# Log converter circuit

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.

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.