# Astable multivibrator using BJT transistors

#### omar-rodriguez

Joined Jun 24, 2015
67
Hi, I'm trying to analyze this circuit

As you can see the base of Q2 and Q4 is at ground and the emiters looks to be at higher voltage than base, so it looks like Q2 and Q4 are always in active region, supposing that the current gain is much greater than 1, then base current of Q2 and Q4 is ~0A and (Ic2=Ie2, Ic4=Ie4), so basically Q2 and Q4 are like current sources.

- First I assume that Q1 is on active region and Q3 is off so:

I with this inequality, I find the capacitor voltage for which Q3 turns on

When the capacitor voltage is larger than 2.7V then Q3 turns on and Q1 turns off

I do not know how to find the capacitor load voltage and the tau=RC of the capacitor, what should I do? find the thevenin circuit over the capacitor?

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#### Jony130

Joined Feb 17, 2009
5,514

#### MrAl

Joined Jun 17, 2014
11,565
Hi, I'm trying to analyze this circuit

As you can see the base of Q2 and Q4 is at ground and the emiters looks to be at higher voltage than base, so it looks like Q2 and Q4 are always in active region, supposing that the current gain is much greater than 1, then base current of Q2 and Q4 is ~0A and (Ic2=Ie2, Ic4=Ie4), so basically Q2 and Q4 are like current sources.

- First I assume that Q1 is on active region and Q3 is off so:

I with this inequality, I find the capacitor voltage for which Q3 turns on

When the capacitor voltage is larger than 2.7V then Q3 turns on and Q1 turns off

I do not know how to find the capacitor load voltage and the tau=RC of the capacitor, what should I do? find the thevenin circuit over the capacitor?

Hello there,

I took a quick look and it looks like the only way for the cap to charge is through one of the pseudo current sources. That is because one transistor is always off, so that current source is put in series with the cap, and the current in a series circuit is the same through every element and one of those elements is the cap.
If this works then the cap charges through a constant current source and has an equation that is not exponential but is linear with a straight line ramping voltage. That would also give merit to the reason for the constant current sources in the first place.
Try that and see what you get. Hint: dv=i*dt/C.
The only catch is that the other transistor (the one that is 'on') must be able to support 2x the current source current which i think it can, but you could double check that too.
This observation would make the analysis exceptionally simple to perform.

Last edited:

#### omar-rodriguez

Joined Jun 24, 2015
67
mmmm got it thanks

V= (I/C)*t so capacitor is charged linearly

V= 11764.7*t

So it reaches 2.07V at t=175us

#### MrAl

Joined Jun 17, 2014
11,565
mmmm got it thanks

V= (I/C)*t so capacitor is charged linearly

V= 11764.7*t

So it reaches 2.07V at t=175us
Hi,

Ok sure. I did not check anything else though such as that 2.07v you calculated.