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I am learning electronics and I am looking at a clap switch. I don't understand how the collector voltage is controlled by the base voltage.

When there is no signal, the collector voltage is high and when there is a signal, the collector voltage is low. How does this happen?

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  • \$\begingroup\$ I believe a typical electret (this obviously isn't the other kind of condenser mic requiring a separate DC supply) requires about 2 V of overhead voltage. That would put the base-emitter voltage "very high." Instead, the overhead voltage across the condenser mic would be perhaps 900 mV at most. I suppose that may work. But it's not designed well, I suspect. The supply resistor only provides about 200 uA, at most. Typical electrets use 500 uA, I think. Also, I'd normally expect a capacitor to the BJT base from the electret. But then the BJT would need additional biasing. \$\endgroup\$
    – periblepsis
    Commented Feb 3, 2023 at 19:52
  • \$\begingroup\$ I think the transistor will be partially conducting but all the time, but the circuit designer hopes it keeps pin 2 above the threshold voltage (1/3 of 9V) when there is no sound. Eww. At least if they're going to do it that way, R1 or R2 should be adjustable since not every BC547 transistor has the exact same amount of gain. I think. Not fully sure. Maybe I misunderstood how the circuit works. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Feb 3, 2023 at 20:09
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    \$\begingroup\$ The short answer to the question in your title is that it's not. Bipolar transistors are current operated devices, not voltage operated. \$\endgroup\$
    – SteveSh
    Commented Feb 3, 2023 at 20:10
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    \$\begingroup\$ @periblepsis just because there's a model that takes voltage at the base to control the collector current, doesn't mean there isn't another model that takes base current to do the same. So, simply being a model doesn't make for reality. It's reasonable to say that collector current is controlled exponentially by the base voltage. However the base current that's induced by the transistor operation has to be compensated externally so quickly and precisely by a matching base terminal current, that it's also reasonable to say to, especially noobs, that current controls it. \$\endgroup\$
    – Neil_UK
    Commented Feb 3, 2023 at 20:41
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    \$\begingroup\$ @periblepsis If we were at a physicists' convention, I would agree with you. As we are answering a question from a noob (quote "I am learning electronics", and look at the level of his question), then I would prefer Steve's approach. I am not sure if you are familiar with the concept of lies to children, but this is the situation for one if ever there was one! I reckon most hobbyists could work with current controlled models and a nominal VBE for ever, only needing voltage control for professional work or higher education. \$\endgroup\$
    – Neil_UK
    Commented Feb 4, 2023 at 8:53

3 Answers 3

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When the condenser mic is off, R1 provides about 186 uA of current into the base of Q1. If Q1 has a beta (current gain) of at least 50, then Q1 will be close to saturation, meaning that it's collector voltage (which is TRI input of the '555) is close to 0 V. In this condition, the base voltage of Q1 is around +0.7 V.

Updated this section

OP did not provide any information on the microphone, so I'm going to have to make some assumptions. When a sound is detected by the mic, it generates an AC waveform whose frequency is that of the incoming sound wave.

This should alternately pull the base to close to 0 V assuming the mic can sink the 186 uA from R1, and then let it rise to 0.7 V . This actions turns Q1 off and on at the sound frequency, which subsequently causes the collector (TRI input of the '555) to swing between 9 V and 0 V, which triggers the timer.

All this said, it looks like the sense of the TRI input is backwards from what it should be. I also think there should be some filtering - a simple RC network - as I don't think you want the '555 triggering at the sound frequency rate. Or maybe you don't care for this application.

Flip R1 and the mic?

If we do this, I think the circuit makes a bit more sense. Q1 would then be OFF, and the TRI input to the '55 high in the absence of sound, since the base of Q1 would be pulled to ground by R1.

Sound would then turn on Q1, which would trigger the timer, creating a pulse that causes the 7474 to toggle. So long as the pulse duration is longer than the period of the sound wave, I don't think any additional filtering is needed.

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  • \$\begingroup\$ This doesn't make sense because pin 2 of the 555 is active-low. When the pin 2 voltage is close to 0V it turns the output on, therefore the output would be on all the time. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Feb 3, 2023 at 20:30
  • \$\begingroup\$ Right. But pin 2 is only low for a relatively short period of time - the duration of the "clap" picked up by the microphone. Your conclusion would be correct if the mic was constantly being bombarded with sound. \$\endgroup\$
    – SteveSh
    Commented Feb 3, 2023 at 23:02
  • \$\begingroup\$ your answer says it's only high for a short period of time. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Feb 3, 2023 at 23:50
  • \$\begingroup\$ Your right, I have that backwards. Let me re-think my answer. \$\endgroup\$
    – SteveSh
    Commented Feb 4, 2023 at 12:41
  • \$\begingroup\$ I updated my answer. I do not think the existing circuit works as intended. \$\endgroup\$
    – SteveSh
    Commented Feb 4, 2023 at 13:37
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Q1 is an NPN transistor, raising the base voltage relative to the emitter causes a small current to flow into the base and allows a much larger current to flow into the collector (10s to 100s of x).

R2 is connected between +9V and the collector. We assume the current into the Tri input is small relative to the other currents (datasheet value < 1 μA), so the current through R2 is approximately equal to the current into the collector of Q1, Ic1. The voltage difference across R2 is therefore approximately equal to Ic1 * R2. Since the side of R2 connected to the battery is by definition fixed at +9V, the voltage at the collector of Q1 is 9V - (Ic1 * R2).

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Q1 gets base voltage Vbb, from R1.Q1 gets collector voltage Vcc, from R2, IC555 pin2 also gets voltage from R2. Current through base emitter junction, will cause a voltage drop at Collector. Microphone can cause base emitter current to vary, that will cause collector emitter current to vary. When Q1 Ibe is high, Q1 Vcc will be low, which means a lower voltage at pin 2.

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