Hello, I've been working on learning Electricity & Electronics and using Forrest Mims' Electronics Learning Lab. I was doing good with only a few struggles that I slowly worked out but now I'm hit by a truck. I've been trying to figure out for weeks now, precisely how the astable multivibrator works. Attached is the schematic for the curcuit used in the lab.

I'm going to tell you what I think is happening, where I'm confused, etc. I'd like for someone to tell me where I'm thinking wrong and what I need to study more. I feel like I'm just forgetting some basic rule or never understood a particular component in the first place. Hopefully with a little guidance, I'll see the error in my thinking and be able to move forward in my studies.

In this curcuit, I understand that inconsistancies in manufacturing will cause one transistor to conduct before the other. Like all the other websites, I'll assume Q1 conducts first. I *think* that as Q1 is conducting, C1 is charging, that charge creating a negative potential on it's other plate. When that negative potential is strong enough, Q2 will turn on. I don't get what turns Q1 off at this point. Does C2 draw all the current once Q2 starts to conduct, not allowing anything to go through Q1?

I also don't understand how either Q1 or Q2's base dosen't see themselves as grounded through R2/R3 and just both be on.

I just don't understand what's going on and what I missed while studying capacitors & transistors.

Thanks for anyones help!

 

Hi,

I don't get what turns Q1 off at this point.

C2 is uncharged at this point, so when Q2 comes on, the base of Q1 will follow high (until C2 charges where the whole thing reverses).

I also don't understand how either Q1 or Q2's base dosen't see themselves as grounded through R2/R3 and just both be on.

It's just that a discharged capacitor will act like a short circuit, overcoming the resistors.

 

Thanks!

I think I get a little more. What do you mean "the base of Q1 will follow high" ?

The capacitor like a short-circuit -- is that always how they work... if there's a path to a capacitor, everything is going to flow there first?

Thanks again!

 

Hi,

What do you mean "the base of Q1 will follow high" ?

When one side of a discharged capacitor changes potential, so will the other. It will immediately begin to charge of course, but for an instant it will be a total short, so what's connected to the other side will follow suit.

The capacitor like a short-circuit -- is that always how they work... if there's a path to a capacitor, everything is going to flow there first?

A cap with a charge of zero has 0V over it, that's how they allways work.
But as mentioned above, it's only for an instant - then the charge will raise the voltage (with a speed determined by the current going into it).

 

BTW, the capacitors are shown with wrong polarities. When Q1 is on and Q2 is off and C2 has its maximum charge, the LH side of C2 will be at +5.3V (due to Vbe) while the RH side will be at approx +2V(voltage drop of LED2) . So the LH side is more positive than the RH side.

Similarly for C1.

 

Ok, I do believe I'm actually more confused now. Thanks for trying, I just need to reinforce some basics.