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In the following circuit:

enter image description here

We have

$$V_{S_1} = -V_T \ln\left(\frac{V_1}{RI_s}\right), V_{S_2} = -V_T \ln\left(\frac{V_2}{RI_s}\right)$$

then $$ \frac{V_{S_1}}{R_1} + \frac{V_{S_2}}{R_1} = I_s e^{\frac{-V_{S_3}}{V_t}} $$

I'm confused on how to calculate the input voltage for the middle op-amp.

Is it \$V_{S_3} = -(V_{S_2}+V_{S_1})\$?

I found that $$V_s =- \frac{V_1 V_2}{R I_s}$$

Is this valid?

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3 Answers 3

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You have a log amplifier, followed by a summer, followed by an antilog amplifier.

Since ln(A) + ln (B) = ln(AB), the configuration is a multiplier, giving you the product of the two inputs.

So, yes, your answer that gives a linearly scaled version of the product of the inputs, that is a reasonable answer.

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The Shockley diode equation, gives the relation between the voltage across and the current through a diode:

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\space\Longleftrightarrow\space\text{V}_\text{D}=\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{D}}{\text{I}_\text{S}}\right)\tag1$$

Where \$\text{I}_\text{D}\$ is the diode current, \$\text{I}_\text{S}\$ is the reverse bias saturation current, \$\text{V}_\text{D}\$ is the voltage across the diode, \$\text{q}\$ is the electron charge, \$\text{k}\$ is the Boltzmann constant, \$\text{T}\$ is the temperature and \$\eta\$ is the ideality factor.

So, we can see that for the two mos left OPAMP's, we get:

$$\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}}=\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\left(0-\text{V}_{\text{S}_1}\right)}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)\space\Longleftrightarrow\space\text{V}_{\text{S}_1}=-\frac{\displaystyle\eta\text{k}\text{T}}{\displaystyle\text{q}}\cdot\ln\left(1+\frac{\displaystyle\text{V}_1}{\displaystyle\text{R}\text{I}_\text{S}}\right)\tag2$$

For the third OPAMP, we get:

$$\frac{\displaystyle\text{V}_{\text{S}_1}-0}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_{\text{S}_2}-0}{\displaystyle\text{R}_1}=\frac{\displaystyle0-\text{V}_{\text{S}_3}}{\displaystyle\text{R}_1}\space\Longleftrightarrow\space\text{V}_{\text{S}_1}+\text{V}_{\text{S}_2}=-\text{V}_{\text{S}_3}\tag3$$

So, we get:

$$\text{V}_{\text{S}_3}=\frac{\displaystyle\eta\text{k}\text{T}}{\displaystyle\text{q}}\left(\ln\left(1+\frac{\displaystyle\text{V}_1}{\displaystyle\text{R}\text{I}_\text{S}}\right)+\ln\left(1+\frac{\displaystyle\text{V}_2}{\displaystyle\text{R}\text{I}_\text{S}}\right)\right)\tag4$$

So, when using the last OPAMP we end up with:

$$\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\left(\text{V}_{\text{S}_3}-0\right)}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)=\frac{\displaystyle0-\text{V}_\text{S}}{\displaystyle\text{R}}\space\Longleftrightarrow\space\text{V}_\text{S}=-\text{R}\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\text{V}_{\text{S}_3}}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)\tag5$$

So, we end up with:

$$\text{V}_\text{S}=-\text{R}\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}}{\displaystyle\eta\text{k}\text{T}}\left(\frac{\displaystyle\eta\text{k}\text{T}}{\displaystyle\text{q}}\left(\ln\left(1+\frac{\displaystyle\text{V}_1}{\displaystyle\text{R}\text{I}_\text{S}}\right)+\ln\left(1+\frac{\displaystyle\text{V}_2}{\displaystyle\text{R}\text{I}_\text{S}}\right)\right)\right)\right)-1\right)\tag6$$

Which, when simplified is:

$$\text{V}_\text{S}=\text{R}\text{I}_\text{S}-\left(1+\text{V}_1\right)\left(1+\frac{\displaystyle\text{V}_2}{\displaystyle\text{R}\text{I}_\text{S}}\right)\tag6$$

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  • \$\begingroup\$ the only mistake i think you made is that you didnt use the forward bias approximation of shockley equation \$\endgroup\$
    – HellBoy
    Oct 30, 2023 at 21:44
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There is an error.

In your second line equation you have assumed that the current through the second stage feedback resistor will be equal to the last stage diode current. This is an incorrect assumption because the last stage diode current is set by the diodes forward voltage drop (VS3) resulting in the second stage op amp's output sinking or sourcing some current.

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