# How does this circuit *really* work: 2 transistor metronome

by rp55
Tags: circuit, metronome, transistor, work
 P: 389 At first glance you can treat capacitor as a some kind of a voltage source. The capacitor act very similarly to a voltage source. The key equation is this I = C*dV/dt The capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing). The faster the voltage change (frequency of a AC signal is high) the large the current flow through capacitor (short at high frequencies). For example for this circuit We have a switch in "B" position. So we apply a 1.5 volts to the circuit. At the beginning of the charging phase the capacitor is empty, and so Vc1 = 0V The voltage on capacitor cannot change suddenly form 0V to 1.5V. We need time to voltage on capacitor to grow (t≈5*R*C). So immediately after we connect the supply voltage a current will begins to flow. And all voltage from power supply appears across R1 resistor, VR1 = Vsupply. Because of a II Kirchhoff's law, 1.5V = Vc1 + VR1 and since Vc1 = 0V( empty capacitor) VR1 must be equal to 1.5V. So we can say that empty capacitor act just like a short circuit because there is no voltage across it. As the capacitor begins to charge, the voltage across the capacitor begins to increase. At the same time the current in the circuit begins to decrease. Why? Because the voltage across the capacitor begins to increase. So voltage across resistor must decrease (Vsupply = Vc1 + VR1). And this is why charging current begins to decrease. The capacitor eventually charges to the full voltage source potential. When this happens, all circuit current stops, because now Vc1 = 1.5V and VR1 = 0V If so no current can flow (Ohms law I = V/R = 0V/R = 0A). And Vc1 in red show voltage across capacitor during the charging phase. And VR1 in red show resistor voltage during the charging phase. Can you describe what will happen in this circuit after we flip the switch into A position? And can you tell me what is the voltage across the LED immediately after we flip the switch? Attached Thumbnails
P: 82
 Quote by Jony130 We need time to voltage on capacitor to grow (t≈5*R*C).
Right... 1/2 second in this case. So instead of time constant formula R*C (63% charge) you use 5 for full charge? Is that what you did there? I didn't know that, very useful.

 Quote by Jony130 Because of a II Kirchhoff's law, 1.5V = Vc1 + VR1 and since Vc1 = 0V( empty capacitor) VR1 must be equal to 1.5V.
Oh I didn't realize this. This is very useful... since you can't give the cap a resistive value. Then you can use ohm's law to determine current across each component (since voltage varies but current same in series components).

One question I have about your circuit though... doesn't the led start to light once the cap charges to .7v (forward bias or whatever it is 1-2v for the specific led) and then the cap stops charging at that voltage? Perhaps you were first talking about the cap charging as if the led wasn't there?

After thinking about it, you're probably implying that the led has a forward bias of > 1.5v ... so the cap fills before it ever lights. Then it lights when the switch is thrown via the extra cap voltage stored.

 Quote by Jony130 And Vc1 in red show voltage across capacitor during the charging phase. And VR1 in red show resistor voltage during the charging phase.
I'm not sure what you're pointing to here, sorry.

 Quote by Jony130 Can you describe what will happen in this circuit after we flip the switch into A position?
Basically it would cause the cap to discharge thru the led (due ?partially? to Vsupply) along with Vsupply powering the led thru R1. I guess I get a bit confused with parallel circuits still so that may throw a wrench into things also that I'm not seeing.

Then I think the cap would eventually fill again and block and the light would go out (assuming its forward bias is > 1.5v).

 Quote by Jony130 And can you tell me what is the voltage across the LED immediately after we flip the switch?
Well here's my guess: 2.5 volts.

Obviously Vsupply = VR1 + Vled.

First you would get the Vled.
Vled = Vsupply - VR1.

VR1 = Vsupply - Vled. If we assume the led voltage drop is 1v then VR1 drop would have to be .5V.

So the voltage across the led would be 1v (Vsupply - VR1) PLUS the capacitor voltage of 1.5 volts. So 2.5 volts? That's probably wrong lol but its my best attempt right now.
P: 389
Quote by rp55
 Quote by Jony130 View Post We need time to voltage on capacitor to grow (t≈5*R*C).
Right... 1/2 second in this case. So instead of time constant formula R*C (63% charge) you use 5 for full charge? Is that what you did there? I didn't know that, very useful.
Yes, capacitor is "fully" charged after a period of five time constant.

 One question I have about your circuit though... doesn't the led start to light once the cap charges to .7v (forward bias or whatever it is 1-2v for the specific led) and then the cap stops charging at that voltage? Perhaps you were first talking about the cap charging as if the led wasn't there?
I use LED diode and the red LED has the lowest forward bias voltage around 1.5..1.6V.
So for sure LED won't start to light.

 After thinking about it, you're probably implying that the led has a forward bias of > 1.5v ... so the cap fills before it ever lights. Then it lights when the switch is thrown via the extra cap voltage stored.
Exactly, you got this part right.

 Basically it would cause the cap to discharge thru the led (due ?partially? to Vsupply) along with Vsupply powering the led thru R1. I guess I get a bit confused with parallel circuits still so that may throw a wrench into things also that I'm not seeing.
This diagram explain everything.

So immediately after we flip the switch the voltage across the LED is equal to 3V.
Exactly in the same time discharge current will start to flow. So the the voltage across to cap will start to drop. So the LED will blink.
Attached Thumbnails

P: 82
 Quote by Jony130 Yes, capacitor is "fully" charged after a period of five time constant. So immediately after we flip the switch the voltage across the LED is equal to 3V. Exactly in the same time discharge current will start to flow. So the the voltage across to cap will start to drop. So the LED will blink.
Doesn't there have to be a voltage drop (even if in opposite direction) on R1? That's what I was calculating in. But I definitely missed C1 discharging thru R1 the way you show. Good stuff. I've learned a lot.
P: 389
 Quote by rp55 Doesn't there have to be a voltage drop (even if in opposite direction) on R1? That's what I was calculating in.
Why do you think that in this circuit any voltage across R1 resistor will have any effect on diode voltage?
P: 82
 Quote by Jony130 Why do you think that in this circuit any voltage across R1 resistor will have any effect on diode voltage?
So perhaps you're saying that since R1 and C1 are in parallel now, and so both parallel branches should see the same voltage which would be per led's limiting drop of 1.5v (or whatever we said)? So they both get different current but the same voltage since they're parallel? Why does the led get 3v then?
P: 389
 Quote by rp55 So perhaps you're saying that since R1 and C1 are in parallel now, and so both parallel branches should see the same voltage
Yes, in parallel circuits the voltage is the same across components.
And this is why VR1 = VC1 = 1.5V.

 Quote by rp55 Why does the led get 3v then?
Why? Well because II Kirchhoff's law says so.
To measure the voltage we need two point in the space. One of this point is treat as a reference point (GND). We have a very similar situation when we try to measure a height of an object. We need a reference point. The most common reference point is "above mean sea level". But when you measure the height of your kitchen table in your house you pick the floor as a your reference point.

So for our circuit we have this situation.

As you can see I pick GND at the bottom as we usually do.

And now we can tell the voltage at point A and B with reference to gnd.
As you can see the voltage at point A is 1.5V higher then ground. And voltage at point B is 3V higher then ground. And the voltage across R1 is equal to VB - VA = 1.5V also notice that capacitor is also connect between this two point ( VC1 = VB - VA = 1.5V).
I hope that now you see why in parallel circuits the voltage is the same.
Also notice that diode is connect between point B and GND. And this is why Vdiode = VB - Vgnd = 3V - 0V = 3V.
Attached Thumbnails

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