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?
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.
0
JasonMewesAuthor Commented:
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?
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.
0
Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.
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
0
JasonMewesAuthor Commented:
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.
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.
Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.