Regulating High Voltage

[Mike]

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,

 

[Jonathan Kirwan]

On Tue, 16 Nov 2004 17:33:21 -0000, "Mike" <mike@notemail.com> wrote:

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.

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.)

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?

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.)

This system then supplies current for both your zeners and your load. I'm also
gathering that your topology is simply a resistor connected in series to the
zener-pair paralleled with your load -- no emitter follower.

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
your load plus a little something for your pair of zeners at this V_bottom.
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.)

Okay, so now you have R=9.1k. (Your load is assumed to be about 40k, by
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
be about 17.6mA*7.1ms/(560V-500V) which works out to about 2.1uF. But that's
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
ripple you are likely to see on your load. Assuming your zeners stay cool and
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

 

[John Fields]

On Mon, 15 Nov 2004 11:55:27 -0000, "Mike" <mike@notemail.com> wrote:

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.

---
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.

--
John Fields

 

[John Fields]

On Wed, 17 Nov 2004 08:54:52 -0600, John Fields
<jfields@austininstruments.com> wrote:


Ooops...

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
^^^^^^^^^^^^^^^^^^^^^

a 10VRMS input to the transformer

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.

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


--
John Fields

 

[Mike]

"Mike" <mike@notemail.com> wrote in message
news:cna5g4$c7o$1@news6.svr.pol.co.uk...

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,

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.

 

[ND]

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.

"Mike" <Mike@whatever.com> wrote in message
news:cnla6e$taa$1@news8.svr.pol.co.uk...

"Mike" <mike@notemail.com> wrote in message
news:cna5g4$c7o$1@news6.svr.pol.co.uk...
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,




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.