How to measure capacitance with high series resistance

Thread Starter

dwood0403

Joined Mar 23, 2019
4
I have a sensor that can be modeled as a capacitor and resistor in series. The problem with this resistor is the resistance is on the order of 10's MOhm and the capacitance (we believe) is single pFs.

I have tried using a resonant circuit to find the capacitance, but the results end up super noisy and with an extremely low q factor. Does anybody have any ideas of how to determine the capacitance?
 

KeithWalker

Joined Jul 10, 2017
895
Normal measuring devices have input capacitance large enough to mask the value you are trying to measure. Do know the value of the effective series resistance? Having two unknowns certainly complicates things. What does the sensor sense? Can a level change at the input be used in any way to characterize the device?
Regards,
Keith
 

Thread Starter

dwood0403

Joined Mar 23, 2019
4
Normal measuring devices have input capacitance large enough to mask the value you are trying to measure. Do know the value of the effective series resistance? Having two unknowns certainly complicates things. What does the sensor sense? Can a level change at the input be used in any way to characterize the device?
Regards,
Keith
I have been passing a DC current through the sensor to characterize the resistor component. I have to measure the capacitance to be sure to prove that we can negate the effects of the internal capacitance if we use the device at certain frequencies. But I need to figure out what those frequencies are, and I have to determine the capacitance to do that.
 

Analog Ground

Joined Apr 24, 2019
397
If the resistance and capacitance are in series (as stated in your first post), how are you passing a DC current to characterize the resistance component?
 

drc_567

Joined Dec 29, 2008
844
Do you have a scope and a signal genrrator?
Try using a square wave at some arbitrary frequency as an input to the sensor.
Observe the resulting output waveform and note the length of time that it takes for the step input of the square wave to return to the baseline voltage. There should be an exponential curve present with a characteristic time constant of 5*τ. Next place a known resistor in series with the sensor and determine the new time constant.
The procedure here should allow you to have two equations and two unknowns, which can be solved. It may be necessary to repeat the procedure using an auxiliary capacitor instead of a resisfor, in order to confirm the numerical results. If the intrinsic magnitudes of the base components are too extreme, then difficulties may arise in obtaining measurable output waveforms.
 

BobTPH

Joined Jun 5, 2013
2,391
I have not worked out the math, but, if you drive it with a sin wave through a known resistance and measure the amplitude and phase, wouldn’t that be solvable for the complex impedance? And, if you know that and the frequency, you can separate the resistive and capacitive components, no?

Bob
 

vanderghast

Joined Jun 14, 2018
37
Vc = Vt0 + (Vs-Vt0)(1-exp(-t/RC))
where Vc= voltage (difference of tension) between the pins of the capacitor,
Vt0 = voltage at the capacitor at some arbitrary time that we will label as time t=0
Vs= CONSTANT voltage at the end of the group Resistance, R, in serie with the capacitor, C.
This is the basic equation, that we can use. R is the total resistance in series, unknow, and C is also unknown.
The fact that we have two unknown values seems a problem, but since they appear only once, we can replace RC by X.
Measure the charge of the capacitor through time on an oscilloscope ( using a small external known resistor, R0, in serie, with C). Use a cursor following the curve to select a starting point, that will be t=0, and ending at some "nice" point on the curve, read Vc at this point. In fact, the cursor gives you Vc-Vt0 (vertical difference, and t (as Delta t with the cursor, horizontally) with the precision of the scope (and not by visual interpolation), at least, on digital scopes. So (Vc-Vt0)/(Vs-Vt0)-1 = exp(-t/X); take the natural log on each side, and since t is known (the interval of time between the two cursors), you get X after elementary manipulation: X0 = (R0 +Rc) *C. Repeat the experiment with another small known resistor, R1, giving now X1 = (R1+Rc)*C. Rc is the internal serial resistance for the capacitor. You get: X0/X1 = (R0+Rc)/(R1+Rc), where just Rc is unknown. Rc found, you can get C using the definition of X0 or of X1.
The computation is really sensible to errors on the inputs. I strongly suggest that you use Excel like formula to comput X0_min, X0_max, X1_min and X1_max given the precision of your oscilloscope.
 
Last edited:

Janis59

Joined Aug 21, 2017
1,179
Cant immagine better method how to use a low pwm meander peak generator thus to measure the echoe on the system. However oscilloscope capacitance and input resistance here are supercritical. Probably 1:100 probe with it 100 MOhm and 0,1 pF may be solution. If the task was 10 MOhm and 1 pF then tau=1E-12*10E6=10E-6=10 microseconds. Thus the meanders must come out once at 20...100 microsec and have length of 1...10 microseconds. Note to the value of RC corresponds the 63,2% fall down in voltage what is 36,8% in rest of voltage.
 
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