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PRR

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..The preamp is only pulling 4ma and I plan to use a 100va toroid transformer.

4mA at 260V is about one VA. Why use a PT one hundred times larger than you need?

*Good* Zener regulation probably needs raw DC about 1.5X the required voltage. 390V. Voltages over 450V discourage the use of cost-effective Electrolytic caps. So you want around 280V AC winding.
 
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Hi PRR,
Dropping resistor? Zeners are appropriate for that current and voltage depending on standing current. Around 8 mA for the zener stack?

I'm only asking because there is next to zero detail provided except for transformer output voltage for B+ (which I agree with). So the power supply isn't really designed without some important details. You are usually bang on with your figures, so I was surprised when you didn't provide the extra details.

-Chris
 
I am using two 130v 5w zener diodes for voltage regulation in my power supply. How much over the 260 volts that I need should I size the transformer? The preamp is only pulling 4ma and I plan to use a 100va toroid transformer.

Thats a very odd design methodology you have going on there.

Typically, you'd start with the voltage required by the circuit (presumably 260V @ 4mA), choose a traffo that comes in about 1.25 to 1.5 times that voltage at a reasonable headroom of current, design your filter and smoothing circuit, and then regulate appropriately if required.

Is there a particular reason for the big transformer?
 
I'd also ask what you'd like to put between raw B+ and the zener diodes. I would treat a CCS and a dropping resistor a bit differently.

You've specified 4mA for the preamp, so the next step will be to look at the datasheet for your zener diode and see what its actual voltage drop is at 4-12mA of current. The 1N5381 specs a max zener voltage of 136.5V and the voltage vs. current curve actually looks pretty nice, with the diode being a bit under 130V for currents under 10mA.

So the max regulated voltage could actually be 273V if you get unlucky with the diodes. I'd set your target voltage at about 370V and use a 10K resistor between B+ and the zener stack. This will give you 10mA total if your diodes are out of spec on the high side and 12.3mA if your diodes are on the low side. I wouldn't want to run more than about 10mA through a 130V/5W zener for very long. They also seem to survive better if you leave their leads kinda long.

If you plan to use a CCS, 300V is going to be enough for most of them. You'll also dissipate less heat (the 10K resistor will only be cooking off 1.5W though, so we are only talking about a 5W resistor here).

I hope this helps.

(Also, the comments about the 100VA transformer are spot on)
 
Off on a different track - let's just look at the question again. OP wants to regulate a tiny power supply requirement (4 ma) with a couple of Zeners and we should assume, a moderately large valued ballast resistor in series with the Zener stack.

Parenthetically, the OP is going to have a step-up transformer that is wildly oversized for the application. But that part doesn't matter, if one is designing for the goal, rather than the unused source portion.

So the goal is clearly "260 volts more or less" and "regulated with Zeners" (which is do-able because of the small nominal operating current).

5 watt Zeners should not be operated at 5 watts continuously. At least not if we want them to survive for decades. 3 watts, just fine. 4 watts, OK too … but a little close for comfort.

P = IE → 3 = I⋅130 V
I = ³/₁₃₀
I = 0.0231 amps (× 1000 → 23.1 mA)
There's our set point. 23 milliamps. It'll figure into the next (and unfortunately unspecified) part of design for Zeners: figuring the size of the ballast resistor.

Let's say that one allows for the 33:66 rule. 66% of the voltage across the Zener stack and 33% or so across the ballast resistor.

(130 + 130) ÷ ⁶⁶/₁₀₀ → 390 v total on the B+UNREG side. GREAT! Now we know what the ballast resistor will be handling: 390 - (130+130) = 130 V. Remembering Ohm's Law:

E = IR → 130 = 0.023 R
R = 130 ÷ 0.023
R = 5,600 Ω
… and to calculate its power dissipation …
P = IE
P = 0.0231 × 130
P = 3 watts
Now let's also use the real-world: 3 watt resistors are hard to find. 5 watt is easy, but often wire-wound, which is neither really a benefit nor a deficit. Perceived by many to be a not-so-good choice.

But let's go the Whole Hog. We will design for 5 watts of nominal headroom dissipation. We need 5.6 kΩ. And entirely arbitrarily, we want to use a bunch of ½ watt resistors to do the job.

5 W total ÷ 0.5 W each = 10 resistors​

We will run them in SERIES (as this is normally safer - getting the job done while also dividing the resistive drop-voltage by N resistors), so that's ¹/₁₀ of the total per resistor.

5,600 Ω × ¹/₁₀ → 560 Ω ea​

OK! Swinging over to Mouser shows some Vishay metal film ½ watt 560 Ω 5% 200 V resistors for 3.8¢/ea quantity 50. ("Why not…? Stock them!"). So all-in, this is a 38¢ solution.

Working further backwards, what size transformer? Shall we say “semiconductor rectified?” Sure, why not. 1.0 V or so per diode voltage drop. Let's use a full-wave bridge in the single-winding (not center tapped), 2 diodes-in-series for-both-phases configuration. A reasonable peak-holding capacitor (in this case 47 μF, arbitrarily chosen, 450 V can) is chosen to follow the rectifiers. 390 / √(2) = 275 VAC RMS as the secondary winding of the power transformer. Since (1 + 1) volts of voltage drop is near-nothing, and we can't even guarantee what the input power plug voltage will range between, let's just ignore it.

The transformer will provide RMS 275 volts and 23 ma. Remembering

VA = V • A (E•I RMS)
VA = 275 × 0.0231
VA = 6.4​

OK!!! Now I'd think about the regulation chain. Perhaps it makes sense to do a CRCRC pre-regulation filter chain … and we can use up some of those series 560 Ω resistors to the task as well. Hey, it works.

FWB rectifier
→ 47 μF / 450 V
→→ (560 + 560) Ω
→→→ 47 μF / 450 V
→→→→ (560 + 560) Ω
→→→→→ 47 μF / 450 V
→→→→→→ (6 × 560) Ω
→→→→→→→ (130 V + 130 V zeners). …
→→→→→→→→ yet another 47 μF “reservoir” cap
→→→→→→→→→ the preamplifier, as B+ supply.

Reasonable enough? Yep. $1.62 per 47 μF ÷ 450 V CDE (Illinois Capacitor brand). FWB is 80¢ for a 1 amp, 1 kV unit. Transformer? It already is in the OP's hands. The costs of the passive parts isn't even $10.00 all in. That's my kind of PS.

Done.
GoatGuy

A very important PS: ... you could just as easily use any value from the 560 ohms demonstrated above, to 3x that or 1.5 kohm and realize an important free-win. Since all that changes in a static sense is quiescent current, as long as quiescent current is reasonably above the expected (4 mA) current drain, well ... there's still headroom.

But, dropping quiescent current increases ripple suppression by the CRCRC stack. The power supply residual noise DROPS when you use less overhead current draw. Indeed... I might just use 12 ea., 1 kohm, metal film resistors instead of 10 of them. 1 kohm resistors are as cheap-as-dirt. And your box should have a bunch in any case. {end PS}
 
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And, of course, there is always the venerable pmillett Maida regulator: High Voltage Regulator

yah, that (them?) too.

However, in a way there's a certain elegance out of using non-active components to get the job done, especially when the power consumption of the target circuit is so modest.

I could be encouraged into a gas-discharge Zener design. But… since they have a decidedly unappealing hysteresis (i.e. a fairly higher strike voltage, then a non-instantaneous droop, a kind of negative resistance), the use of a post-clamping HF damping capacitor has significant limitations. And lets be fair: the sharpness of their conduction isn't really all that sharp at all.

They were best before Zeners.
Now Zeners are palpably better.

Just no light show.
Oh, darn.

As I've done (only for customers), I've put gas discharge regulator tubes onboard, powered by a few milliamps thru a high-ohmage resistor. To LIGHT UP, mostly. I let zeners do the actual voltage-reference thing, then use an old-fashioned, but actually quite effective series MOSFET regulator and parallel-to-reservoir-capacitor trickle drain resistor (safety!!!) to keep parts count low, regulation excellent and ripple rejection spectacular.

It works.
I like things that work.

GoatGuy
 
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Too complicated. Running the zeners at 8 mA and the load wants 4 mA gives you a total of 12 mA. Lower power and less heat radiated. Nice and neat.

The resistor will drop about 130 VDC (for 390 VDC that PRR suggested), so 12 mA will generate 1.6 watts of heat. So an 1,100 ohm, 3 watt or 5 watt resistor will do fine. If the resistor is mounted on a terminal strip, you can use a 3 watt since it will have much better air flow. Each zener will dissipate about 1 watt, so they will run warm with good ventilation. That setup will deliver reliable power around the 260 VDC required. Don't forget the filter caps at this point as well as the first one. If you want, you can divide the 1,100 ohm resistor into two parts with another filter capacitor to really clean the DC up.

It's that simple.

-Chris
 
Too complicated. If you want, you can divide the 1,100 Ω resistor into two parts with another filter capacitor to really clean the DC up. -Chris

Yah, sure… there definitely are 100 ways to make an omelette. As an old and ornery electronics designer who hates endlessly stocking and buying “yet another different part”, I like to use relatively standard parts that I might actually have. It simplifies repairing things in the future too.

However, you made an error in the 1.1 kΩ recommendation. Dropping 130 volts (which we agree upon, and PRR) at 12 ma (4 for the circuit, 8 for headroom) is:

E = IR →
R = E/I
R = 130 ÷ 0.012
R = 10,800 Ω​

Or about 10× what your recommendation was. You definitely do not want a 1,100 Ω resistor. You'd burn up the Zeners (and the resistor). If you like the idea - as I do - of a CRCRC chain, well … do that resistor splitting!

In fact since the exact values are not critical, you might as well say “10 kΩ”. The wattage?

P = IE
P = 0.012 A × 130 V
P = 1.56 W​

Again, since we're really not talking about something radically different, you could plan for a lower resistor power dissipation (say 2 watts of capacity) and use 4 resistors in series to get you to 10 kΩ. 4 ea, 2.7 kΩ ½ watt el-cheapos does the trick.

Whether you think this is complicated or not really isn't the issue. My recommendation was specifically leveraged off the notion that “if the design criteria is to use up a bunch of ½ watt resistors”.

Just saying.
We're agreeing more than disagreeing.
And I like CRCRC.

GoatGuy
 
yah, that (them?) too.

They were best before Zeners.
Now Zeners are palpably better.
Too bad they are so fragile. I use VR tubes for regulating the screen grids in one of my amps that a Zener would simply fail shorted almost instantaneously. If only they made a high power zener...


I also came across this circuit to get around the fact you can't use a large cap after the VR tube...
voeding_vr_tubes.GIF
 
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Too bad they are so fragile.

voeding_vr_tubes.GIF

They are, aren't they? (Zeners, fragile). Such a small package (and following that, such a small bit-of-a-chip of silicon) is the issue. A VR tube glow-cylinder is maybe what 35 mm tall, 15 mm diameter?

A = 2πRH
A = 2π7.5×35
A = 1,650 mm²​

Compared to the little, conductively useless plastic package for a Zener of 12 mm long, 3 mm diameter (same calc = 113 mm²) … 1650 ÷ 113 = 14.5× the surface area. Granted, that surface is in the near vacuum of whatever magic VR gas mixture makes it light up. Still … over current just causes them to glow brightly and get hot. And wear out fast. Not so, Zeners.
_______

I think the author of that schematic (just above) made 3 errors in transcribing.

(1) 1 μF 600 V peaking cap.
(2) C.T. secondary winding connected to heater
(3) pissy, but a Hy is a henry, not He … which is helium…​

OK, (3) was a bit tetchy. A catch is a catch. (2) really shouldn't have the HV center-tapped winding (at the transformer's secondaries graphic) extended to the filament 5 V winding. Doesn't work that way. And (1) the 1 μF cap is not sized right for a total of (30 + 25 + 20 = 75) ma of current demand. Remembering…

ΔV = IΔt/C

Δt = 1 / (2 F) … for full wave rectification
Δt = 1 / (2×50) … Europe 50 Hz input
Δt = ¹/₁₀₀ = 0.01 sec … now plugging in

ΔV = 0.075 A × 0.01 s ÷ 10⁻⁶ F
ΔV = 750 V​

You can see that the purpose of the 1 μF cap (peak holding) is overcome by the ΔV. (Of course on the bench you wouldn't get ΔV=750 to show up on the oscilloscope. The actual inter-peak voltage would just be 'behind' the rectified pulse stream, collapsing to near-zero inter-pulse.)…

BUT that equation (ΔV = IΔt/C) is useful for estimating what size front-end capacitor you would want to specify. Since it isn't hard to grok that the following inductors tend to settle on the RMS average of whatever 'shaped' AC is presented to their front end, well … then having the initially leveled DC+AC be VPEAK-to–10% less than VPEAK seems reasonable, yes?

The HV secondary is spec'd at 270 VAC RMS. We know that VPEAK = RMS × √(2) = 270 × 1.414 → 381 V. 10% of that is 38 V. So, we might want our ΔV to be 38 V.

ΔV = IΔt/C
C = IΔt/ΔV
C = 0.075 × 0.01 ÷ 38
C = 0.00002 F (… × 1,000,000)
C = 20 μF​

And the closest 'standard sized' one would be 22 μF.
_______

Now - I completely understand that the original designer was NOT trying to peak-hold. Clearly this is so in the annotated voltages. The first DC rail is 270 VDC. The inter-choke rail is 225 VDC. The final is 180 VDC.

That being the case, all of the above is misguided on my part. The 1 μF capacitor (non-electrolytic, to be sure), serves only to act as a 'buzz squasher'. In may not even be required at all, but is OK to have for that purpose. I tend to use lower HV windings and peak-holding initial electrolytics per the above calculations.

I guess it is a matter of style.
(After writing all this, I didn't feel like capitulating and just not posting it.)

In any case, remember (№ 2) error on the winding thing.

GoatGuy
 
A very informative post as usual, GoatGuy.



No. 2 Looks to me like it's not connected to anything but ground though. No. 3 might be pissy, but it's true, and annoying to me as well.

And I agree, I usually design with the peak hold, too. Usually to get the extra voltage from the transformer, and that caps are cheaper than chokes. When 47uF was the largest affordable value we needed to use a 10Hy choke. Now that we can use a 470uF we can get away with a 1Hy choke for the same filter, saving money and adding peak current capability. This is one of the reasons I no longer use tube rectification, too.
 
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