Problem with RC circuit

[Johnny_YU]

HI all
I'm trying to analyse a simple RC circuit.
Condition: the initial energy stored in this simple RC circuit is ZERO.
AT t=0s ,a DC voltage source of 5 Volts is applied to the circuit.
I need to find the Transient Response of Vd.
I did the calculations, the result indicates that when t=0s , Vd approaches to 5Volts.
And then I use 'Multisim 12' to do a Transient Response simulation for Vd, the result indicates that when t=0s, Vd approaches to 2.5Volts.
So, which One is Correct? Is there anything wrong with the Calculations or the Simulation?
Thanks!

 

[esp1]

hi Johnny,
This is what LTSpice shows for that circuit.
As the Vsrc impedance is zero and the two capacitors are the same value, at t= 0, Vout will be Vsrc/2

The bottom cap will then charge up to Vsrc exponentially

E

 

[Johnny_YU]

hi Johnny,
This is what LTSpice shows for that circuit.
As the Vsrc impedance is zero and the two capacitors are the same value, at t= 0, Vout will be Vsrc/2

The bottom cap will then charge up to Vsrc exponentially

E

HI, ESP1
Thank for your reply.
Our simulations lead to the same result .
Now I think maybe there is some mistakes in my calculations, but I still can't find it . I have checked the calculations for about 4 times.....
So, I don't know what to do next....

 

[LvW]

Hi Johnny, Let me try a more or less descriptive/intuitive explanation:

At t=0 both capacitors have no charge.
After closing the switch the initial current through both capacitors must be equal (both capacitors equal and "empty"). More than that, at t=0 the resistor has not yet any influence.
Because of equal currents also both initial voltages across the capacitors must be equal. This is possible only if the voltage across C2 is 0.5Vcc.

 

[crutschow]

The two capacitors in series act as a voltage divider to an AC (or varying) voltage, similar to two resistors in series.
Thus two equal value capacitors will divide the initial input transient voltage by 2.

 

[Johnny_YU]

Hi Johnny, Let me try a more or less descriptive/intuitive explanation:

At t=0 both capacitors have no charge.
After closing the switch the initial current through both capacitors must be equal (both capacitors equal and "empty"). More than that, at t=0 the resistor has not yet any influence.
Because of equal currents also both initial voltages across the capacitors must be equal. This is possible only if the voltage across C2 is 0.5Vcc.

HI，LvW
Thank for your exact explanation.
I can understand your explanation, and i can imagine what happen in that circuit when closing the switch.
I'm writing a report. this RC circuit is in the report.
So, without the simulation and the explanation as you gave , i need calculations..... it's also important...
But i still can't find anything wrong in my calculations...

 

[albbg]

I tried to solve your problem. Here is my solution:




I didn't check accurately your calculation, but from what I've seen you forgot the step function, that is, in the calculation you must consider Vcc*1(t), with 1(t) step function.

 

[D.A.(Tony)Stewart]

The transient Analysis is correct. The capacitive divider transforms an initial voltage = 50% of the step input.

 

[Johnny_YU]

I tried to solve your problem. Here is my solution:




I didn't check accurately your calculation, but from what I've seen you forgot the step function, that is, in the calculation you must consider Vcc*1(t), with 1(t) step function.

Hi albbg
Thank you for your reply and your Calculations.
It's awesome.
I chuck your Z(all of the circuit),and check mine again,then i find the mistake....I did a miscaculation...
finally, i updated my calculations and get the correct answer.
Thank all guys helped me!thanks

 

[Johnny_YU]

A new Problem with RC circuit

HI all
I'm trying to analyse a simple RC circuit.
Condition: the initial energy stored in this simple RC circuit is ZERO.
AT t=0s ,a DC voltage source of 5 Volts is applied to the circuit.
I need to find the Transient Response of I of c2.
I did the calculations, the result indicates that when t=0s , I(c2) approaches to 125uA.
And then I use 'Multisim 12' to do a Transient Response simulation for I(c2), the result indicates that when t=0s, I(c2) approaches to 125mA.
Is there anything wrong with the Calculations or the Simulation?
Thanks!
P.S. VD is V(c2)

 

[FvM]

Strictly speaking, your problem description is physically impossible.

Condition: the initial energy stored in this simple RC circuit is ZERO.
AT t=0s ,a DC voltage source of 5 Volts is applied to the circuit.

A DC voltage can't be applied at a specific time, in analysis terms it's either present since infinite time, or you have a pulsed voltage. You can fool the SPICE analysator by omitting the initial transient solution, but in this case, the result doesn't fully conform with laws of physics.

A switched voltage source will however cause infinite initial current when it tries to charge the capacitors in no time, which is physically impossible as well.

 

[Johnny_YU]

Strictly speaking, your problem description is physically impossible.



A DC voltage can't be applied at a specific time, in analysis terms it's either present since infinite time, or you have a pulsed voltage. You can fool the SPICE analysator by omitting the initial transient solution, but in this case, the result doesn't fully conform with laws of physics.

A switched voltage source will however cause infinite initial current when it tries to charge the capacitors in no time, which is physically impossible as well.

HI FvM
thank you for your reply!
I know maybe it's impossible to find I(0),so in my calculations i use I(0+).
I just want to try discussing about this situation, maybe it's meaningless!:-(
And I'm still trying to find a way to explain it.
Then,I find another way to calculate I(0)
Maybe it's still not correct and it's meaningless.......

 

[FvM]

The "physically impossible" comment was related to your simulation waveform in post #10 and the question about the actual initial current.

You can of course describe the theoretical waveforms generated when switching a voltage source to a capacitor in no time with a symbolic δ operator. But you can't see the waveforms in a circuit simulator, because it's not able to handle infinity.

 

[albbg]

In t=o- all currents and voltages in the circuits will be zero: no energy. In t=0+ when the circuit is energized, ideally there will be a current pulse through C1 a C2 that istantaneously will charge them. This because at t=0- C1 and C2 are discharged so they will act as a short circuit. Of course in real word ESR will limit the current as well as the finite rise time of the generator (it generates more or less a ramp, not a step).
To see this on a simulator you have to substitute the battery with a pulse generator in oder to approximate the step function. You'll see that decreasing the rise time the current pulse in t0 will increase in amplitude.

 

[albbg]

In case you want a numerical calculation, just start from the solution of Vo(t) and apply, as also you said:

Ic2(t)=C2*dVo(t)/dt

remembering that Vo(t)=Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*1(t) then

Ic2(t)=Vcc*C2*C2/(C1+C2)]*1/[R1*(C1+C2)]*exp[-t/(R1*(C1+C2))]}*1(t)+
+Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*δ(t)

for t=0+:

Ic2(0)=Vcc*C2²/[R1*(C1+C2)]²*1(0)+Vcc*{1-C2/(C1+C2)}*δ(0)

this means there will be an (ideally) infinite pulse in 0, than the current will decreases starting from 125 uA

 

[Johnny_YU]

In case you want a numerical calculation, just start from the solution of Vo(t) and apply, as also you said:

Ic2(t)=C2*dVo(t)/dt

remembering that Vo(t)=Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*1(t) then

Ic2(t)=Vcc*C2*C2/(C1+C2)]*1/[R1*(C1+C2)]*exp[-t/(R1*(C1+C2))]}*1(t)+
+Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*δ(t)

for t=0+:

Ic2(0)=Vcc*C2²/[R1*(C1+C2)]²*1(0)+Vcc*{1-C2/(C1+C2)}*δ(0)

this means there will be an (ideally) infinite pulse in 0, than the current will decreases starting from 125 uA

HI,albbg
you are so great!:thumbsup:
here I checked your calculations,and I find something maybe wrong.
Ic2(t)=Vcc*C2*C2/(C1+C2)]*1/[R1*(C1+C2)]*exp[-t/(R1*(C1+C2))]}*1(t)+Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*δ(t)
in the upside expression which colored red,maybe miss a C2.
It should be :Ic2(t)=Vcc*C2*C2/(C1+C2)]*1/[R1*(C1+C2)]*exp[-t/(R1*(C1+C2))]}*1(t)+C2*Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*δ(t)
and here: Ic2(0)=Vcc*C2²/[R1*(C1+C2)]²*1(0)+Vcc*{1-C2/(C1+C2)}*δ(0)
It should be: Ic2(0)=Vcc*C2²/[R1*(C1+C2)²]*1(0)+C2*Vcc*{1-C2/(C1+C2)}*δ(0)

Am I right? waiting for your reply~:???:

 

[Johnny_YU]

The "physically impossible" comment was related to your simulation waveform in post #10 and the question about the actual initial current.

You can of course describe the theoretical waveforms generated when switching a voltage source to a capacitor in no time with a symbolic δ operator. But you can't see the waveforms in a circuit simulator, because it's not able to handle infinity.

HI FvM
You are so kind and helpful!:thumbsup:

 

[albbg]

HI,albbg

Am I right? waiting for your reply~:???:

Yes, you are right. Sorry I forgot to multiply by C2 the second term.

 

[Johnny_YU]

Thank you all the people who makes me progress!