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Subject: Astable Multivibrators - again!

Date: 02/20/03 at 12:02 PM
Posted by: Franklin DeMatto
E-mail: not@available.com
Message Posted:

I asked for help in understanding astable multivibrators (like )

I got the following explanation (thanks, Ron Harrison):
When Q2 is off, it is off because the base is below ground. Take this on faith for a moment. Remember that Q1 is saturated (collector near gnd, low on resistance). C1 charges through R2 until Q2 starts to turn on (base at 0.7v). When this happens, the Q2 collector voltage starts to drop. This voltage change couples through C2, pulling down the base of Q1 and stealing its base current. Q1 starts to turn off, so its collector voltage starts to rise. This couples through C1, turning Q@ on even harder. As you should be able to see, a regenerative process occurs, with the transistors changing state very rapidly. The base of Q1 at (-VCC+0.7v), so Q1 is now cut off. The base of Q1 charges with the R3*C2 time constant until the voltage reaches 0.7 volts, and the circuit changes state as before.

I'm still stuck with the following:

Why can't R2 turn on Q2 directly?
Also, how can C1 turn on Q2 - shouldn't the moment C1 charges to 0.7V, and a little current can flow through Q2, shouldn't C1's charge drop and hence it's voltage drops a bit and it is no longer at 0.7V. So shouldn't Q2 go only for a moment?

"When this happens, the Q2 collector voltage starts to drop. This voltage change couples through C2, pulling down the base of Q1 and stealing its base current."

Could you explain this coupling? I understand that C2 will now discharge through Q2, but I don't see why this will steal all the current flowing through R3. Why doesn't C2 share that current with Q1?

I really appreciate your help. It may be that I don't have a basic concept of capacitors clear - if that's the case, please point this out.

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