Log converter circuit

Question

I am reading Arts of Electronics and they show these 2 circuits (p. 38):

1. Why \$V_{in}\$ has to be much larger than 0.6V?

2. How does the diode drop compensation work exactly? (I am looking for some really simple explanations, as I am new to electronics.)

Thanks.

Answer 1

It is not true that the forward voltage must be near 0.6 for exponential behavior (current vs voltage) to occur. Particularly at low voltages, where self-heating and intrinsic resistance is not a factor, diodes generally do behave exponentially. Unfortunately, at these low current levels changes in ambient temperature become significant sources of error, and the non-linearities inherent in exponential behavior also cause problems. The standard Schockley Equation is $$I = I_s (e^\frac{qV}{kT}-1) $$ or$$V = \frac{kT}{q}ln(\frac{I}{I_s}+1) $$ Note that, if it weren't for that pesky 1 on the right, the equation would become $$V = \frac{kT}{q}ln(\frac{I}{I_s}) = \frac{kT}{q}lnI - \frac{kT}{q}lnI_s$$ and the second term would be a constant, making the output linear with respect to lnI.

Alas, this isn't so. If you plot the equation, you'll see that inclusion of that value of 1 causes the output to be non-linear. In principle the effect can be compensated for, but it's generally not worth it. When you combine the errors in both high and low input voltages, the sweet spot is typically about 3 orders of magnitude wide, and typically occurs with a forward Vf of about 0.6 for silicon.

The second circuit you've provided produces an offset of about 0.6 volts, so it operates reasonably well, or at least as well as Art of Electronics needs to. You should note R1 and -V combine to set the value of Vin which produces a zero output. Specifically, for large Vin, if the voltage drops on the diodes are matched, both diodes will have the same current, and the current through D2 will be approximately (-V +0.6)/2R).