bench said:

Hi everyone,

I am having trouble simulating an astable multivibrator and

I was wondering if anyone can help me. I am using

a package called Circuit Maker 6 and the circuit has

already been inputted and all you need to do is download

the file

http://www.geocities.com/benchdocs/astable.CKT.txt
; rename it without the .txt extension (as it is a .ckt file) and

see if you can get it to oscillate. I am not sure what is wrong

with the circuit, maybee the resistors or capacitors have been

assigned wrong values. I initially used generic npn transistors,

but then I tried the 2n2222, but nothing seems to help.

tia.

CM6 has an example circuit which is almost exactly the circuit you

posted... Look for 'Astable.ckt' amongst the examples.

With regards to your circuit, your timing values are pretty big. If

the oscillator worked with your values, it would oscillate at about

3.5Hz, which would be much too slow to see with the simulation values

you have set. Also, you may need initial conditions for some of the

nodes to get the oscillation going in the first place. I got your

circuit going by changing the cap values to C1=500pF and C2=1nF,

R2=R3=47k, and putting initial conditions on the collectors of the

transistors as 0V (on the left) and 9V (on the right). Also, go to the

transient setup, and change the stop time to 500us, and the step time

and max step to 1us.

Here follows far more information than you possibly want to hear about

the oscillator.

The oscillation frequency is near -ln(.5) * (R1*C1 + R2*C2). Thats

because the fall in the collector on one side (when the transitor

turns on) drags the base of the other side down 9V through the

capacitor, so it has to increase back to 0.7 to turn on again. It does

this by charging through the cap/resistor pair. The equation governing

the time of recharge of a cap C through a resistor R can be shown to

be

Vx -Vs = (Vt - Vs) exp(-t/RC)

Where Vs is the starting voltage, Vx is the 'target' voltage, Vt is

the voltage at the other side of the resistor, t is the time required,

and R and C are of course the values for those components. Since the

voltage is dragged down from 0.7 to -8.3, and has to climb back up to

0.7 again to turn on the transistor and complete that part of the

cycle, you have

Vs = -8.3V

Vx - Vs = 0.7 - (-8.3) = 9

Vt - Vs = 9 - (-8.3) = 17.3V

C1 = 500pF, R2 = 47k

C2 = 1nF, R3 = 47k

So (solving the above for t, and noting that there are two RC

oscillators going, one after the other...)

t = -ln(9/17.3) * (500pF * 47k + 1nF * 47k)

= 0.653 * (70.5u)

= 46u

Therefore, the frequency of the oscillator with the values given above

is

f = 1/46u = 21kHz

This matches pretty well with the simulated value, which is around

20kHz. It doesn't match exactly, because the Vs above isn't

really -8.3; its something a bit less negative, because of various

effects, so it doesn't go down 9V, more like 7.5V... We could take

that into account, but these values are close enough, given that the

caps and resistors you'll use are probably 5% to 10% tolerance.

By the equations above, its easy to see that you can control the 'duty

cycle' of the oscillator by simply changing cap or resistor values. By

keeping the sum constant, the frequency will be kept constant.

Regards,

Bob Monsen