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I need to turn off an LED when the voltage reaches 10V. My source is maximum 12.8V (three lithium batteries.) The (yellow) LED is rated 2.1V and 30mA forward current, but the resistor I chose is 3.3k because I need it dim and low consumption. When powered directly from the source, it marked 2.84mA @ 11.16v. Also 0.033mW.

The Zener diode is rated 10V 500mW.

I tried different values for the diode resistor and the one that came closest to producing the same overall 2.84mA current (and LED intensity) was 11k.

I don't know why and couldn't find a formula that explains my result. The transistor's measured Hfe is 140.

I would really appreciate if someone could explain me a bit of the theory behind, since I need to replicate the circuit under different conditions.

  • \$\begingroup\$ When you say "diode resistor" do you mean the zener diode resistor? An LED is also a diode. \$\endgroup\$
    – TimWescott
    Oct 4, 2021 at 18:47
  • \$\begingroup\$ This circuit will turn the LED on when the supply reaches around 10.6 V (10 V zener diode + Vbe of the transistor). I don't know why and couldn't find a formula that explains my result Circuits aren't defined by formulas. You need to think in terms of: when the voltage reaches 10 V, can the zener diode conduct? No because there's also a Vbe of the PNP. Only when that Vbe is larger than 0.6 V will any current flow through the Emitter-Base. What will happen when when current flows through the Emitter-Base of the PNP? Again: no formulas! \$\endgroup\$ Oct 4, 2021 at 19:05
  • \$\begingroup\$ Thank you Bimpelrekkie, I'll try to think that way and now I understand actually why the zener was not cutting at 10v exactly but almost 11v. Appreciated. But still, given the case of a diode with a 3.3k resistor (per schematic), how can you know the appropriate base (zener) resistor value for it? So no power is dissipated and the same power arrives to the led? I figured it out at 11k trying different values with a potentiometer, but how can one know in advance? @TimWescott I meant the zener diode resistor (see diagram). \$\endgroup\$
    – Domingo
    Oct 4, 2021 at 19:24
  • \$\begingroup\$ Here is another question about a circuit with a similar function: electronics.stackexchange.com/questions/547417/… \$\endgroup\$
    – Theodore
    Oct 4, 2021 at 20:12

1 Answer 1


If you plan to use a zener, then the key idea to hold in your head is that the zener will start conducting when sufficiently reverse-biased. You also need to keep in your mind that you'll need a series, current-limiting resistor to protect the zener as well as to operate it at the correct (datasheet) current (which varies depending on the zener's nameplated voltage.)

So, you have two parts in mind now. Just use your imagination about how to keep a transistor off when there's no current and then on when there is current. That's the only problem to solve in your mind.

A BJT transistor is on when it's \$V_{\text{BE}}\$ is forward-biased, sufficiently. It's off, otherwise. (That's a gross simplification, but so long as you imagine "less than \$400\:\text{mV}\$ means off and more than \$550\:\text{mV}\$ it is starting to be on" then you won't be too far afield.)

Hmm. If the zener conducts then there must be a voltage drop across that current-limiting resistor. And if not, then there will be no voltage drop across it. Sounds like a fit -- maybe, anyway.

So what about:


simulate this circuit – Schematic created using CircuitLab

It almost looks okay. Using the overly-simplistic argument I just made, this circuit looks like it may work well. When the zener is 'on' then there will be a voltage drop across \$R_1\$ and that should turn on the BJT. And it does. So, technically, it will turn on the LED at "some determined voltage," now.

The problem with it (well, the bigger problem anyway) is that once on and as the supply voltage rises still further then the BJT base will start supplying insane levels of current into the zener. That will probably not comport with your:

I need it ... low consumption

So, another cheap idea is to just move the LED resistor to the other side of the BJT. That should limit the ability of the BJT base to jam current into the zener, because the emitter resistor will be in the way of it.


simulate this circuit

Now, if you try this arrangement, it will much better meet your requirement as well as turning the LED on at a particular threshold.

The problem with this arrangement is that now it is a voltage-dependent current source to the LED. Let's say \$R_1=R_2=1\:\text{k}\Omega\$. Then the max LED current will be about \$2\:\text{mA}\$. But that LED current will decline linearly with lower voltages. So the brightness will be less when the voltage is less. This may be acceptable. I cannot say.

But the point of the above is to use your imagination and just think, a little. Turn things over in your head in different ways.

Here's another simple approach that minimizes the voltage-dependent current source period and makes it more of an on/off circuit -- through it still has its "behaviors."


simulate this circuit

That might perform acceptably. Now, \$R_1\$ sets the LED current and because this is a voltage-independent current source (once sufficient voltage overhead is reached), it behaves more like on/off than the earlier circuit.

These aren't by any means the only ideas that come to mind. There are countless ways to do this. For example, the very first thing I may want to do for a low power unit is to avoid the zener. You already may only want about \$1\:\text{mA}\$ or maybe \$2\:\text{mA}\$ in the LED. That's bad enough. You don't want to have to waste about the same amount (or still more) when feeding a zener diode. So you may want to look for a lower-current "zener-like" circuit for the voltage comparison part of this circuit.

You cannot do anything much about the LED, except to buy as high-efficiency LED as possible. But once that's decided, you are stuck with its requirement. The only thing left is to worry about the voltage comparison part. And much can be done on that front.

For example, Diodes Inc has the AP431S, which is very low current (tens of microamps) and can be set to a desired voltage, as well.

In addition, BJTs require recombination currents. These can be quite modest -- absolutely no more than 10% but can be 1% or so. And that isn't usually a big deal. But a FET can use even less. So if you get to the point of really wanting to squeeze the last few current-pennies out of the circuit, then you might consider using more expensive FETs and modifying the design once again.

  • \$\begingroup\$ A mind opener jonk... brilliant. but how to decide the value of R1 in your first two examples? given that I want the led to receive 2.84mA @ 11.16v? I ask because I also want to turn on a relay (45ohm) instead of a led, but can't figure out the base resistor value when a zener is involved. \$\endgroup\$
    – Domingo
    Oct 4, 2021 at 21:05
  • \$\begingroup\$ @Domingo Then you don't want a constant current circuit. You want a switch circuit for the relay. That's a different thing. Did I miss it? Or did you fail to include this requirement in your original question? \$\endgroup\$
    – jonk
    Oct 4, 2021 at 21:42
  • \$\begingroup\$ I failed to include the requirement in the original question, because I assume that a relay would work the same way than a led, ie. when the transistor is closed current flows through the collector-emitter junction and bias the coil of the relay. I don't see the difference with a circuit where a led or a relay is turned on; as long as the transistor is closed, once the base is charged via the zener. Probably I'm missing out something. \$\endgroup\$
    – Domingo
    Oct 4, 2021 at 21:54
  • \$\begingroup\$ @Domingo The LED has an exponential response to voltage and therefore requires some means of limiting the current. Besides, an LED's brightness is linearly related to its current. So the current in an LED is what everyone cares about. A relay is basically a coil. And you cannot jam a current into a coil. Just doesn't work like that. A relay requires a minimum voltage to engage and then a different voltage to hold, once engaged, and when released there is a significant kick-back voltage they generate that has to be accommodated. These are about as different as night and day. \$\endgroup\$
    – jonk
    Oct 4, 2021 at 21:58
  • \$\begingroup\$ @Domingo In general, it's just good advice that you write the truth about what you are doing. Even those of us who know enough to replace one thing with another, as a valid simplifying equivalent, have enough sense to state what we are doing and why we are doing it. It's perfectly fine to replace a circuit that is only a modest approximation of a current source with an ideal current source if the difference doesn't matter to the question and if using the simpler, idealized form makes it easier for others to focus on the section where the question is at. \$\endgroup\$
    – jonk
    Oct 4, 2021 at 22:07

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