chase it down, but I can supply an LTSpice simulation that
closely matches their spec sheet. It performs in LTSpice as advertised.
If you want it, let me know.
Jeff Liebermann
Jul 27, 2013, 5:18:45 PM7/27/13
to
On Sat, 27 Jul 2013 14:44:34 -0500, John S <Sop...@invalid.org>
wrote:
>I don't know exactly what the problem is with your analysis
My analysis is fine. However, my arithmetic is atrocious (as usual).
Slight change in numbers below.
>and I really
>don't want to chase it down, but I can supply an LTSpice simulation that
>closely matches their spec sheet. It performs in LTSpice as advertised.
>If you want it, let me know.
Sure I'm interested. I want to see if it can produce 10 watts of
output with only a 23v P-P output swing. No need for LTSpice for that
calculation.
23 v p-p / 2 * 0.707 = 8.2 v rms
With a 50 ohm load (i.e. no xformer):
Power = Volts^2 / Load = (8.2)^2 / 50 = 1.3 watts rms
However, the LT1210 can handle loads below 50 ohms, so with a
transformer, more power can be delivered. The data sheet suggest a 10
ohm load, which might be convinced to deliver 6.6 watts rms with an
impossible broadband power xformer or 5.3 watts, with a 4:1 impedance
transformer.
Another line in the data sheet suggests that it can deliver 1.1 amps
peak output current. Converting to rms, that's:
1.1 * 0.707 = 0.78 amps rms
Power = E * I = 8.2v rms * 0.78 amps rms = 6.4 watts rms output.
That's maximum voltage and maximum current simultaneously, which is
generally considered a bad idea.
I'm sure the model closely matches the data sheet and applications
schematic. However, that's not the problem. I'm questioning whether
it meets the OP's vague requirements. Methinks not.
My abilities with LTSpice are similar to my arithmetic and spelling
abilities, but I still would like to see the model. I could use the
practice. My email in the signature below works. However, I won't
have time to play with it for a few days as I just found myself with a
very strange intermod problem on a local ham radio repeater and need
to build and install an FM broadcast trap filter to eliminate the
garbage.
wrote:
>I don't know exactly what the problem is with your analysis
Slight change in numbers below.
>and I really
>don't want to chase it down, but I can supply an LTSpice simulation that
>closely matches their spec sheet. It performs in LTSpice as advertised.
>If you want it, let me know.
output with only a 23v P-P output swing. No need for LTSpice for that
calculation.
23 v p-p / 2 * 0.707 = 8.2 v rms
With a 50 ohm load (i.e. no xformer):
Power = Volts^2 / Load = (8.2)^2 / 50 = 1.3 watts rms
However, the LT1210 can handle loads below 50 ohms, so with a
transformer, more power can be delivered. The data sheet suggest a 10
ohm load, which might be convinced to deliver 6.6 watts rms with an
impossible broadband power xformer or 5.3 watts, with a 4:1 impedance
transformer.
Another line in the data sheet suggests that it can deliver 1.1 amps
peak output current. Converting to rms, that's:
1.1 * 0.707 = 0.78 amps rms
Power = E * I = 8.2v rms * 0.78 amps rms = 6.4 watts rms output.
That's maximum voltage and maximum current simultaneously, which is
generally considered a bad idea.
I'm sure the model closely matches the data sheet and applications
schematic. However, that's not the problem. I'm questioning whether
it meets the OP's vague requirements. Methinks not.
My abilities with LTSpice are similar to my arithmetic and spelling
abilities, but I still would like to see the model. I could use the
practice. My email in the signature below works. However, I won't
have time to play with it for a few days as I just found myself with a
very strange intermod problem on a local ham radio repeater and need
to build and install an FM broadcast trap filter to eliminate the
garbage.
John S
Jul 28, 2013, 6:39:54 AM7/28/13
to
On 7/27/2013 4:18 PM, Jeff Liebermann wrote:
> On Sat, 27 Jul 2013 14:44:34 -0500, John S <Sop...@invalid.org>
> wrote:
>
>> I don't know exactly what the problem is with your analysis
>
> My analysis is fine. However, my arithmetic is atrocious (as usual).
> Slight change in numbers below.
>
>> and I really
>> don't want to chase it down, but I can supply an LTSpice simulation that
>> closely matches their spec sheet. It performs in LTSpice as advertised.
>> If you want it, let me know.
>
> Sure I'm interested. I want to see if it can produce 10 watts of
> output with only a 23v P-P output swing. No need for LTSpice for that
> calculation.
> 23 v p-p / 2 * 0.707 = 8.2 v rms
> With a 50 ohm load (i.e. no xformer):
> Power = Volts^2 / Load = (8.2)^2 / 50 = 1.3 watts rms
It is a bridged amp configuration. The 23V is _peak_. Do not divide by
> On Sat, 27 Jul 2013 14:44:34 -0500, John S <Sop...@invalid.org>
> wrote:
>
>> I don't know exactly what the problem is with your analysis
>
> My analysis is fine. However, my arithmetic is atrocious (as usual).
> Slight change in numbers below.
>
>> and I really
>> don't want to chase it down, but I can supply an LTSpice simulation that
>> closely matches their spec sheet. It performs in LTSpice as advertised.
>> If you want it, let me know.
>
> Sure I'm interested. I want to see if it can produce 10 watts of
> output with only a 23v P-P output swing. No need for LTSpice for that
> calculation.
> 23 v p-p / 2 * 0.707 = 8.2 v rms
> With a 50 ohm load (i.e. no xformer):
> Power = Volts^2 / Load = (8.2)^2 / 50 = 1.3 watts rms
two. That means your power calculation is 1/4 the actual.
> However, the LT1210 can handle loads below 50 ohms, so with a
> transformer, more power can be delivered. The data sheet suggest a 10
> ohm load, which might be convinced to deliver 6.6 watts rms with an
> impossible broadband power xformer or 5.3 watts, with a 4:1 impedance
> transformer.
> Another line in the data sheet suggests that it can deliver 1.1 amps
> peak output current. Converting to rms, that's:
> 1.1 * 0.707 = 0.78 amps rms
> Power = E * I = 8.2v rms * 0.78 amps rms = 6.4 watts rms output.
> That's maximum voltage and maximum current simultaneously, which is
> generally considered a bad idea.
>
> I'm sure the model closely matches the data sheet and applications
> schematic. However, that's not the problem. I'm questioning whether
> it meets the OP's vague requirements. Methinks not.
> My abilities with LTSpice are similar to my arithmetic and spelling
> abilities, but I still would like to see the model. I could use the
> practice. My email in the signature below works. However, I won't
> have time to play with it for a few days as I just found myself with a
> very strange intermod problem on a local ham radio repeater and need
> to build and install an FM broadcast trap filter to eliminate the
> garbage.
SHEET 1 880 680
WIRE -176 -224 -208 -224
WIRE -32 -224 -112 -224
WIRE -208 -208 -208 -224
WIRE -288 -160 -608 -160
WIRE -32 -128 -32 -224
WIRE -32 -128 -144 -128
WIRE 144 -128 -32 -128
WIRE 528 -128 384 -128
WIRE -288 -96 -320 -96
WIRE -32 -64 -32 -128
WIRE 144 -64 144 -128
WIRE 384 -64 384 -128
WIRE -608 0 -608 -160
WIRE -320 48 -320 -96
WIRE -32 48 -32 16
WIRE -32 48 -320 48
WIRE -800 64 -800 32
WIRE 528 64 528 -128
WIRE -32 80 -32 48
WIRE 384 80 384 16
WIRE -800 96 -800 64
WIRE -608 96 -608 80
WIRE 144 96 144 16
WIRE 528 160 528 144
WIRE -32 208 -32 160
WIRE -32 208 -320 208
WIRE -800 224 -800 176
WIRE -800 224 -880 224
WIRE -32 240 -32 208
WIRE 144 240 144 160
WIRE 384 240 384 160
WIRE -880 256 -880 224
WIRE -800 272 -800 224
WIRE 384 336 384 320
WIRE -320 352 -320 208
WIRE -288 352 -320 352
WIRE -32 384 -32 320
WIRE -32 384 -144 384
WIRE 144 384 144 320
WIRE 144 384 -32 384
WIRE -800 416 -800 352
WIRE -288 416 -320 416
WIRE -800 432 -800 416
WIRE -320 448 -320 416
WIRE -208 480 -208 464
WIRE -144 480 -208 480
WIRE -32 480 -32 384
WIRE -32 480 -80 480
FLAG -320 448 0
FLAG -608 96 0
FLAG 528 160 0
FLAG 384 336 0
FLAG -880 256 0
FLAG -800 64 +15
FLAG -256 304 +15
FLAG -800 416 -15
FLAG -256 464 -15
FLAG -256 -48 +15
FLAG -256 -208 -15
SYMBOL res -48 -80 R0
SYMATTR InstName R1
SYMATTR Value 680
SYMBOL res -48 64 R0
SYMATTR InstName R2
SYMATTR Value 220
SYMBOL res -48 224 R0
SYMATTR InstName R3
SYMATTR Value 910
SYMBOL voltage -608 -16 R0
WINDOW 123 24 124 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value2 AC 1
SYMATTR InstName V1
SYMATTR Value SINE(0 2.5 100k)
SYMBOL cap 128 96 R0
SYMATTR InstName C1
SYMATTR Value 100n
SYMBOL res 512 48 R0
SYMATTR InstName R4
SYMATTR Value 50
SYMBOL ind2 128 -80 R0
SYMATTR InstName L1
SYMATTR Value {L}
SYMATTR Type ind
SYMBOL ind2 128 224 R0
SYMATTR InstName L2
SYMATTR Value {L}
SYMATTR Type ind
SYMBOL ind2 368 224 R0
SYMATTR InstName L3
SYMATTR Value {L}
SYMATTR Type ind
SYMBOL ind2 368 64 R0
SYMATTR InstName L4
SYMATTR Value {L}
SYMATTR Type ind
SYMBOL ind2 368 -80 R0
SYMATTR InstName L5
SYMATTR Value {L}
SYMATTR Type ind
SYMBOL Opamps\\LT1210 -256 -128 M180
WINDOW 3 -125 69 Left 2
SYMATTR InstName U1
SYMBOL Opamps\\LT1210 -256 384 R0
WINDOW 3 -209 -1 Left 2
SYMATTR InstName U2
SYMBOL voltage -800 80 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 15
SYMBOL voltage -800 256 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V3
SYMATTR Value 15
SYMBOL cap -112 -240 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C2
SYMATTR Value 10n
SYMBOL cap -80 464 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C3
SYMATTR Value 10n
TEXT -800 -224 Left 2 !;tran 0 100u 20u
TEXT 616 0 Left 2 !k1 l1 l2 l3 l4 l5 .996
TEXT -800 -264 Left 2 !.ac dec 10000 10k 100e6
TEXT 656 104 Left 2 !.param L=33u
Jeff Liebermann
Jul 28, 2013, 1:37:01 PM7/28/13
to
On Sun, 28 Jul 2013 05:39:54 -0500, John S <Sop...@invalid.org>
wrote:
>It is a bridged amp configuration. The 23V is _peak_. Do not divide by
>two. That means your power calculation is 1/4 the actual.
Oops(tm). My mistake. I was looking at a single amplifier section
and missed the bridge amp which is further down the page.
<http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf>
Pg 6 thru 8 and 13.
>The transformer shown is configured as 2:3 ratio.
I was trying to eliminate the transformer.
>> I'm sure the model closely matches the data sheet and applications
>> schematic. However, that's not the problem. I'm questioning whether
>> it meets the OP's vague requirements. Methinks not.
>It does. Again, see page 16 of the data sheet.
The single LT1210 version does not. The bridged amp comes very close.
9 watts instead of 10. Your LTspice model shows the output -3dB down
at 10MHz. No word from the OP as to what waveforms he's going to be
feeding this amplifier but if it's a 10 MHz square wave, the amp will
need much more bandwidth.
Looks like Coiltronics no longer makes the VERSA-PAC CTX-01-13033-X2
xformer. I couldn't find a data sheet. Plenty of possible
substitutes, but I can't determine which one might work best:
<http://www.mouser.com/catalog/645/usd/1180.pdf>
<http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
I'm worried about the comment "Frequency range to over 1MHz".
At full output (9 watts), the voltage swing across the 50 ohm load
will be 21 V rms or 0.42 A rms. With a 2:3 turns ratio, that's
0.42 a rms * 1.5 = 0.63 a rms
Peak current in the xformer primary is:
0.63 * 1.414 = .89 A peak
Looking at the Mouser catalog page for the transfomer, anything with
more than about 1A peak saturation current should work.
In order to have gain at 0.1 KHz, the transformer primary reactance
must be high enough so that it doesn't reduce the load amplifier
impedance very much. Generally, that means 4 to 10 times the
amplifier output impedance (10 ohms) or about 50 ohms reactance.
XL = 50 = 2*Pi*f*L = 2.828 * 100*10^3 * L
L = 200 uH
Since there are two windings in the primary, that's 100 uH per
winding. Looking at the Mouser catalog page, none of the transformers
qualify with about 100 uH winding inductance and >1A Isat.
Assuming I haven't screwed up the math again (an all too common
occurrence), my guess(tm) is that LT simply ignored the saturation
current, which only affect amplifier linearity on power peaks, and
used whatever had an acceptable inductance. Even so, there's still
nothing on the Mouser page that has enough inductance and that will
handle the current.
Where did you find the 33uH winding inductance shown on the LTspice
model?
Thanks for doing the model. I'll play with it for a day or two and
see how it works.
wrote:
>It is a bridged amp configuration. The 23V is _peak_. Do not divide by
>two. That means your power calculation is 1/4 the actual.
and missed the bridge amp which is further down the page.
<http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf>
Pg 6 thru 8 and 13.
>The transformer shown is configured as 2:3 ratio.
>> I'm sure the model closely matches the data sheet and applications
>> schematic. However, that's not the problem. I'm questioning whether
>> it meets the OP's vague requirements. Methinks not.
>It does. Again, see page 16 of the data sheet.
9 watts instead of 10. Your LTspice model shows the output -3dB down
at 10MHz. No word from the OP as to what waveforms he's going to be
feeding this amplifier but if it's a 10 MHz square wave, the amp will
need much more bandwidth.
Looks like Coiltronics no longer makes the VERSA-PAC CTX-01-13033-X2
xformer. I couldn't find a data sheet. Plenty of possible
substitutes, but I can't determine which one might work best:
<http://www.mouser.com/catalog/645/usd/1180.pdf>
<http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
I'm worried about the comment "Frequency range to over 1MHz".
At full output (9 watts), the voltage swing across the 50 ohm load
will be 21 V rms or 0.42 A rms. With a 2:3 turns ratio, that's
0.42 a rms * 1.5 = 0.63 a rms
Peak current in the xformer primary is:
0.63 * 1.414 = .89 A peak
Looking at the Mouser catalog page for the transfomer, anything with
more than about 1A peak saturation current should work.
In order to have gain at 0.1 KHz, the transformer primary reactance
must be high enough so that it doesn't reduce the load amplifier
impedance very much. Generally, that means 4 to 10 times the
amplifier output impedance (10 ohms) or about 50 ohms reactance.
XL = 50 = 2*Pi*f*L = 2.828 * 100*10^3 * L
L = 200 uH
Since there are two windings in the primary, that's 100 uH per
winding. Looking at the Mouser catalog page, none of the transformers
qualify with about 100 uH winding inductance and >1A Isat.
Assuming I haven't screwed up the math again (an all too common
occurrence), my guess(tm) is that LT simply ignored the saturation
current, which only affect amplifier linearity on power peaks, and
used whatever had an acceptable inductance. Even so, there's still
nothing on the Mouser page that has enough inductance and that will
handle the current.
Where did you find the 33uH winding inductance shown on the LTspice
model?
Thanks for doing the model. I'll play with it for a day or two and
see how it works.
Jeff Liebermann
Jul 28, 2013, 3:05:11 PM7/28/13
to
On Sun, 28 Jul 2013 10:37:01 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
><http://www.mouser.com/catalog/645/usd/1180.pdf>
><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
(...)
VPH5-0155-R. That's 22.3 uH and Isat = 1.6 A.
k = (1 - Leak_Induct / Inductance)^0.5
k = (1 - 0.235/22.3)^0.5 = 0.946
At 100 KHz, 2 * 22.3 uH = 28 ohms which should be good enough with
only a slight drop in gain at 100KHz.
wrote:
><http://www.mouser.com/catalog/645/usd/1180.pdf>
><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
(...)
>Looking at the Mouser catalog page, none of the transformers
>qualify with about 100 uH winding inductance and >1A Isat.
The closest approximation on the Mouser and Cooper pages is the
>qualify with about 100 uH winding inductance and >1A Isat.
VPH5-0155-R. That's 22.3 uH and Isat = 1.6 A.
k = (1 - Leak_Induct / Inductance)^0.5
k = (1 - 0.235/22.3)^0.5 = 0.946
At 100 KHz, 2 * 22.3 uH = 28 ohms which should be good enough with
only a slight drop in gain at 100KHz.
Jeff Liebermann
Jul 28, 2013, 3:11:04 PM7/28/13
to
On Sun, 28 Jul 2013 12:05:11 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
>On Sun, 28 Jul 2013 10:37:01 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>><http://www.mouser.com/catalog/645/usd/1180.pdf>
>><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
>(...)
>>Looking at the Mouser catalog page, none of the transformers
>>qualify with about 100 uH winding inductance and >1A Isat.
>
>The closest approximation on the Mouser and Cooper pages is the
>VPH5-0155-R. That's 22.3 uH and Isat = 1.6 A.
> k = (1 - Leak_Induct / Inductance)^0.5
> k = (1 - 0.235/22.3)^0.5 = 0.946
Sorry, type. That should be k = 0.995
>On Sun, 28 Jul 2013 10:37:01 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>><http://www.mouser.com/catalog/645/usd/1180.pdf>
>><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
>(...)
>>Looking at the Mouser catalog page, none of the transformers
>>qualify with about 100 uH winding inductance and >1A Isat.
>
>The closest approximation on the Mouser and Cooper pages is the
>VPH5-0155-R. That's 22.3 uH and Isat = 1.6 A.
> k = (1 - Leak_Induct / Inductance)^0.5
> k = (1 - 0.235/22.3)^0.5 = 0.946
Jeff Liebermann
Jul 28, 2013, 4:03:56 PM7/28/13
to
Here's were I am now with the amplifier.
<http://802.11junk.com/jeffl/10-watt-amp/LT1210-Bridge-01.jpg>
<http://802.11junk.com/jeffl/10-watt-amp/LT1210-Bridge.asc>
I tweaked the transformer to resemble a Coilcraft VPH5-0155-R and
added the 6th winding.
I managed to expand the -3dB bandwidth to 50 KHz to 25 MHz. The way
it works is that C4 resonates with the primary winding inductances to
provide a parallel resonant peak at about 20 MHz.
After I do what I'm suppose to be doing today, instead of playing on
the computer, I'll see if I figure out how to model the linearity,
power handling, etc.
<http://802.11junk.com/jeffl/10-watt-amp/LT1210-Bridge-01.jpg>
<http://802.11junk.com/jeffl/10-watt-amp/LT1210-Bridge.asc>
I tweaked the transformer to resemble a Coilcraft VPH5-0155-R and
added the 6th winding.
I managed to expand the -3dB bandwidth to 50 KHz to 25 MHz. The way
it works is that C4 resonates with the primary winding inductances to
provide a parallel resonant peak at about 20 MHz.
After I do what I'm suppose to be doing today, instead of playing on
the computer, I'll see if I figure out how to model the linearity,
power handling, etc.
John S
Jul 28, 2013, 10:11:15 PM7/28/13
to
On 7/28/2013 12:37 PM, Jeff Liebermann wrote:
> On Sun, 28 Jul 2013 05:39:54 -0500, John S <Sop...@invalid.org>
> wrote:
>
>> It is a bridged amp configuration. The 23V is _peak_. Do not divide by
>> two. That means your power calculation is 1/4 the actual.
>
> Oops(tm). My mistake. I was looking at a single amplifier section
> and missed the bridge amp which is further down the page.
> <http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf>
> Pg 6 thru 8 and 13.
>
>> The transformer shown is configured as 2:3 ratio.
>
> I was trying to eliminate the transformer.
That is a noble effort.
> On Sun, 28 Jul 2013 05:39:54 -0500, John S <Sop...@invalid.org>
> wrote:
>
>> It is a bridged amp configuration. The 23V is _peak_. Do not divide by
>> two. That means your power calculation is 1/4 the actual.
>
> Oops(tm). My mistake. I was looking at a single amplifier section
> and missed the bridge amp which is further down the page.
> <http://cds.linear.com/docs/en/lt-journal/LTMag_V06N2_May96.pdf>
> Pg 6 thru 8 and 13.
>
>> The transformer shown is configured as 2:3 ratio.
>
> I was trying to eliminate the transformer.
>>> I'm sure the model closely matches the data sheet and applications
>>> schematic. However, that's not the problem. I'm questioning whether
>>> it meets the OP's vague requirements. Methinks not.
>
>> It does. Again, see page 16 of the data sheet.
>
> The single LT1210 version does not.
> The bridged amp comes very close.
> 9 watts instead of 10.
not ask for a complete design.
Your LTspice model shows the output -3dB down
> at 10MHz.
> No word from the OP as to what waveforms he's going to be
> feeding this amplifier but if it's a 10 MHz square wave, the amp will
> need much more bandwidth.
> Looks like Coiltronics no longer makes the VERSA-PAC CTX-01-13033-X2
> xformer. I couldn't find a data sheet. Plenty of possible
> substitutes, but I can't determine which one might work best:
> <http://www.mouser.com/catalog/645/usd/1180.pdf>
> <http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/Bus_Elx_DS_PM%204301_VERSA-PAC(VP1_2_3_4_5_VPH1_2_3_4_5).pdf>
> I'm worried about the comment "Frequency range to over 1MHz".
> At full output (9 watts), the voltage swing across the 50 ohm load
> will be 21 V rms or 0.42 A rms. With a 2:3 turns ratio, that's
> 0.42 a rms * 1.5 = 0.63 a rms
> Peak current in the xformer primary is:
> 0.63 * 1.414 = .89 A peak
> Looking at the Mouser catalog page for the transfomer, anything with
> more than about 1A peak saturation current should work.
>
> In order to have gain at 0.1 KHz, the transformer primary reactance
> must be high enough so that it doesn't reduce the load amplifier
> impedance very much. Generally, that means 4 to 10 times the
> amplifier output impedance (10 ohms) or about 50 ohms reactance.
> XL = 50 = 2*Pi*f*L = 2.828 * 100*10^3 * L
> L = 200 uH
> Since there are two windings in the primary, that's 100 uH per
> winding. Looking at the Mouser catalog page, none of the transformers
> qualify with about 100 uH winding inductance and >1A Isat.
incorrect at the start, it makes the rest of your analysis useless.
My simulation uses 33uH per winding. Did you notice? Two of those
(coupled) in series is 4x33 or about 132uH for the primary. Also, if you
actually read the paper you referenced, you will gain some insight into
their thinking about all this.
> Assuming I haven't screwed up the math again (an all too common
> occurrence), my guess(tm) is that LT simply ignored the saturation
> current, which only affect amplifier linearity on power peaks, and
> used whatever had an acceptable inductance.
However, I seriously doubt LT made as big a mistake as we ourselves make
here.
> Even so, there's still
> nothing on the Mouser page that has enough inductance and that will
> handle the current.
provided some indication that it is over your requirement.
> Where did you find the 33uH winding inductance shown on the LTspice
> model?
inductance to match the response curve supplied by LT. I had no other
way to deduce those without the data on the transformer that they used.
It was an interesting and educational exercise for me.
> Thanks for doing the model. I'll play with it for a day or two and
> see how it works.
George Herold
Jul 29, 2013, 8:26:34 AM7/29/13
to
On Saturday, July 27, 2013 12:37:26 AM UTC-4, Jeff Liebermann wrote:
> On Fri, 26 Jul 2013 10:37:02 -0700 (PDT), George Herold
>
> <ghe...@teachspin.com> wrote:
>
>
>
> >Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>
> >It looked something like those pictures.. not at all easy to do.
>
> >(Well at least it took me a while. I was learning RF at the same time.)
>
> We may have been competitors. I did an NMR driver and power amp
> system using then new VMOS devices. My guess(tm) was about 1979.
> Getting ultra low distortion of -60dB 3rd order IM at rated power gave
> me headaches and nightmares.
>
Certainly *not* competitors. If we'd had any competition, we'd have thrown in the towel, or not picked up the towel in the first place. Or just purchased your towel. :^)
> On Fri, 26 Jul 2013 10:37:02 -0700 (PDT), George Herold
>
> <ghe...@teachspin.com> wrote:
>
>
>
> >Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system.
>
> >It looked something like those pictures.. not at all easy to do.
>
> >(Well at least it took me a while. I was learning RF at the same time.)
>
> We may have been competitors. I did an NMR driver and power amp
> system using then new VMOS devices. My guess(tm) was about 1979.
> Getting ultra low distortion of -60dB 3rd order IM at rated power gave
> me headaches and nightmares.
>
>
> >> If OP wants 10 watts of distortion and
>
> >> intermod free output, which is likely if he's feeding it with a
>
> >> function generator, he's likely to get only about 15% amplifier
>
> >> efficiency with class A. That means he has to burn 67 watts of heat
>
> >> in order to get 10 watts of clean output.
>
>
>
> >Hmm is it that bad?
>
>
>
> It's always worse, never better.
>
>
>
> >(the efficiency) I was thinking about just
>
> >biasing the transistor at the mid current point. And then the
>
> >sine wave turns it on or off.. but the average power loss still
>
> >looks to be about 1/2.
>
> >So ideally 50% efficiency.. but of course it will be a bit worse
>
> >than that in practice. (I may be making some huge blunder here,
>
> >but then (hopefully) I'll be corrected and learn something.)
>
>
>
> The problem is linearity. If I setup the amplifier for best
>
> efficiency, I end up driving the output from rail to rail. That's
>
> fine for getting the maximum output from the device, but not so fine
>
> for getting the best linearity, lowest intermod, and minimum
>
> distortion. Generally, the output swing needs to stay away from the
>
> rails. A safe guess(tm) would be about 50% voltage swing, which is
>
> 1/4th of the maximum output power. For example, for 10 watts of clean
>
> output, I would need an amp capable of producing 40 watts. Of course,
>
> if one doesn't care about distortion, intermod, and such, a 10 watt
>
> amp would do just fine.
OK with the opamp->transistor configuration I haven't look too hard at distortion (inter mod or otherwise.) At low frequencies the negative feedback seems to cure a lot of the problems..(and you can run pretty close to the rails.).. of course once the frequency gets higher and the opamp looses gain it gets worse.
> >> If OP wants 10 watts of distortion and
>
> >> intermod free output, which is likely if he's feeding it with a
>
> >> function generator, he's likely to get only about 15% amplifier
>
> >> efficiency with class A. That means he has to burn 67 watts of heat
>
> >> in order to get 10 watts of clean output.
>
>
>
> >Hmm is it that bad?
>
>
>
> It's always worse, never better.
>
>
>
> >(the efficiency) I was thinking about just
>
> >biasing the transistor at the mid current point. And then the
>
> >sine wave turns it on or off.. but the average power loss still
>
> >looks to be about 1/2.
>
> >So ideally 50% efficiency.. but of course it will be a bit worse
>
> >than that in practice. (I may be making some huge blunder here,
>
> >but then (hopefully) I'll be corrected and learn something.)
>
>
>
> The problem is linearity. If I setup the amplifier for best
>
> efficiency, I end up driving the output from rail to rail. That's
>
> fine for getting the maximum output from the device, but not so fine
>
> for getting the best linearity, lowest intermod, and minimum
>
> distortion. Generally, the output swing needs to stay away from the
>
> rails. A safe guess(tm) would be about 50% voltage swing, which is
>
> 1/4th of the maximum output power. For example, for 10 watts of clean
>
> output, I would need an amp capable of producing 40 watts. Of course,
>
> if one doesn't care about distortion, intermod, and such, a 10 watt
>
> amp would do just fine.
George H.
timt...@hotmail.co.uk
Nov 4, 2017, 7:10:22 AM11/4/17
to
I tried VPH5-0155-R as an impedance matching transformer for an RF amp with a 50R output into a 300R load at 16MHz. It didn't work.
I suspect this transformer is not much use above 1MHz (as specified in the data sheet).
Is there a defect in the model or am I doing something wrong?
I'm an experienced engineer but a novice at RF.
Regards
Tim.
I suspect this transformer is not much use above 1MHz (as specified in the data sheet).
Is there a defect in the model or am I doing something wrong?
I'm an experienced engineer but a novice at RF.
Regards
Tim.
Jeff Liebermann
Nov 4, 2017, 1:39:25 PM11/4/17
to
On Sat, 4 Nov 2017 04:10:18 -0700 (PDT), timt...@hotmail.co.uk
wrote:
That transformer is made for a switching power supply, not a broadband
transformer. The data sheet on the VPH5-0155-R shows that it's only
good up to about 500KHz. See graphs on Pg 8.
<http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/BUS_Elx_PM_4301_VERSA_PAC.pdf>
The description also includes:
"Frequency range to over 1 MHz"
100khz to 10Mhz is about 6.5 octaves of frequency range. Getting that
kind of bandwidth is not easy, especially if you're doing high RF
power, low losses, and a flat frequency response. This might help:
"Designing Wideband RF Impedance Transformers"
<http://www.mwrf.com/components/designing-wideband-rf-impedance-transformers>
or Google for "wideband RF transfomer design".
Good luck.
wrote:
transformer. The data sheet on the VPH5-0155-R shows that it's only
good up to about 500KHz. See graphs on Pg 8.
<http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/BUS_Elx_PM_4301_VERSA_PAC.pdf>
The description also includes:
"Frequency range to over 1 MHz"
100khz to 10Mhz is about 6.5 octaves of frequency range. Getting that
kind of bandwidth is not easy, especially if you're doing high RF
power, low losses, and a flat frequency response. This might help:
"Designing Wideband RF Impedance Transformers"
<http://www.mwrf.com/components/designing-wideband-rf-impedance-transformers>
or Google for "wideband RF transfomer design".
Good luck.
John Larkin
Nov 4, 2017, 2:08:53 PM11/4/17
to
On Sat, 04 Nov 2017 10:39:21 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
>On Sat, 4 Nov 2017 04:10:18 -0700 (PDT), timt...@hotmail.co.uk
>wrote:
>
>>I tried VPH5-0155-R as an impedance matching transformer for an RF amp with a 50R output into a 300R load at 16MHz. It didn't work.
>>I suspect this transformer is not much use above 1MHz (as specified in the data sheet).
>>Is there a defect in the model or am I doing something wrong?
>>I'm an experienced engineer but a novice at RF.
>
>That transformer is made for a switching power supply, not a broadband
>transformer. The data sheet on the VPH5-0155-R shows that it's only
>good up to about 500KHz. See graphs on Pg 8.
><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/BUS_Elx_PM_4301_VERSA_PAC.pdf>
>The description also includes:
> "Frequency range to over 1 MHz"
>
>100khz to 10Mhz is about 6.5 octaves of frequency range. Getting that
>kind of bandwidth is not easy, especially if you're doing high RF
>power, low losses, and a flat frequency response. This might help:
>
>"Designing Wideband RF Impedance Transformers"
><http://www.mwrf.com/components/designing-wideband-rf-impedance-transformers>
>
>or Google for "wideband RF transfomer design".
>
>Good luck.
I wind transmission-line transformers with a 1000:1 frequency range.
It's really not difficult.
Sevick's book is OK up to 16 MHz. I don't trust his high frequency
measurements.
At a fixed frequency, it's easier to use a conventional LC tuned
matching network.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
wrote:
>On Sat, 4 Nov 2017 04:10:18 -0700 (PDT), timt...@hotmail.co.uk
>wrote:
>
>>I tried VPH5-0155-R as an impedance matching transformer for an RF amp with a 50R output into a 300R load at 16MHz. It didn't work.
>>I suspect this transformer is not much use above 1MHz (as specified in the data sheet).
>>Is there a defect in the model or am I doing something wrong?
>>I'm an experienced engineer but a novice at RF.
>
>That transformer is made for a switching power supply, not a broadband
>transformer. The data sheet on the VPH5-0155-R shows that it's only
>good up to about 500KHz. See graphs on Pg 8.
><http://www.cooperindustries.com/content/dam/public/bussmann/Electronics/Resources/product-datasheets/BUS_Elx_PM_4301_VERSA_PAC.pdf>
>The description also includes:
> "Frequency range to over 1 MHz"
>
>100khz to 10Mhz is about 6.5 octaves of frequency range. Getting that
>kind of bandwidth is not easy, especially if you're doing high RF
>power, low losses, and a flat frequency response. This might help:
>
>"Designing Wideband RF Impedance Transformers"
><http://www.mwrf.com/components/designing-wideband-rf-impedance-transformers>
>
>or Google for "wideband RF transfomer design".
>
>Good luck.
It's really not difficult.
Sevick's book is OK up to 16 MHz. I don't trust his high frequency
measurements.
At a fixed frequency, it's easier to use a conventional LC tuned
matching network.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
timt...@hotmail.co.uk
Nov 11, 2017, 12:40:14 PM11/11/17
to
I'll have a stab at winding my own transformer on a T200-2 core.
I'll try 5 turns with the input across turns 2-4 to give a 2:5 turns ratio.
Is this likely to work?
The RF amp I'm using goes from 5-30MHz so not a huge range to match across.
I'm guessing I don't need to use transmission line winding techniques to cover this small frequency range?
Kind regards,
Tim Tanner.
Tim Williams
Nov 12, 2017, 12:06:47 AM11/12/17
to
<timt...@hotmail.co.uk> wrote in message
news:f5058430-e771-4556...@googlegroups.com...
How much magnetizing inductance will 5 turns yield? How much inductance can
you tolerate?
Ferrite is fantastic for transformers.
Note the semantic distinction between transformers (converts voltage/current
by ratio with a minimum added current draw) and inductors (stores energy as
magnetizing current, may have multiple windings).
You want powdered iron, and gapped ferrite, for inductors, and ungapped
ferrite for transformers.
If you need matching reactance as well as transformation, powder cores are
good. But then, you'll only be getting a match over a relatively narrow
frequency range, if that's the case.
Tim
--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Contract Design
Website: https://www.seventransistorlabs.com/
news:f5058430-e771-4556...@googlegroups.com...
> Many thanks for the advice John.
> I'll have a stab at winding my own transformer on a T200-2 core.
> I'll try 5 turns with the input across turns 2-4 to give a 2:5 turns
> ratio.
> Is this likely to work?
> The RF amp I'm using goes from 5-30MHz so not a huge range to match
> across.
> I'm guessing I don't need to use transmission line winding techniques to
> cover this small frequency range?
>
Doesn't sound like a very good combination. What's the A_L on that core?
> I'll have a stab at winding my own transformer on a T200-2 core.
> I'll try 5 turns with the input across turns 2-4 to give a 2:5 turns
> ratio.
> Is this likely to work?
> The RF amp I'm using goes from 5-30MHz so not a huge range to match
> across.
> I'm guessing I don't need to use transmission line winding techniques to
> cover this small frequency range?
>
How much magnetizing inductance will 5 turns yield? How much inductance can
you tolerate?
Ferrite is fantastic for transformers.
Note the semantic distinction between transformers (converts voltage/current
by ratio with a minimum added current draw) and inductors (stores energy as
magnetizing current, may have multiple windings).
You want powdered iron, and gapped ferrite, for inductors, and ungapped
ferrite for transformers.
If you need matching reactance as well as transformation, powder cores are
good. But then, you'll only be getting a match over a relatively narrow
frequency range, if that's the case.
Tim
--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Contract Design
Website: https://www.seventransistorlabs.com/
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