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:
- C1 and C2 are fully discharged.
- 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!
- 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.
- 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.
- 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.
- 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.
- As a result of the previous event:
- Q1 shuts off completely due to the negative voltage on its base from C2.
- C2 begins discharging via now conducting Q2.
- C2 also starts reverse charging its B plate via R3.
- Once plate B of C2 reaches 0.7V, it turns on Q1 and:
- Discharges C1 via Q1.
- Causes Q2 to turn off because of the negative charge on C1 due to its plate A now connected to ground.
- 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.