Astable multivibrator using transistor

I have started learning transistors few weeks ago and I hit a roadblock (again). I have been following this tutorial on Sparkfun -https://learn.sparkfun.com/tutorials/transistors

Now I have moved to building an Astable Multivibrator using transistor. But I couldn't figure out the logic on how the transistor switches itself ON and OFF over and over again since as far as I can see both the base of transistors have been biased with 0.6v as soon as the power is turned ON via R2 and R3 resistors.

This will make both the transistor to be in saturation mode and transistor will turn ON. Am pretty confused with this circuit diagram.

1) Can you help me to understand figure out the logic behind this circuit?

2) How one transistor remain in cut off as stated in explanation when both the transistors are biased to 0.6v which is sufficient for a transistor to turn ON?

Any help will be much appreciated, thanks in advance.

• Nice diagram and statement of your problem
– RoyC
Mar 13 '18 at 9:23
• ...but there are also the capcitors. The base bias of one transistor will be shorted (for a while) via the capacitor if the other transistor is turned on.
– Curd
Mar 15 '18 at 10:19

As Tony Stewart has explained one of the transistors will turn on first. When it does the voltage on its collector drops causing the voltage at the other end of the capacitor connected there to drop well below 0.6 V. The voltage across the capacitor cannot be changed instantaneously.

This point is also the base of the other transistor so it will remain turned off until the base end of the capacitor charges to 0.6 V through one of the 47 K resistors.

Lets assume Q2 turns on first. When Q2 is off the voltage on its collector the capacitor + plate is about VCC. The voltage on the - plate is about .6 V . When it turns on the collector voltage drops to 0V and the negative plate of the capacitor drops by the same amount (VCC) so the voltage on the base of Q1 is -VCC+.6 V this will firmly turn off Q1 until the negative plate charges to 0.6 V through R3.

When this happens the transistor turns on pulling down the Voltage on its collector and the base of the other transistor connected through the capacitor. Causing that transistor to turn off.

Rinse and repeat.

• Ok I think now am moving somewhere. So let's say Q2 turns ON first so voltage at the collector junction of Q2 will be of 0v at this instant. What will be the voltage with respect to ground in negative plate of capacitor at this instant? Is it 0.6v or it will drop to 0v pulling the base of transistor to Q1 to low? Similarly at this instant Q1 will be off and collector voltage will be of +Vcc and does the negative plate of capacitor exhibit -VCC ?? Sorry am a beginner with these concepts struggling to understand, kindly forgive my ignorance. Mar 14 '18 at 11:52
• Answer edited to clarify these points.
– RoyC
Mar 15 '18 at 9:10
• I absolutely loved your explanation. I've been trying to understand this for a few days and this is the clearest I've seen it! Sep 12 '18 at 21:17

In theory if both side components were identical both sides would reach Vbe threshold at the same time and amplify the collector current and pull down each other's side. But nothing is every perfectly matched so the race is always won by one side first with the smallest RC time constant on the base and/or largest hFE.

• Hi Tony, thanks for your answer. But am also looking to understand how transistor turns ON and OFF alternatively as a multivibrator and cap voltages influence the switching of these transistors. Can you please help? Mar 14 '18 at 11:55
• tinyurl.com/yc7r3d2n play with this Mar 14 '18 at 14:21