This is a circuit extracted from the book "Electronic Circuits Analysis" by Phillip Cotler.


simulate this circuit – Schematic created using CircuitLab

The potentiometer R1 can be adjusted to control the polarization voltage across the "Si Capacitor". My question is how can this circuit work if the voltage E is connected to the positive input of the diode and not to the negative (direct polarization)?

I'd also like to ask how does C1 restrict the coil to put the DC tension in short circuit through Cv ?.


2 Answers 2


RV1 adjusts the negative bias on the diode. If you turn the pot all the way 'up', you get voltage E across the diode (negative bias) through R2. All the way down, and the diode has zero bias.

This varies the capacitance of D1 (by varying the width of the junction depletion zone) so that the resonant frequency of the inductance and the D1 capacitance can be changed. Capacitors C1 and C2 are large enough that they appear as short circuits to the frequencies of interest. C1 and C2 isolate the DC voltage at the diode from the signal so that the only current through R2 is the leakage current through the two capacitors (and diode leakage in reverse bias, all of which should be tiny).

Since the only current flowing through R2 is leakage, that allows the value of R2 to be relatively high- since it shunts the parallel-resonant tank circuit it reduces the 'Q' of the tank, so it's desirable to have the resistance reasonably high.

Normally variable capacitance diodes are operated in reverse bias (the higher the voltage the lower the capacitance) but it's possible to slightly forward bias them to get higher capacitance than at 0V bias (another effect- diffusion capacitance- also comes into play). The lower the bias is in relation to the signal, the more the capacitance will vary with the signal itself, which causes distortion, so it's more common to have a fairly high reverse bias voltage nominally.

Here is what the capacitance vs. voltage curve of a typical varactor diode- a 1SV280 looks like:

enter image description here


The diode (known as varicap, tuning diode, varactor) needs to be reverse-biased to work as a capacitor. Reverse bias creates a charge-carrier depletion zone around the p-n junction of the diode. The depletion zone serves as the dielectric, while the rest of the p- and n-regions around it serve as plates.

If the diode were forward-biased, it would conduct (which is not its purpose here) and effectively work as a resistor, shunting the signal from the transformer.

C1 will charge to Cv and stop conducting current, and that's how the DC voltage at the coil will be zero.

I doubt that in any practical circuit C2 and C2 would need to be polarized, but I'm guessing it's just the symbols you used in CircuitLab. And the C2 polarity is probably wrong.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.