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.

  • 2
    \$\begingroup\$ Leakage current depends (highly) upon temperature, even if it is said (by fiat) that it does not vary by impressed reverse voltage. The rule of thumb I use is that it varies by about a factor of 2 for each change by \$10^\circ\text{C}\$. Your problem statement appears to ignore this particular behavior. Also, why do you assert that both diodes are off? \$\endgroup\$ – jonk Feb 11 '18 at 0:10

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:


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?

  • \$\begingroup\$ I had assumed that both diodes were OFF. Because the thought that a large reverse current of D1 could be the forward current for D2 didnot occur to me. After considering that, I found out I (current in the circuit). Since I was 1 microampere, it seemed that indeed the diodes were in reverse bias. I am aware of the rule of thumb that every 10℃ the reverse current doubles. So I found out reverse current for 0℃ and 40℃ which were 1/4uA and 4uA. \$\endgroup\$ – Mohammed Arshaan Feb 12 '18 at 1:33
  • \$\begingroup\$ @MohammedArshaan Did what I write then fit your understanding of things? Or do you feel I failed in some fashion? \$\endgroup\$ – jonk Feb 12 '18 at 1:35
  • \$\begingroup\$ After that I used diode law to find voltage. V2 - V1 = nVt.ln(I2/I1). With V1 and I1 being the condition for 20℃. Now my doubt was that if I consider tbe 2mV/℃ my answer was exactly what you are getting. But if not then it was slightly off. \$\endgroup\$ – Mohammed Arshaan Feb 12 '18 at 1:37
  • \$\begingroup\$ No I understand now that the reason we can use the 2mV/℃ and doubling of reverse current per 10℃ rise is because one diode is forward biased and other is reversed. Earlier I believed both to be reverse, hence, the use of 2mV/℃ drop which is observed in the forward region seemed incorrect. Just how did you arrive at the conclusion that D1 is OFF and D2 is ON? \$\endgroup\$ – Mohammed Arshaan Feb 12 '18 at 1:41
  • 1
    \$\begingroup\$ @MohammedArshaan Your problem statement. Imagine no current, then the resistor has no voltage drop. D2 is arranged forward with respect to the voltage source. So all of the voltage applies to D1. Which as you say "leaks." Since there is a leakage current, then there is no difficulty. D2 will be forward biased and allow the leakage and R1 will drop some voltage due to that. It's a simple fact that D2 is polarized forward with respect to the voltage source. \$\endgroup\$ – jonk Feb 12 '18 at 1:53

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