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