Transfer function of a Gilbert cell multiplier

A Gilbert cell multiplier is a circuit that takes two inputs as differential currents and produces a differential current that is proportional to the product of the two inputs

Let X and Y be arbitrary values between 1 and -1. Let I_x and I_y be arbitrary non-zero currents. The combined emitter current of the left differential pair, \(I_{le}=I_y\left(1+Y\right)\), and the combined emitter current of the right differential pair, \(I_{re}=I_y\left(1-Y\right)\). Let V₁ be the collector voltage of the leftmost transistor and V₂ be the collector voltage of the rightmost transistor. Let I_s be the saturation current of the base-emitter junction. Let V_t be the thermal voltage, about 0.025865 volts.

Part 1

Consider the leftmost transistor. Note that its base and collector are connected together. Recall that the base current is \(I_s\left(\frac{e^{V_1/V_t}-1}{\beta}\right)\). Therefore the total current flowing through through the transistor is \(I_s\left(\frac{\beta+1}{\beta}\left(e^{V_1/V_t}-1\right)\right)\)

Part 2

\(I_{le}=I_y\left(1+Y\right)\) and \(I_{re}=I_y\left(1-Y\right)\), therefore: $$ I_{le} + I_{re} = I_y\left(1+Y\right) + I_y\left(1-Y\right) $$ $$ I_{le} + I_{re} = I_y\left(1+Y+1-Y\right) $$ $$ I_{le} + I_{re} = 2I_y $$

Part 3

Consider the left differential pair. The combined emitter current is \(I_{le}\). Let \(I_{llc}\) be the collector current of the left side, and \(I_{lrc}\) be the collector current of the right side. $$ I_{llc} - I_{lrc} = \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{le}\beta}{\beta+1}\right) $$ Since \(I_{llc} + I_{lrc} = \frac{I_{le}\beta}{\beta+1}\), \(I_{llc} = \frac{I_{le}\beta}{2\left(\beta+1\right)} + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{le}\beta}{2\left(\beta+1\right)}\right)\) and \(I_{lrc} = \frac{I_{le}\beta}{2\left(\beta+1\right)} - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{le}\beta}{2\left(\beta+1\right)}\right)\). The base current is equal to the collector current divided by the gain. Therefore: $$ I_{llb} = \frac{I_{le}}{2\left(\beta+1\right)} + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{le}}{2\left(\beta+1\right)}\right) = \frac{I_{le}}{2\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right)$$ $$ I_{lrb} = \frac{I_{le}}{2\left(\beta+1\right)} - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{le}}{2\left(\beta+1\right)}\right) = \frac{I_{le}}{2\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right)$$ $$ I_{rlb} = \frac{I_{re}}{2\left(\beta+1\right)} + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{re}}{2\left(\beta+1\right)}\right) = \frac{I_{re}}{2\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ $$ I_{rrb} = \frac{I_{re}}{2\left(\beta+1\right)} - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_{re}}{2\left(\beta+1\right)}\right) = \frac{I_{re}}{2\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ By KCL, $$ I_x\left(1+X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + I_{llb} + I_{rlb} $$ $$ I_x\left(1+X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_{le}}{2\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) + \frac{I_{re}}{2\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right)$$ $$ I_x\left(1+X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_{le} + I_{re}}{2\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ $$ I_x\left(1+X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_y}{\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ Similarly, for the other side: $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + I_{lrb} + I_{rrb} $$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_{le}}{2\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) + \frac{I_{re}}{2\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right)$$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_{le} + I_{re}}{2\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) + \frac{I_y}{\left(\beta+1\right)}\left(1 - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ Now, this equation for \(I_x\left(1+X\right)\) is solved for \(V_2\) $$ I_x\left(1+X\right) - \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) = \frac{I_y}{\left(\beta+1\right)}\left(1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right)\right) $$ $$ \frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right) = 1 + \tanh\left(\frac{V_1-V_2}{2V_t}\right) $$ $$ \frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1 = \tanh\left(\frac{V_1-V_2}{2V_t}\right) $$ (this result will be important later) $$ V_2 = V_1-2V_t\operatorname{artanh}\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\right) $$ Now we can use this to solve for \(V_1\): $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_2/V_t}-1\right) + \left(1-\tanh\left(\frac{V_1-V_2}{2V_t}\right)\right)\left(\frac{I_y}{\beta+1}\right) $$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_2/V_t}-1\right) + \left(1-\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\right)\right)\left(\frac{I_y}{\beta+1}\right) $$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_2/V_t}-1\right) + \left(2-\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right) \left(\frac{I_y}{\beta+1}\right) $$ $$ I_x\left(1-X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_2/V_t}-1\right) + \frac{2I_y}{\beta+1} - I_x\left(1+x\right) + \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right) $$ $$ I_x\left(1-X\right) + I_x\left(1+X\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_2/V_t} + e^{V_1/V_t} - 2\right) + \frac{2I_y}{\beta+1} $$ $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_2/V_t\right) + \exp\left(V_1/V_t\right) - 2\right) + \frac{2I_y}{\beta+1} $$ $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t - 2\operatorname{artanh}\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\right)\right) + \exp\left(V_1/V_t\right) - 2\right) + \frac{2I_y}{\beta+1} $$ $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right)\exp\left( - 2\operatorname{artanh}\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\right)\right) + \exp\left(V_1/V_t\right) - 2\right) + \frac{2I_y}{\beta+1} $$ Let \(u = \frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\) $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right)\exp\left( - 2\operatorname{artanh}\left(u\right)\right) + \exp\left(V_1/V_t\right) - 2\right) + \frac{2I_y}{\beta+1} $$ \(\operatorname{artanh}\left(x\right)\) is defined to be \(\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\). Thus \(e^{-2\operatorname{artanh}\left(x\right)} = \frac{1-x}{1+x}\) $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) \left(\frac{1-u}{1+u}\right) + \exp\left(V_1/V_t\right) - 2\right) + \frac{2I_y}{\beta+1} $$ $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) \left(\frac{1-u}{1+u}+1\right) - 2\right) + \frac{2I_y}{\beta+1} $$ $$ 2I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) \left(\frac{2}{1+u}\right) - 2\right) + \frac{2I_y}{\beta+1} $$ $$ I_x = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) \left(\frac{1}{1+u}\right) - 1\right) + \frac{I_y}{\beta+1} $$ $$ I_x\left(1+u\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) - \left(1+u\right)\right) + \frac{I_y\left(1+u\right)}{\beta+1} $$ Expand u: $$ I_x\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) - \left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right)\right) + \frac{I_y\left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right)}{\beta+1} $$ $$ \left(\frac{I_x\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(\exp\left(V_1/V_t\right) - \left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)\right)\right) + \left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right) $$ $$ \frac{\left(\beta+1\right)I_x^2\left(1+X\right)}{I_y} - \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta}\left(e^{V_1/V_t}-1\right) = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}\right) - \frac{I_x I_s\left(\beta+1\right)^2\left(1+X\right)}{I_y \beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2}\left(e^{V_1/V_t}-1\right) + I_x\left(1+X\right) - \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t} - 1\right) $$ $$ \frac{\left(\beta+1\right)I_x^2\left(1+X\right)}{I_y} - \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta}\left(e^{V_1/V_t}\right) + \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} = \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}\right) - \frac{I_x I_s\left(\beta+1\right)^2\left(1+X\right)}{I_y \beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2}\left(e^{V_1/V_t}\right) - \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} + I_x\left(1+X\right) - \frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}\right) + \frac{I_s\left(\beta+1\right)}{\beta} $$ $$ \frac{\left(\beta+1\right)I_x^2\left(1+X\right)}{I_y} + \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} - \frac{I_s\left(\beta+1\right)}{\beta} - I_x\left(1+X\right) + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} + \frac{I_x I_s\left(\beta+1\right)^2\left(1+X\right)}{I_y \beta} = \left(e^{V_1/V_t}\right)\left( \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} + \frac{I_s\left(\beta+1\right)}{\beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} - \frac{I_s\left(\beta+1\right)}{\beta} \right) $$ $$ I_x\left(1+X\right)\left( \frac{I_x\left(\beta+1\right)}{I_y} + \frac{I_s\left(\beta+1\right)^2}{I_y \beta} - 1 \right) + \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} - \frac{I_s\left(\beta+1\right)}{\beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} = \left(e^{V_1/V_t}\right)\left( \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} \right) $$ $$ \frac{ I_x\left(1+X\right)\left( \frac{I_x\left(\beta+1\right)}{I_y} + \frac{I_s\left(\beta+1\right)^2}{I_y \beta} - 1 \right) + \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} - \frac{I_s\left(\beta+1\right)}{\beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} }{ \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} } = e^{V_1/V_t} $$ $$ \frac{ I_x\left(1+X\right)\left( \frac{I_x\left(\beta+1\right)}{I_y} + \frac{I_s\left(\beta+1\right)^2}{I_y \beta} - 1 \right) - \frac{I_s\left(\beta+1\right)}{\beta} }{ \frac{I_x I_s\left(\beta+1\right)^2}{I_y \beta} + \frac{I_s^2\left(\beta+1\right)^3}{I_y\beta ^2} } = e^{V_1/V_t} - 1 $$ $$ \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) - I_y I_s\left(\beta+1\right) }{ I_x I_s\left(\beta+1\right)^2 + \frac{I_s^2\left(\beta+1\right)^3}{\beta} } = e^{V_1/V_t} - 1 $$

Part 4

Find the output voltage given V₁, V₂, and I_y Note that the two differential pairs are wired in reverse parallel. Therefore, the currents subtract. $$ I_{out} = \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_y\left(1+Y\right)\beta}{\beta+1}\right) - \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_y\left(1-Y\right)\beta}{\beta+1}\right) $$ $$ I_{out} = \tanh\left(\frac{V_1-V_2}{2V_t}\right)\left(\frac{I_y\beta}{\beta+1}\right)\left(2Y\right) $$ $$ I_{out} = \left(\frac{\left(\beta+1\right)}{I_y}\left(I_x\left(1+X\right)-\frac{I_s\left(\beta+1\right)}{\beta}\left(e^{V_1/V_t}-1\right)\right)-1\right) \left(\frac{I_y\beta}{\beta+1}\right)\left(2Y\right) $$ $$ I_{out} = 2Y\left( \left(\beta I_x\left(1+X\right)-I_s\left(\beta+1\right)\left(e^{V_1/V_t}-1\right)\right) - \frac{I_y\beta}{\beta+1} \right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-I_s\left(\beta+1\right)\left( \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) - I_y I_s\left(\beta+1\right) }{ I_x I_s\left(\beta+1\right)^2 + \frac{I_s^2\left(\beta+1\right)^3}{\beta} } \right) - \frac{I_y\beta}{\beta+1}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-\left( \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) - I_y I_s\left(\beta+1\right) }{ I_x\left(\beta+1\right) + \frac{I_s\left(\beta+1\right)^2}{\beta} } \right) - \frac{I_y\beta}{\beta+1}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-\left( \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) - I_y I_s\left(\beta+1\right) }{\left(\beta+1\right)\left( I_x + \frac{I_s\left(\beta+1\right)}{\beta} \right)} \right) - \frac{I_y\beta\left(I_x + \frac{I_s\left(\beta+1\right)}{\beta}\right)}{\left(\beta+1\right)\left(I_x + \frac{I_s\left(\beta+1\right)}{\beta}\right)}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-\left( \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) - I_y I_s\left(\beta+1\right) }{\left(\beta+1\right)\left( I_x + \frac{I_s\left(\beta+1\right)}{\beta} \right)} \right) - \frac{I_x I_y\beta + I_y I_s\left(\beta+1\right)}{\left(\beta+1\right)\left(I_x + \frac{I_s\left(\beta+1\right)}{\beta}\right)}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-\left( \frac{ I_x\left(1+X\right)\left( I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta \right) }{\left(\beta+1\right)\left( I_x + \frac{I_s\left(\beta+1\right)}{\beta} \right)} \right) - \frac{I_x I_y\beta}{\left(\beta+1\right)\left(I_x + \frac{I_s\left(\beta+1\right)}{\beta}\right)}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)-\beta I_x\left(1+X\right)\left( \frac{ I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta }{\left(\beta+1\right)\left( \beta I_x + I_s\left(\beta+1\right) \right)} \right) - \frac{I_x I_y\beta ^2}{\left(\beta+1\right)\left(I_x\beta + I_s\left(\beta+1\right)\right)}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)\left(1- \frac{ I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 - I_y\beta }{ I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 } \right) - \frac{I_x I_y\beta ^2}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right) $$ $$ I_{out} = 2Y\left(\beta I_x\left(1+X\right)\left( \frac{ I_y\beta }{ I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2 } \right) - \frac{I_x I_y\beta ^2}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right) $$ $$ I_{out} = 2Y\left(\beta I_x X\left(\frac{I_y\beta}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right) + \left(\frac{I_x I_y\beta^2}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right) - \frac{I_x I_y\beta ^2}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right) $$ $$ I_{out} = 2Y\left(\beta I_x X\left(\frac{I_y\beta}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2}\right)\right) $$ $$ I_{out} = \frac{2XYI_xI_y\beta^2}{I_x\beta\left(\beta+1\right) + I_s\left(\beta+1\right)^2} $$ For transistors with a large gain, \(\beta\left(\beta+1\right) \approx \left(\beta+1\right)^2\), and if \(I_x \gg I_s\), then the second term in the denominator can be ignored with little effect $$ I_{out} = \frac{2XYI_xI_y\beta^2}{I_x\beta\left(\beta+1\right)} = \frac{2XYI_y\beta}{\beta+1} $$ As one can see in the above equation, the output current is proportional to the product of the input values X and Y.