Electronic – Why would you attach a Diode to the base of a BJT

bjtcurrent-sourcediodes

I was looking at a DC BJT setup for sourcing current and came accross this

I have never seen a diode attached to the base of BJTs before and was wondering what it might be used for? I believe it might be used for compensation due to effects in temperature, but I haven't seen much info on this or why you wouldn't bridge the voltage at the base of Q1 with a resistor instead. Does anyone have any suggestions to why you might do something like this?

Best Answer

It is there to keep the transistor's current less susceptible to temperature changes.

In the case of Q1:

Suppose that instead of having R1 and D1, Q1base was connected directly to ground.
Emitter current would be: $$ I_{e} = \frac{20V - V_{be}}{R_{2}}$$

You can see Ie is susceptible to variations in Vbe, which has a known dependency on temperature (T), so you might as well express it as: $$ I_{e}(T) = \frac{20V - V_{be}(T)}{R_{2}}$$

But with the diode, if they are matched and thermally bonded: $$ V_{diode}(T) = V_{be}(T) $$

So now: $$ I_{e} = \frac{20V+V_{diode}(T)-V_{be}(T)}{R_{2}} $$ Which simplifies to: $$ I_{e} = \frac{20V}{R_{2}} $$ Independent of Vbe, and its variations with temperature.

The diode is effectively providing the little voltage offset that would be needed to compensate for Vbe changes with T, in order to maintain a constant current.