any easy way to stepdown voltage from 12v dc to 9 v dc?
Mike Maas
Hi,
I've got a problem that has cost me several plants this summer. I
have a piece of land that has no grid power. I am powering a 24v ac
sprinkler controller through a voltage converter that is turned on
with a dc timer. While there is no ac power to the timer I depend on
the 9v battery backup to preserve the programming. Problem is that
the 9v batteries don't last long enough. The last battery I tried, a
nickel hydride that was supposed to last 6x the time of an alkaline
only lasted two weeks. I am considering putting a dozen or so
alkaline batteries in a box connected to one another but if I can put
together a good stepdown converter I could let the backup run off a
battery that will be recharged daily by the solar system.
So, is there any easy way to step voltage down to 9v dc from 12v dc?
Thanks for any advice.
Mike Maas
Nick Hull
The easy way is to get a LM317 adjustable voltage regulator. With 2
resisters you can control any voltage down to 1.2 out, 37 max in. With
one a pot, it is adjustable. I usually get 317's for $1, radio shack is
more but they have schematic & instructions on the bubble pAck.
--
nick...@usit.net
Free men own guns-slaves don't
Shaw Moldauer
: I've got a problem that has cost me several plants this summer. I
: have a piece of land that has no grid power. I am powering a 24v ac
: sprinkler controller through a voltage converter that is turned on
: with a dc timer. While there is no ac power to the timer I depend on
: the 9v battery backup to preserve the programming. Problem is that
: the 9v batteries don't last long enough. The last battery I tried, a
: nickel hydride that was supposed to last 6x the time of an alkaline
: only lasted two weeks. I am considering putting a dozen or so
: alkaline batteries in a box connected to one another but if I can put
: together a good stepdown converter I could let the backup run off a
: battery that will be recharged daily by the solar system.
: So, is there any easy way to step voltage down to 9v dc from 12v dc?
: Thanks for any advice.
: Mike Maas
I have used the LM317 method and it works great, but wastes a small amount
of power. Another idea is to get a 9V solar battery charger from someone
like Edmonds Scientific.
-Shaw
Duane C. Johnson
Hi Mike;
> like Edmond Scientific.
>
> -Shaw
I have found that most things that run on 9 volts also run on 12 volt
battery power. This means 14.6 volts as a lead acid battery at full
charge and still charging lightly will be this voltage.
If you still don't trust this then make a voltage dropper using
about 8 to 10 1 amp rectifier diodes in series. These diodes don't
regulate preceisly but the timer surely dosn't need much regulation.
I have used these voltage droppers many times with good results.
It is a low tech solution to the problem.
--
CUL8ER
Stupid is Forever
Ignorance can be Fixed
Duane C. Johnson
Ziggy
WA0VBE
Red Rock Energy
1825 Florence St.
White Bear Lake, MN, USA 55110-3364
(612)635-5065 w
(612)426-4766 h
red...@pclink.com
dc...@PO8.RV.unisys.com
http://www.geocities.com/SiliconValley/3027/
un...@iap.net.au
mas...@dnai.com (Mike Maas) wrote:
>Hi,
>I've got a problem that has cost me several plants this summer. I
>have a piece of land that has no grid power. I am powering a 24v ac
>sprinkler controller through a voltage converter that is turned on
>with a dc timer. While there is no ac power to the timer I depend on
>the 9v battery backup to preserve the programming. Problem is that
>the 9v batteries don't last long enough. The last battery I tried, a
>nickel hydride that was supposed to last 6x the time of an alkaline
>only lasted two weeks. I am considering putting a dozen or so
>alkaline batteries in a box connected to one another but if I can put
>together a good stepdown converter I could let the backup run off a
>battery that will be recharged daily by the solar system.
>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>Thanks for any advice.
>Mike Maas
Seems a bit funny that the battery would run down
that quick. Did you put an ammeter in the line to
see what current it's taking? An extraordinary
amount may indicate something amiss.
Easiest way to drop the voltage down from 12v
would be to use either a 7809 (up to 1 amp w/
heatsink) or any TO-3 transistor as a pass regulator.
(Suggest PNP configuration, such as a 2N3054,
but probably any xstr in TO-3 pkg would work.)
Uncle Brian VK6BQN
John Lundgren
un...@iap.net.au wrote:
: mas...@dnai.com (Mike Maas) wrote:
: >Hi,
: >Thanks for any advice.
: >Mike Maas
: Uncle Brian VK6BQN
Hey, unk! The 2N3054 is NPN, just like the 2N3055 but at 1 amp.
Stick with the LM317, it's adjustable. I agree tjat the battery should
last longer than that, but then I don't know what the circuit is. But I
would not try to parallel 9V bats. I would put 6 D cells in series.
You can also buy solar cells that put out a few hundred mA for $20 or $40
or something like that. It depends on the area or size.
--
#===================================================================#
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#======P=G=P==k=e=y==a=v=a=i=l=a=b=l=e==u=p=o=n==r=e=q=u=e=s=t======#
Aaron Lung
>In article <masimoDw...@netcom.com>, mas...@dnai.com (Mike Maas)
>wrote:
>>
[..snip...]
>> So, is there any easy way to step voltage down to 9v dc from 12v dc?
>>
The easiest is to wire up 4 or 5 silicon diodes in series. Rectifier
types with sufficient current ratings should work just fine. I assume
your min. current draw is enough to forward bias the guys to get approx
.6V-.7V drop across each. That'll bring you down to just over 9V.
Another alternative is to stick a linear regulator there. An LM7809
type 9V regulator will work just fine. It's a 1 part solution, and
it's good for up to an amp or so. I'm not sure if Radio Shack still
sells these things or not...at least they used to. In either case, it
shouldn't cost more than a buck.
aaron
Harry H Conover
Mike Maas (mas...@dnai.com) wrote:
: Hi,
:
: I've got a problem that has cost me several plants this summer. I
: have a piece of land that has no grid power. I am powering a 24v ac
: sprinkler controller through a voltage converter that is turned on
: with a dc timer. While there is no ac power to the timer I depend on
: the 9v battery backup to preserve the programming. Problem is that
: the 9v batteries don't last long enough. The last battery I tried, a
: nickel hydride that was supposed to last 6x the time of an alkaline
: only lasted two weeks. I am considering putting a dozen or so
: alkaline batteries in a box connected to one another but if I can put
: together a good stepdown converter I could let the backup run off a
: battery that will be recharged daily by the solar system.
:
:
:
: Mike Maas
:
The easiest way I can think of is simply to put either a string of
silicon diodes (totaling a voltage drop of 3V) or a 3V Zener Diode
in series with the supply.
Either method will give you a relatively stable and current independet
voltage drop, and likely will be more than adequate for your application.
Harry C.
Shaw Moldauer
: Even simpler: Use a _voltage regulator_. What a concept!
I think the common theme was to find a solution which was easily
implemented. 7809 regulators are not a common item in many places. If they
can be found in an appropriate package for the current load, great. I
still contend they will waste some power, as will the diode solution.
A TO3 is difficult for most people to mount. A TO220 is easier, but a TO39
or TO42 package may not be able to dissapate the power for the specific
load. I contend that a seperate solor 9V battery charger requires the
least design. The solution of just using 12V instead of the 9V also has
merit.
: You then screw the heatsink tab to some ground voltage metal (the
: timer itself?) to help cool it off, so's it'll last next-to forever.
I would preface this with "for a TO220............" Don't try
this with a 7909 though............
: Three connections! And probably less than a dollar for the 7809,
If you can find it. Does Radio Shack sell anything for less than a
dollar any more? (except maybe the four D cell flashlight, which they give
away).
: Paul.
: (Yes, I do this for a living...)
So do most of the rest of us. And we make a fine living doing it.
Solutions to problems are a very subjective thing, luckily, we all can
usually understand each others solutions.
-Shaw
un...@iap.net.au
>Hey, unk! The 2N3054 is NPN, just like the 2N3055 but at 1 amp.
Oooosht, I did flkt up!
Unc.
Jim Barr
In article <masimoDw...@netcom.com>, Mike Maas <mas...@dnai.com>
writes
>Hi,
>
>I've got a problem that has cost me several plants this summer. I
>have a piece of land that has no grid power. I am powering a 24v ac
>sprinkler controller through a voltage converter that is turned on
>with a dc timer. While there is no ac power to the timer I depend on
>the 9v battery backup to preserve the programming. Problem is that
>the 9v batteries don't last long enough. The last battery I tried, a
>nickel hydride that was supposed to last 6x the time of an alkaline
>only lasted two weeks. I am considering putting a dozen or so
>alkaline batteries in a box connected to one another but if I can put
>together a good stepdown converter I could let the backup run off a
>battery that will be recharged daily by the solar system.
>
>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>
>Thanks for any advice.
>
>Mike Maas
>
Hi Mike,
Most of the solutions to yor problem have vied for acceptance on the
basis of simplicity or cost, and I just want to make sure we are all
addressing the *problem* before putting in my $2cents.
I have a couple of questions
1. What is the mark space ratio of the timer?
2. Is it possible to use a different type of feeder that could
deliver in 24 hours, a continuous flow (no timer) that amounted to the
same *total* flow that was acomplished via the timed sprinkler during
the same period?
I hope the questions made sense!
I have to say that some of the suggested solutions have elegance.
Best wishes, Jim
Jim Barr Machine Conversation, Bedfordshire England
Best is the enemy of good enough
Leaves Rustle....Blades turn..... Water moves
Paul Renault
"Duane C. Johnson" <red...@pclink.com> wrote:
>Hi Mike;
>Shaw Moldauer wrote:
>>
>> Mike Maas (mas...@dnai.com) wrote:
>>
...snip, snip...
>> : together a good stepdown converter I could let the backup run
off a
>> : battery that will be recharged daily by the solar system.
>>
>> : So, is there any easy way to step voltage down to 9v dc from 12v dc?
>>
>> : Thanks for any advice.
>>
>> : Mike Maas
>>
>> I have used the LM317 method and it works great, but wastes a small amount
>> of power. Another idea is to get a 9V solar battery charger from someone
>> like Edmond Scientific.
>>
>> -Shaw
>I have found that most things that run on 9 volts also run on 12 volt
>battery power. This means 14.6 volts as a lead acid battery at full
>charge and still charging lightly will be this voltage.
>If you still don't trust this then make a voltage dropper using
>about 8 to 10 1 amp rectifier diodes in series. These diodes don't
>regulate preceisly but the timer surely dosn't need much regulation.
>I have used these voltage droppers many times with good results.
>It is a low tech solution to the problem.
>--
>CUL8ER
Even simpler: Use a _voltage regulator_. What a concept!
Depending on how you wish to wire it up, you can use a 7809 regulator,
or a 7909 regulator. The 7809 is a +12VDC regulator, the 7909 is a
-12VDC regulator. (78XX pos. regulator, 79XX neg. regulator...)
For the purposes of this explanation, I'll use the 7809:
It comes in a TO-220 package, a little plastic box with three legs on
the bottom, and a heatsink tab with a convenient screw hole on top.
When you're facing the regulator (so you can read the stuff printed on
the plastic part). the legs are, from left to right, INPUT, GROUND,
OUTPUT. You connect your +12DC to the INPUT leg, the return (the
negative of both your 12V power supply and your 9V circuit) to the
GROUND leg, and the OUTPUT leg, obviously goes to the +9VDC of your
timer circuit.
You then screw the heatsink tab to some ground voltage metal (the
timer itself?) to help cool it off, so's it'll last next-to forever.
Three connections! And probably less than a dollar for the 7809,
Paul.
Fowler_Barry
Why not just place a 3 volt Zener diode in series with the 12 volt supply?
Andy Hill
It looks like you've had quite a few answers to your question. However, why not
take another approach: find a timer that doesn't suck down so much power. My
sprinkler controller (Nelson brand) is battery powered, and has run for nearly
four years on the same batteries.
Duane C. Johnson
Fowler_Barry wrote:
>
> Why not just place a 3 volt Zener diode in series with the 12 volt supply?
Strictly speaking this is true. A zener will do the job.
First the zener if used I would use a 4.7 volt one as the "12 volt"
battery is really more like 13.6 volts and sometimes more when
charging.
I usually use the seriesed rectifiers becouse they are more robust
than the zeners and allow voltage adjustment by changing the tap.
John J. McGarvey
Mike Maas wrote:
>
> Hi,
>
> I've got a problem that has cost me several plants this summer. I
> have a piece of land that has no grid power. I am powering a 24v ac
> sprinkler controller through a voltage converter that is turned on
> with a dc timer. While there is no ac power to the timer I depend on
> the 9v battery backup to preserve the programming. Problem is that
> the 9v batteries don't last long enough. The last battery I tried, a
> nickel hydride that was supposed to last 6x the time of an alkaline
> only lasted two weeks. I am considering putting a dozen or so
> alkaline batteries in a box connected to one another but if I can put
> battery that will be recharged daily by the solar system.
>
> So, is there any easy way to step voltage down to 9v dc from 12v dc?
>
> Thanks for any advice.
>
> Mike Maas
Hi Mike,
You can drop the voltage a couple of ways. The cheapest and easiest is to put a few
forward biased silicon diodes in series with the 12 volt battery. A typical silicon
diode has a voltage drop of 0.7 Volts. So to drop ~3 volts use 4 or 5 silicon diodes.
The type of diodes you want are 1 amp, 50 - 100 PIV rectifier diodes. Try Radio Shack
for a supply.
To get a more accurate voltage you can use a voltage requlator chip. Unfortunetly, 9
Volts is not a common value so you would have to construct one fron a zener diode and
pass transistor. This is difficult unless you have some electronics background. Also
simple series regulators waste some power even when they are not loaded.
Note also that battery voltages are not as accurate as you might think. A cheap 9V
battery may be 11 Volts when new. A 12 Volt battery may run 14 Volts when well charged.
Another way around your problem is to optain 9 Volts from larger alkaline cells. It's
better to put 6 - AA cells in series then to parallel 6 - 9 Volts batteries. Radio
Shack sells the snap connectors for 9 Volt batteries (be careful with polarity) and
battery holders for AA, C and D cells. Wire 6 - Alkaline cells in series in a box and
you can get a replacement for a 9 Volt battery with a lot more capacity.
Good Luck,
Big John
Mungo Henning
Can someone help me out with the calculations here... would it not
be more "efficient" to chain a series of silicon diodes together where
each diode drops 0.7 volts (or so)?
I guess there's got to be a reason against this, so enlighten me please? :-)
Mungo
--
Mungo Henning - it's a daft name but it goes with the face...
mun...@itacs.strath.ac.uk
Steve Work
Jim Barr (Jim...@wandana.demon.co.uk) wrote:
: 2. Is it possible to use a different type of feeder that could
: deliver in 24 hours, a continuous flow (no timer) that amounted to the
: same *total* flow that was acomplished via the timed sprinkler during
: the same period?
This would not really be practical. Most of the solenoids on sprinkler
systems are just on/off, they can't reliably restrict the water flow
partially. Also, if you did have reduced water flow, the sprinkler heads
themselves wouldn't work properly. They wouldn't pop up or rotate, and
they would just dribble the water out nearby without spreading it out.
Steve Work
Fowler_Barry (ba...@rose.hp.com) wrote:
: Why not just place a 3 volt Zener diode in series with the 12 volt supply?
Better yet, four or five regular diodes, forward biased, in series. Each
one will drop about 0.7 volts. These will be easier to obtain and will
better handle high currents.
Tu and Bob Myers
Radio Shack has voltage converters to use less than 12vdc appliances in
cars. I think one of the outputs is 9vdc.
Bob
of Tu and Bob
Fowler_Barry
One more suggestion. I have some land and I have automatic drip irrigation
on a small orchard and a garden. I use battery-powered hose bib valves
to do the job. That way, I don't have to string wires (under cement
walkways and a driveway). I bought the valves at a local (Home Depot)
home supply store. They're going two years on the same batteries (AA cells).
Beats messing with photo cells and wire.
Mike Maas
mas...@dnai.com (Mike Maas) wrote:
>Hi,
snip...
Wow, ain't the internet great! Thanks for all the fine suggestions.
To answer a few of the questions. This is the third controller I tried
but the first one with the 9v battery problem. The others had other
problems. Last year I used individual timers on spigots for each
irrigation line but this year decided to step up the level of
sophistication a little -- considering all the problems it probably
wasn't such a good idea. Incidentally last year I used a gasoline
powered pump and had to go up at a minimum of every two weeks to fill
the tank. This year I'm running a 12v Shurflo pump with a pressure
switch and power supplied by solar panels. Getting that to work right
was another adventure.
As to the plants -- they are various ornamental trees. I plan to build
at the site eventually and am trying to get some landscaping and
gardening done first.
Strangely enough there is a power pole on the property but it is
$3,000+ worth of trenching away from where I need the power right now
and I'm not quite ready to have that work done.
Thanks to all of you for the responses. I'm not sure if I'll simply
go to larger cells and hook them up to create a 9 volt source or if
I'll hook the clock up to a big 12v wet cell with a circuit to reduce
the voltage and keep the wet cell charged off the solar panels.
Thanks again,
Mike
F.P.Harrison
> > I have used the LM317 method and it works great, but wastes a small amount
> > of power. Another idea is to get a 9V solar battery charger from someone
> > like Edmond Scientific.
> >
as you have to have at least three volts drop across the regulator for
it to remain in regulation. Chances are your 12v is > 12v anyway.
> If you still don't trust this then make a voltage dropper using
> about 8 to 10 1 amp rectifier diodes in series. These diodes don't
> regulate preceisly but the timer surely dosn't need much regulation.
>
> I have used these voltage droppers many times with good results.
> It is a low tech solution to the problem.
>
are all forward biased in this application.
Frank
Don Parker
> Mike Maas (mas...@dnai.com) wrote:
> : Hi,
>
> : I've got a problem that has cost me several plants this summer. I
> : have a piece of land that has no grid power. I am powering a 24v ac
> : sprinkler controller through a voltage converter that is turned on
Since there's already been a whole bunch of excellant solutions suggested
for his problem, I was just wonderin - what kind of plants you got out
there in the boonies that need all this TLC, Mikey??????? 0 ;))
Don
Greg Ebert
In article 9...@netcom.com, mas...@dnai.com (Mike Maas) writes:
>Hi,
>
>I've got a problem that has cost me several plants this summer. I
>have a piece of land that has no grid power.
>
What kind of plants ?? Maybe I shouldn't ask......
[...]
>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>
How about 4 series-connected diodes ? They drop about 0.7 volts each and are
very cheap.....
altavoz
The LM317 needs more than 3 vdc to operate. See above subject..
"Re: any easy way to stepdown voltage from12v to 9 v dc ?
Since you're using 9 v alkaline batts , you can use a 9v low drop out
regulator . But be careful of the minimum load ! you may need to put
a resistor from reg' out to ground to maintain this min'.
It sounds like the load is very small ( 9v lasts 2 weeks ) so you may
try a very much less expensive "shunt" regulator , which is just 2
$.05 transistors and a resistor ( 200 ohm). measure the current draw
if its < .01 amps , then use the shunt reg'
BTW 1 transistor IS the zener ( reverse base to emmitter).
______End of text from alt...@azstarnet.com___________
altavoz
Mike Maas wrote:
> Hi,
> I've got a problem that has cost me several plants this summer. I
> sprinkler controller through a voltage converter that is turned on
> the 9v battery backup to preserve the programming. Problem is that
> the 9v batteries don't last long enough. The last battery I tried, a
> nickel hydride that was supposed to last 6x the time of an alkaline
> only lasted two weeks. I am considering putting a dozen or so
> alkaline batteries in a box connected to one another but if I can put
> together a good stepdown converter I could let the backup run off a
> battery that will be recharged daily by the solar system.
> Mike Maas++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
PPPPPPLLLLEEEEAAAASSSSSEEEEEE let this guy get back to raising
his marijauna !
1) The zener idea will work but zeners at 3 v are very soft.
2) Lm317 will likely run out of voltage room ( needs 3.25v min, it
will still work, but it will surprize you with its volt error)
3) diode strings work but are very soft.
4) 7809 is probably the worst idea ( no volt room, high cost,Iq in mil amps
5) my idea is the cheapest but its probably no good cause he may not be
able to do electronics . A shunt reg using 1 xistor as a zener and the
other as a super diode.
Here's the results of the contest :
Barry Fowler zener
Paul Renault 7809
Shaw LM317
Uncle Brian 7809
John Lundgren 7809
Aaron Lung 7809 , diodes
Harry Conover zener,diodes
John McGarvey diodes or batt
Greg Ebert diodes
One thing you probably didnt notice is that this is one of the longest
useful , productive threads in all news groups ( most long threads are
nothing more than FLAME ! ha ha ha ha ha ha ha )
Joseph & Anna Orgeron
mas...@dnai.com (Mike Maas) wrote:
>Hi,
>I've got a problem that has cost me several plants this summer. I
>have a piece of land that has no grid power. I am powering a 24v ac
>sprinkler controller through a voltage converter that is turned on
>with a dc timer. While there is no ac power to the timer I depend on
>the 9v battery backup to preserve the programming. Problem is that
>the 9v batteries don't last long enough. The last battery I tried, a
>nickel hydride that was supposed to last 6x the time of an alkaline
>only lasted two weeks. I am considering putting a dozen or so
>alkaline batteries in a box connected to one another but if I can put
>together a good stepdown converter I could let the backup run off a
>battery that will be recharged daily by the solar system.
>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>Thanks for any advice.
>Mike Maas
It sounds like your timer is powered by the 9 volt battery. What you
needf is a timer that uses the 9 volt battery as a back up for its
programing. If that is what your timer is suspose to have on it. then
you are having lost of power. Check with the manufacture.
Joe
Joseph & Anna Orgeron
Anna,
USAF Retired, Wildlife Rehabber, Hunter, Housewife
Joe
USAF Retired, NRA Life Member, Hunter, HVAC Tech.
"E" Mail Address: e7...@tesser.com
"I am a half-bleed Cajun, if I was full-bleed, I could not stood Myself."
Justan Wilson
altavoz
LM431 ! If you can get it for $.35 like i did . An adjustable
shunt regulator 2.5v to 40 v , 3 resistors .
Do i win the prize ? huh huh ? do i win the prize ?
Harry H Conover
Kirk Kerekes (kke...@iamerica.net) wrote:
: For the ultimate cheapness-- measure the current flow out of the 9VDC
: batteries when in use (an inexpensive DVM ought to do the trick). Compute
: the resistor necessary to obtain a 3V drop at that current [R=
: 3000/current(in mA)].
:
: Select a potentiometer (variable resistor) that has this value in the
: middle of its range. Use the DVM to set the pot to about 20% higher than
: this value, and wire it in series with your 12VDC source. Measure the
: voltage on the 9VDC side, and adjust the pot for 9.0 volts.
:
: Of course, if you have to buy the meter, this ceases to be cheap...
:
Of course, this method is only good for a fixed load.
Harry C.
altavoz
They would work but poor regulation , and a zener in that
range is also very soft, and if you can get surplus like
i do , an LM431 3 term is the neatet cause it regulates
very well as a shunt reg , it does something like .1 amps.
He said he was using a transistor batt , so LM431 would
be perfect ( especially since it's adjustable )
Duane C. Johnson
Of course this is the best solution to use in this application.
1. There is no continuous power wasted in a regulator. In fact
there is preceisly the correct power disipation in the
diode drops. All the current delivered to the timer passes
through the diode string.
2. The timer is very tolerent of variations in voltage as
it was designed to run on a 9v battery which can vary
from 7.5v to as high as as 11v.
3. The source will be a 12v lead acid battery which can vary
from 11v to as high as 14v.
4. The timer may draw several hundred milleamps when it switches
power in the externam circuit.
5. A bunch of 1 Amp diodes in series can be adjustable if a small
alegater clip is used. This way if one setting is wrong you can
change it.
6. 1 Amp diodes are less fragile than the zener diodes. With the
exception of shorting them across the battery terminals they
are indistructable.
7. BTW diodes don't have a set voltage drop, such as .7v . Their
voltage is basicly dependent on their temperature and the
current level. Generally you will get
.7v-.8v when the current in at the rated value.
.6v-.7v at 1/10 rated current.
.5v-.6v at 1/100 rated current. And so on.
I would sugest the use of 8 to 10 diodes in the string. I just checked
Mouser for a currennt priceand found 1N4002 diodes $.07 . This would
push the total cost 10 less $.70 for the string. Pretty cheap huh.
Mouser part# = 583-1N4002 1amp 200 volts
BTW you can get 100 for $5.40 in case you have other projects that
need 1 amp rectifiers. This is one of those staple parts one should
have on hand anyway.
Mouser Electronics
1(800)346-6873
sa...@mouser.com
They take plastic.
No minimum order except for export it is $100 minimum.
This is of the subject but I use diodes often as temperature sensors.
When the current is kept constant, say at 1 mamp, the voltage drop
varies inversly with temperature. One big plus is the voltage vs
temperature is linear, to less than 1% error, over a wide temperature
range of -40C to 175C, -40F to 347F. I generally use 1 amp rectifiers
diodes but where the higest of temperatures are needed you should use
glass encapsulated signal diodes.
Duane C. Johnson
Hi Mungo;
Hey do you like beans? Oh that was a movie not the internet.
Mungo Henning wrote:
>
> Can someone help me out with the calculations here... would it not
> be more "efficient" to chain a series of silicon diodes together where
> each diode drops 0.7 volts (or so)?
> I guess there's got to be a reason against this, so enlighten me please? :-)
> Mungo
> --
> Mungo Henning - it's a daft name but it goes with the face...
> mun...@itacs.strath.ac.uk
I apologize for this multiple post but somehow I lost mungo's question
in the first post. Sorry.
Harry H Conover
Duane C. Johnson (red...@pclink.com) wrote:
: Of course this is the best solution to use in this application.
: 1. There is no continuous power wasted in a regulator. In fact
: there is preceisly the correct power disipation in the
: diode drops. All the current delivered to the timer passes
: through the diode string.
What? Back to the EE course for you! The wasted power is exactly
equal to the voltage drop times the currnent. It doesn't matter if
this power is wasted in a series regulator, a string of forward
biased diodes, or a Zener diode. The bottom line is identical.
: 5. A bunch of 1 Amp diodes in series can be adjustable if a small
: alegater clip is used. This way if one setting is wrong you can
: change it.
Or you can figure it out in advance -- a simple calculation.
: 6. 1 Amp diodes are less fragile than the zener diodes. With the
: exception of shorting them across the battery terminals they
: are indistructable.
True, but they have a somewhat more stable, predictable, and documented
voltage drops than does a string of forward biased silicon diodes.
: 7. BTW diodes don't have a set voltage drop, such as .7v . Their
: voltage is basicly dependent on their temperature and the
: current level. Generally you will get
: .7v-.8v when the current in at the rated value.
: .6v-.7v at 1/10 rated current.
: .5v-.6v at 1/100 rated current. And so on.
Which is exactly what makes a Zener diode preferable.
:
: I would sugest the use of 8 to 10 diodes in the string. I just checked
: Mouser for a currennt priceand found 1N4002 diodes $.07 . This would
: push the total cost 10 less $.70 for the string. Pretty cheap huh.
Can't argue with that, except for the fact that he wants to drop
3-Volts, not 7. Four should do the job nicely.
Harry C.
Frank Dinger
> ba...@rose.hp.com (Fowler_Barry) writes:
> Why not just place a 3 volt Zener diode in series with the 12 volt supply?
===========
Although a 1 component solution ,depending on the current required ,a
3V Zener diode 'could' be more expensive.
Suppose the current required is say 800mA. You then will need a 1A or
3Watts zener diode. Any 78xx regulator for less than 10V would be
suitable.If for example a 7805 is used ,add 2 resistors (one could be
made variable)
_______
| |
input +12V -----| 78xx |--------- output +9 V
| | |
-------- ----
| | | R1 for 7805 R1=Variable 0-4700 Ohms
| | | R2=fixed 3900 Ohms
| ----
| | alternative :no R1 and
| | R2 replaced by 6 or
7 low current
|--------| silicon diodes in series
---- (forward bias)
| | R2
| | It all depends what can be found
---- in the 'junkbox'
|
0 --------------------------
Good Luck
Frank Dinger , Inver by Tain , Ross-shire - Scotland UK
e-mail : gm0csz...@ukrs.org
Packet : GM0CSZ @ GB7NOS.#76.GBR.EU
Jerry Ponko
In article <masimoDw...@netcom.com>, mas...@dnai.com says...
>
>Hi,
>
>I've got a problem that has cost me several plants this summer. I
>have a piece of land that has no grid power. I am powering a 24v ac
>sprinkler controller through a voltage converter that is turned on
>with a dc timer. While there is no ac power to the timer I depend on
>the 9v battery backup to preserve the programming. Problem is that
>the 9v batteries don't last long enough. The last battery I tried, a
>nickel hydride that was supposed to last 6x the time of an alkaline
>only lasted two weeks. I am considering putting a dozen or so
>alkaline batteries in a box connected to one another but if I can put
>together a good stepdown converter I could let the backup run off a
>battery that will be recharged daily by the solar system.
>
>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>
>Thanks for any advice.
>
>Mike Maas
>
to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
0.6-0.7 volt drop across it and so the total voltage drop will be about 3
volts. The benefit of doing it this way is that you don't have energy wasted
(due to IR drop) as you do when using a voltage regulator. Typically You can
use something like 100 PVI 1A diodes commonly available at Radio Shack or
whatever.
--
==========
Jerry Ponko mailto:jerry...@software.rockwell.com
Network Specialist, Be Developer
Rockwell Software Inc.
2424 S. 102nd Street West Allis WI 53227
Rockwell Software Home Page: http://www.software.rockwell.com
Be Inc. Home Page: http://www.be.com
The one what?
Harry H Conover (con...@tiac.net) wrote:
>Kirk Kerekes (kke...@iamerica.net) wrote:
>: For the ultimate cheapness-- measure the current flow out of the 9VDC
>: batteries when in use (an inexpensive DVM ought to do the trick). Compute
>: the resistor necessary to obtain a 3V drop at that current [R=
>: 3000/current(in mA)].
>Of course, this method is only good for a fixed load.
And a regulated supply.
TheOne
Shaw Moldauer
: Here's the results of the contest :
: Barry Fowler zener
: Paul Renault 7809
: Shaw LM317
No, I recommended a solar 9 volt battery charger. I just commented
that the 7809 is not easily picked up at any electronics store. Neither is
the charger, but it is the better solution :-)
: Uncle Brian 7809
Harry H Conover
Jerry Ponko (jerry...@software.rockwell.com) wrote:
: If the 12 volt source is regulated well enough, the least expensive way would
: to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
: 0.6-0.7 volt drop across it and so the total voltage drop will be about 3
: volts. The benefit of doing it this way is that you don't have energy wasted
: (due to IR drop) as you do when using a voltage regulator. Typically You can
: use something like 100 PVI 1A diodes commonly available at Radio Shack or
: whatever.
Jerry, with an in series voltage dropping element, you have identical
power dissipation independent of the method chosen. In this case, power
dissipation in the dropping element will be 3-Volts * Load Current,
regardless of the chosen method.
Harry C.
Duane C. Johnson
Hi Harry;
Not al regulation methods disipate the same amount of power.
Of course the series element of the regulators would be identicle
and unavoidable. Electronic regulators also waste an additional
amount of power in their control circuitry.
In the special case of the timer running on a battery the electronic
regulator would represent a constant and I bieleve larger load the timer.
In this case the timer is not sensitive to voltage variations, I know
this becouse it was designed to be used with a 9V transister radio
battery. The use of purly passive series voltage droping diodes is the
ideal solution to the problem as they don't waste power from the
battery unnessassarily in control circuitry.
They are cheap too.
I would sugest the use of 8 to 10 diodes in the string. I just checked
Mouser for a currennt price and found 1N4002 diodes $.07 . This would
push the total cost for 10 to $.70 for the string. Pretty cheap huh.
Mouser part# = 583-1N4002 1amp 200 volts
1(800)346-6873
sa...@mouser.com
Paul Mathews
<snip>
> 1. The power supply is outputting 12 Watts.
> 2. The load is dissipating 9 Watts.
> 3. If the switching mode regulator "in theory dissipates
> no power at all, where is the remainder of the power
> being output by the supply (3 Watts) being dissipated?
>
...
A properly designed stepdown switcher will have less input current than
output current. For example, you'd probably see something like 800mA
input at 12V for 1000mA output at 9V. If the switcher is truly
lossless, the product Vin * Iin = Vout * Iout
--
Paul Mathews, consulting engineer
AEngineering Co.
opt...@whidbey.com
non-contact sensing and optoelectronics specialists
Steve Work
Harry H Conover (con...@tiac.net) wrote:
: Jerry, with an in series voltage dropping element, you have identical
: power dissipation independent of the method chosen. In this case, power
: dissipation in the dropping element will be 3-Volts * Load Current,
: regardless of the chosen method.
Not if you use a "switching" regulator, these things in theory dissipate
no power at all, in in real operation they dissipate far less than ones
which simply offer resistance to the current.
Kevin AstirCS 1U KO0B
con...@tiac.net (Harry H Conover) wrote:
>Let me toss out a puzzle to all of you: If I have a 12-V dc power
>supply supplying 1-Amp to a load consisting of a switching mode
>regulator (which here drops 3-Volts) and a 9-Volt load:
> 1. The power supply is outputting 12 Watts.
NO! The power supply is "outputting" (sorry, Noah) only 9 W.
The (average) input current is LESS than the output current. YES, it
can so be.
The source supplies (Mr. Webster is much happier now, and would be
ecstatic if I could spell) 1 A or 12W 75% of the time, for 9W
average. I am of cource neglecting losses, which will raise that to ~
1.1 A, or around 9.9 W for a good (not best) switcher.
Folks have no problem trading volts for amps when a transformer is
involved, but drop one winding, and they start raving about
impossibilities.
BTW, a boost regulator has higher input than output current.
> 2. The load is dissipating 9 Watts.
Yup. Lower the source voltage, and the regulator draws more input
current, raise it and it draws less. Switcher has negative input
impedance. Can cause nasty problems...plan on it.
> 3. If the switching mode regulator "in theory dissipates
> no power at all, where is the remainder of the power
> being output by the supply (3 Watts) being dissipated?
Its neither dissipated, nor "output" and every time I read input and
output (ab)used as verbs, I feel like overkeeling and upthrowing
'till I outpass. AND, if output was a verb, it still wouldn't need
"being" for help!
And while your at it tell the damned stewardesses "de-plane" makes
them sound as stupid as they probably are. The word they seek is
"disembark".
-KF-
-MUCH grumpier than usual this AM.... so pass the damned coffee.
Kevin AstirCS 1U KO0B
con...@tiac.net (Harry H Conover) wrote:
>Jerry Ponko (jerry...@software.rockwell.com) wrote:
>: The benefit of doing it this way is that you don't have energy wasted
>: (due to IR drop) as you do when using a voltage regulator. Typically You can
>: use something like 100 PVI 1A diodes commonly available at Radio Shack or
>: whatever.
>Jerry, with an in series voltage dropping element, you have identical
>power dissipation independent of the method chosen. In this case, power
>dissipation in the dropping element will be 3-Volts * Load Current,
>regardless of the chosen method.
If Jerry had said "due to Iq * Vsupply" he would have been correct.
If load is fairly large, then you can ignore Iq losses. If load is
small, use a MAX666 and ignore them anyway.
-KF-
Coffee working, less grumpy now.
Steve Offiler
In article <50438s$9...@news-central.tiac.net> con...@tiac.net (Harry H Conover) writes:
(snip)
>Let me toss out a puzzle to all of you: If I have a 12-V dc power
>supply supplying 1-Amp to a load consisting of a switching mode
>regulator (which here drops 3-Volts) and a 9-Volt load:
> 1. The power supply is outputting 12 Watts.
> 2. The load is dissipating 9 Watts.
> 3. If the switching mode regulator "in theory dissipates
> no power at all, where is the remainder of the power
> being output by the supply (3 Watts) being dissipated?
Easy. The power supply only supplies 12 watts 75% of the time. Voila, 9
watts. ("In theory")
Steve O.
Harry H Conover
Steve Work (slw...@netcom.com) wrote:
: Harry H Conover (con...@tiac.net) wrote:
:
: : Jerry, with an in series voltage dropping element, you have identical
: : dissipation in the dropping element will be 3-Volts * Load Current,
: : regardless of the chosen method.
: Not if you use a "switching" regulator, these things in theory dissipate
: which simply offer resistance to the current.
Agree, but I hadn't noticed them previously enter into this discussion.
There are IC switching mode regulators on the market, but I didn't
think the LM309 (previously mentioned) was one of them (I don't have
the LM309 data sheet available). I do have data sheets on the LM185
through LM385, and circuitry for that series is just plain old series
regulator circuitry.
Let me toss out a puzzle to all of you: If I have a 12-V dc power
supply supplying 1-Amp to a load consisting of a switching mode
regulator (which here drops 3-Volts) and a 9-Volt load:
1. The power supply is outputting 12 Watts.
2. The load is dissipating 9 Watts.
3. If the switching mode regulator "in theory dissipates
no power at all, where is the remainder of the power
being output by the supply (3 Watts) being dissipated?
Harry C.
:
:
Duane C. Johnson
Kirk Kerekes wrote:
>
> For the ultimate cheapness-- measure the current flow out of the 9VDC
> batteries when in use (an inexpensive DVM ought to do the trick). Compute
> the resistor necessary to obtain a 3V drop at that current [R=
> 3000/current(in mA)].
>
> middle of its range. Use the DVM to set the pot to about 20% higher than
> this value, and wire it in series with your 12VDC source. Measure the
> voltage on the 9VDC side, and adjust the pot for 9.0 volts.
Now you are just being silly.
Electronic loads rarely are constant.
I venture that the electronic timer that started this thread
may have an activation curent as much as 100 times the steady
state load maybe even 1000 times the steady state load.
Resisters work best into steady resistive loads.
Duane C. Johnson
Kirk Kerekes wrote:
>
> In article <32270C...@pclink.com>, "Duane C. Johnson"
> <red...@pclink.com> wrote:
>
> - > For the ultimate cheapness-- measure the current flow out of the 9VDC
> - > batteries when in use (an inexpensive DVM ought to do the trick). Compute
> - > the resistor necessary to obtain a 3V drop at that current [R=
> - > 3000/current(in mA)].
> - >
> - > Select a potentiometer (variable resistor) that has this value in the
> - > middle of its range. Use the DVM to set the pot to about 20% higher than
> - > this value, and wire it in series with your 12VDC source. Measure the
> - > voltage on the 9VDC side, and adjust the pot for 9.0 volts.
> -
> - Now you are just being silly.
> - Electronic loads rarely are constant.
> - I venture that the electronic timer that started this thread
> - may have an activation curent as much as 100 times the steady
> - state load maybe even 1000 times the steady state load.
>
> If you will read the original post, the 9v is being used to preserve the
> programming in an electronic timer -- not only is this likely to be a
> constant load, it is also not terribly sensitive to minor voltage
> fluctuations -- in most designs, the internal programming will be retained
> without any battery at all for several minutes, or at least 10 sec or so
> (enough time to change batteries).
>
> I wasn't being silly, I was paying attention to the problem being solved.
I apologize for speaking to harshly. Sorry.
I am still going to say that a resistor is an unwise solution to the
problem on several counts.
1. Yes the clock portion of the timer has a resonably stable current
draw but the Solid State Relay, SSR, section of these timers does
not. I have one of these and it takes several milleamps of current
to activate the SSR. When the SSR is off the current is in the
microamp range. Most of these, I would venture all, SSRs have an
isolation berior from the control side to the power side. This
isolation circuitry takes some power to operate. This power comes
from the input control signal and ultimatly from the battery.
2. Voltage regulators by there nature are non-linear devices. What I
mean by this is they have at least some degree of voltage stability
when the current load is changed. In my timer the current load
presented to the battery changes by a factor of about 100 to 1.
I have one of these units in my hands right now. Even as I type this.
Honest I do.
3. When devising awswers to problems presented by persons, unknown,
to problems, not seen and possibly misstated, we must endever to
make the solution one that can be universally used by the largest
number of readers. This includes the readers that are, possibly,
not well versed in the arts we are so familer with. It does the
reader no good to have a solution that may be techically the lowest
cost solution if it requires $100 worth of test equipment, that
the reader may not have, to implement this solution.
4. The "Elegant" solution to a problem often has the virtue of
"simplicity". This solution is not only correct for the set of
requirements presented but encompasses requirements not even perseved
by the person asking the question. The "Elegant" sopluton may be
used in diverse applications not apparent initially.
My first solution to the problem was that there is no problem. I think
that the timer will run on the 12volt, really 14.6volt battery, right
out of the box. The timer I have runs upto 20 volts. I havn't broken it
yet and don't expect to.
The more universally acceptable solution is the string of 1 amp rectifier
diodes. This solution wastes no power unnessassarily and is compliant
with load variations presented by circuits that operate in various modes.
It is cheap. I showed it cost $.70 for 10 diodes from a mailorder company.
It is adjustable.
Roeland Krawl
In article <32235...@moci.mke.software.rockwell.com>
jerry...@software.rockwell.com (Jerry Ponko) wrote:
>>So, is there any easy way to step voltage down to 9v dc from 12v dc?
> If the 12 volt source is regulated well enough, the least expensive way would
> to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
> 0.6-0.7 volt drop across it and so the total voltage drop will be about 3
> volts. The benefit of doing it this way is that you don't have energy wasted
> (due to IR drop) as you do when using a voltage regulator. Typically You can
> use something like 100 PVI 1A diodes commonly available at Radio Shack or
> whatever.
> --
Ohms law still applies to diodes. A diode will dissipate .7 watts at 1 amp
and .7 volts across the diode.
If the load has some capacitance, then be sure to choose a diode that can
handle the surge created by charging the capacitance at initial power on.
You could insert a small series resistance to limit the surge current.
Duane C. Johnson
a few milleseconds.
Dave Zawodny
Lurk mode off.
opinion mode on @ $.02 rate...
"Duane C. Johnson" <red...@pclink.com> wrote:
>Roeland Krawl wrote:
>>
>> In article <32235...@moci.mke.software.rockwell.com>
>> jerry...@software.rockwell.com (Jerry Ponko) wrote:
>>
>> >>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>>
>> > If the 12 volt source is regulated well enough, the least expensive way would
>> > to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
>> > 0.6-0.7 volt drop across it and so the total voltage drop will be about 3
>> > volts. The benefit of doing it this way is that you don't have energy wasted
>> > (due to IR drop) as you do when using a voltage regulator. Typically You can
>> > use something like 100 PVI 1A diodes commonly available at Radio Shack or
>> > whatever.
Excuse me, this is probably the cheapest way, but there
is`wasted energy in the drop across the diodes. diodes will
drop from ~.5-1.5 volts depending on the current draw.
>> > --
>> Ohms law still applies to diodes. A diode will dissipate .7 watts at 1 amp
>> and .7 volts across the diode.
Ohms law does apply. Amps X Volts = Watts and watts = Power.
>> If the load has some capacitance, then be sure to choose a diode that can
>> handle the surge created by charging the capacitance at initial power on.
>> You could insert a small series resistance to limit the surge current.
>BTW the minimum surge for 1 amp diodes is around 20 amps for
>a few milleseconds.
>
>--
>CUL8ER
>
>Stupid is Forever
>Ignorance can be Fixed
>
>Duane C. Johnson
>Ziggy
>WA0VBE
>Red Rock Energy
>1825 Florence St.
>White Bear Lake, MN, USA 55110-3364
>(612)635-5065 w
>(612)426-4766 h
>red...@pclink.com
>dc...@PO8.RV.unisys.com
>http://www.geocities.com/SiliconValley/3027/
>
A more reliable way could be to use the TO-5 case 3 terminal
regulators and a couple of capacitors, but the dissapation
losses are still there.
A more efficient way is to use a swithcing regulator which is
more efficient, but at a significantly higher cost.
With the slight voltage difference (3 volts), and the typical
automatic 2.4 volt loss in the TO-5 3 term regulator, and the
cost factor of the switching regulator factored, it is almost
a lose lose situation unless you are drawing a lot of
current.
Some devices will tolerate 12 volts instead of 9, however the
spec for a 12 volt lead-acid storage battery is really 12.6v.,
and under a charging situation, up to ~14.5 volts.
Depending on what you are powering, you might have to
experiment a little.
Sorry for butting in:\
Z
Lurk mode on...
JEMSTAR
This problem of reducing 12 Volts to 9 Volts brings me to the question of
How does a dimmer switch work? I know there is a variable rate resister
that is turned to increase/decrease the intensity of the lights, but does
that affect the voltage, or current? Also I have a hard time believing that
the power that doesnot go to the light bulbs, is absorbed by the resister,
because it doesnot get as hot as the lights would have been with no
resister.
This is a little off topic, but I would appreciate constructive comments.
Thanks----------J. Cooke
Peter Kesselman
In article <01bbae10$8990d260$350f...@midwest.greynet.net>, "JEMSTAR"
<redb...@greynet.net> wrote:
A dimmer works by varying the conduction time of a triac. Picture a sine
wave, where the "width" of each half of the wave is cut off. This reduces
the RMS voltage value. This is why dimmers cause radio inteference. They
generate tons of harmonics.
The voltage is not "absorbed by the resistor". It's not absorbed by
anything. The waveshape is modified. The relatively small amount of heat a
dimmer dissipates is due to the voltage drop across the triac, but that's
incidental.
--
Peter Kesselman
Uncle
>>> >>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>>>
>>> > If the 12 volt source is regulated well enough, the least expensive way would
>>> > to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
>>> > 0.6-0.7 volt drop across it and so the total voltage drop will be about 3
That's going over the top.
>A more reliable way could be to use the TO-5 case 3 terminal
>regulators and a couple of capacitors, but the dissapation
>losses are still there.
True there's dissipation, but what are we doing here? Are you
talking about powering up a 9v device from a car battery?
What sort of power does the 9v device need?
If it _is_ a battery, the series diode thingy wouldn't do too well.
The voltage of the battery would vary from some low point, up
to over 14v at full charge. Use a series regulator like the LM7909.
Input can be anywhere from about 10 1/2v up to 35v. If you need
more than 1 Amp, go to a pass transistor (virtually ANY PNP in a
TO-3 case. Look in the library (or whatever your source is) under
power supplies, thence to pass transistors.
Uncle Brian VK6BQN
- -
"Once again, this'd be a great world if not for
the fuckin' people."
- Jim Hill in alt.peeves
Ian Halsema
JEMSTAR wrote:
>
> This problem of reducing 12 Volts to 9 Volts brings me to the question of
> How does a dimmer switch work? I know there is a variable rate resister
> that is turned to increase/decrease the intensity of the lights, but does
> that affect the voltage, or current? Also I have a hard time believing that
> the power that doesnot go to the light bulbs, is absorbed by the resister,
> because it doesnot get as hot as the lights would have been with no
> resister.
> This is a little off topic, but I would appreciate constructive comments.
> Thanks----------J. Cooke
Power control methods vary depending on whether you are using AC
(alternating current- typical household power) or DC (such as is
available from a battery). A dimmer switch for AC power uses an
electronic switching system to turn the electricity to the lamp on and
off at different points in the cycle. The ratio of on time to off time
determines the brightness of the lamp. Since the switching system is
either fully off or fully on, there is very little power dissipated in
it.
A similar arrangement (though more complex) can be set up for DC power
control. However, dimmers in automotive applications do in fact often
use resistors because they are cheap and simple. And these resistors
can become quite hot.
If you want to reduce 12VDC to 9VDC, use a Zener (voltage regulating)
diode rated for 9 volts. You will need a resistor in series with the
diode to limit current flow so that the diode doesn't burn out. The
power rating of the Zener, and resistance/wattage of the resistor in
this simple circuit will depend on the power required by your
appliance. Check out Radio Shack. They have the components to build a
voltage regulator, and may also have pre-built units.
--Ian
Steve Work
Organization: NETCOM On-line Communication Services (408 261-4700 guest)
Ian Halsema (ahal...@es.xerox.com) wrote:
: Power control methods vary depending on whether you are using AC
: (alternating current- typical household power) or DC (such as is
: available from a battery). A dimmer switch for AC power uses an
: electronic switching system to turn the electricity to the lamp on and
: off at different points in the cycle. The ratio of on time to off time
: determines the brightness of the lamp. Since the switching system is
: either fully off or fully on, there is very little power dissipated in
: it.
A drawback of the "switching" lamp dimmer is that it generates a lot of
radio-frequency energy, because of the sharp turn-on and turn-off
transients. If there is a nearby radio (especially AM) you'll hear
static on it. I have one of these near a radio. If the lamp is either
off or full brightness, there is no static, but if the lamp is dimmed to
less than full power, there is a considerable amount of static.
Also, I would recommend against using the zener diode approach to
regulation in most cases. The trouble here is that the zener itself is
dissipating (wasting) a considerable amount of power, often more than the
load itself. This is in addition to the power disspated by the resistor,
and the resistor itself will also have its power dissipation increased.
I would recommend using an integrated circuit regulator in series.
Either an "ordinary" one (which will be about 75% efficient),or a
switching one, which will be noisier, more complex, more expensive, but
its efficiency will be much closer to 100%.
Another thing might be just to string about 4 or 5 forward-biased diodes
in series with the load, as each one drops about 0.6 to 0.7 volts.
However, if the load current drops to near zero, the voltage drop across
the diodes will decrease, you need to be aware of this. Perhaps a
resistor of 10K ohms or so across the load would be appropriate.
Tracy Hooker
Uncle wrote:
> snip...
> If it _is_ a battery, the series diode thingy wouldn't do too well.
> The voltage of the battery would vary from some low point, up
> to over 14v at full charge. Use a series regulator like the LM7909.
> Input can be anywhere from about 10 1/2v up to 35v. If you need
7809 please.
T
Sylvan Butler
Uncle (un...@net1.nw.com.au) wrote:
>If it _is_ a battery, the series diode thingy wouldn't do too well.
>The voltage of the battery would vary from some low point, up
>to over 14v at full charge. Use a series regulator like the LM7909.
If the device was originally designed to work from a 9v battery, then the
regulator is way overkill, and the diode series string will work fine.
sdb
Carl H. Hovermale
another way is to just put in a resistor. a simple resistor is one of the
wires in an old oven. Hook the wire to a 12vdc source and take a
multimeter lead down the wire until it reads 9 volts that is the resistor
Aaron Cake
--
This wastes lots of power and generates lots of heat. The voltage drop
would also be dependent on how much current you are pulling through the
resistor.
-----
Aaron Cake
London Ont.
aaron...@blackhole.xg.com
http://www.lmcs.edu.on.ca/jp2/students/aaronc/index.html
------
It is written in thirty-foot-high letters of fire on top of the Quentulus
Quazgar Mountains in the land of Severbeupstry on the planet
Prelimumtarn, third out from the sun Zarss in Galactic Sector QQ7 Active
J Gamma, God's final message to his people, "We apologize for the
inconvienence".
James S. Bassett
un...@net1.nw.com.au (Uncle) wrote:
>>>> >>So, is there any easy way to step voltage down to 9v dc from 12v dc?
>>>>
>>>> > If the 12 volt source is regulated well enough, the least expensive way would
>>>> > to use 5 or 6 silicon rectifier diodes in series. Each diode has about a
>>>> > 0.6-0.7 volt drop across it and so the total voltage drop will be about 3
>That's going over the top.
>>A more reliable way could be to use the TO-5 case 3 terminal
>>regulators and a couple of capacitors, but the dissapation
>>losses are still there.
>True there's dissipation, but what are we doing here? Are you
>talking about powering up a 9v device from a car battery?
>What sort of power does the 9v device need?
>If it _is_ a battery, the series diode thingy wouldn't do too well.
>The voltage of the battery would vary from some low point, up
>to over 14v at full charge. Use a series regulator like the LM7909.
>Input can be anywhere from about 10 1/2v up to 35v. If you need
>more than 1 Amp, go to a pass transistor (virtually ANY PNP in a
>TO-3 case. Look in the library (or whatever your source is) under
>power supplies, thence to pass transistors.
>Uncle Brian VK6BQN
> - -
>"Once again, this'd be a great world if not for
>the fuckin' people."
> - Jim Hill in alt.peeves
Good advice but use the '7809', the '7909' is for negative Voltages.
May the force live long and engage.
good luck
jb