Require help solving Circuit Exercise

[nautilus]

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?

 

[BioniC187]

There are no time values on that graph. Can't determine T. No R value also

 

[panic mode]

as mentioned, no values are shown so one can only reason what is expected.
assume at t=0, Vc=0...

when Vi is 10V, Vd=0, V=5V so Vc is 10-(5+0)=5V
current would be large doing rapid charging of C.

when Vi=-20V diode is reverse biased so that branch is dead (consider it open circuit)
and capacitor will discharge through R. how far it discharges depends on values of R and C as well as time. if the time is long enough Vc will become -20V. depending on R,C and t any value between -20 and +5V is possible.

 

[MrAl]

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.

 

[nautilus]

Sorry. I forgot to write which is the period of time and the values for R and C.

**broken link removed**

So, when Vi is 10V, Vc is 5 V (because the diode is forward biased) and when Vi is -20V the capacitor will discharge through resistor R and the Vc is 0V ?

 

[panic mode]

Sorry. I forgot to write which is the period of time and the values for R and C.

well, you still did not write any of them. also the attachment is not working.

 

[ericgibbs]

hi pm,

gif copy of the OPs image.

E

 

[nautilus]

hi pm,

gif copy of the OPs image.

E

Thank you

 

[MrAl]

Sorry. I forgot to write which is the period of time and the values for R and C.

**broken link removed**

So, when Vi is 10V, Vc is 5 V (because the diode is forward biased) and when Vi is -20V the capacitor will discharge through resistor R and the Vc is 0V ?

Hello,


Look at my previous post, take your numbers and plug them into the formulas and you have your answer. You cant get any simpler than that

 

[nautilus]

Yes, I read your post. But I don't have to find out the output voltage. I only need the capacitor's voltage (Vc) after the first period of time.

 

[MrAl]

Yes, I read your post. But I don't have to find out the output voltage. I only need the capacitor's voltage (Vc) after the first period of time.

Hello again,


Well if you have to find the capacitor voltage then you can use the rule that the sum of voltage drops around a closed circuit equals zero.

This means the cap voltage is:
Vc=-Vout-20

where Vout is the previous equation i had posted for the output voltage.

You also need to know the time of the second half cycle if you want to calculate the voltage at the end of that cycle numerically.

 

[panic mode]

exactly, there is still no mention of time and he just does not get it.

@nautilus:
voltage can be found easily for part where Vi=10V, but after that Vi=-20V and Vc changes exponentially.
if T=infinity, we can use steady state but i doubt this is the case so you need to use the equations as posted by MrAl...
t

 

[panic mode]

nautilus,

how does this look like? can you tell voltage after t=T as pointed by blue arrow?

 

[nautilus]

I don't have any information of time. The problem I have to solve has only the information I have written.

 

[MrAl]

Hello again,


Well not all problems require a numerical solution. Sometimes it is enough to find the function that would provide all the results had we known all of the parameters.

Thanks to panic mode for graphing the function, where you can see that the place where you want the solution is falling exponentially and is given by:
Vc=25*e^(-t/RC)-20, {0 <= t <= t1}, where t1 is the time length of the second half cycle.

Since you were not given a numerical value for t1 you will have to be satisfied with that result which is an equation not a single number like 1 volt. If you find out the time length of the second half cycle t1 then we would be able to calculate a number like that. Without that however, we can not find the value numerically.
What you can do though is plug in the R and C values you gave and that only leaves t1 to be given at a later date by whomever constructed this problem, if in fact they wanted that kind of solution to be found and not just the equation.

 

[nautilus]

Got it Thank you MrAl.

Thank you all for helping me out with this exercise