Question: In the circuit D1 is a large-area, high-current diode whose reverse leakage is independent of applied voltage, while D2 is a much smaller, lower-current diode for which n = 1. At an ambient temperature of 20C, resistor R1 is adjusted to make VR1 = V2 = 520mV. Subsequent measurement indicates that R1 = 520kohms. What do you expect the voltages VR1 and V2 to be at 0C and 40C?
simulate this circuit – Schematic created using CircuitLab
I have some conceptual queries related to the numerical.
The diodes are both OFF. So a reverse leakage current flows through them. My question is what is the implication of providing that D1 is independent of applied voltage?
Another question is that in forward bias the voltage decreases by 2mV per degree rise in temperature. Is this applicable in reverse bias? I haven't found any confirmation nor any denial. The answer however considers a drop, even for reverse bias.
(Note: I solved the question with and without the drop due to temperature increase/decrease and the answer was correct in the first case while it varied from the correct one in the without clause.)
Another question is that does, area difference between diode bring any change other than in its saturation current?
Disclaimer: I know it might feel that these questions (especially the last one) should be asked differently. But to me they seem to be related to the problem at hand. So please excuse that.
Best Answer
I don't know how you arrived at the idea that both diodes are off in the nominal case. It's a stated fact that reverse-biased \$D_1\$ has a leakage current. This leakage current, independent of applied voltage, provides a forward bias for \$D_2\$.
So that part of your statement carries an incorrect assumption.
KVL provides, for the nominal case at \$20^\circ\text{C}\$:
$$10\:\text{V}-I_{LEAK}\cdot R_1-V_{D_1}-V_{D_2}=0\:\text{V}$$
But you know that \$V_{D_2}=520\:\text{mV}=I_{LEAK}\cdot R_1\$ and so it is then very obvious that:
$$I_{LEAK}=\frac{520\:\text{mV}}{520\:\text{k}\Omega}=1\:\mu\text{A}$$
From this, you also now know, from \$V_D= n V_T\operatorname{ln}\left(1+\frac{I_D}{I_{SAT}}\right)\$ and \$n=1\$ and \$V_T\approx 26\:\text{mV}\$ that:
$$I_{SAT_{D_2}}=\frac{I_{LEAK}}{e^\frac{V_{D_2}}{V_T}-1}\approx 2.06\times 10^{-15}\:\text{A}$$
This last part probably doesn't matter for your problem. While \$I_{SAT}\$ is itself highly temperature-dependent and affects the diode drop voltage, you've made another assumption about that variation by instead stating that the change is \$-2\:\frac{\text{mV}}{^\circ\text{C}}\$. So that trumps any discussion about the change in \$I_{SAT}\$ over temperature.
By the way, the above is all at \$20^\circ\text{C}\$.
So the question breaks down into the following three steps.
That's all that can be done with your problem, given how few details were provided.
Since you say you already got the right answers, please do tell what they were. I'm curious how they compare in value and in the reasoning you applied. The above is mine. What's yours?