I have been looking at ways to derive low voltage rails from a higher voltage and current supply, which in practical terms is about 53-0-53V from a linear power supply (toroidal, bridge rectifier and electrolytic caps).

I naively thought that the circuit below should produce a nice 30V across the test load R3, instead I got a dead zener diode and a nice explosion from transistor Q2 which was somewhat unexpected and disappointing. It actually blew its middle leg off, the poor thing.

The idea is to get +15V and -15V rails to power an op amp or two. I expected that R1, D1, and R2 would drop respectively 38V, 30V and 38V and thus, like a pair of standard series regulators, Q1's emitter would stabilise at 15V (relatively to the hypothetical 0V rail which isn't there) and likewise the collector of Q2 would be at -15V.

What have I done wrong? I am wondering if I've misunderstood the current flow through the PNP, they always make my brain fry because of the reverse sort of nature of them. Anyway, what's my mistake?


simulate this circuit – Schematic created using CircuitLab


The zener is now a 1N4751A, 30 V at 8.5 mA, see these specs. The zener resistors are now 4K7 for a zener current of about 8.5 mA.

After adding voltage sources the simulation runs and results in about +/- 2.54 V over the zener and +/- 2.1 V over the output resistor.

Strange! Either the simulator does not know that the zener zeners at 30 V, or the transistors draw a lot of base current, but with such a big load resistor that is unlikely.

simulator screenshot

  • \$\begingroup\$ Q2's collector can only be at -53V, perhaps you meant its emitter? What are the voltage ratings of these? can they stand Vce=106V if the base circuitry is unbalanced for any reason? \$\endgroup\$ – Brian Drummond Sep 11 '16 at 19:45
  • 3
    \$\begingroup\$ The Absolute Maximum collector-emitter voltage for both 2N2222 and 2N2907 is 40 volts, and the 1N4730A is a 3.9 volt zener diode. Things WILL blow up! \$\endgroup\$ – Peter Bennett Sep 11 '16 at 19:53
  • \$\begingroup\$ They're rated at 60 and -75V respectively. I assumed they'd never see more than half the total rail voltage. \$\endgroup\$ – Ian Bland Sep 11 '16 at 19:54
  • \$\begingroup\$ Peter Bennett, oops I read the collector base rating not the collector emitter voltage. That might explain the explosion then! Should this circuit in principle work with higher rated transistors? Also, the zener I used was 30V, i just picked one in the schematic software that I presumed from its part number to be 30V. \$\endgroup\$ – Ian Bland Sep 11 '16 at 19:56
  • \$\begingroup\$ Trouble is, unless the transistors are identical in current gain, one is going to load your zener-resistor network more than the other which means your output won't be centred within the supply rails, it will be closer to one side than the other which means one of the transistors would be seeing more than half the voltage. \$\endgroup\$ – Tom Carpenter Sep 11 '16 at 20:02

You already have an unregulated DC supply. As you say, built from a bridge and some capacitors. Apparently, you have a center-tap on your transformer secondary, too. So you have a ground, too, and \$\pm\:53\:\text{V}\$ measured with your meter for the two other rails. I'll assume that this is probably unloaded, so you probably will have less than that when loaded. How much less is anyone's guess, as it depends a lot on the loading, your toroid design, the capacitors, and other factors. But less, for sure.

I gather you are trying to learn about how to design your own \$\pm\: 15\:\text{V}\$ supply for use with opamps. So you aren't necessarily just wanting to buy a nice supply (they are cheap, these days.) And since this is about learning, it's going to be a linear design and not a switcher. So your power supply will be generally inefficient, power-wise. But you are fine with that.

Perhaps I'm projecting, but I think this is a good idea to start with. It's modest enough that you have every reason to succeed. But there is enough to learn about that it's worth struggling for, too. I think my very first learning experience, where I really learned a few things well, was in trying to design my own power supply like this. At the time, then, I pretty much didn't have a choice. Existing lab supplies were unobtainable for a young teenager. And there was no set of cheap ebay suppliers for fancy switchers based on ICs, either. So I had to do it myself or go without. And faced with that, one learns or one does without.

Your approach is perhaps a little too much like a sink/source output driver used in everything from opamps to audio amplifiers. You could take the approach you are taking, but you'd have to make two of them -- one for \$+15\:\text{V}\$ and one for \$-15\:\text{V}\$. And they are even less efficient, as they can each source from your (+) rail and sink to your (-) rail, and you need to run them in class-AB. You really only need to source from (+) to make the \$+15\:\text{V}\$ rail and to sink to (-) to make the \$-15\:\text{V}\$ rail.

Just as a side note, it may be a good idea to include a pair of bleeder resistors to your existing capacitor bank at the output of your bridge. Something to get rid of the stored charge if you turn things off. Some \$\tfrac{1}{2}\:\text{W}\$, \$10\:\text{k}\Omega\$ resistors? That would only present a \$5\:\text{mA}\$ load, when running.

While you are considering that idea, consider also trying to load down your existing unregulated supply to measure what it does under load. I'd try something like a \$\ge 5\:\text{W}\$, \$1\:\text{k}\Omega\$ resistor to get an idea about a \$50\:\text{mA}\$ load, measuring the voltage with that load present. I'd then try something like a \$\ge 10\:\text{W}\$, \$270\:\Omega\$ resistor to see what happens when I get near \$200\:\text{mA}\$ load. This will test your entire unregulated system and give you an idea about its limitations. Those values were picked at random. If you already know the limitations of your toroid, then try out two different resistor values that hit the maximum load you expect to support and another one to hit perhaps 30% of the maximum load. And just take note of the voltage values measured. It helps to have an idea about your unregulated rail when loaded down a bit.

I'd recommend that you start by focusing on just one side, say creating the \$+15\:\text{V}\$ regulated supply rail from your unregulated (+) rail. You need to consider whether or not you want any current limits, too. I think it would be safer to include them. But that's your decision. It's not hard to include something for that, though. And, just personally, I'd probably want to be able to go to \$+12\:\text{V}\$, too. So perhaps a variable output supply that works over some modest range of output voltages?

You have lots of headroom! This means you can use an NPN emitter follower, a Darlington follower, or just about any configuration you want to have. Things are not tight, so you have room for control structures. Lots of room. The downside is, of course, that you have to dissipate and that your voltage rails are enough to make you have to check datasheets to stay within safe operating parameters for devices.

Finally, you can probably accept having to separately set the two voltage rail values, independently. Some power supplies are designed to provide tracking so that if you set the regulated \$+\text{V}\$ supply to \$+15\:\text{V}\$ then your regulated \$-\text{V}\$ supply will track that and provide \$-15\:\text{V}\$. But you can live without that, for now, I suspect.

If you write up a separate question, or clarify this one better, I may get you started with three or four different discrete (non-IC) topologies to consider analyzing on your own and building. But, for example, I have no idea what kind of current compliance you want to have. And it would help to know what voltage you measure when your unregulated supply is loaded down to the maximum current compliance you want to support (using a high wattage resistor and then taking a moment to measure the voltage with a voltmeter before it gets too hot.) And it would help still more to know if you do want a variable voltage over a range (what range, exactly?) and, if you just want a fixed voltage, how much initial accuracy do you feel you need? And I'd like to know if this is strictly for an opamp supply (suggesting a lower current compliance) or if you will want to use this to actually supply higher currents at still lower voltages, for some projects. Finally, it would be nice to know what BJTs you have, or are willing to get.

EDIT: So. Something simple, not very much current compliance of only \$5\:\text{mA}\$. Let's first focus on the (+) rail side... could go either with NPN or PNP for the pass transistor. It's more a matter of how you want to control it. Do you want to siphon away current from a source, or pull out current as needed? Hmm. Let's try this -- emphasis on simple.


simulate this circuit – Schematic created using CircuitLab

I've written down some design notes on the schematic. The resistor values are standard ones, so the actual output voltage will be a little off. But it should be close. Here's the logic.

I started out using \$Q_1\$ as an emitter follower topology. It's emitter targets \$15\:\text{V}\$. So I wrote down "15V @ 5mA" there. I initially estimated a useful \$\beta_{Q1}=50\$ and computed \$I_{B_{Q1}}=100\:\mu\text{A}\$ and estimated (from memory only) \$V_{BE_{Q1}}=750\:\text{mV}\$. From this, I decided I wanted \$5\times\$ as much from the unregulated supply, so I set \$R_1=\frac{53V-15V-750\:\text{mV}}{500\:\mu\text{A}}=74.5\:\text{k}\Omega \approx 75\:\text{k}\Omega\$. This means that I'll need to pull away between \$400-500\:\mu\text{A}\$ from \$R_1\$ to control \$Q_1\$'s behavior at the output. That's a small enough range, \$450\:\mu\text{A}\pm 50\:\mu\text{A}\$, that variations in a simple circuit won't be too sensitive. Oh, and I chose the BC546, which has a \$V_{CEO}=65\:\text{V}\$. (Could use a 2N5551 for \$V_{CEO}=150\:\text{V}\$.)

I decided to use another NPN down below, with its base nailed to a resistor divider, to pull that current. \$Q_2\$'s collector is nailed to a voltage, so no Early Effect. Fine. Dissipation in \$Q_2\$ is under \$10\:\text{mW}\$, so no problem. (You already know there may be a problem in \$Q_1\$.) A diode and capacitor provides a semi-stable voltage reference, as it is fed a relatively stable \$450\:\mu\text{A}\pm\:50\:\mu\text{A}\$ current. I estimated \$\beta_{Q2}=50\$ (again) and computed \$I_{B_{Q2}}=10\:\mu\text{A}\$ and estimated (from memory only) \$V_{BE_{Q1}}=650\:\text{mV}\$. I also know that the 1N4148 does about \$550\:\text{mV}\$ running at \$500\:\mu\text{A}\$ current. So this told me that the divider node should be guessed at \$1.2\:\text{V}\$. I wrote that down, too.

I chose to make the divider current at least \$10\times\$ the max required base current for \$Q_2\$. One of the problems with this circuit is going to be ambient temperatures, as these affect the base-emitter junction of \$Q_2\$ (and \$D_1\$, too) and this affects our divider point and pretty much everything else. But adding \$D_2\$ and \$D_3\$ in the divider helps here. It provides two more temperature dependent junctions that will track the other two over temperature. The remaining problem being \$R_3\$ and the differing current densities.

\$D_2\$ and \$D_3\$ are running with about \$\tfrac{1}{5}\$ of the current density of \$D_1\$ and \$Q_2\$. I happen to remember that a 1N4148 presents about \$\Delta V \approx 100\:\text{mV}\$ per decade change in current density, so I guess that \$\Delta V = 100\:\text{mV}\cdot \log_{10}\left(\tfrac{1}{5}\right) \approx -70\:\text{mV}\$ per diode for those two. So this means that to reach \$1.2\:\text{V}\$ at the divider, \$R_3=\frac{1.2V - 2\cdot\left(550\:\text{mV}-70\:\text{mV}\right)}{87\:\mu\text{A}}\approx 2.7\:\text{k}\Omega\$ (I used \$87\:\mu\text{A}\$ as the mid-point current value.) So that sets \$R_3\$, at a guess.

I added a speed up cap across divider resistor \$R_2\$ so that short-term load variations might more immediately drive \$Q_2\$. (If the \$15\:\text{V}\$ regulated rail suddenly jumps upward, then \$C_3\$ will pull up immediately on the base of \$Q_2\$ making it pull away more of the drive current going to \$Q_1\$, countering the rise. Similarly, in the other direction, too.)

You should be able to pony up the (-) regulated rail, I think. And keep in mind that you do not want to load this thing down too much! You will definitely cause that poor little TO-92 serious problems. It's dissipating \$5\:\text{mA}\cdot\left(53\:\text{V}-15\:\text{V}\right)\approx 200\:\text{mW}\$ and the package has \$\tfrac{200 ^{\circ}K}{W}\$, so this works out to about \$+40^{\circ}C\$ over ambient, already. You can see just how quickly this thing will heat up if you run much more current through it. You may be able to get away with \$10\:\text{mA}\$, but not much more.

OVERVIEW NOTE: Now that you can see one person's process (other, more experienced designers will apply still more knowledge than I applied), let's take a moment to view this from a distant perspective.

The circuit boils down to:

  1. A pass transistor (\$Q_1\$) which is supposed to stand-off about \$40\:\text{V}\$ between the unregulated (+) rail and the desired \$15\:\text{V}\$ rail. This pass transistor will need a source of base current so that it can be kept in its active region. It is also arranged into an emitter-follower configuration, so that moving its base voltage around moves its emitter around in roughly 1:1 (voltage gain from base to emitter is \$\approx 1\$.)
  2. We can solve all of the needs in (1) above by using a simple resistor (\$R_1\$) to the unregulated (+) rail. This not only can provide the needed base current, but it also makes it very easy to control the base voltage of \$Q_1\$, by just pulling more or less current through it. For design purposes, we do not want variations in \$Q_1\$'s base current to seriously impact the current stream we are also using to control the voltage at the base of \$Q_1\$. So we need to make this stream of current large, by comparison. Larger is better, and perhaps by default we might choose a factor of \$10\times\$. But we are also constrained by the fact that this is a \$5\:\text{mA}\$ power supply. So we might want to use something that is about \$\tfrac{1}{10}\$th of \$5\:\text{mA}\$ to keep it modest. This means something from \$10\cdot 100\:\mu\text{A}=1\:\text{mA}\$ on the one side to about \$\tfrac{5mA}{10}=500\:\mu\text{A}\$ on the other side. I decided to use the smaller value, since this is just a simple regulator and I can accept a slightly less stiff base source.
  3. Something to control the current being pulled through \$R_1\$, based upon a voltage comparison of some kind. It turns out that a BJT is okay for something like this. (More BJTs would be better, as in an opamp, but one is sufficient here.) It has a collector current that depends upon the voltage difference between its base and emitter. So it compares its' base and emitter and adjusts a current on that basis! Practically made in heaven for this, yes? So we now stick a new BJT (\$Q_2\$) with its collector tied up to \$R_1\$ and the base of \$Q_1\$.
  4. We need a reference voltage. Could use a real reference, like a zener or a more sophisticated IC device, but this is a simple design. Well, a diode with a fixed current density is a voltage reference. (Excepting temperature.) And guess what? We just happen to have a current we can use that is relatively stable! The very current we are using to adjust \$Q_1\$'s base voltage through \$R_1\$. So now, \$R_1\$ provides three services for us -- it provides base current to \$Q_1\$, allows us to control \$Q_1\$'s base by adjusting the current through it, and now that very same current can be used to stabilize the voltage of a voltage reference diode. All we do is stick that diode into the emitter of \$Q_2\$. And add a small capacitor across it o kill high frequency noise there. It's nice when things do multiple duties for you.
  5. We have our current control collector, a voltage reference at the emitter, and now all we need to provide is a comparison voltage, derived from the output voltage, at the base of \$Q_2\$. It's important that if this comparison increases (the output voltage appears to increase for some unknown reason), that we will pull more current through \$R_1\$ to force the base voltage of \$Q_1\$ to decline to oppose this change. Turns out that a simple voltage divider does this job well. All we need to do is to make sure that the current through the voltage divider is a lot more than the required base current of \$Q_2\$, so that when \$Q_2\$ adjusts its collector current and needs more (or less) base current, that this doesn't affect the divider voltage (much.)

That's really the essence of it. I added those two diodes to help stabilize things vs ambient temps. But they aren't strictly necessary if you don't mind your voltage rails shifting around a little more with temperature. As it is, they may still drift around by maybe \$\tfrac{25\:\text{mV}}{^{\circ}C}\$, just doing a short loop-around bit of guess-work. But if you don't mind it being twice as bad then you can replace the resistor and two diodes with a simple resistor, instead:


simulate this circuit

The actual value of \$R_3\$ may need to be adjusted a bit here, as we don't actually know just how much base current is needed (probably less than I guessed -- a lot less.) So perhaps closer to the \$12\:\text{k}\Omega\$ value? But you can use a potentiometer here, I suppose, to make this adjustable, too.

  • \$\begingroup\$ Thanks for your comprehensive reply! The 53V supply is for a 100W FET amplifier I built years ago and am rebuilding and modernising. As you say, this is about learning as much as outcome. The intention is a low current op-amp input stage (currently a single TL072) so I only want a few mA on the 15-0-15 supply and I wanted to avoid another "proper" PSU (linear or SM) to keep down the parts count and avoid more transformers. Hence the high resistor values. The BJTs are ones I happen to have lots of but I'll obtain whatever I need. The 53V PSU has LEDs (in series with 20k) that bleed the caps. \$\endgroup\$ – Ian Bland Sep 12 '16 at 5:55
  • \$\begingroup\$ @IanBland: Thanks. Do you need accuracy for the 15V rails? Do you want a range of voltages? Do you care about wasting a BJT on a current limiter for each side? Anything more you'd want to add? \$\endgroup\$ – jonk Sep 12 '16 at 5:57
  • \$\begingroup\$ I don't mind wasting reasonable amounts of current and the voltage isn't critical, so long as it's stable, I just picked 15V as a nice round number, lots of headroom on the op amps and (ironically) minimising the necessary volt drop through the BJTs. This version was meant as a test of the principle, I can for instance add extra smoothing capacitors etc for a final version where necessary. I'm trying to avoid needing heatsinks on the BJTs as well to keep this small and sweet :) \$\endgroup\$ – Ian Bland Sep 12 '16 at 6:08
  • \$\begingroup\$ @IanBland: Just a few mA? If you have to have a BJT drop away 40V, it only takes 5mA or so to get to a quarter watt. A lot of opamps support 20mA or 30mA outputs. So when you say "a few mA" are we talking 5 or are we talking 30? \$\endgroup\$ – jonk Sep 12 '16 at 6:15
  • 1
    \$\begingroup\$ @IanBland: Okay. So at 10mA would be talking about up to half-watt for the pass BJT. At 5mA would be quarter-watt. TO-92 will do a quarter, but at half watt I'm thinking "not really." I think the TO-92 is like 200C/W. I don't like anything that says +100C over ambient at the die (assuming you even have air flow.) Oh, heck. You are okay with blowing the things up, anyway. So let's stay with a TO-92 and plan on 5mA max. I'll stick a current limit on it. You can remove it if you don't want it and like to watch BJTs fry. \$\endgroup\$ – jonk Sep 12 '16 at 6:29

For one thing, a 2N2222 is only rated for 40 V. The 2907 is good for 60, but that still doesn't leave much margin for things to go wrong, particularly at startup.

I suspect the real problem is that the transistors were wired incorrectly. That could leave a direct path thru Q1, D1, and Q2. Poof!

Added about voltages on the transistors

Even when everything is working perfectly, each half of the circuit sees 53 V. The 1N4730 is a 3.9 V zener diode. That means, when everything is working perfectly, the transistor bases will be held at ±2 V. Even saying the B-E drop of each transistor is only 600 mV, the emitters will be at ±1.4 V. That means each transistor will see 52 V across it when everything is perfect.

Everything is never perfect. How accurate are the ±53 V supplies? What about startup transients? What are the real zener voltages with only half a milliamp thru them? What happens when the load draws some real current, even if only on startup to charge up a capacitor or something?

Did you look up the voltage spec for the transistors you are actually using, not just any datasheet you could find for the generic part number? There are minimum voltage specs somewhere for a 2N2222 and 2N2907, but specific manufacturers sometimes make their parts more capable. You can't use one of those datasheets to tell you the maximum a generic part is good for. To get the numbers I quoted above, I grabbed random datasheets. That means the real specs could be lower than what I quoted.

One transistor is already well out of spec, and the other is close to it. This is not good engineering.

  • 1
    \$\begingroup\$ Thanks. The transistors should only be dropping 38V each though surely, the resistor R3 is dropping the rest? I triple checked the wiring. Maybe my 2907s have a different pinout to my data sheet or something. I agree that the bang was quite big enough to indicate a dead short path through the transistors and zener. \$\endgroup\$ – Ian Bland Sep 11 '16 at 21:13

First, Google is your friend. A 1N4730 is a 3.9 volt zener.

That said, I'm inclined to believe that you either miswired your circuit or you used the wrong values of resistors. I'm especially inclined to think that R1 or R2 might have been 100 ohms, rather than 100k. At any rate, your nominal resistor values are large enough to prevent Magic Smoke Emission, so your circuit in some way was different from your schematic.

  • \$\begingroup\$ Yes, I used the wrong part number on the schematic, the actual zener was a BZX. I thought that about the resistor values too (which are correct) but if the transistors fail short there's a short cirucuit path that bypasses all the resistors, so I guess that's where the electrons went. \$\endgroup\$ – Ian Bland Sep 11 '16 at 20:11
  • 1
    \$\begingroup\$ \$V_{CE} > 50V\$ will kill a lot of small signal BJTs. Not sure what he used, exactly. But that's a risk here. The 2907 comes in a 40V and a 60V variety, for example. And the PN2222 is 30V, with the A variety being 40V. It just seems reckless and I suspect he could have destroyed them even with that circuit. That 1M resistor isn't dropping much \$\Delta V\$. \$\endgroup\$ – jonk Sep 11 '16 at 20:20
  • \$\begingroup\$ Never mind. Just saw his comment about using a 30V zener. \$\endgroup\$ – jonk Sep 11 '16 at 20:23
  • IF Vcemax for Q2 is 40V and beyond in secondary breakdown then Ve max is -12V

  • Vb for Q2 is 1/2 of Vz (D1=3.9) or -2V approx. this Vbe = -10V while spec is -5V ABSOLUTE MAX.

  • due to the catastrophic mode of failure for Vbe reverse ,

  • and your careless design,
  • only you are responsible for it's middle leg getting blown off, perhaps by construction errors.
  • \$\begingroup\$ As said earlier I put the wrong zener part number on the schematic while compiling my question, it should be a 30V BZX85C. Obviously I can't prove I wired the breadboard correctly, but I triple checked the pinouts before and after powering it up. \$\endgroup\$ – Ian Bland Sep 12 '16 at 5:57

This is an easier way of getting +/-15V from your rails:


simulate this circuit – Schematic created using CircuitLab

R1 and R2 allow about 2.5mA to flow to the transistor bases and to the 16V zeners. The voltage at the emitters of the transistors will be about 0.7V less than the zener voltage or about +/-15.3V.

While this is a very simple and reliable circuit, note that it is not short-circuit or overload proof as a 3-terminal regulator would be.

There are a few linear regulators which can operate from your relatively high supply rails but they will not be all that cheap. Do a parametric search on a distributor or supplier web sites to find them. The negative regulator may be more of a problem, especially as your (presumably unregulated) rails might go considerably higher than 53V peak. While you can use the above circuit to drop down the voltage for a 3-terminal regulator you have to consider the worst-case conditions and how much dissipation the transistors will experience.

  • \$\begingroup\$ Thanks Spehro, this was basically what I was trying to do except clever me wondered if I could replace the two zeners with one, and then it all exploded. Shouldn't high resistance connections to the Q bases act as a crude current limit circuit? That was the reason for my 100k resistors in the original. Also, are the values of C1 critical? I haven't got any 100nF caps to hand, but I have got 1uF... \$\endgroup\$ – Ian Bland Sep 12 '16 at 18:01
  • 1
    \$\begingroup\$ @IanBland It will act as a very crude current limit but you don't want to go so low in zener/base current that you start getting rail voltages unduly modulated by the loads (and the beta is not well known). So you might typically get 0.5-1A with my values, which is going to burn up the transistors pretty quickly. There's nothing critical about the cap values. An emitter current sense resistor and a small signal transistor (per rail) could be used to limit the current more accurately, say to 50mA meaning a few watts dissipation so a small heatsink or copper on a PCB would save the transistors. \$\endgroup\$ – Spehro Pefhany Sep 12 '16 at 18:55

Reviewers rejected my latest edits to the question, and suggested to create a new answer, so:

Here is the schematic from the OP, completed with voltage sources and more appropriate zener resistors, for the recommended zener current of about 8.5 mA:


simulate this circuit – Schematic created using CircuitLab

And here is the result of the simulation using the Simulate This button:

screenshot of simulation

The zener is now a 1N4751A, 30 V at 8.5 mA, see these specs. Setting the correct part nr does NOT set the related zener voltage, I did that manualy in the circuit diagram editor. The zener resistors are now 4K7 for a zener current of about 8.5 mA.

After adding voltage sources the simulation runs and results in about +/- 15.0 V over the zener and +/- 14.5 V over the output resistor.

Perfect! This circuit seems to do what is expected from it.

As for the blown parts: that must be something like a wrong connection, as suggested by one of the commenters.


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