# Circuit Problem

Discussion in 'Homework Help' started by hangman88, Sep 26, 2009.

1. ### hangman88 Thread Starter New Member

Sep 26, 2009
2
0
Hello Everyone,

The following is a problem that I have been having trouble with:

A voltage source generating a 20V square wave of period 8ms is connected in series with a resister(100 Ohms), a diode, and a capacitor(10 microFarads). What is the behavior of the voltage across the capacitor, assuming that it is initially uncharged and the diode is ideal.

Help would be greatly appreciated.
Thank you

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Is your source 20V square wave from 0 to 20V or -10V to +10V? That has some bearing on the outcome.

Draw the circuit and think about what is going to happen in principle as time progresses. What effect does the diode have on the circuit behavior?

What is the circuit time constant? How will that influence the rate at which the capacitor charges? How long will it take to fully charge the capacitor to the peak source voltage? Relate this charging period to the period (or rather half-period) of the square wave.

Hope this gives you a start.

3. ### Ataleph Active Member

Apr 20, 2009
31
0
Ideal diode means that if your square wave source has a negative half a period (like in case of -10V to +10V that t_n_k mentioned), it would be cut off, i.e. the voltage in this negative half a period simply will not reach the capacitor.

A 100 Ohms resistor and a 10 microFarads capacitor form charge constant of TAU=R*C=100*10e-6=1msec, for a full charge you need time of aproximatelly 6*TAU=6msec. This means that during half a period (4msec) the capacitor will be almost fully charged. The exact solution of the time domain equation for capacitor voltage is:
V(t)=Vf*(1-exp(-t/TAU) for charging during the positive half a period of clock, where Vf=20V (in case of 0-->20V switching), and
V(t)=Vi*exp(-t/TAU) for discharging during the negative half a period of clock, where the Vi is the voltage on the capacitor after positive half a period charge.
Does this help?

Sep 26, 2009
2
0

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Since it was stipulated that the capacitor is initially uncharged, I can't see how that could matter. The direction of the diode will determine the polarity of the voltage across the capacitor, but that also depends on which end of the cap is chosen as the reference node.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
The discharge time constant is infinite, not R*C.

7. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
512
Here's a hint: remember the voltage across a cap can never change instantly (jump) it can only ramp up or down as driven by the current:

I = C x dV/dt

The term "dV/dt" is called the "voltage ramp rate" of the capacitor and is directly proportional to the charging current, and inversely proportional to capacitance.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Thanks Ron - point taken - my error.