1
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ I'm not understanding your question. The Ebers-Moll equation does a good job in predicting Ic for Vbe. And indeed an increase in the collector current by a factor of 10 (at room temp.) gives an increase in the Vbe voltage of about 60 mV. \$\endgroup\$ – George Herold Sep 2 '15 at 20:05
  • \$\begingroup\$ I am too lazy to dig in to your question. Please check here to see if it is a known erratum: artofelectronics.net/errata \$\endgroup\$ – mkeith Sep 2 '15 at 21:15
  • \$\begingroup\$ @mkeith Thanks for the errata page. I checked it and this issue is not there. \$\endgroup\$ – Rol Sep 3 '15 at 5:47
  • \$\begingroup\$ @GeorgeHerold I was looking for the change in collector current with temperature at constant Vbe \$\endgroup\$ – Rol Sep 6 '15 at 17:06
2
\$\begingroup\$

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).

|improve this answer|||||
\$\endgroup\$
1
\$\begingroup\$

In these formulas it assumed that the voltage across the PN junction is set externally and that it also decreases with -2.1 mV/K. If you would do that indeed the current would change significantly. In practice there would need to be some mechanism in place to control the current. A series resistor will do the trick, it will limit the current.

But then you say: but then V will be lower !

And that is correct, only the V to I relation is exponential so even though I can change a lot, V will not change much except for the -2.1 V/K.

Another complication is that Is(T) is VERY temperature dependent.

Also have a look at this question

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ That 2.1V/K coefficient only applies at a particular bias point. Just FYI. If you haven't seen it before, check out the Bob Pease article: "What's all this Vbe stuff, anyhow?" I linked to it in my comment above. \$\endgroup\$ – mkeith Sep 2 '15 at 21:45
  • \$\begingroup\$ I cannot follow your calculation. I get e^(700/25)/10000 = 1.44 * 10^8 \$\endgroup\$ – Rol Sep 3 '15 at 5:44
  • \$\begingroup\$ Oops, I made a mistake there ! I completely changed my answer :-) \$\endgroup\$ – Bimpelrekkie Sep 3 '15 at 6:15
0
\$\begingroup\$

I was reading some thermodynamics when I saw something that can answer my question (?!)

It holds

$$ \left. \frac{d I}{d T} \right|_V = - \left. \frac{d V}{d T} \right|_I \left. \frac{d I}{d V} \right|_T $$

The second factor can be easily calculated from the Ebers-Moll equation.

I get: $$ \left. \frac{d I}{d T} \right|_V \approx \frac{I}{12~K}$$

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.