Differential Amplifier

Differential amplifier can amplify a small difference $\triangle v=v_1-v_2$ between two voltages $v_1$ and $v_2$. Differential amplification has many applications, such as the first stage of operational amplifiers (Op-amps).

Assume the two transistors $T_1$ and $T_2$ in the circuit are identical with the same properties, and their emitters are connected to a current source with constant current so that $I_E=I_1+I_2$, i.e., if $I_1$ increases, $I-2$ will decrease, and vice versa. Consider the following three cases:

• When the two input voltages are the same $v_1=v_2$, then $I_1=I_2=I_E/2$ and the output voltage is $V_{out}=V_{cc}-I_2 R_C$, which is treated as a reference level corresponding to $\Delta V=0$.
• If $v_1>v_2$, the following changes happen:
  
  $$v_1 \uparrow \Longrightarrow I_1 \uparrow \Longrightarrow I_2 \downarrow \Longrightarrow v_{out}=V_{cc}-I_2 R_c \uparrow$$
  
  

• If $v_1<v_2$, the following changes happen:
  
  $$v_1 \downarrow \Longrightarrow I_1 \downarrow \Longrightarrow I_2 \uparrow \Longrightarrow v_{out}=V_{cc}-I_2 R_c \downarrow$$
  
  

The output voltage $V_{out}$ can be further amplified to indicate the difference and its polarity between the two input voltages $v_1$ and $v_2$.

The current source shared by the two transistors is also shown in the figure. Due to the fact that the forward biased diodes have fixed voltage $V_D=0.7V$, the base voltage of the transistor is also fixed at $2.1V$, so is the current $I_{ce}$, i.e., the circuit can be used as a constant current source.