Is it possible to measure semiconductor capacitance (as in my case, a p-n junction LED) manually using an oscilloscope probe?
I have tried usual AC voltage divider circuits (RC series circuit) with one probe connected to the input signal, and the other probe connected to the device under test DUT.
The results appear similar to the usual RC capacitor voltage response. However, my worry is that the input signal from the AC source would vary also the capacitance, due to the nature of the p-n junction that capacitance is also time-dependent.
Regards,
Yes, the pn junction capacitance is bias-dependent. You should avoid voltage swings in excess of kT (0.03V for room temperature) to get meaningful readings.
Dear James!
I think you can measure capicitance dependence on p-n junction voltage. For doing it, you can summ the bias voltage and small sine signal. And you measure small sine current. And of course you are right, the capicitance have very strong dependence on voltage. For reverse voltage the Area of Volume Charge is big and the capicitance is big.
Best regards,
Dima
Thank you for your answers.
I am currently reading of the work of Shrestha et al. (article link attached) regarding measurement of Capacitance measurements. Maybe this would address the concern of slower measurement from LC meters, but other parameters such as resistances have to be taken into account.
Article Accurate Fast Capacitance Measurements for Reliable Device C...
Dear James,
welcome!
Yes you can use the oscilloscope to measure the capacitance of an element like the reverse biased p-n junction diode. The reverse biased p-n diode small signal model is a series combination of the junction capacitance in series with the diode series resistance Rs.The series resistance is normally very small for conventional diodes.
To measure the diode small signal equivalent circuit you have to DC bias it at the required reverse bis by applying the DC bias voltage through high resistance or large inductance to prevent the ac signal from flowing in the DC source. Then you apply a constant small signal voltage on the diode through a coupling capacitor to prevent the DC current to netter the function generator.
Then you have to a add small resistance Rm in series with the diode to sense the ac current through the diode.
Then you use two channels one to measure the voltage across the diode and the other to measure the voltage drop across the current sense resistance Rm. You have to choose the right frequencies such that the ac current flowing in the DC source is negligible compared to the current in the diode as they are connected in parallel.
By measuring Vd and Vm across Rm= Id Rm one can obtain the magnitude of Vd, the AC diode voltage and Id the AC diode current.
The phase of the Id relative to Vd can be measured by the Lissajous ellipse.
You enter Vm to the y of the oscilloscope and Vd to the x of the oscilloscope. To determine the phase phi , then sin phi= yint/ymax, with yint is the intercept of the ellipse with y axis and ymax== Vm.
Then you can calculate y the admittance of the diode y= (Id/Vd) and phase angle Phi= g + jwC, with g the conductance C the diode capacitance.
Best wishes
Prof Zekry has given a good explanation of the method. I would just add that not all DC sources of bias are low impedance at the measuring frequency so to be sure its impedance is not affecting anything bypass it with a suitable capacitor. Best wishes.
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