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Also known as a mixer, although those usually have inputs that can handle negatives.
The left two OP-amps take the logarithm of their inputs. The average of those two logarithms are found with a voltage divider and then multiplied by two with the middle OP-amp. This gives the sum of the two logarithms. That then goes to the right OP-amp which takes the exponent of its input. This creates the equation b^(log_b(x) + log_b(y)), which simplifies to x*y for values of x and y greater than 0 and values of b greater than 0 and not equal to 1.
For the b in the equation, that’s the base. For the base I chose 2, which requires the special value of 55.48827 (rounds to 55.5) as the emission coefficient of the diodes. For the saturation current I chose 10μA as that was the highest it goes. 10μA simply means that the correct resistance for most of the resistors was 100kΩ so that 1V would create 10μA. Also there are some 10μA current sources which make it so that with infinite negative voltage there will be no current and with no voltage there will be 10μA of current, which is the expected behavior for exponentials.
To find the correct value of the emission coefficient to achieve a given b, use the equation 1/(0.026*ln(b)), where ln is the natural logarithm. 0.026 is the value of the thermal coefficient that EC uses and corresponds to 28.57 degrees Celsius. To find the b given the emission coefficient, use the equation e^(1/(0.026*c)), where e is Euler’s number (2.71828182846) and c is the emission coefficient.
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