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Calculate target voltage in RC circuit

Posted on 2014-10-13
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Last Modified: 2014-10-20
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?
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Question by:JasonMewes
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9 Comments
 
LVL 84

Expert Comment

by:Dave Baldwin
ID: 40376576
It would be difficult to do it with just two measurements because the possible range of values is too great for both the voltage and the time constant.   The voltage could range from almost 0 to thousands of volts and the time constant could range from sub-microseconds to days.  But the charging curve for an RC circuit is well defined and if you took a series of measurements, you could eventually calculate where you were on that curve which will tell you where you will end up.
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Author Comment

by:JasonMewes
ID: 40376656
We can do any number of measurements for this, and in this case we know the voltage is known to be roughly 10V and the time-constant is around 1-2min.

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?
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LVL 84

Expert Comment

by:Dave Baldwin
ID: 40376701
If you can measure the source voltage first, then 63% of that is one time constant.  The time it takes to get to 63% represents the RC values in the circuit.  Resistance is much easier to measure and 1% values are very affordable.  1-2 minutes represents a very large tolerance.  I would measure the capacitors before they are installed.  

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.
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LVL 84

Expert Comment

by:ozo
ID: 40376719
Where are you applying the input voltage?
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LVL 27

Expert Comment

by:aburr
ID: 40379103
I assume that you are measuring the voltage (V) across the capacitor (C)
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
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Author Comment

by:JasonMewes
ID: 40379327
aburr, your assumptions are indeed correct - sorry I did not make this obvious.
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.
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LVL 84

Expert Comment

by:Dave Baldwin
ID: 40379339
What happens when you get the results of the current tests?  Do you have to replace any parts to make it meet specs?
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LVL 27

Expert Comment

by:d-glitch
ID: 40380651
You can solve the equations easily if you have precise control over the timing of the switch closure/initiation and the sampling.

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.
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LVL 27

Accepted Solution

by:
d-glitch earned 2000 total points
ID: 40382153
Here is an Excel spread sheet that demonstrates the calculations I described earlier.
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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