It seems that the circuit bellow acts as a differential comparator. I confirmed it by simulation on LTSpice.


simulate this circuit – Schematic created using CircuitLab

I remark that output flips when VA=VB. I can easily determine Low to High transition by neglecting base current of Q1 (the results are close to simulation value): $$V_{LH}=\left( 1+{\frac{R_1}{ \left( R_4+R_5 \right)//R_2}} \right) *V_{ZD1}$$ How can I determine High to Low transition?

Can someone explain how Q2 is passed/blocked?

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Commented Jun 13, 2020 at 4:14

1 Answer 1


Initial Approach

Ignoring BJT base currents and simplifying the behavior of the long-tailed BJT pair so that I can assume in all cases that \$V_{\text{B }Q_1}=V_{\text{B }Q_2}\$, I'd go through the following steps:

  1. Estimate \$I_Z=\frac{V_\text{CC}-V_Z}{R_7}\approx 125\:\mu\text{A}\$. The 1N4739A datasheet says it should be \$I_Z=28\:\text{mA}\$. Note that the value computed for your circuit is very far from the recommended operating point.

    The datasheet doesn't show the static resistance of a zener diode. Zeners do have some, from fractions of an Ohm to perhaps a few Ohms. The datasheet instead shows the zener's apparent resistance, \$Z_Z\$, which is the local resistance slope near the recommended operating point and it includes both the static (leads, bonding wires, bonding contact points, and doped semiconductor bulk) and dynamic (related to breakdown in this case) resistance co-mingled together. To make this clearer, let's look at a plot I just developed using LTspice and an ORCAD model I have for the 1N4739 zener diode:

    enter image description here

    In the above, you can see the cursors located at approximately the right place for operation of this zener and you can also see the zener voltage. This pretty much confirms that this really is a \$9.1\:\text{V}\$ zener, when operated at the recommended zener current.

    But also take note of the slope of the green line at that operating point. From the datasheet, for this zener, it shows \$Z_Z=5\:\Omega\$. And you may be able to just barely see it curving a bit right at this operating point (keep in mind this is a log-plot, though.) There is a "local slope" you can get by placing a ruler there and drawing a tangent line that just touches the curve at the operating point. This is the \$Z_Z\$ resistance they are specifying in the datasheet.

    From the datasheet, I don't really have "better information" than this slope. However, you can see that the slope isn't fixed, but varies. So the assumption I'm going to make is just that: an assumption. But it's all I have available from the datasheet and it will have to do.

    With the above caveats in mind, I find the following guess:

    \$\text{ }\therefore V_{\text{B }Q_2}=V_Z^{'}=V_Z-I_Z\cdot Z_Z\approx 8.96\:\text{V}\$, or \$9.0\$ in round numbers for use below.

    (LTspice instead computes it as \$8.93\:\text{V}\$, when supplied with the estimated \$125\:\mu\text{A}\$. So I'm not complaining.)

  2. Assume \$600\:\text{mV}\le V_{\text{BE }Q_2}\le 700\:\text{mV}\$, make an initial estimate of \$I_{R_3}= \frac{V_\text{CC}-V_Z^{'}-V_\text{BE}}{R_3}\$ or \$530\:\mu\text{A}\le I_{R_3}\le 540\:\mu\text{A}\$.

    \$\text{ }\therefore I_Q=I_{R_3}\approx 550\:\mu\text{A}\$, in round numbers.

  3. In this configuration, assuming full saturation of either one or the other BJT and ignoring the Early Effects, etc., \$I_{\text{C }{Q_2}}\$ is either all of the current in step #2 or none of it (this is kind of a current-teeter-totter of sorts):

    \$\text{ }\therefore V_{\text{C }{Q_2}}=\left(V_Z^{'}+R_4\cdot I_{\text{C }{Q_2}}\right)\cdot\frac{R_5}{R_4+R_5}=\left.\begin{array}{r|cc} 1.6\:\text{V}\\ 6.1\:\text{V}\end{array}\right.@I_{\text{C }{Q_2}}\left\{\begin{array}{r} 0\:\text{A}\\ 550\:\mu\text{A}\end{array}\right.\$

  4. This is enough, now, to solve for the input voltage thresholds:

    \$\text{ }V_\text{IN}=V_Z^{'}\cdot\left(1+\frac{R_1}{R_2\:\mid\mid\: R_4}\right)-V_{\text{C }{Q_2}}\cdot\frac{R_1}{R_4}\implies\text{ }\left\{\begin{array}{l} V_\text{HI}\approx 105\:\text{V}\\ V_\text{LO}\approx 97\:\text{V}\end{array}\right.\$

There are a lot of assumptions, above. But this would be my "back of the envelope" approach in order to get an initial approximation.

Targeting a Wide Hysteresis

Given the above and a little bit of algebra, the hysteresis width will be something like this:

$$\Delta V = \:\mid V_\text{HI}-V_\text{LO} \, \mid \: = I_Q \cdot \frac{R_1}{R_4} \cdot \left(R_4 \mid \mid R_5 \right) = I_Q\cdot R_1 \cdot \frac{R_5}{R_4+R_5}$$

That provides a few things to consider.

So first off, I'd increase \$I_Q\$ a bit (not a lot, because I don't want to mess with the magnitudes of your resistors too much) by setting \$R_3=6.8\:\text{k}\Omega\$. This boosts things a little, so that \$I_Q\approx 780\:\mu\text{A}\$. (Figure somewhere between \$750\:\mu\text{A}\$ and \$800\:\mu\text{A}\$.)

Then I'd definitely increase \$R_5\$ while also decreasing \$R_4\$. I don't want to increase \$R_5\$ too much, just yet. So I'd shoot for about \$R_5=12\:\text{k}\Omega\$. But I'd drop \$R_4\$ quite a bit, to about \$R_4=22\:\text{k}\Omega\$.

Together, this means I've got about \$\Delta I_\text{IN}\approx 780\:\mu\text{A}\cdot \frac{12\:\text{k}\Omega}{12\:\text{k}\Omega+22\:\text{k}\Omega}\approx 275\:\mu\text{A}\$. Since I want \$\Delta V=200\:\text{V}\$, I'd find I need \$R_1=\frac{\Delta V}{\Delta I_\text{IN}}\$, or something like \$R_1=680\:\text{k}\Omega\$ to \$R_1=750\:\text{k}\Omega\$ (It computes out closer to the larger value, so I'd go with that.)

Now, given that you've increased \$R_1\$ by a factor, I'd increase \$R_2\$ by a similar factor, or \$R_2=68\:\text{k}\Omega\$. This should get you pretty close given that all we are doing is "back of the envelope" calculations.

I've approximately this in mind:


simulate this circuit – Schematic created using CircuitLab

Now, you may just need to tweek \$R_2\$ a bit to set the bracket about where you want it. You may need to make minor adjustments to \$R_4\$, too. But maybe not. Those are the only two resistor values I'd mess with, at this point.

Hopefully, that helps.

Of course, you have to have a \$15\:\text{V}\$ power supply. But you seem to have it, already. So that's good.

The Above Design Values in LTspice

I finally got a moment to try out the above design in LTspice, which already comes with models for your BJTs but didn't come with the model for the zener. I got a zener model from ORCAD and stuffed it into the simulation.

Here are the results:

enter image description here

I'm actually kind of shocked at how close it came. It's just a simulation and there are a lot of simplifying assumptions that were made, above. Yet, not bad at all!

Anyway, I guess my rational thinking process based upon what pops out looking at the algebra got close enough for simulation, at least. Of course, reality will set in and your devices won't be matched like those in the simulation.

Keep in mind I have not done any analysis of realistic variations in BJT parts, the zener, etc. So this is just my attempt to help you analyze the circuit. Not to develop a circuit that may be calibrated and the reproduce repeatable results, one circuit to the next, one environment to the next. Nor any thoughts to protection, isolation, safety, etc. etc.

  • \$\begingroup\$ Correct me if I am wrong. Zener current should be : $$I_Z=\frac {V_{CC}-\left(V_{Z@I_{ZT}}-Z_{Z@I_{ZT}} \cdot I_{ZT}\right)}{R_7+Z_{Z@I_{ZT}}}$$ Then $$V_{B_{Q2}}=V_{CC}-R_7\cdot I_Z$$ \$\endgroup\$
    – hiak
    Commented May 27, 2020 at 16:25
  • 1
    \$\begingroup\$ @Abderrezak The \$Z_{Z@I_{\text{ZT}}}\$ isn't needed in the denominator. That's a dynamic resistance, best seen from a derivative point of view. (The local slope about the operating point.) It is my fault, by the way, in how I wrote above. You've pointed up a problem I need to fix. What I was actually trying to do for you (but did NOT do correctly) is point out that the zener voltage will be different if you don't drive it correctly. This is why current sources are usually used, where possible. I extrapolated the dynamic resistance over a long range. It's all you can do, though. So that's why. \$\endgroup\$
    – jonk
    Commented May 27, 2020 at 17:03
  • \$\begingroup\$ @Abderrezak I'm going to clarify my text. \$\endgroup\$
    – jonk
    Commented May 27, 2020 at 17:07
  • \$\begingroup\$ @Abderrezak I've updated it. Hopefully, that helps you better understand where I was coming from. \$\endgroup\$
    – jonk
    Commented May 27, 2020 at 17:46
  • 1
    \$\begingroup\$ @Abderrezak So, I did the best I could to estimate your zener voltage given the datasheet limitations I'm handed. However, you really cannot use the local slope \$Z_Z\$ value to model the static resistance, which may be anything from about \$\frac14\:\Omega\$ to around \$\frac12\,Z_Z\$ (more or less.) It's a lot better just to make sure that your resistor, if you use one, is very, very much larger than \$Z_Z\$ so that you can ignore the value in the denominator when working out the zener current. It's not worth struggling for that precision nor using a current setting resistance near \$Z_Z\$. \$\endgroup\$
    – jonk
    Commented May 27, 2020 at 18:11

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