Art of Electronics Exercise 2.9

[pastro]

Hi all:

I'm working independently through "The Art of Electronics" 2nd Ed. I have a question about Exercise 2.9 (p. 84)

The exercise states:
"Verify that an 8degC rise in ambient temperature will cause a base-voltage-biased grounded emitter stage to saturate, assuming that it was initially biased for Vc = 0.5Vcc."

My solution.

From text page 81, we have
Ic = Ic0 exp(ΔVbe/25mV) (eqn 1)
ΔVbe = (-2.1mV/degC) ΔT(degC) (eqn 2)

So, substituting 2 into 1:
Ic = Ic0 exp(-ΔT(degC)/11.9) (eqn 3)

Roughly, for the collector biased to half the supply voltage, we will have Vc0 = 0.5 Vcc = Vcc - Ic0 Rc => Ic0 = 0.5 Vcc/Rc
Roughly, for saturation, we have the collector near ground, so Ic = Vcc/Rc

Thus, using eqn 3 with Ic = Vcc/Rc, and Ic0 = 0.5 Vcc/Rc, I get ΔT = -8.2degC. If a temperature rise caused saturation, I would expect ΔT = +8.2degC. However, the negative sign would seem to indicate that a FALL in temperature actually leads to saturation, which is not what the problem statement says I should find. So, what am I doing wrong? Or else is this a mistake in the book?

Thanks!

 

[WimRFP]

Hello,

Delta(Vbe) = -2.1mV/K means that when you want to maintain the same diode current, you have to reduce Vbe with 2.1mV/K.

So if you needed, 0.6V at room temperature, you have to reduce to 0.6-(2.1m*8K) = 0.5832 V to compensate for the temperature rise of 8 degrees. However in your excersise the voltage remains (for example 0.6V). So when you increase Vbe from 0.5832 to 0.6V, the BE diode current almost doubles (factor e^(16.8m*40) = 1.96 ). The collector current will do the same (simple approximation).

So if the voltage drop across the collector resistor was 0.5*Vcc, now it becomes 0.5*1.96 = 0.98*Vcc. So the transistor is saturated as you have 0.02*Vcc CE voltage.

The reason for this behavior is that Is (diode saturation current) doubles every about 8..10 degrees temperature rise.