Transfer function of a BJT differential pair

Suppose there are a pair of transistors with their emitters connected together to a current source of magnitude I. Let \(I_a\) and \(I_b\) be the collector currents, \(I_{ab}\) and \(I_{bb}\) be the base currents, and \(V_e\) be the emitter voltage. Let \(V_a\) and \(V_b\) be the base voltages of the transistors. The following equation represents the base current: $$ I_{bb} = I_s\left(\frac{e^{V_{BE}/V_t}-1}{\beta}+\frac{e^{V_{BC}/V_t}-1}{\beta_r}\right) $$ where \(I_s\) is the saturation current, \(\beta\) is the current gain of the transistor, \(\beta_r\) is the reverse gain, and \(V_t\) is the thermal voltage (about 0.025865 volts)
If the collectors are pulled high and \(V_{BC}\) is negative, then we can ignore the second term because the B-C junction is reverse biased. $$ I_{bb} = I_s\left(\frac{e^{\left(V_b-V_e\right)/V_t}-1}{\beta}\right) $$ The following equation represents the collector current $$ I_b = I_s\left(e^{V_{BE}/V_t}-1\right) - I_s\left(e^{V_{BC}/V_t}-1\right)\left(\frac{\beta_r+1}{\beta_r}\right)$$ Similarly, the second term can be ignored here as well $$ I_b = I_s\left(e^{\left(V_b-V_e\right)/V_t}-1\right) $$ Now solve for \(V_e\) and substitute: $$ \frac{I_b}{I_s} + 1 = e^{\left(V_b-V_e\right)/V_t} $$ $$ V_t\ln\left(\frac{I_b}{I_s}+1\right) = V_b-V_e $$ $$ V_b - V_t\ln\left(\frac{I_b}{I_s}+1\right) = V_e $$ $$ I_a = I_s\left(e^{\left(V_a-V_e\right)/V_t}-1\right) $$ $$ I_a = I_s\left(e^{\left(V_a-\left(V_b - V_t\ln\left(\frac{I_b}{I_s}+1\right)\right)\right)/V_t}-1\right) $$ $$ I_a = I_s\left(e^{\left(V_a-V_b\right)/V_t+\ln\left(\frac{I_b}{I_s}+1\right)}-1\right) $$ $$ I_a = I_s\left(\left(\frac{I_b}{I_s}+1\right)e^{\left(V_a-V_b\right)/V_t}-1\right) $$ $$ I_a = \left(I_b+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s $$ For any transistor, the emitter current is equal to the sum of the base and collector currents: $$ I_e = I_c + I_b = I_c + \frac{I_c}{\beta} = I_c\left(\frac{\beta+1}{\beta}\right) $$ Therefore: $$ I = \left(I_a+I_b\right)\left(\frac{\beta+1}{\beta}\right) $$ $$ \frac{I\beta}{\beta+1}-I_a=I_b $$ Substitute into a previous equation: $$ I_a = \left(\frac{I\beta}{\beta+1}-I_a+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s $$ $$ I_a + I_a e^{\left(V_a-V_b\right)/V_t} = \left(\frac{I\beta}{\beta+1}+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s $$ $$ I_a = \frac{\left(\frac{I\beta}{\beta+1}+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s}{e^{\left(V_a-V_b\right)/V_t}+1} $$ Given \(I_a+I_b = \frac{I\beta}{\beta+1}\), we can express \(I_a-I_b\) as follows: $$ I_a-I_b = 2I_a - \frac{I\beta}{\beta+1} $$ $$ I_a-I_b =2\frac{\left(\frac{I\beta}{\beta+1}+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s}{e^{\left(V_a-V_b\right)/V_t}+1} - \frac{I\beta}{\beta+1} $$ $$ I_a-I_b =2\frac{\left(\left(\frac{I\beta}{\beta+1}+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s\right)\left(\beta+1\right)} {\left(e^{\left(V_a-V_b\right)/V_t}+1\right)\left(\beta+1\right)} - \frac{I\beta\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ $$ I_a-I_b = \frac{2\left(\left(\frac{I\beta}{\beta+1}+I_s\right)e^{\left(V_a-V_b\right)/V_t}-I_s\right)\left(\beta+1\right) - I\beta\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ $$ I_a-I_b = \frac{2\left(I\beta+I_s\left(\beta+1\right)\right)e^{\left(V_a-V_b\right)/V_t}-2I_s\left(\beta+1\right) - I\beta e^{\left(V_a-V_b\right)/V_t} - I\beta} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ $$ I_a-I_b = \frac{\left(I\beta+2I_s\left(\beta+1\right)\right)e^{\left(V_a-V_b\right)/V_t} - 2I_s\left(\beta+1\right) - I\beta} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ $$ I_a-I_b = \frac{\left(I\beta+2I_s\left(\beta+1\right)\right)e^{\left(V_a-V_b\right)/V_t} - \left(I\beta + 2I_s\left(\beta+1\right)\right)} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ $$ I_a-I_b = \frac{\left(I\beta+2I_s\left(\beta+1\right)\right)\left(e^{\left(V_a-V_b\right)/V_t} - 1\right)} {\left(\beta+1\right)\left(e^{\left(V_a-V_b\right)/V_t}+1\right)} $$ By definition, \(\tanh\left(x\right) = \frac{e^{2x}-1}{e^{2x}+1}\). Therefore: $$ I_a-I_b = \tanh\left(\frac{V_a-V_b}{2V_t}\right)\left(\frac{I\beta+2I_s\left(\beta+1\right)}{\beta+1}\right) $$ If \(I \gg I_s\), then we can remove the terms with \(I_s\) with little effect: $$ I_a-I_b = \tanh\left(\frac{V_a-V_b}{2V_t}\right)\left(\frac{I\beta}{\beta+1}\right) $$

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