# The capacitor Filter + RC circuit time constant

#### Dave_

Joined Mar 22, 2007
28
Hi all. Firstly, Love the site!

Now, the problem.
I have a complex RC problem similar to that found here:

except I have just R1, R2 and C. So it's even easier!

That page works out the time constant using the Thevenin equivalent circuit.
The time constant Im after however is different for the charge and discharge times. Ideally I want a capacitor that charges quickly and discharges slowly. Such as suggested here:
http://www.tpub.com/neets/book6/22e.htm
The capacitor filter.

I'm using a Schottky Barrier diode to half wave rectify a 1GHz signal which is being pulse modulated at 1kHz prf and 1µS pulse width. Currently I'm using 3(tore) = 1nS. So the capacitor will be charged to 95% in 3 time constants, however, I'm guessing that it will discharge really quickly too?

So what was suggested in the tpub site was to use a large capacitor and large load in parallel with the capacitor. This combination means the discharge time will be slow. But will that not mean the charge time will be slow also?

So just to summarise my questions:
1. Is it possible for a capacitor to have a fast charge time and slow discharge time?
2. Should I take my time constant value to be 0.333nS assuming 1GHz, or 0.333µS assuming 1µS pulse width.
3. (If question 1 = yes) Is it still accurate to convert this circuit into its Thevenin equivalent circuit?

Many Thanks

Dave

#### Dave_

Joined Mar 22, 2007
28
Right, I've come back to this problem after figuring out to build a LPDA antenna and I have figured out it is a silly question!

Just looking at the RC time constant graph
http://tpub.com/content/neets/14181/css/14181_198.htm

shows me that if I only charge up to 63% each time, then the charge is fast, and the discharge slow! And the time constants are the same for charging and discharging if both going for 5+ time constants.

Can I just confirm that whether you have a resistor in parallel or series, or both, the time constant will remain the same?

As for the time constant, this one is tricky, because the capacitor doesn't have to charge up fully before the sinewave is in the negative cycle, but it has to retain enough charge so that on the next cycle when it is charging up again it has slightly more charge than before, so it can reach a higher charge before the sinewave is in its negative cycle. There will be a time when it reaches its maximum charge where its discharge matches the charge level. This I am unsure of calculating, but will do a few searches and hopefully find some more answers.

Now I just need to work out how much the jitter will be effecting things. I think my measurement of this voltage level would be quick enough, the log amp responds at 0.5µS and then the ADC responds at 0.4µS.

#### thingmaker3

Joined May 16, 2005
5,083
The only way to get different time constants when charging and discharging is to charge and discharge through separate pathways with different resistances. The full wave rectifier described at TPub charges through the diode tubes and discharges through the load - the diode tubes have considerably less impedence than does the load.

#### Dave_

Joined Mar 22, 2007
28
Hi thingmaker3, thanks for the reply So the full wave rectifier described at TPub shouldn't be reduced down to its Thevenin equilivant? As this would assume the same time constant for charge and discharge.

So is it correct to assume that the resistor in series with the capacitor only effects the charge up time constant, and the resistor in parallel only effects the discharge time constant?

#### thingmaker3

Joined May 16, 2005
5,083
Thevenin analysis works fine for a rectifier and filter, as long as one runs separate analysis for when the diode conducts and when it doesn't. This would be the same as an example Thevenin analysis of a circuit with switch closed for charging versus switched open after charging.

Let's have a look at the circuit you are working on and go from there.

#### Dave_

Joined Mar 22, 2007
28
Ah, ok. Then things get alot more difficult!

Here's my circuit: The switch in this case is the schottky barrier diode, and displayed here is its equilivant circuit including its package.
The values are:
Z0 = 50 ohm
Lp = 2nH
Rs = 25 ohm
Rj = 9000 ohm
Cp = 0.08pF
Cj = 0.18pF
C = 0.05pF
RL = 25000 ohm

I have control over the values for C and RL, and I chose these to give me a time constant value of 0.333nS, and a voltage sensivitity reduction of 0.7353

When I reduced it down to its thevenin equilivant, I ignored Z0, Lp, Rs, Cp, Cj. Leaving just Rj in series with the capacitor C, which is in parallel with RL.
This gave me a thevenin equilivant of 6617.6 ohm in series with the 0.05pF capacitor.

#### thingmaker3

Joined May 16, 2005
5,083
I don't recall anything in Thevenin's theorem about capacitors in series... I'll re-read the pertinant chapters in Leach and A.A.C. again and get back to you. Certainly, you shouldn't be able to ignore components in series with the capacitor.

#### thingmaker3

Joined May 16, 2005
5,083
Looks like the frequency is low enough (or the caps and inductor small enough) for Thevenin to work. We need linear values for Thevenin to work. There are both capacitive reactance and inductive reactance at work in the circuit, so impedence will be a root of something. It will be the root of something really small in this case, thouhg.

I concur with your math. Charging TC will be 0.33nSec.

Assuming Rs and Rj are actually the diode when foreward biased, discharging TC will be in the neighborhood of 1.25nSec.

Thank you very much for your time on this subject thingmaker3 