Precision capacitor measurement with op amp, resistors, oscilloscope? SOLVED

Using only op amps, resistors, and a normal lab oscilloscope, how can you make a precise measurement of the capacitance of an unknown capacitor? This is a bonus question on my problem set but it sounds crazy.

by nitabobby64
November 17, 2016

2 Answers

Answer by mrobbins

One way would be to make an oscillator using (known) resistances and the (unknown) capacitor. This uses the op-amp more as a comparator than an op-amp, but I think it should count... Take a look at this simulation:

Open and run it!

C1 is your unknown capacitor, and $\tau = R_1 C_1$ is the RC time constant. If you don't even know what order of magnitude C1 is, you can try a bunch of different R1 values so you get the time period into a place you can measure it accurately enough.

Basically, R2, R3, and R4 form a voltage divider. When $V_A = V_{CC}$ then the non-inverting input of the op amp is at $\frac {2} {3} V_{CC} = 3.33$ volts. And when $V_A = 0$ then the non-inverting input is at $\frac {1} {3} V_{CC} = 1.67$ volts.

Pretend that when the circuit starts, the capacitor is discharged, so $V_B=0$ and therefore since the non-inverting input is higher, the op-amp output $V_A=5$. This will start charging the capacitor through R1, and it will keep charging exponentially until it crosses the non-inverting input at 3.33 volts. At that time, the op-amp inverting input will be higher than the non-inverting input, so the output will swing the other way to $V_A=0$. At that point the capacitor will start discharging, again until it hits 1.67 volts... and at that time it switches again and the cycle starts over! Tada -- oscillator!

How long does it take to switch? Well, how long does it take to charge an RC-combination from 1.67 up to 3.33 volts (or from 1/3rd to 2/3rds of full voltage)? If the charging-up voltage as a function of time looks like $$V_B(t) = 5 (1 - e^{-\frac {t} {\tau}})$$, then we can figure out for $V_B(t_1) = 5 * \frac {1} {3}$ and $V_B(t_1) = 5 * \frac {2} {3}$... that's $$\frac {1} {3} = 1 - e^{-\frac {t_1} {\tau}}$$ $$\frac {2} {3} = e^{-\frac {t_1} {\tau}}$$ $$ln(\frac {2} {3}) = -\frac {t_1} {\tau}$$ $$ln(\frac {2} {3}) = -\frac {t_1} {\tau}$$ $$t_1 = \tau * ln(\frac {3} {2})$$ You can do the same thing for $t_2$ and you'll get $$t_2 = \tau * ln(3)$$ The difference is $$t_2 - t_1 = \tau * \big( ln(3) - ln(\frac {3} {2}) \big)$$ $$t_2 - t_1 = \tau * ln(2)$$ That's our half-period -- the time to go from 1/3rd to 2/3rds VCC. But we have to discharge too to complete the cycle, so I'll call our overall oscillator period $T$. Well, $$T= 2 * (t_2 - t_1)$$ $$T= 2 ln(2) * R_1 C_1$$ If you plug in our values in the schematic above, you get $T =139$ milliseconds.

Does that match the simulation?

You bet it does :)

So if the capacitance was unknown you could just measure the period and reverse the formula: $$C_1 = \frac {T} {2 ln(2) * R_1}$$

ACCEPTED +2 votes
by mrobbins
November 17, 2016

Answer by berniekorrie

@mrobbins perfect. Just wanted to point out that you could also do this with a digital comparator chip (it would look the same as your circuit).

Or if you had an inverter like a 7404 laying around, its even simpler...

Simulation works too!

Same basic idea. The oscillation period $T$ will be different than the op-amp one because it the inverter will have different thresholds (not exactly 1/3rds or 2/3rds of $V_{CC}$), but if you have an oscilloscope you can measure those too and fix your math.

+1 vote
by berniekorrie
November 17, 2016

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