# Differential Amplifier

A differential amplifier is a type of electronic amplifier that amplifies the difference between two input voltages only. The differential amplifier two inputs are the inverting input and non-inverting input. Its output signal is 180° out of phase with inverting input signal and in phase with non-inverting input signal. It is used in an Operational Amplifier (OPAMP).

The different modes of operation of the differential amplifier are:

- Common mode input signal v
_{CM}. Both inputs are connected together to the input signal.
- Large differential input signal.
- Small differential input signal v
_{i}<2v_{T}

The circuit below shows the differential amplifier with a small differential input signal v_{i}=50mVpp @ 1KHz.

Notes:

- Click on the colored nodes V1, VE, VC1, VC2 in the schematic to show/hide that node waveform. To zoom in to the smaller input and emitter (VE) voltages, hide both larger output voltages.
- Select the different R1 and R2,R3 values, press Simulate and observe the change in gain and DC biasing voltages with the component values.
- Move the cursor in the chart to obtain the waveform values to calculate the gains for the inverting and non-inverting outputs. Compare with the calculated values using the formulae derived below.

## DC Analysis

To perform the DC analysis, we need to redraw the schematic by replacing the transistors with its BJT DC model.

The voltage across R1, V_{R1} can be calculated using KVL

\begin{equation}
V_1 = V_{R1} + V_{BE}
\end{equation}

Ignoring I_{B}, assuming circuit symmetry and using KCL, the DC collector current I_{C} is half the current through R1.

\begin{equation}
I_C = I_E = {V_1 - V_{BE} \over 2 R_1}
\end{equation}

The DC collector voltage VC1 (wrt GND and not to VE) can be calculated using KVL

\begin{equation}
V_{CC} = V_{C1} + I_C R_2
\end{equation}

## AC Analysis

Similarly for the AC analysis, we need to redraw the schematic by replacing the transistors with its small signal model.

For the AC analysis, calculate r_{e} using I_{E} from the DC analysis and v_{T}=26 mV. Note that the AC analysis is only valid for v_{be} < v_{T}

\begin{equation}
r_e = {v_{T} \over I_E}
\end{equation}

To simplify the calculation of the amplifier gain, we assume the following

- R1 which is parallel to re can be ignored if
\begin{equation}
R_1 > 10r_e
\end{equation}
- The current ic1 = i
_{re} = -ic2 (KCL)

\begin{equation}
v_1 = i_{re} {2r_e}
\end{equation}

The output voltages at VC1 (inverting) and VC2 (non-inverting) are

\begin{align}
v_{C1} &= {- {R_2 \over {2r_e}}v_1 } \\
v_{C2} &= {R_3 \over {2r_e}}v_1
\end{align}

Finally, the voltage at v_{e} is a simple voltage divider of 2 r_{e}

\begin{equation}
v_e = {v_1 \over 2}
\end{equation}