The transistor as an amplifier and a switch

A small signal made large

A microphone gives a signal of a few millivolts. A loudspeaker needs several volts. Between them an amplifier makes the signal larger. In a transistor amplifier the small signal changes the base current, and the transistor changes a much larger collector current in step with it.

The common-emitter amplifier

The circuit of the previous lesson is used with one change. A small alternating signal $v_i$ is added in series with the base supply.

1. The base supply $V_{BB}$ sets a steady bias, so that the transistor stays in the active region. The steady values are the working point or Q point.
2. The signal changes the base current by $\Delta I_B = \dfrac{\Delta v_i}{r_i}$, where $r_i$ is the input resistance.
3. The collector current changes by $\Delta I_C = \beta\, \Delta I_B$.
4. The extra current in $R_C$ changes the voltage across it. Because $V_{CE} = V_{CC} - I_C R_C$, the output voltage is $\Delta v_o = -R_C\, \Delta I_C$.

Putting the steps together:

• The minus sign says that the output is upside down: a phase change of $180^\circ$. When the input rises, the collector current rises, the drop in $R_C$ rises, and $V_{CE}$ falls.
• The current gain is $A_I = \beta$.
• The power gain is $A_P = A_V \times A_I$ in size, $\dfrac{\beta^2 R_C}{r_i}$. The extra power is not made by the transistor. It comes from the supply $V_{CC}$. The signal only controls how much of it reaches the load.
• Two stages in a row have a gain equal to the product of the two gains. The two phase changes cancel.

Bias and clipping. The output voltage $V_{CE}$ cannot be larger than $V_{CC}$ (the transistor is then cut off) and cannot fall below about $0.2$ V (saturation). If the signal tries to take it past these limits, the peaks are cut off. This is clipping. For the largest swing without clipping, the working point is set near the middle, $V_{CEQ} \approx V_{CC}/2$. The largest input amplitude without clipping is

Here $h$ is the smaller of $V_{CEQ} - 0.2\ \text{V}$ and $V_{CC} - V_{CEQ}$. It is the room that the output has to swing.

The load line $V_{CE} = V_{CC} - I_C R_C$ is a straight line on the output characteristics. The working point is where the line crosses the curve of the base current $I_B$.

An oscillator. If part of the output of an amplifier is fed back to the input in step with the input signal (positive feedback), the amplifier keeps going with no input. A coil and a capacitor in the circuit set the frequency $f = \dfrac{1}{2\pi\sqrt{LC}}$. The energy comes from the supply.

A worked example. An amplifier has $\beta = 60$, $R_C = 3.3\ \text{k}\Omega$ and $r_i = 1.1\ \text{k}\Omega$. An input of $10$ mV amplitude is applied. Find the voltage gain, the output amplitude and the power gain.

1. The voltage gain. $A_V = -\dfrac{\beta R_C}{r_i} = -\dfrac{60 \times 3.3\ \text{k}\Omega}{1.1\ \text{k}\Omega} = -180$.
2. The output. $|v_o| = 180 \times 10\ \text{mV} = 1.8$ V, upside down.
3. The power gain. $A_P = |A_V| \times \beta = 180 \times 60 = 10\,800$.

Make the input $40$ mV. Which part of the output is cut off? Then move the resting voltage, and find the setting that clips the fewest peaks.

The transistor as a switch

Use the same circuit with two states only. The input is either low or high.

• Low input ($V_i$ below about $0.7$ V for silicon): $I_B = 0$, the transistor is in cut-off, $I_C = 0$, and the output is $V_o = V_{CC}$. The switch is open.
• High input: $I_B$ is large, and the transistor is in saturation. The output is $V_o = V_{CE(sat)} \approx 0.2$ V. The switch is closed.

For saturation, the base current must be at least

A design uses a base current a little larger than this, so that saturation is sure even for a transistor with a smaller $\beta$.

The output is high when the input is low and low when the input is high. The circuit inverts. It is a NOT gate. The same circuit with a relay or a lamp in the place of $R_C$ lets a small signal switch a large load on and off.

A worked example. A buzzer of resistance $150\ \Omega$ is switched by a transistor with $\beta = 100$ on a $6.0$ V supply. The control voltage is $3.3$ V. Find the least base current and the largest base resistor.

1. The saturated collector current. $I_{C(sat)} = \dfrac{6.0\ \text{V} - 0.2\ \text{V}}{150\ \Omega} = 38.7$ mA.
2. The least base current. $I_{B,min} = \dfrac{38.7\ \text{mA}}{100} = 0.387$ mA.
3. The largest base resistor. $R_B = \dfrac{3.3\ \text{V} - 0.7\ \text{V}}{0.387\ \text{mA}} = 6.7\ \text{k}\Omega$. A resistor of half this size gives a safe margin.

Move the input from $0$ to $5$ V. At which input does the output start to fall, and at which does it stop falling? Make the base resistor larger. Which of the two points moves?

Practise with new numbers

Try it in Python

The program amplifies a $100$ Hz sine wave with a gain of $-200$, and clips the output at $0.2$ V and $12$ V. It tries three input amplitudes.

Common mistakes

• Forgetting the minus sign of the gain. The output is $180^\circ$ out of phase with the input.
• Saying the transistor makes power. The extra power comes from the supply.
• Using the supply $V_{CC}$ as the input to the formula of the gain. The gain uses $\beta$, $R_C$ and $r_i$.
• Using a bias that leaves no room. A working point near $V_{CC}$ or near $0.2$ V clips one half of the signal.
• Calling the switch closed when the output is high. A high output means the transistor is off, which is the open switch.
• Using $V_{CC}/R_C$ for the saturation current. The transistor keeps $0.2$ V, so use $(V_{CC} - 0.2\ \text{V})/R_C$.

Remember these

• $A_V = -\dfrac{\beta R_C}{r_i}$. The output is upside down. $A_I = \beta$ and $|A_P| = \beta |A_V|$.
• The Q point is set near $V_{CC}/2$. Clipping occurs between $0.2$ V and $V_{CC}$.
• Switch: cut-off gives $V_o = V_{CC}$, saturation gives $V_o \approx 0.2$ V. Saturation needs $I_B \ge \dfrac{V_{CC} - 0.2\ \text{V}}{\beta R_C}$.
• The transistor switch is a NOT gate.

References

• NCERT Physics Class 12, Semiconductor Electronics, the older edition on the transistor as an amplifier, an oscillator and a switch.
• Horowitz and Hill, The Art of Electronics, third edition, Chapter 2, Transistors.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An amplifier has $\beta = 120$, $R_C = 2.0\ \text{k}\Omega$ and $r_i = 800\ \Omega$. What is the voltage gain?

2. An input of $5.0$ mV goes into the amplifier of question 1. What is the output voltage?

3. The input to a transistor switch is $0$ V. What are the transistor state and the output voltage?

4. A transistor switch has $V_{CC} = 12$ V, a load of $600\ \Omega$ and $\beta = 50$. What base current saturates it?