I have to do an exercise and I don't know how to solve it. Can anyone explain in to me?
I have the following circuit:
View attachment 67041
V is 5 volts, diode's voltage drop is 0 volts and V[SUB]i[/SUB] is 10 volts or -20 volts (as shown in the graph).
What is the capacitor's voltage (V[SUB]c[/SUB]) after the first period of time of the input voltage?
Hello,
We will assume this is more of a theoretical question rather than a practical one because no practical circuit would include a capacitor being charged by a zero impedance voltage source as this circuit has. So we have to assume that there is some small but insignificant capacitor ESR, insignificant such that it doesnt prevent the cap from being charged very quickly but prevents an infinite current from flowing at t=0.
First we have the positive portion of the input wave. It goes to +10v and this is relative to the minus input terminal which also connects to the cap left side. The diode conducts but the small battery subtracts 5v from that 10v so we end up with +5v across the cap with the left side negative and right side positive as the diagram shows. The first half cycle can last as long as it does and this voltage will stay at +5v. So the cap has 5v across it with left side negative and right side positive.
Now the input switches to -20v. But looking at the plus input as common now, that puts +20v at the minus input terminal. So with plus input common we have a +20v signal in series with the capacitor that is charged to +5v as mentioned before and so this puts +25v across the resistor R (again with the plus input as common). The general equation for a voltage exciting a capacitor and resistor and looking across the resistor is:
Vout=Vin*e^(-t/RC)
and Vin is simply the +25v at t=0 so we get:
Vout=25*e^(-t/RC)
But we declared the plus input which is also the plus output as common, so this voltage is actually negative so we get finally:
Vout=-25*e^(-t/RC)
And that is the output voltage for any R and C and any time t where t is the time of the second half cycle (and the period is one complete cycle) so t=0 here when the second half cycle starts so this solution is not valid during the first half cycle but only during the second half cycle. The first half cycle would look different.
Stated relative to t=0 where t is the start of the first cycle, we get:
Vout=-25*e^(-(t-t1)/RC)
where t is time and t1 is the time length of the first half of the cycle, but the solution is only valid for t>=t1 which again is the second half cycle.
Note also that both of these are also not valid for the third half cycle or any that follow.