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)\l