# How does this circuit diagram work?

I don't understand this circuit diagram here, but I'm sure that you can help me out. I'll explain what I have figured out so far.

EDIT: It's a metronome circuit. Sorry for being unclear about that. 1. In every circuit, current tries to connect "ground" (-) and "+9 VOLTS" in this situation.
2. From the top down, 9 volts of current is waiting at the emitter side of Q2 (transistor). I'll get back to that.
3. On the left, the voltage is tuned down a bit by the R1 , then after it is a variable resistor (R2).

I am lost here. I know that R2 controls how fast C1 charges, and I know that at a certain point C1 will trigger Q1 and Q2 to release the 9 volts (see 2). This 9 volt burst creates the clicking sound.

What I am confused about is:

1. Why does C1 need to be directly connected to the speaker?
2. Doesn't some of the current from R2 go to Q1?
3. How does the capacitor discharge? I imagine that the capacitor will slowly charge up to a point, then let it out (presumably to Q1). When does it let out? Why does it stop it's "charging" routine to discharge?
4. Doesn't Q1's base run straight to ground? Why is it going to Q2?

CircuitLab allows you to simulate this: simulate this circuit – Schematic created using CircuitLab

Open the circuit, click on "simulate," and figure out how it works yourself! Note that I had to simulate the inductance of the speaker (important in this circuit) by inserting a series inductor, as CircuitLab didn't include this in the speaker model.

Separately, the reason the capacitor is connected to the speaker's "top" terminal, rather than ground, is that it needs to see the voltage drop across the speaker, which fluctuates with both how the speaker conducts, and how Q2 conducts; this is what keeps the oscillation going. If the capacitor was connected directly to ground, there would be a single click, and then it would be silent.

Specifically, when Q2 is open, the voltage across C1 will drop to 0 (it will discharge.) But, as C1 discharges, it turns from an isolator to a conductor (a discharged capacitor works a bit like a wire; a charged capacitor works a bit like an isolator.)

This circuit appears to be an oscillator. Likely it is designed to produce some form of siren.

1. I think C1 is connected in this way to ensure that the base of Q1 can be stimulated by the oscillations that will occur.
2. Correct.
3. As the voltage across the speaker increases, the capacitor will release more energy through Q1. This in turn causes the current through the speaker to decrease repeating the cycle.
4. Remember, transistors can act as 'variable resistors' not just switches.
• Thank you for the quick reply! I have some more questions as I am a bit confused now :) May 17, 2013 at 12:19
• For (1) can you please elaborate more on how this part of the circuit works? I don't really understand what you mean. For (2) can you explain why it needs to be connected in the way that it is and how Q1 functions in this example? Also wouldn't that just let more voltage into Q1, which may cause it to trigger Q2 and a preemptive 9V click? and for (3) I don't understand at all how this works ... Sorry I don't get much of this! If you want to give me links to a site that would explain the behavior of this circuit in simple terms that would be fine. Thanks for the help! May 17, 2013 at 12:26
• For (3) does this have anything to do with when the capacitor is charged, it won't conduct any more? May 17, 2013 at 12:31

Why does C1 need to be directly connected to the speaker?

Don't forget that C1 is is connected to the collector of Q2 too and this is actually critical to the circuit design. The fact the the speaker is connected to that node to is simply to convert the Q2 collector voltage to sound.

How does the capacitor discharge?

When Q2 is "off", the (relatively small) charging current is "down" through the resistors, "down" through C1 and to ground through the speaker voice coil.

When Q2 is "on", the (relatively large) discharge current is "down" through the collector of Q2, "up" through C1 and to ground through the base-emitter junction of Q1.