This is the PCB from an Elmo Sound Monitor 912, marketed sometime back in the 1970/1980's.

The circled component seems to be a diode of some sort. Or, at least the diode mode on a multimeter measures 0V in one direction and 1.1V in the other. The two-tone paint also seems to indicate that the polarity of whatever it is matters. And comparing it to the size of the other nearby components, it is quite a bit smaller than most through-hole components.

Unknown through-hole component

Can anyone identify more information--like a manufacturer that made this style of component or even the name of the package--that might aid my searching?

This question shows a device that appears to be a similar size but has a different paint style. The forward voltage is also in the same ballpark. Would that make this a "stabistor"?

Trying to reverse-engineer the board, it's D1 in this schematic:

reverse-engineered schematic diagram of second half of Elmo Sound Monitor 912 PCB

The signal is coming from a magnetic tape head that passes through two fairly standard BJT amplifier stages and a potentiometer voltage divider before entering this schematic through C4. It's a little unclear what the role of D1 or even T3 is in this situation.


Several people asked: the diode forward voltage was measured in-circuit.

And here is the complete schematic from the magnetic tape head, including the two BJT pre-amp stages:

complete reverse-engineered schematic diagram of second half of Elmo Sound Monitor 912 PCB

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    \$\begingroup\$ When you measured the part on the DVM was it in circuit or out of circuit? \$\endgroup\$
    – Autistic
    Commented Mar 8 at 1:37
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    \$\begingroup\$ Check out this SE amusing self-accepted answer. electronics.stackexchange.com/questions/636786/… Two-junction stabistor would be my bet. \$\endgroup\$ Commented Mar 8 at 6:28
  • \$\begingroup\$ @SpehroPefhany Reading... Okay. A soft low-voltage zener had crossed my mind. But I have never heard the word stabistor before. That one is new to me. \$\endgroup\$ Commented Mar 8 at 9:15
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    \$\begingroup\$ Nicholas - Hi, You said: "the diode mode on a multimeter measures 0V in one direction and 1.1V in the other" Did you really mean that in one direction, the DMM showed "0V" or did you mean "0L"? \$\endgroup\$
    – SamGibson
    Commented Mar 8 at 14:20
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    \$\begingroup\$ The circuit seems to be a standard series-pushpull output stage. The role of D1 is then to create the right bias voltage for the sum of the VBE's of the output transistors, so that must be about 2 diode voltages. D1 probably contains 2 diodes in series, internally. So you should look for components with that kind of description (e.g. "double diode", or "stacked diode") to find what the exact type might be! \$\endgroup\$ Commented Mar 10 at 9:02

3 Answers 3


initial thoughts

Every part in this circuit is doing something useful. Some of them are doing more than one thing. It's a minimal class-AB/class-B output stage, with voltage gain. I'll try and cover most of it.

Assuming this is a line-out, it probably anticipates about \$10\:\text{k}\Omega\$ load at the output and needs to support at most \$2\:{\text{V}_\text{PP}}\$.

Let's assume \$V_{_\text{BE}}=700\:\text{mV}\$ and \$\beta=200\$ for small signal devices.

(Note: Martin Henne has provided a link to a reverse-engineering PDF, in comments below, that specifies the output is designed for headphones at \$15\:\Omega\$. This is not the \$10\:\text{k}\Omega\$ load I had assumed when writing what follows. The new information explains the values for \$R_5\$ and \$R_6\$ as dropping a quiescent \$100\:\text{mV}\$ and qualifies as good emitter degeneration, for example. It will also require several updated equations, the quiescent point and voltage gain at the very least, if I want to update things, later. Just FYI for now.)

initial summary

\$C_4\$ is a DC-blocking cap to avoid DC bias at the signal input from interfering with the operating point of the circuit.

\$T_3\$ is the voltage amplifier stage (VAS.) \$C_{12}\$ must be set larger than the intrinsic BJT capacitance of \$T_3\$ and manages and sets the dominant pole to roll off the gain for higher, unwanted frequencies.

\$T_4\$ and \$T_5\$ are the two-quadrant active driver pair for the output that operate in class-AB/class-B.

\$R_5\$ and \$R_6\$ normally would be emitter degeneration. But see notes on specific detail sections below for more about that.

\$R_3\$ works together with \$R_4\$ to set the voltage gain. The NFB will be \$\approx \frac1{1+\frac{R_4}{R_3}}\$. Each resistor contributes a current and the VAS applies its \$\beta\$ to that. So I'd estimate \$A_v\approx \frac1{\frac1{\beta}+NFB}\approx 29.8\$ (\$\beta=200\$.)

\$C_7\$ is a bootstrap capacitor. It causes \$R_{14}\$ to have a fixed voltage across it and therefore act like a current source. (A good idea that should be studied.)

\$R_{11}\$ is there for \$C_7\$ to work against while maintaining \$R_{14}\$ as a current source.

\$D_1\$ sets the necessary voltage difference between the bases of \$T_4\$ and \$T_5\$ to keep the output stage as a class-AB/class-B with lower cross-over distortion.

\$C_1\$ is another DC-blocking capacitor so that the single-ended line-output appears to be centered around the shared ground reference.

\$C_{10}\$ is there for start-up/power-on. (Maybe helps with pop?)

minimum \$V_{_\text{CC}}\$

To keep \$T_3\$ (the VAS) in active mode, it's collector should never go lower than \$1\:\text{V}\$. Given the maximum swing, the range for the collector voltage should be at least \$V_{\text{C}_\text{VAS}}=2\:\text{V}\pm 1\:\text{V}\$. So the mid-point voltage range, between \$R_5\$ and \$R_6\$, should be no less than \$V_{_\text{MID}}=2.7\:\text{V}\pm 1\:\text{V}\$.

Now, here is where a part does double-duty. \$R_4\$ provides NFB for the circuit, setting the overall voltage gain to about \$A_v\approx 30\$. But at this moment what I care about is that it also provides the recombination current needed by \$T_3\$.

A very simple analysis tells me that if \$k=\beta\cdot\frac{R_{14}+R_{11}}{R_4}\$ then \$V_{_\text{MID}}=\frac{V_{_\text{CC}}-V_{_\text{BE}}}{1+k}+\frac{V_{_\text{BE}}}{1+\frac1{k}}\$.

Solving for \$V_{_\text{CC}}\$ and plugging in \$V_{_\text{MID}}=2.7\:\text{V}\$, I would find that \$V_{_\text{CC}}\ge 6.953\:\text{V}\$.

Rounded to \$V_{_\text{CC}}\ge 7\:\text{V}\$ gives the absolute minimum operating voltage for this circuit.

quiescent collector current for \$T_3\$

Returning to \$R_4\$, it's now possible to work out the quiescent collector current for \$T_3\$, given some \$V_{_\text{CC}}\$.

This is \$I_{\text{C}_\text{VAS}}=\beta\cdot\frac{V_{_\text{MID}}-V_{_\text{BE}}}{R_4}\$. With \$V_{_\text{CC}}= 7\:\text{V}\$, this means \$I_{\text{C}_\text{VAS}}\approx 2.35\:\text{mA}\$ (and \$V_{_\text{MID}}=2.72\:\text{V}\$.)

This is exceptionally hot for a line-out stage, which will never need to supply more than \$100\:\mu\text{A}\$ into its load. But there it is. And it only gets hotter with higher \$V_{_\text{CC}}\$ values.

(It's possible that this current value, or still somewhat larger, is appropriate for the stabistor that Spehro mentions in comments.)

I think this choice should be re-evaluated. But it's not my circuit and I don't know the details of \$D_1\$, either.

\$R_5\$ and \$R_6\$

In arriving at the minimum \$V_{_\text{MID}}\approx 2.72\:\text{V}\$, I discounted any voltage drop across \$R_5\$ and \$R_6\$.

The reason is that both are \$3.3\:\Omega\$. Normally, I'd first think emitter degeneration as an important part of their purpose. But that would suggest \$100\:\text{mV}\$ drop and there's no way on Earth that they would design such a high implied quiescent current of \$30\:\text{mA}\$ through \$R_5\$ and \$R_6\$.

Maybe they did, extreme as that may be, but for now I conclude their values were arrived at for a reason I don't follow.

(Note: \$D_1\$ could drop enough (maybe if more than \$1.6\:\text{V}\$?) to set that particular quiescent current so high. But it just doesn't really make much sense. It's grossly wasteful. Might as well go class-A, if so.)

So, ignorant about \$D_1\$, I'm assuming those resistors won't have a lot of drop across them. Negligible, I believe. So I didn't count them into the above mid-point voltage value.


Here, I just think it should drop two \$V_{_\text{BE}}\$, plus a little. Say, \$1.5\:\text{V}\$ to move it a little out of class-B and into class-AB. There's plenty of base drive, it seems. So I'm not too worried about running it as a hotter class-AB.

(Spehro mentions in comments that this may be a stabistor.)

\$C_7\$ and \$R_{14}\$

\$C_7\$ is just a bootstrap capacitor. It is big and the voltage across it won't move around much. So this means there will be a relatively fixed voltage across \$R_{14}\$, turning it into a cheap current source.

With \$V_{_\text{CC}}=7\:\text{V}\$ then about \$2.35\:\text{V}\$ across \$R_{14}\$, given \$I_{\text{C}_\text{VAS}}= 2.35\:\text{mA}\$ (from above.) This also means that the voltage across \$C_7\$ will be about \$3.1\:\text{V}\$.


This one gets a widely varying voltage across it because of the bootstrap capacitor. With \$V_{_\text{CC}}=7\:\text{V}\$ and the peak \$V_{_\text{MID}}= 3.72\:\text{V}\$ (with the peak output voltage added to it) and \$C_7\$'s voltage being another \$3.1\:\text{V}\$, there can be as little as \$7\:\text{V}-3.72\:\text{V}-3.1\:\text{V}=180\:\text{mV}\$ across it at the peak of a swing. It's just there to give something for the bootstrap capacitor to work against.

It's normal to ensure that, at minimum, it's current doesn't go to zero. But the circuit will work even if that idea isn't followed to the letter.

input signal

Because \$A_v\approx 30\$, this should not exceed a peak of \$33\:\text{mV}\$ (and should normally be substantially less.)


Here's what LTspice says. I set \$D_1\$ to drop \$1.4\:\text{V}\$ and \$V_{_\text{CC}}=7\:\text{V}\$ and let it run. The predictions above were \$V_{_\text{MID}}=2.72\:\text{V}\pm 1\:\text{V}\$, \$I_{R_{14}}=2.35\:\text{mA}\$, and \$A_v=30\$. The .TRAN run appears to show very similar values:

enter image description here

final note

This is actually a very good circuit to study, carefully. I'm impressed by how much has been packed into so few parts, putting some to multiple important duties. It's not often you see something like this. So study it.

I don't agree with all the exact value choices. But I also don't know anything about \$D_1\$'s datasheet. And I'd probably do something different about the emitter degeneration.

But the topology is well worth a thorough study. It uses NFB to set voltage gain and good linearity and immunity to operating temperature changes and power supply ripple. It provides two-quadrant drive. It illustrates bootstrapping. It does all this with economy.

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    \$\begingroup\$ Your analysis is amazing. The audio output is for headphones and has an impedance of 15 ohms according to the manual. The full schematics can be found here: PDF - Elmo_SoundMonitor-912_ReverseEng v17 Eagle - Elmo_SoundMonitor-912_ReverseEng v17 \$\endgroup\$ Commented Mar 8 at 14:25
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    \$\begingroup\$ @MartinHenne So, it's not for line-out but for headphones. That would move some needles. The emitter degeneration for one. I'll to take a look in half a day. Thanks! \$\endgroup\$ Commented Mar 8 at 15:12
  • \$\begingroup\$ Great answer. But can you explain in more detail how did you arrive K factor and Vmid equation? \$\endgroup\$
    – G36
    Commented Mar 8 at 20:44
  • \$\begingroup\$ @G36 You know that (VMID-T3 VBE)/R4 has to be the base current into T3. Active T3 will have BETA times that as its collector current. That collector current times R11+R14 will be a drop from VCC to the base of T4. One more T4 VBE drop from there and you are back at VMID. Do you think that's enough for you to solve for VMID? Or do you disagree with the approach or otherwise need it slow-walked? \$\endgroup\$ Commented Mar 8 at 22:42
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    \$\begingroup\$ @MartinHenne That reverse-engineered schematic definitely says 15 Ohms and sets VCC to 9 V. And now I can readily imagine 100 mV across the 3.3 Ohm resistors as a design goal. So this erases that confusion of mine. Voltage gain will drop to about 20, as well. Thanks for the link. \$\endgroup\$ Commented Mar 9 at 0:31

T3 is the main voltage amplifier for the circuit. The circuit gain is set by R3 and R4 (and the input impedance of T3).

D1 is a dual diode, like two 1N914's in series.


1.1V is rather high for a DVM reading of a single silicon diode, but it might reading of 2 series diodes at a very low current density. Two series diodes suggest Class-AB biasing where a single diode would give you Class B biasing.

Comparing the reading of a known single silicon diode and this unknown device could help you work out what is on the inside.

  • \$\begingroup\$ JkingNH - Hi, Thanks for posting an answer. However, questions to the OP should not be asked in answers, so I removed them. Clarification questions to the OP should be asked using comments, usually below the question. That way: (a) The OP will get notified of your clarification request. (b) Other people will see that it has been asked (in fact one of your questions has already been asked in a comment!). (c) The OP can (and often should) add new info into the question in response to clarification comments, and not leave it under just one answer. \$\endgroup\$
    – SamGibson
    Commented Mar 8 at 13:37
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    \$\begingroup\$ Well, selenium has around 1V drop... \$\endgroup\$
    – PlasmaHH
    Commented Mar 8 at 15:24

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