Differential Amplifiers

Op-amps are differential amplifiers, designed to amplify the difference between two inputs. In an op-amp, the gain $A$ is usually very large, approaching $\approx 10^6$ in practice, and is modelled as infinite.

$$V_{out}=A(V_1-V_2)$$

Op-amps are based on a circuit known as a long-tailed pair:

• $Q_1$ and $Q_2$ are a matched pair of transistors, meaning they have the exact same electrical properties (
  • ${\beta }_{Q_1}={\beta }_{Q_2}$
• The quiescent current $I_Q$ is the current through the shared emitter resistor, $R_{EE}$
• $I_Q=I_{E_1}+I_{E_2}$
• If $\beta$ is large, then $I_C=I_E$

Biasing

When doing bias calculations, the two inputs $V_1$ and $V_2$ are assumed to be grounded, $V_B=0$. As $V_{BE}=0.7$, $V_E=-0.7$. $V_{EE}$, the tail voltage, is always taken as negative. Using this, we can calculate $I_Q$:

$$I_Q=\frac{-0.7-(-V_{EE})}{R_{EE}}=\frac{V_{EE}-0.7}{R_{EE}}=2I_E$$

$R_{EE}$ sets the quiscent collector current.

When the two inputs are grounded, the output at the collectors $V_{C_1}$ and $V_{C_2}$ are the same.

$$V_{C_1}=V_{C_2}=V_{CC}-I_C{R}_C$$

For matched transistors, $V_{C_1}-V_{C_2}=0$.

AC Analysis

The long-tailed pair can operate in two modes, depending upon how input is applied

• Differential mode amplifies the difference between the two input signals
• Common mode works similar to a regular BJT amplifier
• Better amplifiers have a high ratio of differential to common gain, called the Common Mode Rejection Ratio (CMRR)

Differential Mode

The circuit below shows two AC sources connected, $V_d/2$ and $-V_d/2$, to give a differential input signal of $V_d$.

The differential output is the difference between the two outputs:

$$V_{o1}-V_{o2}=V_{od}$$

And the differential mode gain:

$$A_{id}=\left|\frac{V_{od}}{V_d}\right|=g_m{R}_C$$

The way this circuit is usually used, however, is with one output referenced to ground:

This gives a single-ended output, with a gain of:

$$\left|\frac{V_o}{V_d}\right|=\frac{g_m{R}_C}{2}$$

The input and output resistances for differential mode inputs are:

$$R_{id}=2r_{\pi }$$ $$R_{od}=2R_E$$

Common Mode

Common mode input is when the same signal is connected to both input terminals, $V_1=V_2=V_{cm}$. An ideal differential amplifier would reject common mode input, but this is often not the case. The performance of a differential amplifier is defined by it's CMRR, which would ideally be infinite, but is usually just very large in practice.

$$A_{cm}=\left|\frac{V_{o1}}{V_{cm}}\right|=\frac{-\beta R_C}{r_{\pi }+2(1+\beta )R_{EE}}\approx \frac{-R_C}{2R_{E}E}$$

The common mode input resistance:

$$R_{ic}=\frac{r_{\pi }}{2}+R_{EE}(1+\beta )$$

The generalised output of a differential amplifier, factoring in both common mode and differential mode input signals is:

$$V_{out}=A_{id}{V}_{id}+A_{cm}{V}_{cm}$$

Example

Find $R_{EE}$ to give $I_Q=1mA$, and $R_C$ for max AC swing, when $V_{CC}=15V$.

$$I_Q=\frac{15.7-0.7}{1m}=15k\Omega$$

For max AC swing, $R_C\approx 7.5V$:

$$R_C=\frac{7.5}{0.5mA}=15k\Omega$$

Using $\beta =100$, calculate the differential and common mode gains, and the CMRR of the circuit.

$$r_{\pi }=\frac{\beta V_T}{I_C}=\frac{100\times 25}{0.5}=5k\Omega$$

$$A_{id}=\frac{g_m{R}_C}{2}=\frac{\beta R_C}{2r_{\pi }}=\frac{100\times 15k}{2\times 5k}=150$$

$$A_{cm}=\frac{-\beta R_c}{r_{\pi }+2(1+\beta )R_{EE}}=\frac{-100\times 15k}{5k+202\times 15k}=-0.49$$

$$CMRR=\left|\frac{A_{id}}{A_{cm}}\right|=\frac{150}{0.49}=303.5=49.6dB$$

The common mode rejection ratio for this circuit is fairly low, because $R_{EE}$ is low. $A_{cm}\to \infty$ as $R_{EE}\to \infty$, so ideally $R_{EE}$ is as large as possible. Replacing it with an ideal current source with infinite resistance can acheive this.