BJT Amplifiers

• BJTs make excellent amplifiers when biased in the forward-active region
• Transistors can provide high voltage, current and power gain
• DC biasing stabilises the operating point
• DC Q-point determines
  • Small-signal parameters
  • Voltage gain
  • Input & output impedances
  • Power consumption
• DC analysis finds the Q-point
• AC analysis with the small-signal model is used to analyse the amplifier

Hybrid-Pi Model

The hybrid-pi small signal model is what is used for hand analysis of BJTs:

• Intrinsic low-frequency representation of a BJT
  • Does not work for RF stuff
• Ignoring output impedance assumes $V_A$ is large
• Parameters are controlled by the Q-point
• Transconductance $g_m=\frac{I_{CQ}}{V_T}\approxeq 40I_{CQ}$
  • Thermal voltage $V_T=25mV$
• Input resistance $r_{\pi }=\frac{\beta V_T}{I_{CQ}}=\frac{\beta }{g_m}$

For AC analysis, coupling capacitors are replaced by short circuits, and DC voltages replaced by short circuits to ground. The circuit below shows a 4-resistor bias amplifier replaced by it's small signal model.

AC Analysis

The impdance at the base input $R_{ib}$:

$$R_{ib}=\frac{V_{in}}{i_b}=\frac{i_b{r}_{\pi }+R_E(1+\beta )i_b}{i_b}=r_{\pi }+R_E(1+\beta )$$

The impedance at the emitter is reflected back to the base, multiplied by $(1+\beta )$. This makes the overall input impedance of the amplifier:

$$\begin{gathered} R_i=R_{ib}||R_1||R_2 \\ {R}_i=(r_{\pi }+R_E(1+\beta ))||R_{TH} \end{gathered}$$

The output impedance is easy, as lookong into the collector, we can see $R_C$ in parallel with a current source which has infinite impedance, so:

$$R_O=R_C$$

The voltage accross $R_C$ is the output voltage:

$$V_o=-\beta i_b{R}_C$$

The voltage accross $r_{\pi }+R_E$ is the input voltage:

$$V_{in}=i_b(r_{\pi }+R_E(1+\beta ))$$

The overall voltage gain is therefore:

$$A_v=\frac{V_o}{V_{in}}=\frac{-\beta R_C}{r_{\pi }+R_E(1+\beta )}$$

Note that the gain is negative meaning this is an inverting amplifier. If we make the assumption that $r_{\pi }<<R_E(1+\beta )$, and that $\beta$ is large, then:

$$A_v=\frac{V_o}{V_{in}}\approx \frac{-\beta R_C}{R_E(1+\beta )}\approx \frac{-R_C}{R_E}$$

Common Collector Amplifier

The common collector (or emitter-follower) amplifier is another amplifier circuit used with BJTs (as oppose to the common emitter shown above).

The hybrid-pi model of this circuit looks like, as without the collector the circuit can be re-arranged to:

The output voltage is the voltage accross the emitter resistor, and as $i_E=(1+\beta )i_b$:

$$V_o=i_b(1+\beta )R_E$$

The input voltage is the voltage accross both the emitter resisitor and $R_{\pi }$:

$$V_{in}=r_{\pi }{i}_b+i_b(1+\beta )r_E$$

Therefore the voltage gain for this amplifier is:

$$A_v=\frac{V_o}{V_{i}n}=\frac{i_b(1+\beta )R_E}{r_{\pi }{i}_b+i_b(1+\beta )r_E}$$

$$A_v=\frac{R_E(1+\beta )}{r_{\pi }+R_E(1+\beta )}$$

As usually, $r_{\pi }>>R_E(1+\beta )$, $A_V\approx 1$. This amplifier has very low voltage gain, and instead acts as a current amplifier:

$$A_i=\frac{i_E}{i_b}=\frac{(1+\beta )i_b}{i_b}=1+\beta$$

The input impedance is large, as it is the reflected impedance from the emitter resistor again:

$$R_{in}=\frac{V_i}{i_b}=\frac{i_b{r}_{\pi }+i_b(1+\beta )R_E}{i_b}=r_{\pi }+R_E(1+\beta )$$

The output impedance can be calculate by shorting $V_{in}$, and by applying a test current source $i_x$ accross the output terminals. I'm not going to type out all the analysis but:

$$R_o=\frac{r_{\pi }}{1+\beta }=\frac{1}{g_m}$$

The emitter follow has high input and low output impedance with a high current gain, so acts as an impedance transformer and a buffer.

Example

A circuit for a common-emitter amplifier is shown below.

Work out the values of the DC biasing components $R_{B1}$, $R_{B2}$, and $R_E$ for the following conditions:

• $V_{CC}=10V$
• $I_E=5mA$
• $V_E=1.3V$
• Voltage accross $R_C=4V$
• $\beta =100$

Assuming $I_E=I_C$ and $\alpha =\frac{\beta }{1+\beta }=0.99$, we have:

$$R_C=\frac{4}{5mA}=800\Omega R_E=\frac{1.3}{5mA}=260\Omega$$

Calculating the Thevenin equivalent of the biasing resistors:

$$R_{TH}=\frac{\beta R_E}{10}=2.6k\Omega V_{TH}=V_E+V_{BE}+I_B{R}_{TH}=1.3+0.7+50\mu A\times 2.6k\Omega =2.13V$$

Then calculating the bias resistors from the Thevenin values:

$$R_1=\frac{V_{CC}{R}_{TH}}{V_{TH}}=\frac{10\times 2.6k}{2.13}=12.2k\Omega$$

$$R_2=\frac{R_{TH}{R}_1}{R_1-R_{TH}}=\frac{2.6k\times 12.1k}{12.1k-2.6k}=3.3k\Omega$$

To derive an expression for the voltage gain, need to replace the BJT by it's small signal model

$$V_{out}=-\beta i_b{R}_c$$

$$V_s=i_b{r}_{\pi }+i_b{R}_E+\beta i_b{R}_E=i_b(r_{\pi }(1+\beta )R_E)$$

$$A_v=\frac{V_{out}}{V_s}=\frac{-\beta R_c}{R_{\pi }+(1+\beta )R_E}=\frac{-100\times 800}{500+101\times 260}=-2.99$$