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'[PICLIST] [EE] Power supply'
2002\01\27@122415 by

Hi-
After looking at the sample ascii drawings in Dave Dilatush's example, I
came across this....wasn't there a big discussion about a low cost power
supply -- how does this one rate?
David

Subject: Re: power-supply voltage soar - NOT - Win's solution
From: Win Hill <whillmediaone.net>
Date: Tue, 30 Oct 2001 15:03:37 GMT
--------
Hi folks,

Here's my circuit, as promised.  I started this thread to
discuss my solution to the age-old problem of soaring no-

My immediate goal is to upgrade the soaring power supply in
one of my older designs, a low-noise wideband laboratory
power amplifier, by limiting its output to 38V; this is
necessary to protect some components rated at 40V.  I'll
need to do this in an electrically "quiet" fashion and also
avoid generating any additional heat inside the instrument.

45V _|  eliminate objectionable soar
x   /
| x
40V _|    x
o o  o  o      but keep all this
|           o    /
35V _|               o
|                   o
|                       o
30V _|_______________________________o_____
|       |       |       |       |
0       1A      2A      3A      4A load

My solution is to interject a p-channel power MOSFET switch
between the bridge rectifier and a large 10,000uF storage
cap, which remains connected directly to the power amplifier.
A differential amplifier turns the FET off when the output
rises above 39V and off after it drops ~ 1.5V lower, with
hysteresis provided by R6.  Note that the R2/R3 ratio sets
the Q1's ON gate voltage to about 12V.  Note, this circuit
can be used at much higher voltages, by changing R5 and R6,
limited only by the breakdown ratings of Q2 and Q3.  (In my
PCB implementation I use a cascode connection for Q2, etc.,
but that's another story.)

D3  TVS
rectified  ,------|<|----------------.
ac in     | ,--|<|--.   p-channel   |
| |  ___D4|    Q1  FET    |           +38V
--|>|---+--+-+--OOO--+---+-- s   d --+-+----+---o  4A
|    | L1    | \_|_    g       |    |
--|>|---+    |      R2  /_\    | IRF9Z |    | C2
|    |       |   | D2  |  34N  |   ===
C1 |    |       +---+-----+---,   |    |
===   R1      |        ,----|---+   gnd
|    |       |       |     |   |
|    |    Q2 |    Q3 |    R6   R5   R1  12k
gnd   | 5V    c       c     |   |    R2  12k
+---- b           b --+---+    R3  4.7k
\_|_      e --+-- e         |    R4  2.49k
D1 /_\          |             R4   R5  18.2k
| LM336-    R3            |    R6  191k
|  5.0      |            gnd   C1  100uF
gnd         gnd                 C2  10,000uF
L1  10uH 5A

Low noise is an important goal here, which means minimizing
rapid changes in C2's current, which make undesirable voltage
steps on its 0.025-ohm esr.  Q1 may turn off in the middle or
even the peak (up to 12A) of an ac charging cycle, but the
transformer's 2mH leakage inductance insures that the current
into C2 does not abruptly stop; a 15V TVS, D3, provides the
path for this current until it's done.

A similiar issue arises when Q1 suddenly turns on, connecting
C1 and C2.  C1 is a small cap, but it can still put a current
spike into C2, so a small 10uH 5A inductor L1 is added to slow
dI/dt to acceptable levels.  D4 clamps L1 when Q1 shuts off.

A similar circuit with an IRFZ46N FET creates -38V, but it
doesn't need the L1 D4 addition.  I experimented on paper with
using a dual comparator instead of the long-tail pair, this
appeared simple at first, but ended with a higher parts count.
___
Keep the faith,                     /.-.\
(( * ))
- Win                      \\ //
\\\
Winfield Hill                       //\\\
Rowland Institute for Science      /// \\\
Cambridge, MA                      \/   \/

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David Harris wrote...

>After looking at the sample ascii drawings in Dave Dilatush's example, I
>came across this....wasn't there a big discussion about a low cost power
>supply -- how does this one rate?

[Win Hill's design snipped]

One thing to keep in mind when going through the circuits in that file I
lot of what I would call "design by committee" and they also toss a lot
of speculative (i.e., "in theory, this should work") stuff around in
their discussions.
Not all of it is practical.  Any given circuit may work as presented, or
it may not work because it's flawed, or it may be untested, or it may
even be only a partial schematic and unusable without adding the missing
components; so use with caution.

The particular circuit you pointed to is not a general-purpose
regulator; I think Win Hill was presenting it for discussion as one
possible solution for a problem he was having with a power supply, as a
way to keep the voltage on his power supply's main filter capacitor from
rising excessively under low-load conditions.  AFAIK, that's all the
circuit shown was meant to do.

I have a general-purpose "hobby" power supply I built for my own use,
which provides +5 VDC at a couple hundred milliamps for PICs and digital
stuff, plus adjustable +/- 0 to 15 VDC outputs current-limited at 70 mA
for analog stuff.  I use it for all of my bench work.

If there's any interest, I can post the schematic for it tomorrow.

Dave

--

David H

Dave Dilatush wrote:

{Quote hidden}

--

> I have a general-purpose "hobby" power supply I built for my own use,
>... If there's any interest, I can post the schematic for it tomorrow.

Yes please. Should be enough interest to merit it.

RM

--

OK, I'll be putting it up tomorrow evening (US East coast time) and
posting a link to it here.

DD

David Harris wrote...

>
>Dave Dilatush wrote:
>
>> If there's any interest, I can post the schematic for it tomorrow.

--

The "Junkbox" experimenter's lab supply has been posted, at:

http://members.home.net/dilatush/

Dave

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'[PICLIST] [EE] Power supply ignorance'
2002\05\15@133057 by
I have a problem that has got to be a simple fix, and I am already feeling
stupid.

I am trying to build power supply that will give me 40v, 12v, and 5v.
I have the 40v part working fine. My 12v regulator will only take 35v max,
so I got a resistor, put it between the +40 output and my meter, and it
shows about 30v.  I installed the regulator with this same resistor btween
the input and the +40v. Ground to the -40v.

It does not work.  The meter only shows .638 volts at the input of the
regulator. Output is 0.

I planned to then put a 5v regulator off of the 12 volts.

Do I need to put the resistor at ground instead of + ?

Thanks

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Not sure you can use a resistor because of ohms law. V=I/R
Since the current drawn isn't constant, the voltage won't be either. I
suppose that's why we have voltage regulators :)
I have a PSU that does something similar to your requirements.
To drop the 40v to 15v (which is used as input to the 12v regulator) I
use a resistor and zenner diode between 40v + and - rails. Their
junction connects to the base of a 2N3055. Collector goes to the 40v and
emitter gives the 15v ish...
Depending on the current you need, you can vary the transistor and mount
it on a heat sink.

{Original Message removed}
that ohm guy gets in my way alot ;-)

How about a voltage divider? Or would that affect the voltage of the entire
circuit?
Also, my load will fluctuate alot on both the 40v and 12v circuit.

{Original Message removed}
Whenever you use a resistor network with a variable load, your in
trouble.

The transistor/zenner circuit won't suffer too much from the load. I
think from memory that a 2n3055 will handle about 5 amps. You can always
use a bigger tran if needed, a 400mw zenner and standard 1/4 watt
resistor are all you need. If you going to have big load surges you need
some beefy caps on the power rails.

{Original Message removed}
You have not yet stated the output current for each of the supplies. There are several ways to solve your problem. A
voltage divider may work, depending on the load current this may be a high heat and a high cost solution if heat sinks
are necessary. A simple series pass regulator may work.

What kind of load regulation do you require?

A fixed PWM with a PNP transistor can drop the voltage to near the voltage needed, then a good linear regulator will
give you a solid output. This is the lowest heat generating and very inexpensive for a preregulator method of dropping
the voltage. (example: 555 as the PWM, a PNP transistor, a diode, inductor and cap will give you a preregulator for
under a dollar)

Gordon Varney
http://www.iamnee.com

{Quote hidden}

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Having been thru something similar recently, I can
provide some assistance...

Even if your 12V regulator could handle 40 V, it would
generate a lot of heat (this was my problem), and the
solution was to drop the voltage a bit before regulating.
This would spread the heat out to other components as
well.  Of course you may have a large enough heatsink
for this.

However, if your current is fairly constant (I doubt),
then you could calculate the appropriate resistor to
drop the voltage to say 15 or 20V.  But a changing
current will cause a varying voltage drop (V = I x R).
For varying current, you may come up with a resistor
value that will give you say 35V at lowest current,
and say 20V at max current and that would be fine.  Ensure
that you calculate the power requirement for the resistor,
add a little buffer and get the appropriate wattage
resistor.  Or you could use something with a constant
voltage drop, like rectifier diodes.  But to get even
35V from 40V, you would need 7 or 8 diodes in series
(assuming roughly 0.7V drop across each).  You could also
try another regulator like an 18V regulator "ahead of"
the 12V.

However, since 40V to 12V is a big drop, you may want
to look into something more efficient like a DC-to-DC
converter, or split your PS into 2 -- one transformer
that will get you a 40V supply and another that will
start you off at something like 15 - 20 V for the
12V and 5V regulators.

Cheers,
-Neil.

{Original Message removed}
part 1 494 bytes content-type:text/plain; (decoded 7bit)

hello

you could use one of the following circuits
the 1st is by using the LM317,the o/p voltage equation is
Vref is the voltage between the two node of R1
Iadj is the current following from the common node of the LM317
the O/P voltage of the circuit given is approximatly 12.5 V
The second circuit is using the zener diode it will give you an O/P voltage
depending on the two Zener voltage .
the pictures are virus free.

Amin

part 2 3405 bytes content-type:image/gif; (decode)

part 3 2600 bytes content-type:image/gif; (decode)

part 4 136 bytes
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'[PICLIST] [EE] power supply heat'
2002\06\19@144955 by
I have a circuit with a 7805 regulator that was set-up with a 12.6V CT
transformer and two diodes on the input side. This gave it an input voltage
of 8.9V (or, 6.3V x 1.414 = 8.9).

The 12.6V transformer output is rated with the primary at 230V. Now I am
considering using it on 277VAC. This should bring the input to the 7805 up

At 8.9V no heatsink was required with a .5A load. How can I pre-figure how
much power will be dissipated when on the 277V primary, so I can select a
heatsink, or perhaps decide that the whole idea is either OK or NG???

I checked the National datasheet for the 7805, and still came up short.
Is it as simple as saying that I will have 5.7V dropout at .5A, thus will
have 2.85W to deal with?

TIA,
Chris

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At 02:53 PM 6/19/02 -0400, you wrote:

Power dissipation is just:

P= (input voltage - output voltage) * (output current + regulator current)

Regulator current is typically around 5mA for the 78xx05.

You should work out heat sink size based on worst case input voltage,
output voltage tolerance, and ambient temperature.

Best regards,

{Quote hidden}

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At 02:53 PM 6/19/02 -0400, Chris Loiacono wrote:

>The 12.6V transformer output is rated with the primary at 230V. Now I am
>considering using it on 277VAC. This should bring the input to the 7805 up

Don't do that!

I think that you will cook your transformer if you over-voltage the input
that much.  Core saturation.

How much power do you need?  We use several custom transformers for some of
our products - one of those has 120V, 277V, 347V primary taps and supplies
20 Vac @ 4A.  It mates with a little card that has a NS simple switcher
that you could set to whatever voltage you wanted within the range of 5Vdc
- 14Vdc.

dwayne

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(780) 489-3199 voice          (780) 487-6397 fax

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> >The 12.6V transformer output is rated with the primary at 230V. Now I am
> >considering using it on 277VAC. This should bring the input to the 7805
up
>
> Don't do that!
>
> I think that you will cook your transformer if you over-voltage the input
> that much.  Core saturation.

Doesn't core saturation depend on current, rather than voltage? I would
think that you'd have less chance of saturating at 277V than 230V.

Bob Ammerman
RAm Systems

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{Quote hidden}

But no-load current depends heavily on primary coil voltage.

Mike.

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At 11:39 PM 6/19/02 -0400, Bob Ammerman wrote:
> >
> > I think that you will cook your transformer if you over-voltage the input
> > that much.  Core saturation.
>
>Doesn't core saturation depend on current, rather than voltage? I would
>think that you'd have less chance of saturating at 277V than 230V.

Try it and see!  Do it outside so that the stink has doesn't get to you ;)

I'm not sure how best to describe this.  A transformer core can handle only
a certain flux density before it saturates.  This is directly proportional
to applied voltage as follows:

Peak AC Flux (Gauss)   B(AC) = Vrms x 10^8 / 4.44 * N * f * A
B gauss  V volts  N #turns  f Hz  A cm^2

One of many web pages that describe this:
<http://www.coilws.com/magneticandhow.html>

I had a couple of our power supplies returned after failing because the
tap.  They didn't smell nice :(

dwayne

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(780) 489-3199 voice          (780) 487-6397 fax

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>> I think that you will cook your transformer if you over-voltage the input
>> that much.  Core saturation.
>
>Doesn't core saturation depend on current, rather than voltage? I would
>think that you'd have less chance of saturating at 277V than 230V.
>
>Bob Ammerman
>RAm Systems

Nope. The core must standoff the volt*second product. Thus, more volts
needs more core. Likewise, lower frequency means more seconds and also
means more core needed.

This is why higher freq operation is often desirable. If a transformer is
designed for one input voltage and you find you must increase the voltage,
it helps to offset this by raising the operating frequency also if possible.

The core in question *may* work at 277V but this can only be determined by
testing or consulting the manufacturer.  Since excess core usually means
excess expense, the cores are usually made to just barely do the job and
typically run warm at rated load. Raising the voltage will usually raise
the temperature also. Determination of suitability is left as an exercise
to the student.

Tom M.

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This is why I contacted the manufacturer at the same time the original
message was posted here. I don't expec to hear from them any time
soon...Since changing the freq is out of the question, I am most curious

CL

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What sort of wattage are you after?

RS Components have 50/100/200VA kits that comes with a pre wound 115/230
primary on it, and a piece of paper with tables for calculating a secondary.
The unit is not potted, and so it should not be too hard to either remove
the existing primary, and replace it, or else add a supplementary primary
winding before putting a secondary on it. If you did the latter then make
sure you put sufficient insulation in between the primary and secondary.

The kit has a plastic bobbin with primary windings, and a stack of
laminations to put in afterwards, along with the necessary metal pieces to
clamp it all up afterwards.

At rshttp://www.com/ look for catalogue # 182-9919 for 50VA, 182-9925 for
100VA and 182-9931 for 200VA

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On Wed, 19 Jun 2002, Chris Loiacono wrote:

>I have a circuit with a 7805 regulator that was set-up with a 12.6V CT
>transformer and two diodes on the input side. This gave it an input voltage
>of 8.9V (or, 6.3V x 1.414 = 8.9).
>
>The 12.6V transformer output is rated with the primary at 230V. Now I am
>considering using it on 277VAC. This should bring the input to the 7805 up
>
>At 8.9V no heatsink was required with a .5A load. How can I pre-figure how
>much power will be dissipated when on the 277V primary, so I can select a
>heatsink, or perhaps decide that the whole idea is either OK or NG???
>
>I checked the National datasheet for the 7805, and still came up short.
>Is it as simple as saying that I will have 5.7V dropout at .5A, thus will
>have 2.85W to deal with?

Maybe you can remove 1 diode and have more ripple. This will reduce the
power on the regulator. Simulate your circuit with spice. I am amazed that
you ran 9V into it (dissipating 2W) and did not need a heatsink. You will
need a small heatsink for 3W but you can remove the diode as I said or you
way I think. You can put it before the filter cap too but then you will
need another value.

Peter

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If you want to run off of 277VAC and you only have 230V transformers
available, get a 230V to 48V transformer and wire it as a stepdown
autotransformer.  230V + 48V = 277V aprox. so you can input 277 to both
windings in series (correct phasing) and get 230 out, everything will be in
spec.  Ideally you would get a transformer with both 48V and 12.6V windings
so you don't need two chunks of iron.

To test phasing try putting a light bulb in series with the primary on your
first test.  If the phasing is wrong the core will saturate and the light
will light.  If the lamp stays dark you have the phasing right.  This test
is done with no load on the output.

Doug Butler
Sherpa Engineering

> {Original Message removed}
On Fri, 21 Jun 2002 12:41:21 -0500, you wrote:

>If you want to run off of 277VAC and you only have 230V transformers
>available, get a 230V to 48V transformer and wire it as a stepdown
>autotransformer.  230V + 48V = 277V aprox. so you can input 277 to both
>windings in series (correct phasing) and get 230 out, everything will be in
>spec.  Ideally you would get a transformer with both 48V and 12.6V windings
>so you don't need two chunks of iron.
,,and you need to have the full mains spec isolation between the 48
and 12.6v windings, which will often not be the case for standard
transformers, which only have full isolation between primary and
secondary, not between secondaries..

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