# A question about BJT Astable Multivibrators

Discussion in 'General Electronics Chat' started by samy555, Sep 17, 2015.

May 24, 2010
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Feb 17, 2009
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3. ### AnalogKid Distinguished Member

Aug 1, 2013
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Two quickie rules of EE:
1. The voltage across a capacitor cannot change instantaneously.
2. The current through an inductor cannot change instantaneously.

When TR2 conducts, its collector goes from +9V to 0V, and change of -9V. This appears on the other side of C2 as a voltage change from whatever it was to whatever it was minus 9V. Since it was sitting at the Q1 base-emitter voltage (+0.6V), the capacitor drives the base to -8.4V. In the real world the base-emitter junction goes into reverse conduction like a zener diode at around -6V to -8V, "catching" the capacitor spike. If this circuit were run on 24V, this could be a problem because if the capacitors are big enough they could dump damaging currents through the reverse biased junctions.

ak

4. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
OK,,, now we have C1 voltage around 0.6
C2 voltage around 8.4
Q1 was ON
Q2 starts to be ON
OK I get it
You say that when VCE2 = 0 and the right side plate of C1 = 0 (I don't know how right side plate of C1 = 0 !!??)
I can not understand

5. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
OK I know that

Yes of course you are right
Yes, Change = 0-9 = -9 (mathimatically true but I cann't imagine it)
I cann't understand

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Charged capacitor act just like a voltage source. So if we charge a capacitor to say 8.4V and now we "short" a right plate via transistor to GND.
What will be the voltage at the negative plate ??
The situation is exactly the same as in this circuit

Where S1 represent the transistor collector-emitter "junction".

At the beginning the capacitor is empty so Vab = 0V. But at the moment when we connect a power supply the current will start to flow and the situation look like this

The charging current is flowing and the voltage across is rising with the time constant R1*C1.

And after t = 5R1*C1 the capacitor is fully charged which means that I1 = 0A but Vab = 8.4V

And now we push the switch. This means that we short the left plate of a capacitor directly to GND. But our capacitor was previously charged to the voltage around 8.4V. So the capacitor will act just like a battery. And we connect the positive terminal of this battery to GND.
So the voltage at the negative terminal (node B) with respect to ground is now equal to -8.4V.
This pic show the details

and as capacitor discharge the voltage across capacitor drop.

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7. ### crutschow Expert

Mar 14, 2008
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In a nutshell, when the voltage on one end of a capacitor changes instantly (or nearly so) by X volts, the voltage on the other end changes by X volts also.
This is independent of the original voltage at each end of the cap.
Follow that rule and you'll understand the circuit.

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8. ### samy555 Thread Starter Active Member

May 24, 2010
116
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Thank you very much, and in all languages of the world
I understood every letter you said but unfortunately I did not understand the above part, which is the same part that I could not understand all the time.

But wait a minute I see it another way, the voltage between point A and GND when the switch S1 is pushed = zero, so if 8.4V was on the capacitor this means that -8.4V must be on the diode (Total = 0)
Thank you jony130

Your answers always be distinct and I benefited a lot of them

9. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Thank you
please give me a numerical example

10. ### crutschow Expert

Mar 14, 2008
13,501
3,375
OK.
One end of a cap is at 0V and the other end is 9V (capacitor charged to 9V).
If the 9V end is suddenly grounded, the other end of the cap will instantly go to -9V.

Last edited: Sep 21, 2015
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11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But this has nothing to do with capacitors.
What is the voltage at point B with respect to GND for this circuit:

But I'm glad that you finally got it.

Let's take this case as a example

Voltage at point A is equal to VA = 6.2V and VB = 0.6V . And now if we press the S1 switch the VA voltage will changes instantly from 6.2V into 0V (0V - 6.2V = -6.2V), the voltage on the other end of a capacitor will also changes instantly by 6.2 volts. In this case from +0.6V into -5.6V (0.6V - 6.2V = -5.6V )

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12. ### samy555 Thread Starter Active Member

May 24, 2010
116
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Thank you Jony130
Thank you crutschow
With my best wishes