Reg:RC Circuit time constant

[sakthivel.eee]

Hi All,
Please any one can tell what is charging and discharging time constant for the attached circuit. Please give some explanation

Thanks in advance.

Thanks & Regards,
Sakthivel.S

 

[fcfusion]

The circuit's transfer function is:

Vout/Vin = [R2/(R1+R2)]/[jw*C*R1*R2/(R1+R2)+1]

where C*R1*R2/(R1+R2) defines the circuit's time constant

thus time constant = 1 ms

When speaking of charging and discharging time it is usually assume you speak of voltage variation between 10% and 90% of voltage supply (11V).

For example, discharging from VCC (11 V) to 10% of VCC (1.1 V):

Discharge time (10%) = time_constant * (- ln (0.1)) = 0.001*2.3 = 2.3 ms


Read this for detailed info:

https://en.wikipedia.org/wiki/Time_constant

You can calculate charge time constants in time domain (integral or differential equations, too hard !!! ) or using the transfer function (Laplace, easier, it's what I used).

I hope the calculations are right

 

[sakthivel.eee]

HI
How can you considering two resistor in parallel (C*R1*R2/R1+R2). because 1K resistor terminal is connected with positive supply & 10K terminal is connected with negative terminal.

 

[LvW]

HI
How can you considering two resistor in parallel (C*R1*R2/R1+R2). because 1K resistor terminal is connected with positive supply & 10K terminal is connected with negative terminal.

The time constant is a parameter which describes the charging/discharging process of a capacitor. Thus you have to look into the circuit from the viewpoint of the capacitor. And then, it is easy to verify that both resistors work in parallel.

 

[malli_1729]

Hi!

i am just telling what i know in this regard theoritically....

in the circuit at the start of the pulse, when Pulse is going from 0 to 11V as the capacitor will not allow any sudden changes in the voltage across it, at that instant capacitor will be short so, no currnet flows through 10K resistor , as the time progresses voltage builds up across capacitor and current through the capacitor will reduce at the same time current through 10K will increase. when Capacitor fully cahrged current through it is "0" so, full currnet will in 10K which 11V/(1+10K)
So, for charging we have to take the path of 1k and 11nf ...

when dischrging it will discharge through 10K ....so, for discharging we have to take 10K and 11nf. based on these we can calculate the time constants of the circuit....if i am wrong please correct me.........

thanking you....

 

[LvW]

Yes, you are wrong. For input voltage=0 (short circuit) both resistors are in parallel.

 

[sakthivel.eee]

Dear Malli,
You r wrong. I also was thinking like that only. but parallel combination only gives me the correct result(Simulation). But i don't how these two resistor in paralle because both resistors r connected at different battery terminal

Thanks & Regrds,
Sakthivel

Added after 1 minutes:

Dear LVW,
For 0 Voltage it is parallel. But we are applying 11 Volts right?Then how it is parallel. Please tell me

Thanks & Regards,
Sakthivel.S

 

[LvW]

Dear LVW,
For 0 Voltage it is parallel. But we are applying 11 Volts right?Then how it is parallel. Please tell me

*It's an 11 volts pulse which goes back again to 0 volts. Thus, discharging with 1k||10k.
* For the charging process it's not so easy to see by pure inspection that both resistors are in parallel. It's best to calculate (as it was done already). However, as the final voltage across the capacitor will be lower than 11 volts (due to the voltage divider) one can expect that the time constant will be determined by a resistance which is lower than 1k. As a simple example, imagine that both resistors are equal. Then, the time constant is (R||R)*C=0.5RC and the voltage across the cap is 0.5*Vin.

As a general rule - in a linear circuit with only one energy source and only one capacitor you can easily find the time constant by answering the question: What is the effective resistance for the process of discharge of the capacitor if the source is set to zero.

 

[fcfusion]

As LvW said, it is best to calculate the circuits behaviour first and analyse it later.

You shouldn't try to figure out how current flows through the resistor's since both of them always have some current which makes the system complex to understand and may mislead you.

As for the equation, I believe my calculations are rigth. However, you could recalculate them if you want, it is not hard.

You can always confirm results in a simulator.

 

[LvW]

Yes, fcfusion. Your result is correct.

 

[balavinayagam]

For a circuit like this (single time constant (STC) circuits) we can find the time constant by considering the thevenin equivalent across the capacitor. Hence it will be the resistors in parallel.

we have to look from the capacitor end i,e consider a voltage source connected instead of the capacitor ,short the source and find the resistance now.

refer Sedra , Smith "Microelectronic Circuits " appendix for more such circuits

 

[LvW]

Quote FCFUSION:
When speaking of charging and discharging time it is usually assume you speak of voltage variation between 10% and 90% of voltage supply (11V).

Only now I have detected that there was an error in FCFUSION'S interpretation (not in his calculation):
The time constant associated with an RC circuit is NOT the time between 10% and 90% of charge/discharge. These two points are relevant for the rise time (step response) only.
The time constant is the time needed to reach 63.3% of the final voltage (which, in theory, is reached at infinite time).

 

[fcfusion]

You're right, time constant is the time needed to reach 63.3% of final value, and that's how I made my calculations:

Discharge time (10%) = time_constant * (- ln (0.1)) = 0.001*2.3 = 2.3 ms

Perhaps I didn't explained it properly. What I meant is that the voltage in a capacitor never reaches voltage zero or maximum voltage, only converges torward it. So when speaking of charge and discharge times we consider the values 10% and 90% of maximum voltage.

That's just a convention though. Some people use 1% and 99%, or 0.1% and 99.9%. Depends on how much accuracy do you need.

 

[sakthivel.eee]

HI All,
Thanks for your reply. Now i can able to understand. Thanks for your help.

Regards,
Sakthi