Question about Zener diodes in series

[nsummy]

This is a circuit for the antique Electronic Supply farm radio battery eliminator, specifically the B battery circuit.  This is the first "real" circuit I ever soldered together and while my knowledge has came quite far since then, I still have the same question today as I did then: 

Is there a reason to put so many  11v zener diodes in a series like this?  I understand the main reason for the series: dropping down the voltage for the progressively lower outputs, but why so many 11v zeners?  Is there a benefit to doing it this way over using half as many zeners at 22v?  When I built the circuit I used a 39v followed by 3 22's.  It appears to work correct but I have always wondered if it was a bad decision.

[wraper]

More voltage means more power to dissipate.

[floobydust]

It doesn't need to be so many low voltage zeners in the string- unless you like the ability to change the output voltage, say to 90V by simply moving the take off. I think power dissipation is low with the 1.5mA operating point.

In a Perma Power Model A, the B+ voltage was too high after replacing the selenium rectifier with a silicon diode.
So I added a 91V 5W 1N5377B zener as a shunt regulator, it has a 1k8 2W resistor ahead of it. No transistors.

[TimFox]

A good reason to use multiple low-voltage Zeners in series instead of a single high-voltage Zener is that the temperature co-efficient (and therefore susceptibility to self-heating) is lower for the lower-voltage units and their series combination.
A family of Zeners, as a function of nominal voltage, go through zero tempco somewhere around 4.5 V.
Below that are "true Zeners", while above that the diodes are "avalanche diodes" (which also have steeper I-V curves than true Zeners).
When measuring the voltage on a, say 68 V (plastic-cased axial lead) Zeners (with constant current), I could see the voltage change obviously when squeezing the body between thumb and forefinger.

[PGPG]

I think power dissipation is low with the 1.5mA operating point.

Why you assume that supply is designed to only self power itself.
At picture on the right I see 50mA written so I understand supply is expected to be load with 50mA.
So through some Zeners 50mA+1.5mA current flowing should be expected.

51.5mA*11V=0.566W dissipated at 11V Zener diode
51.5mA*22V=1.133W dissipated at 22V Zener diode

[mikerj]

I think power dissipation is low with the 1.5mA operating point.

Why you assume that supply is designed to only self power itself.
At picture on the right I see 50mA written so I understand supply is expected to be load with 50mA.

50mA is the expected maximum current on the 135v output.  This current doesn't go through the zener diodes.

[CaptDon]

Those zeners provide voltage taps for various B+ voltages required for different devices or sometimes even within the same radio. I remember old B batteries actually having a socket to accept a multi-pin plug and indeed the battery provided taps at different points in the series string. The bottom of the zener string is the return for the regulator section. Due to thermal shift and so forth the regulation won't be very tight but then again it wouldn't be very tight coming from a battery either.

[DonKu]

A good reason to use multiple low-voltage Zeners in series instead of a single high-voltage Zener is that the temperature co-efficient (and therefore susceptibility to self-heating) is lower for the lower-voltage units and their series combination.
A family of Zeners, as a function of nominal voltage, go through zero tempco somewhere around 4.5 V.
Below that are "true Zeners", while above that the diodes are "avalanche diodes" (which also have steeper I-V curves than true Zeners).
When measuring the voltage on a, say 68 V (plastic-cased axial lead) Zeners (with constant current), I could see the voltage change obviously when squeezing the body between thumb and forefinger.

Thank you. True Zeners have negative tempcos while avalanche diodes have positive tempcos. Theoretically you can reduce overall tempco by replacing each 1N4741A with a 1N4728A and 1N4737A combo. But the combo's Zener voltage level's a little lower.

[PGPG]

50mA is the expected maximum current on the 135v output.  This current doesn't go through the zener diodes.

I don't agree.
As in transistors there are lacking lines than I assume the arrows saying +90V, +67V, +45V, +22V have wires from between Zener diodes.
If at 22V output you take 50mA than 51.5mA will be going through all Zeners in serie.

[floobydust]

Regardless of the zener (or number of) chosen for the regulation, Q2 requires 22.9V across R5 15k to turn on and back off output. So the entire zener string is always kept at around 1.5mA - it's not a shunt regulator circuit, it is a series-pass regulator. Does this make more sense?

Powering vacuum tube radios, they don't care if B+ moves around several %.

I worked with General Transformer Corp. Model A Perma-Power  "For sets using 4, 5 or 6  1.5V tubes" circa 1947.
It provides A and B battery output: 1.4VDC 350mA and 90VDC 13mA.
Need 0.01uF cap across the (new silicon) rectifier diode to prevent RF switching hash.

[PGPG]

Q2 requires 22.9V across R5 15k to turn on and back off output. So the entire zener string is always kept at around 1.5mA - it's not a shunt regulator circuit, it is a series-pass regulator. Does this make more sense?

Where from (in your opinion) is the output current when someone will not use this supply +135V output, but its +22V output if in all Zeners there is only 1.5mA that can't go to the output as is needed to drive R5?

[floobydust]

I see the regulator output is always taken from the Q1-E, and the zener string gets jumpered or the chosen one (arrow) connected to the output.
If you wanted 90V output, you could either short D4-D7, or move D8-K to the output.
If you wanted a 23V output, you would jumper the entire zener string D4-D13. That leaves only zener D14 22V on Q2 to give you the minimum 23V. This is how I see it working.

[PGPG]

That leaves only zener D14 22V on Q2 to give you the minimum 23V. This is how I see it working.

How you interpreted arrows suggesting outputs as switch shorting selected set of Zeners?
No switch is shown to short some Zeners.

(190-135)*0.05=2.75W
(190-22)*0.05=8.4W

Much easier is to distribute this power between tranistor (2.75W) and Zeners (11*0.05=0.55W each) than to dissipate all 8.4W in transistor.

[mawyatt]

Why not use a Zener with a TC of ~+2.2mv/C for D14 (~6.2V, BZX79-6V2, 1N5234) which will produce an overall TC of ~0 with Q2 (Vbe ~-2.2mv/C) and replace the zener diode string with a resistor string so the "taps" are about the same as the with the diode string?

This would allow a minimum output voltage of ~ 7V (Vbe + Vz (D14)) and have an output voltage of;

Vout ~ (Vz+Vbe)(Rx+Ry)/Ry, where Ry (R5 in schematic as 15K) is the effective resistance at the base of Q2 to ground (no need for R8) and Rx is the resistor value replacing the diode string.

For example for 130V output with ~6.2V Zener for D14 and Ry (R5) at 15K, then Rx~274K. Use Spice to home in on the exact value if needed as this was just a "back of the envelope calculation". Also consider adding a small shunt capacitor around Rx to improve loop phase margin, and no need for C5 as it just reduces phase margin by delaying feedback effect and slows loop. Could consider adding a shunt cap across D14 to reduce the zener noise if needed.

Best

[PGPG]

Why not use a Zener with a TC of ~+2.2mv/C for D14 (~6.2V, BZX79-6V2, 1N5234) which will produce an overall TC of ~0 with Q2 (Vbe ~-2.2mv/C) and replace the zener diode string with a resistor string so the "taps" are about the same as the with the diode string?

Do you assume output from "taps" between resistors?
If so then how it will work if for example 45V output will consume 50mA?

Or
Do you assume shorting some resistors to get other voltages?
If so then add a transistor cooling fan to the project.

[mawyatt]

Why not use a Zener with a TC of ~+2.2mv/C for D14 (~6.2V, BZX79-6V2, 1N5234) which will produce an overall TC of ~0 with Q2 (Vbe ~-2.2mv/C) and replace the zener diode string with a resistor string so the "taps" are about the same as the with the diode string?

Do you assume output from "taps" between resistors?
If so then how it will work if for example 45V output will consume 50mA?

Or
Do you assume shorting some resistors to get other voltages?
If so then add a transistor cooling fan to the project.

Yes "taps" are between resistors. You can simply add an emitter follower for the tapped voltages for the 50ma output. The base connects to the desired tap voltage, collector to the 130V and emitter as output. If you want higher tapped voltage precision, you can add a diode in the tapped resistor string (at the bottom) for the emitter follower Vbe compensation. Use Darlington (and diode pair) to reduce base current effects.

For a single output just use a pot to set the desired voltage as shown.

Best

[PGPG]

You can simply add an emitter follower for the tapped voltages for the 50ma output. The base connects to the desired tap voltage, collector to the 130V and emitter as output.

I'm not sure as it was simply not my design, but I understand that idea behind this schematic was to distribute power among many elements (transistor+Zener diodes).
Dissipating 11*0.05=0.55W in single element is possible without even using radiator (element have to be not the smallest one).
With your solution the 22V output made as emitter follower will dissipate (130-22)*0.05=5.4W in a single element.

Use Darlington (and diode pair) to reduce base current effects.

And you came to replacing one Zener with:
- resistor,
- diode pair,
- Darlington,
- radiator.

[mawyatt]

You can simply add an emitter follower for the tapped voltages for the 50ma output. The base connects to the desired tap voltage, collector to the 130V and emitter as output.

I'm not sure as it was simply not my design, but I understand that idea behind this schematic was to distribute power among many elements (transistor+Zener diodes).
Dissipating 11*0.05=0.55W in single element is possible without even using radiator (element have to be not the smallest one).
With your solution the 22V output made as emitter follower will dissipate (130-22)*0.05=5.4W in a single element.

Use Darlington (and diode pair) to reduce base current effects.

And you came to replacing one Zener with:
- resistor,
- diode pair,
- Darlington,
- radiator.

Many different ways to approach this. The OP ten zener approach might experience some output "tap" variation (as 1% Zeners are expensive) and with temperature variation as shown. The approach we've shown should yield better results over temperature and "tap" voltages, as 1% resistors are cheap.

The single output version should show overall better temperature performance as everything is "inside" the control loop.

Anyway, the OP can decide which is acceptable, and as always YMMV.

Best

[mawyatt]

You can simply add an emitter follower for the tapped voltages for the 50ma output. The base connects to the desired tap voltage, collector to the 130V and emitter as output.

I'm not sure as it was simply not my design, but I understand that idea behind this schematic was to distribute power among many elements (transistor+Zener diodes).
Dissipating 11*0.05=0.55W in single element is possible without even using radiator (element have to be not the smallest one).
With your solution the 22V output made as emitter follower will dissipate (130-22)*0.05=5.4W in a single element.

Use Darlington (and diode pair) to reduce base current effects.

And you came to replacing one Zener with:
- resistor,
- diode pair,
- Darlington,
- radiator.

The 11V Zeners have a TC of 8~9mv/C and since they are in series this is cumulative, so the top of the Zener ladder will see a variation of 80~90mv/C just due to the Zener ladders. Here's a quick spice simulation showing the Multiple Zener approach and Resistive Feedback at load currents of ~1 and 50ma.

We've spent enough time on this, you and others can run your own simulations and judge for yourself.

Best

[floobydust]

I think we've misunderstood that lousy schematic in OP.
Antique Electronics K-101A battery eliminator (wood board kit) replaces multiple batteries. The A, B, C all at once.
Problem I think it only has B batt high current "50mA" available for the +135V rail, while the 90V,67,45,22V outputs are all lower current- through small zeners which could roast. You would not get 90V at 50mA for example, the 22V rail is also very weak.
This must be why they used so many zeners in the string.

[nsummy]

I think we've misunderstood that lousy schematic in OP.
Antique Electronics K-101A battery eliminator (wood board kit) replaces multiple batteries. The A, B, C all at once.
Problem I think it only has B batt high current "50mA" available for the +135V rail, while the 90V,67,45,22V outputs are all lower current- through small zeners which could roast. You would not get 90V at 50mA for example, the 22V rail is also very weak.
This must be why they used so many zeners in the string.

Correct.  Sorry I probably didn't explain the circuit very well or maybe didn't communicate that I was looking for the reasoning of the approach (rather than alternative approaches, which I am sure there are better ways to accomplish this).

So if I understand what you said correctly, the multiple zener approach can dissipate heat better?  From what I have read, adding zeners in series will not increase the power rating, but I learned long ago theory doesn't always match real life.  They claim the entire B circuit can supply a combined total of 50 ma between the taps, which always baffled me (especially when I was new), because as you pointed out the zeners couldn't handle that sort of output at the 1w rating. I guess its also worth noting that they no longer sell this kit or the plans.

Also for those wondering, this is how they describe the circuit (sorry in advance for the lack of paragraphs here.) 

The B or plate outputs can supply up to a combined total
of 50 milliamperes at 22, 45, 67, 90, and/or 135 volts.
This section of the supply uses a two-transistor regulator
and a string of zener diodes that divide the output
voltage. Ql in the accompanying schematic diagram is a
series pass transistor whose collector-to-emitter voltage
changes to maintain a constant output potential, regardless
of changes in line voltage or load current. Ql gets
its base current through R3 and R4, which also furnish
collector current to Q2. In operation, if the output voltage
as sensed at the base of Q2 tries to rise, Q2 is turned on
harder, drawing more current through R3 and R4. The
resulting increase in voltage drop across these resistors
lowers the base voltage at Ql, correcting the output
voltage. Conversely, if the output voltage as sensed at
the base of Q2 tries to fall, Q2 is moved closer to cutoff,
drawing less current through R3 and R4. The consequent
reduction in the voltage dropped across these
resistors tends to increase the voltage appearing at the
base of Ql, correcting the output voltage. The thermistor
in series with the collector lead of the pass transistor
has a positive temperature coeffi cient, and protects
the B regulator against overload. This device has a cold
resistance of about 100 ohms. Excess current heats the
thermistor, which causes its resistance to increase, limiting
current to a safe level.

[TimFox]

Zener power ratings:
Within a family of, say, 1 watt Zeners, obviously four 6.2 V diodes in series will be rated for 4 W, while a 24 or 25 V single diode is rated for 1 W.

[floobydust]

I think the series zener string increases their power and allows you flexibility to change output voltage by jumpering them or moving the take-off point.
Many of these battery eliminators are not a great design. Using 10 zeners is a bit much and constructing it like a crystal radio on a wood board is not OK when there are hazardous voltages present, and there is no fuse.

[CaptDon]

Tim has the right idea. As current is drawn from the zener string the voltage at the pickoff point will begin to sag but will be compensated for because the bottom end of the string will sag an equal amount and the regulation circuit will increase the voltage at the top of the string to maintain the 135 volts. The zener current at the bottom will always try to be the 1.5ma predicted and the total current at the top of the string will be the load current taken from the pickoff point plus 1.5ma The idea works but does waste a lot of power as heat. The higher the physical number of zeners carrying the load current the higher is the wasted power.

[floobydust]

There are repair threads implying these battery eliminators eat zeners. It could be the (radio) capacitor inrush currents, not sure. They were not reliable enough.