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In an RC circuit, the voltage across the capacitor will equal the input voltage after infinity.

If the input voltage is unknown, as well as the RC time constant,

is it possible to calculate or approximate this "target" voltage from two or more careful measurements during early charge, given that these measurements are sufficiently precise in both time and amplitude?

How would one go about this?

If the input voltage is unknown, as well as the RC time constant,

is it possible to calculate or approximate this "target" voltage from two or more careful measurements during early charge, given that these measurements are sufficiently precise in both time and amplitude?

How would one go about this?

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The variance in C and R between units (this is for testing equipment) are causing us long wait times during testing, for the voltage to stabilize to within certain bounds of the target voltage. We wish to reduce this time, as well as increase accuracy by early measurements and a mathematical approach.

I know the curve is well defined, and it seems to me there should be a simple mathematical solution to the problem (at least if the measurements were ideal). But my math knowledge is failing me.

Do you have a mathematical or algorithmical approach in mind?

One of my customers hand selects capacitors for a precision 25Hz oscillator before he installs them. He might have a test fixture to do that. While that's a much lower time constant, measuring the capacitor value is a much quicker thing to do than measuring that long time constant in circuit.

I assume that you have the unknown voltage (E) in series with the unknown resistor (R) in series with the unknown C

You then have two unknowns E and RC

The equation is given below. You have two unknowns so you need two measurement at two known times.

The equation

V(t) = E (1 - e^(t/RC))

Problem solved

If you do not know how to extract E and RC from the equation, ask

I know the equation you are describing well, but no, I don't know how to extract E or RC.

I can measure t[1] and t[2] (two times of measurements) as well as all the corresponding V(t[1]), V(t[2]).

Solving E and RC will allow us to determine that the system is providing correct voltage (E) and that the input filtering is adequate (RC). This is exactly what I am looking for.

Dave Baldwin, this is for existing equipment and I do not control the design or specifications.

For example, if you can set the sampling times at 200 and 400 ms after the start:

Your data would be

V1(t=0.1) = E (1 - e^(0.1/RC))

V2(t=0.2) = E (1 - e^(0.2/RC))

Take the ratio of the equations. The E's cancel out, and you wind up with something of the form

V1/V2 = (1- x)/(1 - x²)

In theory, you can solve for x, and then RC [ = t1/ln(x) ], and then E.

In practice, you will need to choose the sample times carefully to insure accuracy.

Times on the order of RC/2 would probably work best.

Note that the initial behavior of a circuit with parameters E and RC is nearly the same as the that of a circuit with each value doubled.

I assume sample times of 60 and 120 seconds.

I show the exact calculations and the effect of a 1% timing error.

You can change the E and RC to see how the errors behave as the RC gets larger.

Voltage-Calculations-for-ExEx-2.xlsx

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