I was reading the datasheet for the LT1161, and dicovered the below construction in an example application toward the end of the document (Page 11). It appeared to be powering some Logic ICs.

This clearly seems to be clamping the supply voltage to something the logic IC can handle, so how does this work, and how is it better than just using a zener with a resistor?


simulate this circuit – Schematic created using CircuitLab

Context, lower left, click to enlarge: enter image description here


5 Answers 5


This is quite easy. Remember that BE voltage of a bjt is around 0.6V when active. The Zener has a 5.6V drop.

So, the zener sets the base voltage of the transistor to be 5.6V, this makes the Emitter voltage 5V.

As long as the Zener is active and the BJT is active, the emitter voltage will be nearly 5V(depending on tolerances). This is regardless of all other conditions. So, regardless of whatever else the circuit is doing, we have a stable voltage point of 5V. We just need to make sure the zener is active and BJT is active, which requires setting R1 low enough(mainly for the BJT).

The reason why it is better than just the zener, is that it has very little output impedance. The zener would be R1, by itself. With the BJT, it is much lower and can drive much larger loads, depending only on the BJT's current carrying capacity/internal resistance.

One can add a cap to the base and ground to reduce any fluctuation further stabilizing the emitter voltage. The cap would be smaller than using the same on the emitter to stabilize.

So, such sources are good because they provide a cheap(time, cost, real estate, etc) way to produce a different rail voltage with a low internal resistance. If one just used the zener and resistor, and the circuit required a large amount of power for some event, it would be under supplied creating anomalous behavior.


D1 creates a fixed voltage by acting as a shunt regulator. R1 supplies enough current thru D1 so that it is run where its voltage is fairly constant over a wide range of current.

That is all you need if you just want a reference voltage, like into one side of a comparator for example. This reference can only deliver at most the amount of current coming thru R1 minus the minimum D1 needs to stay at its stated voltage. In other words, just R1 and D1 together make a steady voltage, but with rather low current capability.

Q1 is used in the emitter follower configuration to provide current gain. This creates a power supply with much more current capability. Actually the current capability is (gain + 1) higher than just R1 and D1 by themselves.

There is a drawback though. Q1 will drop some voltage from its base to emitter. The B-E junction looks like a diode to the circuit, and will have a typical diode drop, usually somewhere in the 600-750 mV range. The net result is that the output at the emitter of Q1 is a power supply with a voltage around 700 mV less than the zener voltage, and a lot more current capability than just the R1 and D1 "supply".


how is it better than just using a zener with a resistor?

The 2N3904 can output a current up to 200 mA to drive a load. It is connected as an emitter follower so in simple terms, whatever voltage is fed to the base, is outputted on the emitter minus the VBE (Typically 0.7V for a simple Silicon transistor like the 2N3904). In more complex terms, the 0.7V is a little load dependant so there is a small degredation of regulation offered by this type of circuit compared with a straight zener-resistor.

The really big advantage is when you consider the alternative i.e. just a resistor and zener diode. Ask yourself what value of resistance is needed to be able to output up to 200 mA when the emitter follower is removed and the output is taken straight from the zener diode.

To achieve an output of 200 mA (without damaging zener regulation) means that the resistor has to drop 19 volts (to produce a 5V output) at 200 mA - that's a power dissipation of 3.8 watts and it's continuous.

If the "200 mA" required by the load is only very intermittent then 3.8 watts continuous is an awful lot of waste heat. The emitter follower is basically supplying only the demand asked by the load and sure, it will dissipate 3.8 watts every time the load needs 200 mA but, when the load only needs 10 mA then the power wasted in the 2N3904 is only 190 mW.

The straight zener-resistor doesn't have this option but it is a little better at regulating. However, if you need this much voltage regulation a 7805 would be a better option but, as always, taking too much current continuously from the output means a sizable power dissipation and a heatsink.

  • \$\begingroup\$ Why would the load dissipate 190mW when needing 10mA? I thought the current at the emitter is the summation of collector current (Beta * base current) plus base current, and assuming base current is small, the load at output needing 10mA would imply current into the base is 10mA divided by Beta (e.g. 100) hence the power dissipated in 2N3904 is in orders of magnitude less than 190mW. I think I'm confused and wish you can help clarify... \$\endgroup\$
    – KMC
    Apr 3, 2020 at 11:03
  • \$\begingroup\$ I said when the load only needs 10 mA then the power wasted in the 2N3904 is only 190 mW. \$\endgroup\$
    – Andy aka
    Apr 3, 2020 at 11:11

I guess no one has yet explained how it works. Someone did identify the BJT as acting like an emitter follower. But that's not a how, but rather a what topology answer.

Since you ask, "how is it better than just using a zener with a resistor," I'll assume you already understand the resistor and zener section. That leaves just the added BJT to explain.

When a BJT is in its active region, the Vbe drop doesn't change very much regardless of the collector current. So if you know the voltage at the base of the BJT, then you know also the voltage at the emitter. For a silicon NPN BJT (most of them), that will be near 0.7V less than the base voltage. So fixing the base, fixes the emitter to a value somewhat lower (We're ignoring the collector for now.) In short, the emitter is about a "diode drop" less than the base.

Now, let's add in the collector. This is tied to the power supply, directly. All this means is that the collector current (if any) can come straight from the low impedance power supply rail and NOT from your zener supply. It's not very important to this circuit how high of a voltage exists there -- it works about as well with 12V as it would with 20V, for example. (Barring the Early Effect.) The only limit here is the ability of the BJT to "stand off" the voltage across its collector and emitter pins. Most BJTs can stand off 30V and more, but you can always look it up on a datasheet to be sure -- look for the VCE breakdown voltage (or VCEO.) So the collector is just tied there so that it has a ready supply of current from a power supply that has a large current compliance.

Now, the load is attached at the emitter and then to ground. This completes a circuit from the resistor, to base, then to emitter, through the load there, and then to ground, which starts to supply a tiny current into the load. But that current also supplies the needed recombination current to allow a much larger (due to the BJT's beta value) collector current to flow, which now ALSO adds to the current moving out the emitter and into the load. So the load now has much more current compliance from the circuit, getting most of the extra from the collector and almost none of it from the base. This can be a difference by a factor of 100 or more. So the nice thing here is that the zener+resistor circuit is barely loaded, even though the load requires a lot of current.

Suppose the load needs 100mA. In a zener+resistor circuit only, all of this would have to come through the resistor. And this would add an additional voltage drop across it or else nearly or completely starve the zener for its needed current to hold its voltage. But with the BJT in place, perhaps 99mA or even 99.9mA might come from the collector, bypassing the zener's resistor. Still, perhaps 1mA or perhaps 0.1mA would still be needed by the zener's resistor to supply the BJT's base recombination current. But that's a LOT less and it almost certainly won't starve the zener. (They are often operated at 10mA, which makes supplying the BJT's base current of 0.1mA to 1mA far less complex than if you were considering having to supply 100mA from the zener's resistor!)

In effect, the BJT's collector-emitter path supplies most of the needed current, leaving a much, much smaller part of it to load down the zener's resistor. And this is almost always a good thing here.

Also, note that the voltage difference of 0.7V I mentioned earlier DOES depend a bit on the collector current. So the regulation isn't perfect. If the load requires 1mA and then demands 100mA later, the voltage it accesses at the emitter will shift a little bit. But the nice thing here is that the change is only about 60mV for each factor of 10 change in the load's current. So even with a change from 1mA to 100mA, the voltage at the emitter will only shift about 120mV (because there are two factors of 10 between 1mA and 100mA.) This is often quite tolerable and makes an okay power supply for a lot of uses.

EDIT: @DewaldSwanepoel:

This really should be the accepted answer.

Hehe. Thanks. I tried to actually answer the question, not just prove that I can recognize a pattern. Bit of a difference. But the OP chooses.

Would you mind expounding on how one would go about to choose the value of the resistor and the specific BJT for your application?

Each zener will have a specification in the datasheet for the desired operating current. (Or a range of operating currents.) Diverting from the recommended values means that the zener voltage itself may no longer meet the other specifications, so it's best to operate it as intended.

Suppose you picked the 1N4735A, which is a \$6.2\:\textrm{V}\$ zener. It is tested at \$41\:\textrm{mA}\$, but a chart shows curves for currents from \$5\:\textrm{mA}\$ to \$20\:\textrm{mA}\$. That suggests you should operate it at one of those. The zener impedance usually improves a bit (good thing) if you are running them hotter, or with a little more current.

There will also be a base current for the BJT, when fully loaded. Or none, if not. Suppose the BJT had to support a load of \$200\:\textrm{mA}\$. And let's say the voltage rail is \$12\:\textrm{V}\$, so the BJT might have to drop \$7\:\textrm{V}\$ across its \$V_{CE}\$. Then it may have to dissipate up to \$1.4\:\textrm{W}\$. And such BJTs may only exhibit \$\beta=50\$, or so. So we'd plan on a base current of up to \$4\:\textrm{mA}\$.

So we need to handle a base current varying from \$0\:\textrm{mA}\$ to \$4\:\textrm{mA}\$. I'd recommend selecting perhaps the \$20\:\textrm{mA}\$ current for the zener. That way, the total current through the resistor will be \$24\:\textrm{mA}\$, with all \$24\:\textrm{mA}\$ going through the zener when the BJT is not loaded, to just \$20\:\textrm{mA}\$ through the zener when the BJT is loaded, maximally. This variation isn't much, so the voltage reference the BJT sees should hold up, okay.

So in this case the resistor would be:

$$ R= \frac{12\:\textrm{V} - 6.2\:\textrm{V}}{24\:\textrm{mA}} \approx 242\:\Omega$$

You might select either \$220\:\Omega\$ or \$270\:\Omega\$. Either would be okay.

I've seen some circuits use a Darlington but what exactly are the considerations?

The Darlington will drop more voltage getting to the emitter, so your zener voltage selection needs to take that into account. The main other detail is that the Darlington will require less base current.

Suppose in the above circumstance, the maximum current required was \$1\:\textrm{A}\$, instead. In this case, we might plan on \$\beta\approx 40\$ and the base current then of \$25\:\textrm{mA}\$. This is way more than we can expect to design in, for that zener I mentioned. It's just too much to deal with. So here, we might choose a Darlington with a \$\beta=500\$ here, meaning we are back to a base current that is about \$2\:\textrm{mA}\$, which is now perfectly fine. Of course, we would have to use a higher zener voltage so that would be a different zener choice. But that's the idea.

I'd imagine that for the diode you simply want to choose a diode of which the zener breakdown voltage is 0.7V higher than your desired output voltage?

Roughly speaking, yes. You do need to figure out what the base-emitter drop will be at the BJT or the Darlington. Higher current compliance requires a larger drop. You may see anything from \$600\:\textrm{mV}\$ to over \$900\:\textrm{mV}\$, depending on the BJT and circumstances. And significantly more than that from a Darlington. Use the datasheets to help estimate this.

Also, are there advantages to this circuit over simply using a voltage regulator like the LM317?

For most purposes? Probably not. But there are issues of availability, number of part suppliers, cost, number of through-holes (if that's what you are doing), and other factors that may get you to choose one over another. One thing you do get here is the control over the BJT you pick. But you can add BJTs around the LM317, too. So maybe the LM317 still wins out.

  • 1
    \$\begingroup\$ This really should be the accepted answer. Would you mind expounding on how one would go about to choose the value of the resistor and the specific BJT for your application? I've seen some circuits use a Darlington but what exactly are the considerations? I'd imagine that for the diode you simply want to choose a diode of which the zener breakdown voltage is 0.7V higher than your desired output voltage? Also, are there advantages to this circuit over simply using a voltage regulator like the LM317? \$\endgroup\$ Sep 23, 2016 at 22:09
  • \$\begingroup\$ @DewaldSwanepoel See added edits. \$\endgroup\$
    – jonk
    Sep 23, 2016 at 22:56

Simple resistor with zener is suitable for low currents or for very specific and known level of current. If you want to use power output, you have to use something like that circuit you mentioned. It can provide stabilized voltage for wide range of current levels.

  • \$\begingroup\$ This circuit doesn't work very well because of the current dependent voltage drop across the transistor's base-emitter junction. Use it only for low currents. \$\endgroup\$
    – Bart
    Aug 1, 2016 at 13:55
  • 2
    \$\begingroup\$ Bart, you are right. But it is MUCH more stable than single resistor with zener. This part of question I tried to answer. \$\endgroup\$ Aug 1, 2016 at 14:02

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