Rectifier wiring
Ignoramus14602
It has four terminals, but they are not marked. The datasheet is pretty
sparse, too. Which of them are AC and which are DC?
i
Ned Simmons
You can figger it out in a minute or two with your DMM.
--
Ned Simmons
Winston
Put your multimeter in 'junction test' continuity mode.
(Normally indicated by a diode symbol.)
Test across every terminal combination of both polarities
until you locate the diodes as they conduct in forward
direction (Red + to Anode, Black - to Cathode) and as
they block in reverse direction (Red + to Cathode, Black -
to Anode). Write down each reading on a large sketch
of the part and all will be revealed.
It'll prolly end up looking a lot like:
http://www.fairchildsemi.com/ds/GB/GBPC12005.pdf
--Winston
Spehro Pefhany
The one on the chamfered corner that's at 90° to the rest is the +,
the one _diaagonally_ opposite is the -, and the other two are the AC
input terminals.
Check it on 12VDC with a small light bulb in series to be sure. You
should measure 11VDC or so out with the + terminal as I said,
regardless of the polarity of the input (and, of course, the series
light should never light up (!)).
As Ned said, this is easily figured out with a DMM that has a diode
check range, but compare it with a regular diode to be sure of the
polarity. Most digital DMMs have the red lead source positive voltage,
and most analog VOMs are the opposite.
http://www.repairfaq.org/sam/semitest.htm#sttdj
Obviously if you get it wrong you'll probably kill the bridge or cause
the capacitor to vent its lifeblood.
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
BQ340
Usually the clipped corner (or terminal that is not oriented the same as
the other 3) is the positive & the diagonal opposite is negative. The
other 2 are the AC.
<http://www.softwareforeducation.com/electronics/notes/AS/Diode/bridge-rectifier.jpg>
MikeB
ne...@jecarter.us
The GBPC case seems to use the angled corner as the + indicator. See
this datasheet: http://www.vishay.com/docs/88612/gbpc12.pdf
John
Lloyd E. Sponenburgh
news:7NSdneQgVuYjvLDR...@giganews.com:
They ARE marked, Iggy. The corner miter usually denotes the + output lead.
LLoyd
Pete C.
Have you looked on the *top* of the case?
Karl Townsend
> Obviously if you get it wrong you'll probably kill the bridge or cause
> the capacitor to vent its lifeblood.
You guys are taking all the excitement out of Iggy's life. Just hook it up,
the smoke will come out if its wrong, then do something else. Like the time
I tried a start cap in the run cap position on my three phase converter.
Now, I know the difference.
Karl
Ignoramus14602
I figured it out. Thanks to all. The power supply is now working
nicely.
i
Ignoramus14602
I know the fun feeling. In about 0.1 seconds after a loud BANG I know
that something went very wrong.
i
cl...@snyder.on.ca
<ignoram...@NOSPAM.14602.invalid> wrote:
Use your ohm meter to check the polarity of the 4 diodes. The AC
goes to the points where the positive and negative diodes join.
You get DC+ where the 2 positive diodes joinn, and DC- whre the
negative diodes join. GENERALLY the cut off corner, if present, is one
of the DC terminals.
DoN. Nichols
The terminal nearest the corner which is cut off is the positive
output. (Also, it is at right angles to the other three) The diagonally
opposite one is the negative. The other two are AC input.
You can verify this with a multimeter in the ohms or the
diode-check positions. Positive probe on the negative terminal will
show conduction to the other two adjacent terminals. Negative probe on
the Positive terminal will show conduction to the other two adjacent
terminals.
The diodes are arranged pretty much like the schematic symbol
for a bridge rectifier, and surround the center mounting hole as shown
here. Always positive and negative are opposite corners, and the AC
inputs are the other two corners. Something will mark the positive
terminal as different -- here the cut corner and the blade being at
right angles to the rest of the blades. Sometimes there is a red dot by
the positive terminal, black near the negative (except on black potting
epoxy), and yellow on the two AC inputs.
I guess that they did not spell this out on the data sheet
because it is so common a construction -- and has been for at least
forty years or so. Or -- it might be spelled out somewhere else in the
Comchip products manual -- and you only have two pages out of that
manual.
Enjoy,
DoN.
--
Email: <dnic...@d-and-d.com> | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---
LSMFT
--
LSMFT
I haven't spoken to my wife in 18 months.
I don't like to interrupt her.
PeterD
>Karl Townsend wrote:
>> ...
>>> Obviously if you get it wrong you'll probably kill the bridge or cause
>>> the capacitor to vent its lifeblood.
>> `...
>>
>> You guys are taking all the excitement out of Iggy's life. Just hook it up,
>> the smoke will come out if its wrong, then do something else. Like the time
>> I tried a start cap in the run cap position on my three phase converter.
>> Now, I know the difference.
>>
>> Karl
>>
>>
>>
>I did that once; KaBang it went.
Da KaBang I can handle, it's cleaning up the mess afterwards that is a
PITA!
Karl Townsend
"PeterD" <pet...@hipson.net> wrote in message
news:rclr265v7c4bdt9mu...@4ax.com...
Are you talking about the mess in your shorts?
Karl
Jeffrey D Angus
Try this data sheet:
http://www.nteinc.com/specs/53000to53099/pdf/nte53016_20.pdf
If you part has the same internal connections, you should be
able to use an ohm meter to see a diode between AC & + and
AC & -, and an open between AC & AC.
Jeff
--
“Egotism is the anesthetic that dulls the pain of stupidity.”
Frank Leahy, Head coach, Notre Dame 1941-1954
Uffe Bærentsen
"Karl Townsend" <karltown...@embarqmail.com> skrev i en meddelelse
news:4c2e0047$0$77568$892e...@auth.newsreader.octanews.com...
*LOL*
That's a good one Karl :-)
--
Uffe Bærentsen
cl...@snyder.on.ca
If you don't get that mess cleaned up pronto, it IS a PITA.
PeterD
<bg> Ever cleaned up a blown cap? Fluff and foil everywhere!
beecrofter
so that the battery can go in either way .
William Sommerwerck
> devices so that the battery can go in either way.
Yes... if the device can tolerate a constant voltage drop of 1.2V (or
greater).
Joe Pfeiffer
> A neat use for bridge rectifiers is to put in battery powered devices
> so that the battery can go in either way .
Coincidentally, comp.robotics.misc has just been having a lively thread
regarding these battery contacts that allow a battery to be inserted in
either orientation, completely through the geometry of the contacts.
http://www.pcmag.com/article2/0,2817,2365995,00.asp
--
As we enjoy great advantages from the inventions of others, we should
be glad of an opportunity to serve others by any invention of ours;
and this we should do freely and generously. (Benjamin Franklin)
William Sommerwerck
>> so that the battery can go in either way .
> Coincidentally, comp.robotics.misc has just been having a lively thread
> regarding these battery contacts that allow a battery to be inserted in
> either orientation, completely through the geometry of the contacts.
> http://www.pcmag.com/article2/0,2817,2365995,00.asp
I was about to say something ironic/sarcastic, about how this has already
been done with battery shape -- but was pleased to read that this system
works with common existing cells. Great idea -- if it works.
> As we enjoy great advantages from the inventions of others, we should
> be glad of an opportunity to serve others by any invention of ours;
> and this we should do freely and generously. (Benjamin Franklin)
"Giving away" inventions is not necessarily a good thing. The patent system
actually encourages invention by increasing the likelihood an inventor will
profit from his ideas.
DoN. Nichols
> A neat use for bridge rectifiers is to put in battery powered devices
> so that the battery can go in either way .
Presuming that the device can accept the loss of two diode
forward drops from the battery to the load. (About 1.4V at reasonable
current levels and with silicon diodes.) Some things are close enough
to their minimum voltage threshold so this would not work well.
If the battery and diode are followed by a regulator, you
probably are all right.
Karl Townsend
"DoN. Nichols" <dnic...@d-and-d.com> wrote in message
news:slrni37h3q....@Katana.d-and-d.com...
> On 2010-07-06, beecrofter <beecr...@yahoo.com> wrote:
>> A neat use for bridge rectifiers is to put in battery powered devices
>> so that the battery can go in either way .
>
> Presuming that the device can accept the loss of two diode
> forward drops from the battery to the load. (About 1.4V at reasonable
> current levels and with silicon diodes.) Some things are close enough
> to their minimum voltage threshold so this would not work well.
>
> If the battery and diode are followed by a regulator, you
> probably are all right.
>
> Enjoy,
> DoN.
I'm curious. I use a 2:1 transformer, bridge rectifier and caps to power 80
volt AMC drives. This comes out about 4 volts high but hasn't seemed to
cause trouble. For a cheap volt drop could you just run the DC out through a
couple more diodes on a bridge rectifier?
Karl
William Sommerwerck
> 4V high but hasn't seemed to cause trouble. For a cheap
> voltage drop could you just run the DC out through a couple
> more diodes on a bridge rectifier?
Yup. Each silicon diode gives 0.6V to 0.7V drop. Make sure the diode's
current rating is sufficient.
DoN. Nichols
Yes -- but it might be cheaper to use individual diodes since
you would need several bridges to get 4V. Let's see -- somewhere
between 0.600 V and 0.700 V per diode, so 6.6 diode drops (3 + bridges)
and 5.7 diode drops (3 - bridges). I would put the extra diodes between
the output of the bridge and the first capacitor.
And that 4V high is pretty close to right on for 120 VAC input.
We start by dividing the 120 VAC by 2 for the transformer ratio, then
multiplying by the square root of 2 (1.414) and it comes out to 84.84 V
peak.
But any diode drop will soften the supply's stiffness somewhat,
because there is some variation in voltage drop with current. Not
nearly as much as with a resistor, but some.
Better is to add a small filament transformer in series with the
primary to adjust the voltage into the bridge to the right value. 80V /
1.414 requires 56.57 VAC into the bridge -- hmmm ... adjust for two
diode drops at 0.650 V per diode requires 57.87 VAC into the bridge, so
multiply by 2 for the transformer ratio and you get 115.75 VAC. If your
line voltage is really 120 VAC, you need to drop 4.24 VAC -- so a 6.3
VAC center tapped filament transformer would give you a choice of 3.15
VAC for half the winding, or 6.3 VAC for the full winding -- wired to
oppose the line voltage. The secondary of the filament transformer
needs to be rated to handle the worst case current into the main
transformer.
But really -- since line voltage varies with neighborhood load
-- or state wide load -- (and lots of air conditioning load where I am
today) you are as likely to get 115 VAC as 120 VAC -- and that would be
just about right for your transformer. Right now, a line voltage
monitor is telling me that I'm getting 115 VAC. It usually runs between
117 VAC and 120 VAC here.
So -- unless the drive says 80 V *MAX* -- don't sweat it. You
will be right on some of the time, and a bit high some of the time. But
your speed will be controlled by the output from the tach generator and
what the amplifier (drive) does with it, not the supply voltage, which
will simply set the limiting speed of the motor -- and it will almost
never actually run at that speed. Remember that the computer feeds an
analog voltage to the amplifier (drive) which tells it how fast to go,
and the amplifier compares that voltage to the voltage from the tach
generator to adjust what voltage goes to the motor's primary windings.
Your speed will be right as long as the amplifier (drive) does not break
down from the applied voltage. So -- look up what it can safely accept.
If it says that the absolute maximum voltage is 80 VDC -- then you want
to either lower the voltage a bit so at 120 VAC you will have no more
than 80V DC, or set up a regulator as part of the power supply to make
sure that the output voltage is never anything but 80 VDC. (This
requires the transformer to produce a bit more voltage so the regulator
can toss some away in the process of regulating -- say perhaps 100 VDC
before the regulator.
I hope that I haven't told you too much. :-)
Good Luck,
Winston
(...)
> "Giving away" inventions is not necessarily a good thing. The patent system
> actually encourages invention by increasing the likelihood an inventor will
> profit from his ideas.
I'm sorry to inform you that you are in violation of
my rights to those thoughts.
My lawyer discovered that any practitioner in the
field *could* have easily originated those thoughts
before you did; they are insufficiently
novel.
Additionally, with further work, I believe
we will be able to uncover historical instances
of each of those thoughts occurring as 'prior art'.
You are to cease expression of those thoughts or any
thought that could be considered derivative of them.
Either of those thoughts are now available for license
from me for a single - use fee of US $5.00.
See how this really works? :)
--Winston