Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

prove of Capacitor charging

Status
Not open for further replies.
H

Highlander-SP

Member level 3
Joined
Sep 21, 2005
Messages
64
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,884
laplace+capacitor+charging

How can be proved the formula of the capacitor charge in a RC series circuit:

Vc = Vin * [1 - e^(-t/(R*C))]

where:
Vc = capacitor voltage
Vin = power source
R = resistor
C = capacitor
t = tempo

and how about the discharce: Vc = Vcmax * [ e^(-t/(R*C))]
 

capacitor charging s domain equation solution

Treat the RC circuit as a voltage divider with R as the top element
Write transfer function Vo/Vin =Xc/(Xc+R).
Xc = 1/jwc (Capacitive Reactance)
Using the Laplace operator in place of jw,
Xc = 1/sC
So
Vo/Vin = (1/sc)/(R + (1/sc)) = 1/(RCs+1)
Laplace transform of a step input is 1/s
LaPlace transform of the output response to the step is the product
of the input times the transfer function (multiplication in s domain is
equivalent to convolution in the time domain)
vout(s)=[1/s] [1/RCs+1]
Expanding by partial fractions
vout(s) = 1 - 1/(s+(1/RC))
taking the inverse Laplace transorm:
vout(t) = 1-e^(-t/RC).
~
A similar approach can be taken for the discharge condition.
Regards,
Kral
 

The proof is very easy. You can find the proof from any physics textbook, like 'Fundamentals of physics, by Halliday, 6 or 7th edition', or ' Physics by James walker, 2nd edition'.
I can help you to prove it, but I cannot type it on here because it requires integration symbols. Feel free to contact me should you have any physics problems
 

U need to know that the capacitor doesn't really charge, the total charge in a capacitor is preserved and what happens is that one plate accumulates electrons which reppels the electrons on the other plate.
if u get this then u must know that this will genetrate an electrical field that will fade through time. the recombination rate of negative and positive charges is explained by maxwells equations.
 

The solution is very very very very easy :
from the relation q(t)=Q(1-exp(-t/RC)
you can substitude instead of charge Q=VC then you can get this relation :
VC=εC(1-exp(-t/RC)) , but the capacitance of capacitor doesn't change then you can get the relation V = ε(1-exp(-t/RC)) , where :
V : Voltage of capacitor
ε : electromotive force or the voltage of the source .
t : time
R: Resistor .
C: Capacitor ,
best wishes ,
Yousef Omran .
 

you can draw the circuit with source resistor and capacitor all in series. then

voltage across the resistor is V1=IR, capacitor V2= C*integral(Idt);

If source is Vin then Vin=V1+V2; from this you ger a differential equation and hence get I; V2=Vin-IR get the voltage across the capacitor
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top