The astable multivibrator I found in wikipedia and also my college material are the same, which is on the left hand side of the picture. While for the schematic at the right hand side, I found it from the example of LTSpice.

For the left hand side, when Q1 is turned on, C1 is charged through R2 right?
How about the circuit from LTSpice?
When Q1 is turned on, C1 is charged through R1 and R3? Should be correct.

I just wonder why I couldn't find the same schematic of LTSpice from the web? Is it copy right reserved?

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I would say because it's drawn wrongly?, the correct circuit is the one on the left. It may perhaps work?, but it's not a good example, and it's not something I would consider.

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Wrongly? But it is working, just the frequency is different. Any other example on astable multivibrator formed by R, C and transistor?

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Hi AND_ECE
Post a new thread for your topic.

 

The circuit on the left is prone to not starting to oscillate, especially in simulations. The circuit on the right has DC negative feedback, which keeps the transistors out of saturation except right after they switch ON. This allows oscillation to start more readily, especially in simulations.
There are ways of getting the circuit on the left to start in simulation. In practice, it will generally start, due to component mismatches and the startup transients introduced by the power supply turn-on step.
I agree that the circuit on the right is not as useful as the other one. Since the transistors don't saturate, you won't get good logic "0" levels, although there are relatively simple ways to get around this.
I think the circuit on the right was presented as a simulation exercise, and not necessarily as a useful circuit.
There are other types of multivibrators. Google "emitter-coupled multivibrator".

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Ron (aka Rube)

 

Hi Banana,
In the left circuit, R2 discharges C1 slowly. C1 charges through R1 fairly quickly.

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Uncle $crooge

 

All the while I'm thinking that C1 charges through R2 when Q1 is turned on; and discharges through the base of Q2.
From what you've said, C1 charges through R1 across the base of Q2; and discharges through R2 to the collector of Q1? Is that how it works?

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The wire on the capacitor that connects to a transistor's base is driven to a negative voltage when the other transistor conducts. Then the capacitor discharges into R2.

Since the max reverse voltage on a silicon transistor's emitter-base junction is only about 5V then a small supply voltage is needed or protection diodes should be used in series with the emitters.

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Uncle $crooge

 

Yes, I have just simulated, it is -ve. I'm still figuring out why is it negative.
When C1 is fully charged, Q1 is turned on. C1 acts as a battery and its -ve side is connected to the base of Q2, that's why the potential at the base of Q2 is -ve when C1 starts discharging, right?

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Start with Q2 turned on. C1 is charging quickly into its base through R1. When C1 is fully charged then R2 keeps Q2 turned on.
Then R3 charges or discharges C2 enough to turn on Q1 a little and the negative end of C1 drives the base of Q2 negative which turns off Q2. Then both ends of C2 go positive which quickly turns on Q1 completely.
Then C2 charges quickly into the base of Q1 through R4. At the same time, C1 is discharged slowly by R2 and when the voltage at the base of Q2 rises up to about +0.6v then Q2 begins to turn on a little and the charge in C2 turns off Q1.

When a capacitor is charged and a transistor drives its positive wire to ground, then its negative wire is a negative voltage.

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Uncle $crooge