voltage of capacitor

Thread Starter


Joined Apr 1, 2007
hi, i am currently trying to understand how a astable circuit works. The point i don't understand is why when Q1 is ON, the negative terminal of the charged capacitor, C1, will be negative voltage (~-5). The explanation i got from wiki is that the voltage across a capacitor cannot change suddenly. However, i am sure there is other explanation.

Any explanation will be appreciated.




Joined Mar 26, 2007
dear fugrammer
the first thing you must consider in your circuit is the Q1 and Q2 unlike
so one of them turn on before other and cause turn off the other
the capacitors in the circuit employ to respons which is ON and start charging

Thread Starter


Joined Apr 1, 2007
okay. I don't understand why does the positive lead of C1 is pulled down to 0.35V? I know that the transistor has a low voltage drop of a bout 0.35V, but why did the postitive lead of C1 drop from +5V to +0.35V?

John Luciani

Joined Apr 3, 2007
The voltage of the capacitor cannot change instantly. If the +terminal has a voltage
+5V greater than the -terminal and you lower the +terminal to 0V the +terminal
is still +5V greater than the -terminal. Since the voltage at +terminal is now 0V
the voltage at the -terminal has to be 5V less which means the -terminal is now
at -5V.

The +teminal of the capacitor drops to 0.35V because that is the saturation voltage
of the transisitor (Vce(sat) on the datasheet). The transistor is being used as a switch
but cannot get any closer to a short circuit (0V) than the saturation voltage.

(* jcl *)



Thread Starter


Joined Apr 1, 2007
when measuring the voltage negative lead of the capcitor, are we finding the potential different between it and the - terminal of the battery? Since it is a negative voltage, does it mean electric charges flow from the - terminal of the battery to the negative lead of the capacitor?