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## Summary:

- log-antilog opamp question

## Main Question or Discussion Point

Firstly, this is not a homework question. I found a worksheet online with an example of a square law circuit built using log-antilog operational amplifiers. I tried to derive the transfer function but I can't seem to eliminate the reverse saturation current term ##I_S##. I would really appreciate some help with this problem.

I have included the schematic below:

Here is my attempt:

Assuming all resistors are equal value, all transistors are perfectly matched and at equal temperature and all op-amps are ideal.

Output of first stage:

##V_{O1} = - V_T ln(\frac{V_{IN}}{R I_S})##

Output of second stage:

##V_{O2}= 2V_{O1}##

Output of third stage:

##V_{O3} = - RI_Se^{\frac{V_{O2}}{V_T}}##

Sub 1 into 2:

##V_{O2} = -2V_Tln(\frac{V_{IN}}{RI_S}) = - V_Tln(\frac{V_{IN}^2}{R^2I_S^2})##

Sub into 3:

##V_{O3} = - RI_Se^{-ln(\frac{V_{IN}^2}{R^2I_S^2})} = -RI_S\frac{R^2I_S^2}{V_{IN}^2} = - \frac{R^3I_S^3}{V_{IN}^2}##

I have included the schematic below:

Here is my attempt:

Assuming all resistors are equal value, all transistors are perfectly matched and at equal temperature and all op-amps are ideal.

Output of first stage:

##V_{O1} = - V_T ln(\frac{V_{IN}}{R I_S})##

Output of second stage:

##V_{O2}= 2V_{O1}##

Output of third stage:

##V_{O3} = - RI_Se^{\frac{V_{O2}}{V_T}}##

Sub 1 into 2:

##V_{O2} = -2V_Tln(\frac{V_{IN}}{RI_S}) = - V_Tln(\frac{V_{IN}^2}{R^2I_S^2})##

Sub into 3:

##V_{O3} = - RI_Se^{-ln(\frac{V_{IN}^2}{R^2I_S^2})} = -RI_S\frac{R^2I_S^2}{V_{IN}^2} = - \frac{R^3I_S^3}{V_{IN}^2}##