Rc circuit questions?

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flak88

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Couple of beginner questions about RC circuits:

How do you calculate what value capacity + charging resistor you need at 10v dc supply to take 30 sec from 0v to reach the supply voltage 10v and vice versa for discharge.

To find the second time constant do you find 63% of the first time constant ?

Can someone explain and give an example of how to use the vc = v(1 -e -t/RC) to get the voltage across the capacitor at any time constant ?
 

hi.
There are lots of RC tools on the web.eg: Resistance, Capacitance, Voltage, And Time Calculator

Consider a full charge/discharge at 5 times the 'time constant' R*C...
 
Hello,

To get close to the supply voltage in 30 seconds you can assume 5 time constants does it. In other words, divide 30 seconds by 5 and set RC to that value.

To use Vc=Vs*(1-e^(-t/RC)) simply set RC=R*C and t equal to the time you want to solve for, and also set Vs equal to the source voltage. The result you get for Vc will be the voltage across the cap after time t, assuming the initial cap voltage was zero volts.
 
Five time-constants will get you to within 99.3% of the final voltage (or 9.93V with 10V supply). The same five time-constants will get you to 0.67% of 10V (or 67mV) above ground for discharge.
 
So to calculate the required rc and c value for 30 sec it would be RC 3k x 2uf ?

To calculate the value for 2 time constant do i just get 63% of the first constant ?

To calculate the voltage at time 3sec of rc of 3k and c of 2uf i set Rc to 6 and divide by the time 3sec then what steps?

To get the rc current at any time is it just a case of getting the voltage at that time then dividing it by rc resistor ?
 

hi,
Say you wanted to have a charge/discharge of 99% from it starting value and the 99% value occurs at 30secs, which is 5 * R*C
So 30/5 = 6secs = R * C, say C=1uF and R = 6megohms, that would give the 1st time constant at 6secs [63%] and the 5th time constant at ~30secs.
 
flak88,

Couple of beginner questions about RC circuits:

OK, go ahead.

How do you calculate what value capacity + charging resistor you need at 10v dc supply to take 30 sec from 0v to reach the supply voltage 10v and vice versa for discharge.

You first of all determine what time constant you need. For a RC circuit, the time constant TC=RC. Thereafter, choose the product of R and C that will give you the desired time constant.

To find the second time constant do you find 63% of the first time constant ?

No, time constants are measured in seconds, not percent. Two time constants are 2*TC .

Can someone explain and give an example of how to use the vc = v(1 -e -t/RC) to get the voltage across the capacitor at any time constant ?

Yes I can. The general formula for energizing and de-energizing a capacitor is:
V = Vf + (Vo-Vf)*exp(-t/RC) , where Vf is the final voltage and Vo is the initial voltage. As everyone has been telling you, 5 or 6 time constants will get you very close to the final voltage. Assuming 6 time-constants, then the time-constant will be 5 seconds. Further assuming that R=1 Meg and C = 5 uF, then TC will be RC = 5 seconds. Plugging and chugging into the formula I stated above, we get after 5 secs, 30+(-30*exp(5/5)) = 18.96 volts, 10 secs = 30+(-30*exp(10/5)) = 25.94 volts, 30 secs is 30+(-30*exp(30/5)) = 29.93 volts.

So to calculate the required rc and c value for 30 sec it would be RC 3k x 2uf ?

The time-constant for the RC values you gave is 3000*2uF = .006 secs. The is way too short compared to 5 or 6 secs. Did you do the arithmetic?

To calculate the value for 2 time constant do i just get 63% of the first constant ?

I assume you mean the capacitor voltage value. As shown above, after the first time-constant, the capacitor voltage will be 18.96 volts. The difference between 30 and 18.96 is 11.04 . 63% of 11.04 is 6.96 volts . Adding 18.96 + 6.96 is 25.92 volts which compares favorably with the 10 sec value calculated above.

To calculate the voltage at time 3sec of rc of 3k and c of 2uf i set Rc to 6 and divide by the time 3sec then what steps?

Your specified a time constant of 0.006 secs. 3.0/0.006 = 500 time-constants. The voltage will have long reached its final value within 3 secs with such a short time-constant.

To get the rc current at any time is it just a case of getting the voltage at that time then dividing it by rc resistor ?

I = C*(dE/dt) , but you will find it easier to subtract the voltage across C from the supply voltage to get the voltage across R. Then using the resistance formula I = v/R , the current is obtained.

Ratch
 


No. Supply is only 10V. After the 1st TC, Vc=6.3V. After the 2nd TC, Vc=6.3+2.3 = 8.6V. After the 3rd TC, Vc = 8.6+.8=9.5V. After the 4th TC, Vc=9.5+.3=9.8V. And finally after the 5th TC, Vc=9.9V.
 
flak88,

Sorry I got the supply voltage wrong. I was thinking of 30 volts instead of 10 volts. The voltages wiill be proportionately smaller.

Ratch
 
I think what you are looking for is how to calculate each TC (1,2,3 etc.) That is your formula TC (in %) = (1-1/e^)X100%.
Where e is a constant (2.718). So first TC is 1-1/2.718 = 63%. Second TC is 1-1/2.718^2 or 86.5%.
 
I think what you are looking for is how to calculate each TC (1,2,3 etc.) That is your formula TC (in %) = (1-1/e^)X100%.
Where e is a constant (2.718). So first TC is 1-1/2.718 = 63%. Second TC is 1-1/2.718^2 or 86.5%.

Think that formula needs a little work.
 
r/c

Works OK for me.

1 TC e^1 -- 63.2%
2 TC e^2 -- 86.5%
3 TC e^3 -- 95%
4 TC e^4 -- 98.2%
5 TC e^4 -- 99.3%
 

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Your formula is missing an exponent. You couldn't get those values using it. The correct formula is:

Vn(tc)(%) = (1 - 1/e^n)X100%,

n = 1, 2, 3, 4, 5...
 
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Can you rearrange that formula so that you put time require for 0% vs to 99% vs in and get required capacitor + rc value out ?
 
I can, but it would involve taking natural logs. As many posters have already said, for 0% to 99% is 5 RC time constants. So, TC = 30s/5 = 6s. Now you have to play with different R's and C's to get to 6s. ie

R= 1MEGOHM, C= 6uF - R*C = 1 * 10^6 * 6 * 10^-6 = 6s.

If you absolutely have to have the formula rearranged...

n = log(e)(1/1-99%) ~= 5, as previously stated.
 
BrownOut-
I can, but it would involve taking natural logs.


Sounds like flak88 just wants to enter a time, and be given the RC values?

T = 5RC

so;

RC = T/5
 
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