Let's say I am given the following circuit: along with $$dV_p/dT=0.139mV/K$$:

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Defining V1 the voltage above Diode D1 and V2 the voltage above diode D2 , we use the expression for the current of a diode for both diodes :

$$I_{D1}=I_{S1}e^{\frac{V1}{V_T}}$$ $$I_{D2}=I_{S2}e^{\frac{V2}{V_T}}$$

If we solve with respect to the voltages, we get :

$$V_1=V_T\ln{\frac{I_{D1}}{I_{S1}}}$$ $$V_2=V_T\ln{\frac{I_{D2}}{I_{S2}}}$$

Subtracting, we finally get Vp :


What I know for sure, is that the thermal voltage VT depends on temperature and it is given by :


I think that the saturation currents depend on the temperature as well. However, will these currents be different? Or can I just cancel them out in the logarithm above?

Cancelling them out, the result doesn't make sense:


The result is different form the given value of 0.139 mV/K and does not depend on T so I get no information from here.

So, it seems I can't cancel out the saturation currents .

After looking it up, saturation currents seem to depend on :


However, even if I use the long expression for the saturation currents, ni1 and ni2 will still cancel out in the logarithm. Therefore the dependence of Vp on T vanishes even if the saturation currents depend on it.

Long story short, this was my attempt at a solution. How do I find the temperature?


1 Answer 1


Obviously you can't ignore the Is temperature dependence or the diode temperature coefficient would be of the opposite sign (let alone the magnitude) of what we know it to be (~-2mV/K).

But basically you have the answer.

Vp = \$ \frac{n KT}{q} (\ln{(\frac {Id1}{Is}}) - \ln{(}{\frac{Id2}{Is}})) = \frac{n KT}{q} \ln{(\frac {Id1}{Id2}}) \$

dVp/dT = \$\frac{n K}{q} \ln(5) \$= 0.139mV/K so n = 1 (ideality factor)

(if you got a different number, check your calculations here)

So the temperature T = \$\frac{Vp}{0.139mV/K}\$ (it's simply proportional to absolute temperature of the junctions, assuming they're the same temperature, of course)

eg, If Vp = 50mV then T = 56.6°C


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