# Given a diode circuit and some of its values, how do I find the temperature?

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

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

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

$$V_T=\frac{kT}{q}$$

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:

$$\frac{dV_p}{dT}=\frac{k}{q}\ln{\frac{I_{D1}}{I_{D2}}}$$

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 :

$$n_i^2=(5.2\times10^{15}T^{\frac{3}{2}}e^{\frac{-E_g}{2kT}})^2$$

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?

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