Electrical – maximum reverse voltage around diode

My, that's a lot of questions.

Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.

Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.

Two such diodes in series behave like one 1.4V diode with 8 ohms in series.

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.

1. Does \$I_{LEAK}\$ change over temperature? Either it does, or it does not. You failed to state your assumption here. However, I happen to know there is a rule of thumb for diodes, regarding leakage currents. (I know this well from working with photodiodes.) The leakage currents increase by a factor of \$2\$ for each \$+10^\circ\text{C}\$ change. (You can work out the implication in the other direction of temperature change.) So, we have the following: $$\begin{align*}I_{LEAK_{20^\circ\text{C}}}=1\:\mu\text{A}\cdot 2^\frac{20^\circ\text{C}-20^\circ\text{C}}{10^\circ\text{C}}&=1\:\mu\text{A}\\\\I_{LEAK_{0^\circ\text{C}}}=1\:\mu\text{A}\cdot 2^\frac{0^\circ\text{C}-20^\circ\text{C}}{10^\circ\text{C}}&=250\:\text{nA}\\\\I_{LEAK_{40^\circ\text{C}}}=1\:\mu\text{A}\cdot 2^\frac{40^\circ\text{C}-20^\circ\text{C}}{10^\circ\text{C}}&=4\:\mu\text{A}\end{align*}$$
2. What is voltage across \$D_2\$ at \$0^\circ\text{C}\$ and at \$40^\circ\text{C}\$? Since you've already stated the variation elsewhere, assuming the current through the diode remains the same, this seems easy. But you need to include the difference that is also due to any change in leakage current. So At \$0^\circ\text{C}\$, it is $$V_{D_2}=520\:\text{mV}+\left(-2\:\frac{\text{mV}}{^\circ\text{C}}\right)\cdot \left(0^\circ\text{C}-20^\circ\text{C}\right)+V_T\operatorname{ln}\left(\frac{250\:\text{nA}}{1\:\mu\text{A}}\right)\approx 524\:\text{mV}$$ and at \$40^\circ\text{C}\$ it is $$V_{D_2}=520\:\text{mV}+\left(-2\:\frac{\text{mV}}{^\circ\text{C}}\right)\cdot \left(40^\circ\text{C}-20^\circ\text{C}\right)+V_T\operatorname{ln}\left(\frac{4\:\mu\text{A}}{1\:\mu\text{A}}\right)\approx 516\:\text{mV}$$
3. What is voltage across \$R_1\$ at \$0^\circ\text{C}\$ and at \$40^\circ\text{C}\$? Well, you have the leakage currents from #1, above. So at \$0^\circ\text{C}\$ you have, $$V_{R_1}=520\:\text{k}\Omega\cdot I_{LEAK_{0^\circ\text{C}}}=130\:\text{mV}$$ and at \$40^\circ\text{C}\$ it is $$V_{R_1}=520\:\text{k}\Omega\cdot I_{LEAK_{40^\circ\text{C}}}=2.08\:\text{V}$$

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?