Regulating High Voltage

Discussion in 'Electronic Basics' started by Mike, Nov 15, 2004.

1. MikeGuest

Hi,

I'm trying to design a high voltage DC power supply, which will be used for
charging up a capacitor. To start with I only want an output of 400V DC.
The supply voltage to the supply will be around 10V AC.

My current design consists of a simple step-up transformer (10V AC to 400V
AC), a bridge rectifier (AC to DC), and a smoothing capacitor (reduce the
ripple). With this design (I think) I should be able to get a high voltage
of something around 400V DC. As it's used to charge a capacitor I don't
want the voltage to be any higher than 400V. But I don't want the voltage
to be much lower than 400V. So I guess I need someway of regulating the
output. It seems a zener diode is the way to go, but I can't find any that
will regulate 400V (only 200V).

So my question is, does anyone know a way to regulate the output voltage
without using zener diodes? I could add / remove turns from the transformer
until I get the correct voltage, but this is far from ideal (not to mention
tedious).

If it makes any difference, the current requirements of the supply are 10mA
at 400V.

Thanks for any help,

2. Roger JohanssonGuest

Take two.

Two diodes in series, add the zener voltages.

3. Jonathan KirwanGuest

I'm not sure how good that will be. The 1N5281A is rated at 200V @ 650uA but
it's tolerance is 10% with some .11%/C temperature drift, on top. I suppose the
OP could hand select them to get the right voltage, though.

Also, since the Izt of the 1N5281A isn't anything like the 10mA desired, a BJT
emitter follower would be needed that only requires some percentage of that
650uA biasing, say 100uA (it's .16V change for 65uA change in bias, which is
probably fine.) This means beta of 100, which may be achievable so long as
there is at least a volt or two drop across V(CE). But how much the V(CE) will
have to stand off could be a problem. The MJ12005 is mentioned in the Art of
Electronics, 2nd ed., as an NPN with Vceo of 750V, though. Also, that same book
includes a couple of high voltage circuit examples on pages 369 and 370 that
could be used for something like this. The second example regulates the ground
return.

Jon

4. Roger JohanssonGuest

Two of them in series will still have the same tolerance.
If his 200V zeners have the precision he needs then two of them will have
the same precision.

5. Jonathan KirwanGuest

10% tolerance, yes. At 400V, put in absolute terms, this is +/-40V, yes?
This still begs the question, though.

Is 10% okay?

And it still means that an emitter follower BJT is required as the 650uA Izt
rating simply cannot be used for a regulated 10mA supply without one.

Jon

6. peterkenGuest

bad idea using zeners, since those high voltage zeners have large tolerance
(upto 10%)
also, if they are "regulating" constantly at say 5mA then every zener must
dissipate 1W....
bad efficiency I think, also the switching primary side will get in trouble
eventually

so the way to go is by using feedback
IF common of primary and secondary may be connected, then use a voltage
divider (resistors), with say a ratio of say 1:100 (proposal 1M/10k, close
enough for trimming)
this way you get 4V from the divider, which can be used to "measure" the
secondary voltage and regulate the 10Vac in.

IF common of primary and secondary may NOT be connected, then use a linear
opto-coupler to create primary side "measuring"

7. Roger JohanssonGuest

If he was going to use one 200V zener, he can just as well use two, the
parameters are very similar for two in series and one single zener of the
same type.

8. Jonathan KirwanGuest

Agreed. But I think another part of the post, "I don't want the voltage to be
any higher than 400V. But I don't want the voltage to be much lower than 400V,"
bears on this question. I don't think the OP knew about the 10% regulation of
such high-V zeners in the first place. But we can see what the OP says. We're
past the useful point of "intent mining" of the OP's post, I think.

Jon

9. CFoley1064Guest

Subject: Regulating High Voltage
Hi, Mike. Hmmn. Quick 'n' dirty cap charger. 400VDC, 10mA max, 10VAC from a
transformer available.

Before you start, remember that you're dealing with potentially lethal voltage
here, and you should observe appropriate safety precautions.

Also, you should be aware that there are good isolated, current-limited high
voltage power supplies available at hamfests, ebay and such that will give you
exactly what you want with no hassle and at good prices. Just provide the 40K,
10 watt series resistor (use 2 ea. 20K, 5 watt in series) and you're good to
go.

Also, if you happen to have a constant voltage transformer handy, you might not
need a regulator at all. Just set up your transformers and rectifier, and
tweak a voltage divider in using a "Dividohm" power resistor and another diode
to make it so your peak no load voltage (cap fully charged) is 400V.

Having said that, the prior post mentioning zeners is also a good idea,
although I'd probably use a string of 10V and 12V 1 watt zeners (1N4740,
1N4742) to do the job. Of course, I happen to have a small drawer full of
these. Use any 1W zeners below 40V. That would allow me to tweak in final
voltage to within a couple of volts of where I wanted to be, and these zeners
are commonly available. If you've got access to It would look something like
this (view in fixed font or M\$ Notepad):

____ T1 ___ ___ ___ +
o-|_--_|---o--. ,---. .--|___|-o-|___|-o-|___|-o-->|--o
FU1 | )|( | | R1 | R2 | R3 | D1
| )|( | | | | |
.----' '-. | .-------. | | | | 400V
| | | | | | | | | |
| | T2 | '-o ~ + o-' | VZ1/-/ |
| o--. ,-' | | +| ^ .-.
120VAC | | )|( | BR1 | --- | | |
| | )|( | | --- | R4| |
o----' '-. | | C1 | VZ2/-/ '-'
| | | .-o ~ - o-. | ^ |
| | T3 | | | | | | | |
| '--. ,-' | '-------' | | | |
| )|( | | | | |
| )|( | | | | | -
o--------o----' '---' '--------o-------o-------o------o
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

T1 - T3 120VAC pri, 10VAC sec, 20 VA plus transformers
R1 100 ohm, 10 watt
R2 5000 ohm, 5 watt
R3 40,000 ohm, 10 watt (or 2 ea. 20K 5W)
R4 330K 1 watt (bleeder resistor)
C1 10uF 600V or greater electrolytic
BR1 -- 800V bridge rectifier, 6A or more
D1 1N4007
VZ1 - VZ2 String of 1N4740 (10V) and 1N4742 (12V) 1 watt zeners on a perfboard
(you'll need 33 to 37 total -- stack up the '42s to get over 400V, then replace
'42s with '40s to get down to 400V+/-2V)

Out of the above components, you'll probably have the most trouble finding a
cap. You might want to scrounge one out of some older toob stuff (check the
cap first to avoid misery).

The idea is that you get the 10VAC from your first transformer, and backfeed
three more of the same type of transformer, applying the 10VAC to the
secondary. You'll get around 120VAC out at the primary of each (make sure to
drastically oversize the transformers for this job), and putting them in series
will give you 360VAC, which should rectify to a little more than 500V. R1
limits the surge current at turn-on, saving your bridge rectifier and possibly
the transformers. C1 accepts the charge, and then R2 gives you a voltage drop
for the shunt zeners. The 5K value sets around 20mA steady state, out of which
10mA will be stolen from the shunt by the load when cap is initiating charge.
20mA (no load) can be handled easily by the 10V and 12V zeners. The 40K R3
limits the current charging the cap, and R4 is the bleeder resistor necessary
to discharge C1 when you turn off.

One thing I used to do was have a microswitch interlock, so that if somebody
opened the door to remove the cap or otherwise placed themselves in peril, the
power would automatically be turned off, and the bleeder resistor (which could
be made much lower to discharge more quickly) was switched into the circuit.
Think carefully about your application, and how to minimize the danger, before
you start. After somebody gets hurt is too late. Don't forget the fuses, but
remember that they're not personal protection devices.

Good luck
Chris

10. BanGuest

He said he wanted 400V not 40.

11. MikeGuest

Thanks for all the replies.

I've had a look at the circuit in the Art of Electronics book and it looks a
bit complicated for me. I think using zener diodes in series is the way to
go. I didn't take the tolerance into account in my original design. The
zener diodes I've seen have a tolerance of (+/-) 5%. So I think I'll try
and get a voltage of around 380V. 5% of 380V is 19V and 380+19 = 399V (or
it could be 380-19 = 361). It's not the original 400V I wanted, but it
seems fairly easy to implement.

If I use zeners in series, how do I calculate the value of the resistor?
For a single zener I would use:

R = (Vs - Vz) / Imax

Do I just add up the individual diodes Vs and Vz?

Thanks again.

12. Roger JohanssonGuest

Yes, two 200V zeners become a 400V zener when connected in series.
Decide what current you want, and find the resistor value.

But what are you trying to use the zener for?

What kind of circuit are you talking about?
Where is the schematics?

13. Jonathan KirwanGuest

I'm not sure if this is a confused view or not. Basically, when you use two
200V zeners in series to get 400V, but with 10% tolerance, you can expect to see
anywhere from 360V to 440V at the node when you wire it up. This means it won't
necessarily be 399V, at all. It might be. It probably won't be.

The 1N5281 may come in five flavors:

--1 unit-- --2 units--
1N5281 +/-20% +/-40V +/-80V
1N5281A +/-10% +/-20V +/-40V
1N5281B +/-5% +/-10V +/-20V
1N5281C +/-2% +/-4V +/-8V
1N5281D +/-1% +/-2V +/-4V

I've honestly NO idea which of these you can find or what you can live with.
But there is the list. Also keep in mind that the voltage is spec'd at those
tolerances when EXACTLY I(zt) is flowing. Variances in I(zt) will add
additional error, as will temperature-caused variations (and time, too.)
Hmm? "diode Vs"? I know what Vz is, but do you mean your supply voltage?

I gather that you read Art of Electronics on those mentioned pages. Take a look
at pages 68 and 69, also.

You mention a 10VAC input and a 400VAC output, fed to your bridge. This 400V is
reasonably assumed RMS, unless you state otherwise. This output is a pulsing
full wave output with peaks at about SQRT(2)*400V or about 560V peaks. This is
followed by capacitors for "smoothing," as you say. With no load at all, these
will largely just sit at the high voltage. But with a load applied, the
capacitors will be supplying the current during significant portions of the AC
cycle (the frequency of which you haven't mentioned, if memory serves.)

gathering that your topology is simply a resistor connected in series to the

In any case, the peak filter capacitor voltage of about 560V reached at the peak
of each cycle will decline as the capacitors supply the current, once the
transformer voltage declines faster and separates away towards zero. At some
point towards the next cycle, this transformer voltage will catch back up with
the falling voltage on the caps and will then supply both the caps and the load,
causing the voltage to follow the sine curve on the way back up to 560V. This
bottom point, let's call V_bottom. V_top will be 560V, let's say.

You have to set your resistor so that it will supply up to a maximum of 10mA to
Let's say that we set this "little something" for the zener pair as 250uA
(somewhat below the stated 650uA, but don't worry -- they will get a lot more,
later.) Let's say your zeners are somehow exactly tweaked so that they zener
right at 400V. Also, let's pick (arbitrarily, for now) that V_bottom will be
500V. So, we have: R= ((V_bottom-400V)/(10mA+250uA)) = (500-400)/(10.25mA) =
9760 Ohms. Call it 9.1k. (V_bottom is used because R must be set to still work
at the lowest source voltage it will be presented with.)

comparison, 400V/10mA.) The rise from 500V to 560V takes about 3ms at 60Hz and
the half-cycle time at 60Hz would be about 8.3ms, so the droop time that the cap
is supplying current will be roughly 7.1ms (a little less, probably -- I'm using
this: {[PI/2-arcsin(V_bottom/V_top)]/(2*PI*f)}, as an estimate.) Using this
figure as a rough guess, your filter capacitor should be about I*dt/dV. Your
peak current will be based on V_top, and will be about 17.6mA. So the cap will
more worst case, so let's call it 2uF for now.

Now, getting back to the rest. At V_top=560V, your current will be about 17.6mA
(160V/9.1k), assuming the zener voltage is stable (which it isn't.) Since your
load is taking up 10mA, hypothetically, this means 7.6mA is pouring through the
zeners. At V_bottom=500V, your current will be about 11mA (per design for
100V/9.1k) and 1mA will be pouring through the zeners.

What if there is no load? Then there will be anywhere from 11mA to 17.6mA
pouring through the zeners.

The voltage slope of the 1n5291 is about 2500 ohms at 650uA. This suggests (as
a first order guess) that fluctuations of from 1mA to 7.6mA should yield
variations of (7.6mA - 1mA)*2500 or about 16.5 volts variation. This is the
can handle the current load. But at 400V across them and some 5mA average
current flow, that's two watts of power. These are 500mW devices, so this
definitely exceeds their capabilities, assuming you could tolerate to voltage
variations.

Assuming your load is constant, you can make this better by sizing your
capacitor upwards, so the droop is less and then the resistor can be better
sized and the average current through the zeners closer to their design range.
Roughly, I get something like:

V_delta = (V_top-V_bottom) * Rzt * I_total / (V_bottom-V_out)

As you can see, this arrives at (560-500)*2500*(10.25mA)/(500-400) or over 15V
in the case we mentioned before. Since I arbitrary adjusted R down to 9.1K,
though, that's why the slight variation here.

As you can see, moving V_bottom closer to V_top improves things in two ways.
With V_bottom set to 540V, for example, you get:

(560-540)*2500*10.25mA/(540-400)

Which is about 3.7V peak-to-peak ripple voltage.

R= ((V_bottom-400V)/(10mA+250uA)) = (540-400)/(10.25mA) = 13660 Ohms.

If you actually used a 2% resistor, you can get a 13.7k value. Very close.
This would be (160+140)/(2*13.7k) or about 11mA. That's an average of 1mA for
the zeners, or about .4 watt for both (.2 watt each.) So that is within their
specification. But now your capacitor needs to be:

(160V/13.7k)*(PI/2-arcsin(540/560)/(2*PI*60)/(560V-540V)

or about 4.5uF. Call it 4.7uF ... @ 600V, or so.

However, the zeners will be experiencing as little as (540V-400V)/13.7k - 10mA =
220uA and as much as (560V-400V)/13.7k - 10mA = 1.7mA, which at Rzt of 2500 Ohms
means: (2500)*(1.7mA - 220uA) = 3.7V -- which is what we calculated for the
estimated ripple above. Good.

Now what happens when you remove your load? Well... all of some 11mA goes
through the zeners for about 11mA*400V, plus some more for the Rzt -- probably
about 5 watts total. More than the two zeners are likely to handle.

Or, what happens if your load doesn't actually use all of the 10mA? Same thing
-- the zeners pick up the current your load doesn't use. And that translates to
dissipation. This is a reason why an emitter follower is often used.

There are a lot of factors I didn't worry about, but this is a napkin kind of
thing to give a rough sense. I again need to point out that I'm not an expert
and not even trained on these things -- I am a hobbyist who has read some stuff
and built one or two things and that's all. So take this with a grain of salt.

Jon

14. John FieldsGuest

---
If all you're using the supply for is to charge a cap, then you could
do something like this:

+HV
+------+ +-----+ |
MAINS>----|P1 S1|--|~ +|---[16K]--+--[1N4007>]---<<---+
| | | | | |
| | | | [40K] [CAP]
| | | | | |
MAINS>----|P2 S2|--|~ -|----------+--------------<<---+
+------+ +-----+
XFMR FWB

Assuming no copper losses in the transformer and no leakage current
through the cap and based on 120VRMS nominal mains, 10mA out of the
secondary, and the resistances shown, +HV will be about 400V. the 16k
resistor will dissipate about 1.6 watts and the 40k will dissipate
about 4 watts after the cap is charged, so a 5 watter and a 10 watter
would probably be OK.

15. CFoley1064Guest

Subject: Re: Regulating High Voltage
Hi, Ben. Sorry -- the ASCII art isn't correct, even though the explanation is.
I'm assuming he has 10VAC. If he gets three more 10VAC transformers and
backfeeds them (applying 10VAC to the secondaries, using them as the primaries
in step-up configuration), he can get 120VAC out of the primary of each one,
giving him 360VAC if he puts them in series to go to the rectifier and shunt
regulator.

Sorry it wasn't clear. Slinging ASCII. My bad.

Chris

16. John FieldsGuest

On Wed, 17 Nov 2004 08:54:52 -0600, John Fields

Ooops...
^^^^^^^^^^^^^^^^^^^^^
a 10VRMS input to the transformer
and the schematic should look like this:
+HV
+------+ +-----+ |
10VRMS>----|P1 S1|--|~ +|---[16K]--+--[1N4007>]---<<---+
| | | | | |
| | | | [40K] [CAP]
| | | | | |
10VRMS-----|P2 S2|--|~ -|----------+--------------<<---+
+------+ +-----+
XFMR FWB

17. MikeGuest

That looks simple enough, but what if the 10VRMS has some noise? Wouldn't
that cause more than 400V to be put across the capacitor?

18. MikeGuest

I've been looking at the high voltage regulators in the 'Art of Electonics'
book again. On page 371 it has an opto-isolated high voltage regulator,
which looks pretty straightforward. I think this is what I'll use. By
altering the Vref signal I should be able to fine tune the output.

Does anyone have any opinions on this circuit? The book is pretty popular,
so I guess most of you have it. If not I could post the schematic on one of
the binary newsgroups.

My only question on the circuit is how fast should the comparator be? If it
was slow, then I guess high voltages could be output.

Thanks again for all your replies.

19. NDGuest

You could put 2 of your 200V zeners in series, or use a resistor network and
take your output across a portion of the network.
There are circuits which use lower voltage zeners to control a transistor
with a resistive load also. Look around you can probably find something
like one of the options above for your needs.