Newb trying to understand astable multivibrator

I've been trying to understand one thing about the astable multivibrator below. According to https://www.build-electronic-circuits.com/astable-multivibrator/, at the very beginning of the operation, the capacitors will start developing a positive charge on both plates. The left plate of C1 will reach +8V and so will the right plate of C2. That will happen before the other plates will reach 0.7V. Here's my question:

How can +8V develop there if the transistors are turned off? I must be missing something basic. Trying to get into electronics and this is one of the questions that are driving me crazy.

======== UPDATE AFTER STARING AT THIS CIRCUIT FOR DAYS ========

I've come up with my own explanation based on reading and re-reading multiple descriptions of this crazy but really cool circuit. I also have soldered together a physical model of this.

So here we go, please correct anything is that is utterly wrong.

A note on the types of capacitors needed for this circuit

Some circuits show polarized capacitors which causes additional difficulty in understanding the operation of the circuit but regular capacitors can be used. The reason polarized transistors can be used is because even though both sides of it will be charged positively, their negative side will reach a max of 0.7V which will not destroy it, but why use them then? I don’t know? Just use non-polarized transistors.

Operation according to my understanding

First, it's useful to imagine this circuit without the capacitors. In that case, R2 and R3 are connected to the base of Q2 and Q1 respectively. If there are no capacitors, the base-emitter (BE) voltage of both Q1 and Q2 will have their natural drop of 0.7V and the transistors will turn on immediately after the circuit is powered on. Therefore, the L1 and L2 LEDs will light up without any delay.

But we do have capacitors C1 and C2, which is both crucial to the operation of the oscillator and presents a lot of difficulty for a beginner like me in analyzing the circuit. Here’s what in the circuit and here's what I think happens when the power supply is turned on:

1. C1 and C2 are fully discharged.
2. Due to C1 and C2 being discharged, they are fully conductive for a tiny bit of time (R1C1) and (R4C2). Therefore R1->C1->Q2 and R4->C2-Q1 provide a path for the capacitors to start charging their A plates and while that is happening both LEDs are ON!
3. With time, both capacitors accumulate enough charge to cause more and more resistance turning off the corresponding transistors more and more. The A plates are now close to 9v.
4. At the same time plates B begin to charge the other way via R2 and R3. This is interesting since both sides of the capacitor will have a positive charge with plates A having a much larger charge. From the perspective of the transistors their bases will have 0.7v on their bases but the total charge of the capacitors will be 9v - 0.7v or 8.3v. The full charge of the capacitors is not “visible” to the cross-connected transistors half the time because the corresponding cross connected A plate is hidden from each by the turned off cross-connected transistor. For example, Q1 can’t see the full charge on C2 when Q2 is off. And Q2 can’t see the full charge of C1 while Q1 is off. And when they do see the A plates, the A plates are referenced to 0v and the charge will appear negative in relation to the base. More on negative voltage below.
5. Eventually one of the capacitors reaches 0.7V on plate B and turns on the corresponding cross-connected transistor (C1->Q2 or C2->Q1). As mentioned above, the total charge on capacitor would be around 8.3v.
6. Suppose C1 reaches 0.7V first and turns on Q2. That opens the CE path of Q2 and C2’s plate A becomes referenced to 0V but the charge on the capacitor is still 8.3v with plate B 8.3V lower. So when plate A becomes 0v, plate B will continue to be 8.3v lower in relation to the base of Q1.
7. As a result of the previous event:
1. Q1 shuts off completely due to the negative voltage on its base from C2.
2. C2 begins discharging via now conducting Q2.
3. C2 also starts reverse charging its B plate via R3.
8. Once plate B of C2 reaches 0.7V, it turns on Q1 and:
1. Discharges C1 via Q1.
2. Causes Q2 to turn off because of the negative charge on C1 due to its plate A now connected to ground.
3. C1 starts reverse charging via R2 so its B plate becomes more and more positive until it reaches 0.7V and turns on Q2 - AGAIN!

… and so it continues back and forth

Of note is that if we removed a capacitor C2 right after Q2 turned on, Q1 would never turn on.

• Who says the transistors are turned off? Feb 1, 2022 at 15:52
– G36
Feb 1, 2022 at 16:30
• Ah, those are nice links. I'll definitely review. Thank you.
– Serg
Feb 1, 2022 at 18:48
• @Hearth, you're right, just one of them is turned off at a time. Let me rephrase my question: before any of them turns on and just before one of the capacitors has reached 0.7V on the "inner" plate (plate facing a transistor), what are the charges on the outer plates (connected to the collectors)? I don't think it's zero because in order for negative voltage to develop it needs to be above at least 0.7V, but I don't understand how it gets to +VCC because I can't see the charging path for that to happen -- especially before any of them turn on.
– Serg
Feb 2, 2022 at 0:08

Your explanation seems correct overall in broad strokes. However, the part about removing C2 preventing Q1 from turning on seems wrong. The resistors R2 and R3 are perfectly capable of turning Q1 and Q2 by themselves...the capacitors are there to turn them off.

1. When you first apply power, the capacitors are discharged (0V). When you apply 9V, they don't start charging immediately, since Q1 and Q2 are both off and aren't passing much current. It's now a race to see which base reaches threshold first.

2. Q2 gets to ~0.6V (threshold) and turns on, bringing the collector voltage down. C2 still has 0V across it, so Q1's base voltage is pulled down as well. Q1 won't turn on until C2 achieves ~-0.4V charge, so that with Q2 in saturation (Vce=~0.2V) Q1's base can achieve threshold. (Q2's base will reach 0.7V [saturation] in the process.)

3. C1 may have charged up to ~8V at this time. When Q1 turns on, it applies ~-8V to the base of Q2. C1 has to charge now to ~0.4 from -8, so this is where the real timing starts.

4. Switch C1/Q1 and C2/Q2 and repeat indefinitely.

The thing to remember is that each Q, when turned on, forces the other Q off for a time through the capacitor. When the other one finally turns on, it turns off the first one the same way. Take out the Cs and both LEDs will stay on permanently.

How can +8V develop there if the transistors are turned off?

If the transistor is off, its base-emitter region still behaves like a forward biased diode and so the negative plates of the capacitors are close to 0 volts (OK maybe 0.7 volts above 0 volts).

at the very beginning of the operation, the capacitors will start developing a positive charge on both plates

Each capacitor will develop positive charge on one plate and negative charge on the other plate. Maybe that's what you meant?

• According build-electronic-circuits.com/astable-multivibrator, the charges on the capacitor are +0.7V and +8V. This simulation does confirm that: falstad.com/circuit/e-multivib-bi.html. I'm only passing along what I've read :)
– Serg
Feb 1, 2022 at 18:46
• @Serg so, is your question (as written) fully answered now? Feb 1, 2022 at 20:05
• Almost, I've watched youtube.com/watch?v=1x570l7dbXE which was AWESOME, but I still have 1 question which boils down to this: What is the charging path for the right left plate of C1 and right plate of C2? I.e.: how can it get it +9V in this case?
– Serg
Feb 1, 2022 at 22:19
• @Serg C1's positive plate charges through R1 and L1 until Q1 turns on, then Q1 is a discharge path. Feb 2, 2022 at 8:08
• I think at this point I need to go back to the drawing board and review everything. I think we can close this question. Thank you for the help
– Serg
Feb 2, 2022 at 18:44