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This circuit comes from Hamilton's Handbook of Linear Integrated Electronics for Research, 1977, p. 240. It is similar to the common BJT monostable multivibrator, but it doesn't include the feedback element. I've been trying to work out what the design intent was, out of curiosity. There isn't much explanation in the original text, only that this converts an impulse to a pulse of finite duration.

Work So Far. I've worked out the following. In the quiescent state, assume Q1 is OFF and Q2 is ON. Suppose the base of Q1 is kept just below the ON voltage, and that Q2 is saturated. An impulse on the base of Q1 causes that transistor to turn ON, and capacitor C1 turns Q2 OFF. This is solid so far. But for Q2 to stay OFF for an extended time, we would prefer Q1 to stay ON even after the TRIG signal has gone.

One way of doing this is to give Q1 and Q2 different bias currents, with the bias current of Q1 being lower than the bias current of Q2. This way, when Q1 turns ON and Q2 turns OFF, the emitter voltage is lower, and the base voltage of Q1 (which was just below the threshold) is now sufficient to keep Q1 ON until Q2 turns back ON and the emitter voltage increases again, turning Q1 OFF.

This works, and can be verified with simulation, but there's still a problem. If the impulse (trigger) is large enough to turn Q1 ON and to keep it ON, then it will also couple an impulse on the emitter, and this will cause an impulse on Vout (since Q2 is saturated, Vout tracks the emitter until Q2 turns OFF). Therefore, the output will not be a pure pulse, but will also include an impulse right before the pulse starts.

To verify, with the circuit as I've designed it (based on the above description), SPICE does show this impulse: enter image description here

Zooming in on the rising edge:

enter image description here

The green trace is Vout, the red trace is the base of Q1.

You could imagine filtering this signal, then passing it through another comparator stage to clean up the edge. But I wouldn't imagine this circuit as-shown to have this kind of flaw. Which makes me wonder if my assumptions about how it works are correct?


Edit I've found this circuit in other sources now, called an emitter-coupled monostable multivibrator. The descriptions I've found available online are lacking, and the only values I've found came from a practice problem, but those values seem to be either wrong, or dependent on the transistor beta so much that the simulation gives very poor results.

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    \$\begingroup\$ There is a feedback; it is implemented by connecting the emitters. \$\endgroup\$ Nov 27, 2023 at 16:53
  • \$\begingroup\$ Emitter resistor R3 “copies” the trigger pulse for a while until trigger pulse persist. Then the voltage across R3 goes almost to zero volt. This happen so quick the Q2 dynamics doesn’t catch up (Q2 is still conducting). \$\endgroup\$ Nov 27, 2023 at 17:32
  • \$\begingroup\$ I have not this behaviour with 2N2222. I have it with 2N3904. \$\endgroup\$
    – Antonio51
    Nov 28, 2023 at 8:56

2 Answers 2

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First Response

In the quiescent state, assume Q1 is OFF and Q2 is ON.

I would have been a little more descriptive. Assuming Q1's base doesn't draw much, it's base sits at \$1\:\text{V}\$ due to the divider.

Assume \$Q_2\$ is saturated, then the collector/emitter current will be about \$\frac{12\:\text{V}}{R_3+R_6}\approx 2.8\:\text{mA}\$. That would put its base at \$3.5\:\text{V}\$, so \$R_5\$ would be supplying \$\approx 26\:\mu\text{A}\$, which just happens to be about \$100\times\$ less. So, yes, it's confirmed to be likely saturated. But a very edgy saturation because we're assuming a lowish active mode \$\beta\$. So this isn't by any means deep saturation. That's why, in fact, \$R_5\$ is set \$100\times\$ the value of \$R_6\$.

Suppose the base of Q1 is kept just below the ON voltage,

No, this is where we separate company. Instead, the shared emitter voltage is about \$2.8\:\text{V}\$. Which means \$Q_1\$ is not kept "just below the ON voltage". There's a reverse voltage on the base-emitter of \$Q_1\$ of about \$-1.8\:\text{V}\$. I'd say that is well-below.

and that Q2 is saturated.

Yes, but again just barely so. Lightly saturated. In a twilight zone between active and saturated, really.

An impulse on the base of Q1 causes that transistor to turn ON,

Look, you need to turn \$Q_1\$ on long enough to matter. What exactly must that be? You've basically got a source impedance in the divider of about \$18\:\text{k}\Omega\$. And a tiny capacitor of \$10\:\text{pF}\$ which is likely on the order of the 2N2222's internal capacitances, which themselves need to be supplied as it turns on.

So let's say that you have to overcome about \$20\:\text{pF}\$ in the BJT (and it is likely more than that) and do that while working against \$18\:\text{k}\Omega\$. All that through \$10\:\text{pF}\$, to cause at least a voltage change of \$1.8\:\text{V}+700\:\text{mV}\$ or \$2.5\:\text{V}\$. That's \$140\:\mu\text{A}\$ just for the resistance.

And then some more for the BJT, once it starts to turn on. Whatever \$\frac{\text{d}V}{\text{d}t}\$ you are supplying into \$C_2\$ is going to pull about twice as much, likely more than twice, for the BJT's capacitance.

So in round figures, I'd guess you need about \$150\:\mu\text{A}\$ for the resistance load. And once the BJT turns on, more than another \$300\:\mu\text{A}\$ on top of that. So we're already at \$450\:\mu\text{A}\$ and if you want to handle various BJTs you need still more. I'd probably figure at least \$800\:\mu\text{A}\$, which I'd round to \$1\:\text{mA}\$ to be more sure.

So this means \$\frac{1\:\text{mA}}{10\:\text{pF}}\approx 100\:\frac{\text{V}}{\mu\text{s}}\$. With a \$12\:\text{V}\$ trigger that means a rise time faster than \$120\:\text{ns}\$. Granted, I more than doubled things to be safe. So simulation might work near two or three times that long. But not much longer than that.

So this means that whatever triggers this must also have a very fast edge. Bear that in mind.

Okay. So with all that said, we've got \$Q_1\$ turning on. But for how long??

Well, that \$18\:\text{k}\Omega\$ Thevenin resistance and \$Q_1\$'s base current won't take long to blunt \$C_2\$. So it will be the rest of the circuit that determines the duration. \$C_2\$ won't have much to do with it.

and capacitor C1 turns Q2 OFF. This is solid so far. But for Q2 to stay OFF for an extended time, we would prefer Q1 to stay ON even after the TRIG signal has gone.

Assume \$Q_1\$ is momentarily saturated from the earlier logic (\$\frac{\text{d}V}{\text{d}t}\$ is large enough.) It's collector/emitter current will be about \$200\:\mu\text{A}\$, so the left side of \$C_1\$ is pulled down to about \$200\:\text{mV}\$. But with \$12\:\text{V}-3.5\:\text{V}\approx 8.5\:\text{V}\$ across it, this yanks the right side definitely well into negative territory. So yes, \$Q_2\$ not only is firmly off but if \$C_1\$ were any larger it might also hold its voltage long enough to be avalanching the base-emitter, as well. In any case, yeah. \$Q_2\$ is turned off.

With the left side of \$C_1\$ held close to ground and the right side likely around \$-6\:\text{V}\$ or more for a moment, all that's left to consider is how long it takes \$100\:\text{pF}\$ to charge back up via \$R_5\$ so that \$Q_2\$ can be back on. That's an RC curve but without going nuts about it I'd guess that it will take about \$22\:\mu\text{s}\$ to yield \$\Delta V\approx 10\:\text{V}\$ with an average of \$15\:\text{V}\$ across \$R_5\$.

So that's where I'd finger the output pulse duration.

Added Per Comments Below Question

Considering the circuit doesn't function properly (doesn't produce a clean pulse) I would say this is not a tertiary issue! The behavior should be mostly independent of transistor parameters, so the design values will need to change. The question is, how drastically and in what way?

Try stuffing a \$22\:\text{pF}\$ capacitor, going from the base of \$Q_2\$ to your \$12\:\text{V}\$ rail. That will blunt the impact of the follow-on effects due to the sudden, rapid rise in \$Q_1\$'s emitter, propagating through CJE and CJC of \$Q_2\$ (and even backwards over to \$Q_1\$'s collector via \$C_1\$.)

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  • \$\begingroup\$ Thanks for the answer, most of your comments were relating to the values I'd designed in (and you're right, I was tinkering with values, so e.g. "just barely below conduction" was no longer true, and that's also why Q2 isn't hard saturated anymore). And getting an expression for the pulse duration is no issue, but eliminating the coupled impulse is. The question is really about that awkward rising edge issue. Any thoughts there? \$\endgroup\$ Nov 29, 2023 at 18:32
  • \$\begingroup\$ @SamGallagher Sorry I didn't earlier understand your question. Seems I just blew past the fact that you cared about some tertiary issue and not about how the thing worked. Had I realized that fact, I'd not wasted time. What you care about are the CJC, CJE, and TR parameters of the BTJs. Try setting them to something very small, like well below the fempto- area and re-run the simulation. \$\endgroup\$ Nov 30, 2023 at 7:11
  • \$\begingroup\$ Considering the circuit doesn't function properly (doesn't produce a clean pulse) I would say this is not a tertiary issue! The behavior should be mostly independent of transistor parameters, so the design values will need to change. The question is, how drastically and in what way? \$\endgroup\$ Nov 30, 2023 at 18:00
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    \$\begingroup\$ The 22pF cap does the trick! I still wonder if this was the original circuit intent, but I'm not sure it's worth it to keep working on it. Thanks for the help \$\endgroup\$ Dec 4, 2023 at 21:41
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I found an explanation in J. Watson's Semiconductor Circuit Design (1970) (link to first edition) which I think adequately explains the circuit behavior, and can be considered a more authoritative source. Here's the description, slightly edited for brevity:

Here, Tr1 is normally ON and Tr2 is normally OFF, and under these conditions the steering diode is unbiased. Consequently, a negative trigger pulse will be allowed to pass to the base of Tr1 and the collector of Tr2, so initiating a change of state. When Tr1 switches OFF, its emitter voltage goes negative and this also being the emitter voltage of TR2, a forward bias is applied to the base-emitter junction of that transistor, so turning it ON. This is because the base voltage of Tr2 is held fairly constant by resistors R1 and R2. Consequently, Tr2 switches ON and the negative-going transient at its collector is applied to the base of Tr1 via the capacitor C, so forcing Tr1 hard OFF. enter image description here Because Tr2 is now saturated, its base will clamp the left-hand side of C very near to the emitter voltage V2. The right-hand side of C will charge towards +Vcc via \$R_{B1}\$ and \$R_E\$ along with the saturated Tr2 in series. If \$ R_{B1} \gg R_E\$, the time constant will be approximately \$R_{B1} C\$.

The emitter voltage is now V2, or \$I_{E2} R_E\$, and when the right-hand side of C reaches a voltage a little higher than this, switch-over will again take place.

Watson goes on to derive an approximate timing law for the pulse duration, giving:

\$ T = R_{B1} C\, \ln \left( 1 + \frac{V_{CC}-V_1}{V_{CC}-V_2} \right) \$

And he goes on to caution:

This equation is rather less accurate than is desirable, for it depends on many approximations, the most important being the infinitely rapid switching of Tr1 and Tr2. However, it does show that T can be controlled by both RB1 and C and by the relative levels of V1 and V2. The stability of V2 is determined by the combination R1 and R2, and for very stable operation, R2 may be replaced by a Zener diode, or shunted by a speed-up capacitor.

The recovery of the circuit after an excursion is not immediate as the foregoing treatment suggests, for C will recover via \$R_{C2}\$ and the parallel combination of \$R_{C1}\$ and \$R_E\$. This time constant is normally small, but it should be ensured that a further trigger pulse does not appear until two or three recovery time constants have elapsed.

Watson concludes with a discussion of the biasing required for the circuit to operate:

Finally, the bias situation must be considered. In order to maintain Tr1 ON and Tr2 OFF during standby, then V1 must be more positive than V2... This assumes that TR1 is fully saturated when in the standby condition, which may be assured by making the base current at saturation large enough. ... During the dwell time, Tr2 must be saturated, which means that by similarity with equation (7.19), \$ h_{FE(0)}R_{C2} > R1 \$. R2 may now be found by noting that the voltage at the junction of R1 and R2 should be positive with respect to the base of Tr 1 if the base lead were open-circuited and Tr1 continued to conduct.

This confirms my analysis above, and due to the way the trigger input is coupled in, the trigger coupling into the output might be reduced. Otherwise, a small 22pF capacitor in parallel with \$R_{B1}\$ can be added.

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