circuit analysisdiodeskirchhoffs-laws
I am trying to calculate the current of the circuit but I am unable to find the correct answer. I was able to solve the same circuit without the diode witch gave me 2mA. I am unsure of what to do with the diode added to the circuit. Here is my circuit.
simulate this circuit – Schematic created using CircuitLab
The diode polarization is direct with a voltage of 0,7V. How do I solve the current of this circuit? Any help would be greatly appreciated.
The circuit as shown is not viable - or you could analyze it in two phases, if you must:
Phase 1:
Phase 2.a: If D2 burns out and becomes an open circuit:
Phase 2.b: If D1 burns out and becomes an open circuit:
Then there are the possibilities of D1 or D2 burning out to become short. That resultant analysis is left for you to do :-)
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:
Best Answer
For problems like this, the first step is to assume the state of the diode. Is it conducting or not?
Once you assume a state, then you redraw your circuit to match the assumption. For example, if you assume the diode is not conducting, what can you replace it in the schematic with? How about if you assume it is conducting (hint: remember you said it has a voltage of 0.7V)?
After making these assumptions, re-drawing and solving, one of the solved circuits will result in a contradiction with the original assumption. Perhaps you calculate the voltage drop across the diode as something smaller than 0.7V by using KVL around the loop formed by R2 and R3. This contradicts the diode being at 0.7V when conducting, which means that assumption was wrong.
The circuit that doesn't result in any contradictions is the correct assumption, and the values from that circuit are the "right" answer.