# DC Transients

#### Steve1992

Joined Apr 7, 2006
100
A capacitor, plates labelled X and Y.

1. If +6V were applied to plate X, Y being at 0V.
2. +6V then applied to plate Y,

What would the voltage be at plate X?

Will it be the sum ie +12V?

Steve

#### beenthere

Joined Apr 20, 2004
15,819
Hi,

Depending on the timing of the application of the voltages, some charging would take place. That would also depend on the current soucring capability of the charging source.

But, very shortly after + 6 was applied to both plates, the voltage on either would also be +6, and no net voltage difference would obtain. To get the 12 volts difference, plate Y would have to be charged from -6 volts.

#### Steve1992

Joined Apr 7, 2006
100
The reason I ask this question is, I read this in an electronics text book, referring to a Triggered Flip-flop (attached):

...When square wave pulses (6V amplitude) are applied to the trigger input, voltage rises of both L and M (C1) to +6V and of W and X (C2) to +6V and +12V respectively since the charges on C1 and C2, and therefore the voltages across them, cannot change instantly.

Is this right?

Steve

#### n9352527

Joined Oct 14, 2005
1,198
Simple answer is yes. If the source impedance is low enough, it will force the voltage on the other side of the capacitance up because of the stated reason you quoted, that the voltage across a capacitor can not change instantaneously.

This principle is used in charge pump circuit that can double voltage using two capacitors.

#### beenthere

Joined Apr 20, 2004
15,819
Hi,

Look at the flip-flop in the stae where Q1 is on (low out) and Q2 off (Q bar high). R3 has 0 volts on the collector end, and so ther is no current through Q2's base. D2 is biased off.

R4, however, "sees" +6 volts, and so Q1 has about .7 volts on the base. D1 is foreward biased.

The trigger pulse amy force point M up in voltage, but D2 will block the current. Point W going high will cause X to go negative. Since D1 is foreward biased, the conduction will pull the voltage off Q1's base, and halt its conduction. This will let R3 conduct to the base of Q2 and turn it on.

The flip-flop will now be set up so that the next trigger pulse will reverse the action. C1 will have a negative pulse that will go through D2 and turn off Q2.