Question

Problem #1 Suppose a Si diode has a forward voltage drop of 1.492V at room temperature, and suppose its IV curve is so steep that the forward voltage drop is independent of current. The diode is a cylinder 0.069mm in radius, 0.053018mm deep and is glass encapsulated. The glass package is 0.09619mm in thickness, and is held at 26°C (room temperature) by a thermally conductive blanket. The diode's forward voltage drop decreases by 21.479 microvolts for every degree C above room temperature (so called "thermal derating"). (use k(Si) 13 W/em/K, k(glass) 0.01 Wiem/K, density rho(Si) 2.328 g/em3, rho(glass)- 2.7g/cm3, Si heat capacity 712 J/Kg°C) a) You want to flow 9.367 mA continuously through it. Estimate the CHANGE in the forward voltage drop across the diode when it reaches quasi-equilibrium. NOTE: Simplify the geometry to a slab by "unrolling" the cylinder about its center, where the heat is produced. (Ignore the heat conduction out the leads, but that turns out to be a very big sink of excess heat.) b) Now you want to send sporadic (and widely spread out) pulses of high current that are so short in duration that all the heat stays in the diode during the pulse. Estimate, as we did in class, the longest (duration) of the pulses in this mode of operation.

Answer 1

Unrolling the geometry about the center we get a slab cross section as above.

Here, heat is generated in the centerline & is transferred to room temperature 26 $\degree$ C sink.

we need to find the quasi-equilibrium temp. & thereafter the voltage drop.

We know, Power generated by diode = I*V = 9.367*10^-3A *(1.492+0.021479(t-26))V, here t= temp.of diode centre.

at equilibrium let us assume the temp. at glass-Si interface as T_x

At equilibrium heat flow will be

$P=I*V = \frac{k_{Si}(T-T_{x})}{x _{si}} =\frac {k_{glass}.(T_{x}-26)}{x_{glass}}$

$q=\frac{k\Delta T}{x}$ taking electrical analogy it is similar to I=V/R

thermal reistance in this case = x/k. now here thermal resistances are in series

therefore total resitance (x/k) = x(Si)/K(si) + X(glass)/K(glass) = $\frac{0.069}{1.3} + \frac{0.09619}{0.01} = 9.672$

therefore I.V = (T-26)/9.672

=> 9.367*10^-3*(1.492+0.021479(T-26)) = (T-26)/9.672

solving for T = 26.135 $\degree$ C, therefore the change in forward voltage drop = 21.479*(T-26)mV = 2.9mV