Let we have a capacitor charged about 300v. we want to deliver energy for one time constant. that means we want to truncate it. does any body has idea how to do so.

 

Can you be a bit more clear? What does truncating energy over here mean?

__________________
"There is no way to peace, peace is the way!"

 

I think what he means is, the cap is charged to 300v, he wants to discharge it to a voltage level determined by 1 time constant....... oh, here we go again..... :lol:


Well 2 things to consider: the cap's value in mfd and the resistor or load.

More info needed ?

 

If you know the initial voltage, theoretically that's all you need to know.

V=Vinit*e^(-t/RC)

He wants to discharge the cap until V=Vinit*e^(-1). If Vinit=300, V=110.36 volts. You don't really need to know the values of R and C.
As is usual, the devil is in the details. You need a comparator that can handle 300 volts, or you can use a normal low voltage comparator, but then you have to use a voltage divider to scale the voltage. The voltage divider will bleed the charge off the cap unless you want to discharge it immediately after charging it, or can use an extremely high value (100 Megohm?) resistor in the divider.
Of course, if you know the values of R and C, you can just use a timer (555?) to time the discharge.

 

Ron, I bow to your superior knowledge of matters concerning all things electronic, however I'd have thought that the value of the capacitor would be vital in the calculation or measurement......purely and simply in terms of energy.......since a 100mfd cap can store more energy than a 1mfd cap..........or is this totally irrelevant....?

 

Well, Chippie, when you have a charged cap, and you connect a resistor across it, the cap discharges at a rate determined by the product of R and C. When it has discharged for a time numerically equal to that product, it will have reached a voltage equal to 36.8% of its original voltage. If you can compare the cap's voltage in real time vs a dc reference equal to 36.8% of the original voltage, and stop the discharge when the two voltages are equal, you will have accomplished the OP's goal. Yes, the amount of energy dissipated is a function of the capacitance, but that doesn't affect the technique.
Like I said, if you know the values of R and C, you can simply discharge for a time equal to the product of R and C.

Ron

 

Ahh now I see......I re read you post and saw the reference to R/C...

As I previously posted, we need to know the value of the cap and the resistor or load.......then I concurr with what you wrote 8)

 

And as I previously posted, you don't need to know the values of R and C f you know the voltage to which the cap is charged. It's an either-or situation: If you know V, you don't need to know R and C. If you know R and C, you don't need to know V (except for voltage breakdown and safety considerations).

Ron

 

I think Petersmith might be looking to discharge for one time constant, then completely disconnect output. That may be what he means by truncate. I see a relay with a timer-controlled coil to remove capacitor output from the load at exactly one time constant.

__________________
BSEE

 

Ron, you have considered t/RC = 1. But Time Constant is RC and t is the time required to charge or discharge.
You should consider RC = 1.

If he says he want to discharge it at unity time constant then RC will be 1.
Here he did not mention that time t should also be 1 second.
And he did not mention upto what voltage he wants to discharge it. The instantaneous voltage on capacitor is given by,

Vc = VF + (VI - VF)*e^(-t/RC)
Where,
Vc = Instantaneous value of cap. voltage
VI = Initial voltage across cap.
VF = Final voltage across cap.

So, RC is needed to find the time required for capacitor to reach upto the desired voltage or initial voltage is needed to find the value of RC to take the final voltage across the capacitor to given value in desired time.

__________________
"There is no way to peace, peace is the way!"

 

Kinjalgp, that's wrong. The units of the product of R and C is seconds. If RC=1, then that is a one second time constant. Petersmith says he wants to deliver energy for one time constant, not one second. Suppose R=1 kohm and C=1 uF. Then RC (one time constant)=1 millisecond. Does RC=1 in this case?
If we want to deliver energy for one time constant, then in the equation you posted, which I agree is correct, t=RC, or, as I said, t/RC=1.
I do want to clarify the meaning of VF. In this case, I have assumed that we are discharging the cap through a resistor whose other end is connected to ground (while the switch is closed). In this case then, VF=0, and the equation simplifies to the equation I posted:

Vc=VI*e^(-t/RC)


Ron

 

It would have helped this discussion if Petersmith has explained his requirements more clearly. It always pays dividends to read and re-read a question before posting it.

 

Hi Ron,
I know how you got Vc=VI*e^(-t/RC).

The question itself is quite confusing. Petersmith has not mentioned clearly what he means by one time constant.

__________________
"There is no way to peace, peace is the way!"

 

Guys, I apologize for being a little rude. I know this is generally a genteel forum.
As far as "time constant" being ambiguous or undefined - see http://ccrma-www.stanford.edu/~jos/m...onentials.html
and
http://www.atis.org/tg2k/_time_constant.html

This definition is universally accepted in the electronics industry and in academia.

Ron