Series-Parallel DC RC circuit

instance.

As I said: "I agree that the thevenin resistance is the correct value to use for the time constant."

I also agree that bj's statement is wrong.

I disagreed with your statement that bg had not specified what resistance he meant to use for the time constant calculation. In his post which I quoted and later posts, he argues that only the series resistance should be used for the time constant calculation. I do not agree with him (and I do not think that you do either).

The charging time constant is C*R' where the value of R' is R1 parallel R2. My numerical example expresses this; try it.

The discharge time constant depends on whether you discharge the cap by jumping V1 to zero volts, or by disconnecting it. In the first case, the discharge time constant is identical to the charge time constant, 500 us in my example. In the second case, R1 is open so only R2 determines tau. Your statement that the cap "discharges through the parallel equivalent of the two resistors" is the first case. Try it.

It's better to do it in real life, but you can also use LTSpice to see this stuff happening.

Interesting: if the cap charge and discharge curves indeed had different time constants, an applied sine wave at V1 would produce harmonics across the cap. That's not possible.

John

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My mistake !! Yes I see what you guys are getting at, the time constant needs to use the parallel eq of both resistors.

A little bit of nit picking: Actually the capacitor charges FASTER to a lower voltage. The thevenin equivalent resistance gives a smaller time constant which means a quicker initial increase in voltage. The final voltage will be the thevenin voltage which is lower since the two resistors form a voltage divider.

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constant

Ah! I couldn't tell from your last sentence. Thanks.

Jon

The initial slope of the capacitor charge, measured in volts/second, is independent of R2 (for nonzero R2 of course). It's just

dv/dt = V1/(R1*C)

John

Since the cap's voltage presents a voltage across the shunt resistor, except at t=0, there is current present in it (except at t=0.)

That's not a very good way to look at it. But maybe I just can't read your English and fit it with my mental models.

Since I already performed the complete calculations using various techniques for solving differential equations, each with the same result -- not to mention that I tested what I wrote with LTSpice simulation to be absolutely certain that a spice program also agreed with me -- Would you care to expose the calculations in detail, starting from basic statements?

Jon

Okay. We are all good, now! We have achieved a shared reality!

Jon

Right ... right ... right. I was focusing on the "All the shunt resistor does is divide the source voltage ..." part & got sloppy. Bob

I mean, V(t) = V_source * [R2/(R1+R2)] * ( 1 - e^( -t/[C1*R1*R2/(R1+R2)] ) )

Jon

Yes, assuming that we are starting with a capacitor voltage of zero (which has been true for all of the discussion so far on this thread). In this case the initial current through R2 is zero which means that for the initial instant we can ignore R2. Your equation then falls out by simple inspection.

A little more nit picking, this time with my own posting. (This is what happens when I have dinner and have some time to think.)

I said that the capacitor charges 'FASTER'. I even capitalized the word 'FASTER'. The change in capacitor voltage is only faster in the circuit with the shunt resistor R2 (versus without R2) if you are looking at change as a percentage of the final value. I.e. it will reach 63% of its final value quicker in the version of the circuit with R2 then in a circuit without it.

However, as Bob points out, the rate of change of the capacitor is lower if the shunt resistor is present. As Bob says, the shunt resistor is taking some of the current that would otherwise be charging the capacitor. This gives us a lower rate of change of the capacitor voltage (when measured in volts per second).

Both the capacitor voltage and the rate of change in the capacitor voltage (in volts per second but not percentage) will always be lower with the shunt resistor than without it. The only exception is at t=0 when the capacitor voltage is zero. At that instant the capacitor voltage is zero in either case and the rate of change is V1/(R1 * C) (as John Larkin posted previously).