Two transistor multivibrator circuit, capacitor question.

[Mint.]

I do not understand why the capacitor gets charged negatively and then all of a sudden drops down to 0v without any time constant and why is there current flowing from one terminal to the other if the current is DC?

http://www.falstad.com/circuit/e-multivib-a.html

[IanB]

Current can flow through a capacitor as long as the voltage is changing with time.

An important factor in understanding how that circuit works is knowing that a capacitor works like a straight piece of wire when you consider short time scales. This means that if you change the voltage on one side of a capacitor, the voltage on the other side will immediately change by exactly the same amount (just the same as a wire--if you change the voltage on one end of a wire the voltage on the other end will follow it).

So for example, if Q1 turns on it will pull the + side of C1 low. This will immediately pull the other side of C1 low at the base of Q2 and turn off Q2. If Q2 turns off it will cause the + side of C2 to go high, which will raise the base of Q1 high and keep it turned on. Therefore you have a state with Q1 turned on and Q2 turned off.

Starting from this position, consider each of the capacitors. Will each one charge or discharge, and what will happen when one or other of them does so?

[Mint.]

I understand perfectly now, thanks Ian!

[IanB]

I have always had trouble getting my head around the capacitor polarity in that circuit, especially understanding why the polarity is the way it is.

So out of curiosity I simulated the circuit and plotted the voltages on each side of the capacitor. It shows clearly that the marked polarity is correct, although there is a momentary reversal of voltage each cycle. Apparently this is not too much of a problem for electrolytic capacitors because it is outweighed by the positive voltage differential during the rest of the cycle.

[Bored@Work]

You can directly measure the voltage difference between two nets in LTspice by clicking on the first net, keeping the mouse button pressed and dragging the probe to the second net.

[IanB]

That's convenient to know, although plotting the two voltages separately clearly shows how Vb stays negative for much of the time and tops out at about +0.6 V when it reaches the turn on threshold of Q2.

[Frenchie]

As I understood it, aluminium electrolytics are not that upset by reverse polarity as long as the voltage is less than a volt or two but they deviate a long way from ideal* when doing so. Not 100% on that so I'd been keen for someone to confirm or deny please.

(* Not that they're ideal even when used with the correct polarity)

[Mint.]

This gives me the impression that electrolytics are a form of cheating in this circuit because we are using them incorrectly. Isn't that also disrespectful and damaging to the components?

[vk6zgo]

This gives me the impression that electrolytics are a form of cheating in this circuit because we are using them incorrectly. Isn't that also disrespectful and damaging to the components?

No!

[MikeK]

I'm confused by the explaination.  Doesn't it work because one side of each capacitor charges up as the other side is discharging because the resistors create two different RC constants?

[vk6zgo]

I'm confused by the explaination.  Doesn't it work because one side of each capacitor charges up as the other side is discharging because the resistors create two different RC constants?

The Falstad simulation is very messy,& doesn't give a good idea of how the thing works.
Ditch any simulations,draw the thing on paper & try to think your way through the operation,or find a good book.
Simulations are good after you have grasped the basic operation,but can be confusing if you get the wrong slant initially.

[Blofeld]

This gives me the impression that electrolytics are a form of cheating in this circuit because we are using them incorrectly. Isn't that also disrespectful and damaging to the components?

There is another potential problem with the circuit: As can be seen in the plot, node b (the base of Q2) goes down to more than -9V. The emitter of Q2 stays at 0V, so there is a reverse bias of more than 9V for the base-emitter junction. But according to the 2N2222 datasheet this reverse bias should not exceed 6V, so strictly speaking you should not do this:

http://media.digikey.com/pdf/Data%20Sheets/ON%20Semiconductor%20PDFs/P2N2222A%20Rev3.pdf
(page 1, maximum ratings, emitter base voltage 6V)

On the other hand, this circuit is so enormously popular (for me it was the first oscillator I built) and people don't seem to complain about damaged transistors. So I guess as long as you are not too concerned about reliability you can get away with it.