In The Art of Electronics 2nd Ed. it is said that:
The Ebers-Moll equation is $$I \approx I_s(T) e^{\frac{V}{25 \, mV}}$$ and the change of voltage with temperature at constant current is
$$\left. \frac{d V}{d T} \right|_I = -2.1 \frac{mV}{K}$$
The way the authors calculate relative changes in the current is
$$\frac{\Delta I}{I} = e^{\frac{- \left. \frac{d V}{d T} \right|_I \Delta T}{25 \, mV}}$$
which indeed they use to state that a for \$\Delta T = 30 \, K\$ you get a factor of \$\approx 10\$ increase in the current.
Is there a valid reason for this or just a mistake in the book?
At first this does not make sense to me, since they are using a coefficient that was found assuming constant current in order to calculate the change in current!
EDIT:
In short, from the definition of \$\left. \frac{d V}{d T} \right|_I\$ I think the right calculation is
$$ \frac{I_s(T + \Delta T)}{I_s(
T)} e^{\frac{\left. \frac{d V}{d T} \right|_I \Delta T}{25 \, mV}} – 1 = \frac{\Delta I}{I} = 0$$
This would mean that they are actually calculating the change in saturation current:
$$ \frac{I_s(T + \Delta T)}{I_s(
T)} = e^{\frac{- \left. \frac{d V}{d T} \right|_I \Delta T}{25 \, mV}} $$
Right?
Best Answer
Just for your understanding: The base emitter voltage does NOT decrease if the temperature (and with it the collector current) are increasing. The sequence is as follows: For a constant VBE voltage the collector current Ic increases for higher temperatures of the transistor body (increased carrier mobility). And the bias voltage VBE must be externally reduced (2mV/K) in order to bring Ic back to the former value (that is the background saying Ic=constant). This VBE reduction should be done automatically by applying voltage feedback (emitter resistorRE).