Analog Electronics I · 120 h · Topic 3 of 8

Small-signal amplifiers

A microphone delivers 2 mV. A loudspeaker needs several volts. In between there is a chain of amplifiers, and the first link is this one: the small-signal stage, where the signal is so small that the transistor behaves linearly.

Amplifiers Transistors Common emitter Common collector Darlington

01DC and AC at the same time

A small-signal amplifier does two things at once, and the secret to understanding it is to separate them:

In DC, the circuit biases the transistor and leaves it at a stable Q point, in the middle of the active region. That is what was calculated in Bipolar transistors.
In AC, the input signal makes the transistor swing around that Q point. The output is an enlarged version of that swing.

Figure 1. AC riding on the quiescent level, animated: with the Q point centered the signal fits entirely, and when it is shifted or too large it clips against the limits.

Superposition theorem

Since the circuit is linear around the Q point, it can be analyzed in two separate parts: first the DC circuit (with the capacitors as open circuits) and then the AC circuit (with the capacitors as short circuits and the DC sources at ground). Then the two results are added. Without this separation the analysis is intractable.

02The amplifier's capacitors

A capacitor blocks DC and passes AC, with a reactance that falls as the frequency rises:

X_{C} = \frac{1}{2 π f C}

Rule of thumb: a capacitor “is a short circuit” for the signal if its reactance at the lowest working frequency is at least 10 times smaller than the resistance that goes with it.

C1 · input coupling

Lets the signal in but keeps the signal source from altering the bias of the base. Typical: 10 µF.

C2 · output coupling

Delivers the signal to the load without passing it the collector's DC. Typical: 10 µF.

C_E · bypass

In parallel with R_E. Keeps R_E in place for DC (stability) but cancels it for AC (gain). Typical: 100 µF.

The bypass capacitor is the most interesting one

R_E stabilizes the Q point but, if it is also allowed to act on the signal, the negative feedback cuts the gain enormously. C_E resolves the contradiction: in DC R_E exists, in AC it disappears. A single component doing two different jobs depending on frequency — it is one of the most elegant ideas in analog electronics.

Polarity

Coupling capacitors are usually electrolytic, which are polarized. The positive terminal goes on the side where the DC level is higher. On C1, the positive faces the base; on C2, the collector; on C_E, the emitter. Connected backwards they swell up and burst.

03Dynamic emitter resistance

For the signal, the base-emitter junction is not a passive component: it behaves like a resistor that depends on the quiescent current. It is called r_e (lowercase r for emitter) and is the constant that governs the entire gain.

r_{e} = \frac{26 mV}{I_{E}}

With I_E in milliamperes, r_e comes out in ohms. The 26 mV comes from the thermal voltage kT/q at 25 °C.

I_E	r_e	Consequence
0.5 mA	52 Ω	Low gain, low power consumption, more noise.
1 mA	26 Ω	The usual compromise value.
2.6 mA	10 Ω	Good gain.
10 mA	2.6 Ω	High gain, but the transistor heats up and the input impedance drops.

It is immediately clear that the Q point is not a free choice: it sets the gain that is possible. And since the 26 mV is really kT/q, r_e changes with temperature, and so does the gain. That dependence is one of the reasons modern amplifiers use feedback.

04Common-emitter amplifier

It is the reference configuration: input at the base, output at the collector, emitter common to both loops.

Figure 2. Common-emitter amplifier, animated: on the right, the input and the output at the same time. The output is seen to be inverted 180° relative to the input.

Voltage gain

A_{v} = - \frac{r_{c}}{r_{e}}, r_{c} = R_{C} ∥ R_{L}

The minus sign indicates phase inversion: when the input rises, the output falls. r_c is the collector resistance seen in AC, that is, R_C in parallel with the load.

Worked example · Complete common emitter

V_CC = 12 V, R1 = 47 kΩ, R2 = 10 kΩ, R_C = 2.2 kΩ, R_E = 470 Ω with C_E, load R_L = 10 kΩ.

DC (calculated in the previous topic): V_B = 2.11 V, V_E = 1.41 V, I_E = 3.0 mA, V_CE = 4.0 V. Correct Q point, in the active region.

AC:

r_e = 26 mV / 3.0 mA = 8.7 Ω
r_c = 2.2 k ∥ 10 k = (2.2×10)/(2.2+10) = 1.80 kΩ
A_v = −1800 / 8.7 = −207

A 5 mV peak input gives 1.03 V peak at the output, inverted. Linearity check: the output swing (1.03 V) must fit comfortably between the Q point (4.0 V) and the limits: downward there is 4.0 − 0.2 = 3.8 V of margin, and upward 12 − 4.0 = 8 V. It fits without clipping.

Impedances

Z_{i n} = R 1 ∥ R 2 ∥ (β \cdot r_{e})

Z_{o u t} \approx R_{C}

In the example, with β = 200: β·r_e = 1.74 kΩ, and Z_in = 47 k ∥ 10 k ∥ 1.74 k ≈ 1.4 kΩ. It is low, and it is the main weakness of the common emitter.

What happens if C_E is removed

Without the bypass capacitor, R_E also acts on the signal and the gain falls to:

A_{v} = - \frac{r_{c}}{r_{e} + R_{E}}

In the example: −1800 / (8.7 + 470) = −3.8. The gain fell by a factor of 54.

It looks like a disaster, but it has an enormous upside: since r_e (8.7 Ω) is negligible compared with R_E (470 Ω), the gain becomes practically −R_C/R_E, that is, it depends only on resistors. It no longer depends on r_e, or on temperature, or on β. Gain is traded for stability and for lower distortion. In practice an intermediate solution is used: split R_E in two and bypass only one part.

Configuration	Gain	Stability	Distortion
R_E fully bypassed	Very high (200)	Poor	High
R_E partially bypassed	Medium (20-40)	Good	Low
Without C_E	Low (4)	Excellent	Very low

05Common collector (emitter follower)

The signal enters at the base and leaves at the emitter. The collector goes straight to V_CC, which for AC is ground — hence the name.

A_{v} = \frac{r_{E}}{r_{E} + r_{e}} \approx 1

Z_{i n} = R 1 ∥ R 2 ∥ β (r_{e} + r_{E})

Z_{o u t} \approx r_{e} + \frac{R_{S}}{β}

r_E is the total emitter resistance seen in AC (R_E ∥ R_L), and R_S the resistance of the signal source.

The voltage gain is less than 1 (typically 0.98). It sounds useless, but it is not: what this circuit does is impedance matching.

What it provides

Very high Z_in (tens or hundreds of kΩ): it does not load the previous stage.
Very low Z_out (a few ohms): it can drive demanding loads.
High current gain (β).
It does not invert the phase.

Where it is used

As an output stage before a loudspeaker or a long cable.
As a buffer between a high-impedance source (a piezoelectric microphone, a probe) and an amplifier with low Z_in.
In the Zener-regulated power supply, which is exactly this circuit.

Why impedance matching matters

If a microphone with a 50 kΩ internal impedance is connected to a common emitter with 1.4 kΩ of Z_in, the divider formed between the two eats 97% of the signal before it reaches the transistor. Putting an emitter follower in front (Z_in = 200 kΩ), 80% gets through. The overall gain of the whole chain rises enormously even though the follower “does not amplify.”

06Common base

The third configuration: input at the emitter, output at the collector, base grounded for the signal. It is seen little in Year 4 but it belongs in the comparison table.

Very low Z_in (r_e, that is, a few ohms).
High voltage gain, the same as the common emitter, and no inversion.
Current gain ≈ 1 (it is α).
Excellent high-frequency response, because it does not suffer from the Miller effect.

Because of that last property it is used in radio-frequency amplifiers and at the input of wideband instruments — a Telecommunications topic, in Year 6.

Configuration	A_v	A_i	Z_in	Z_out	Phase	Main use
Common emitter	High	High (β)	Medium	High	180°	Voltage amplification
Common collector	≈ 1	High (β)	Very high	Very low	0°	Impedance matching, output stage
Common base	High	≈ 1 (α)	Very low	High	0°	High frequency, RF

07Darlington connection

Two transistors connected so that the emitter of the first drives the base of the second. The pair behaves like a single transistor with a gain equal to the product of the two. The interactive transistor publication has a dedicated tab.

Figure 3. Darlington connection. It comes already integrated in packages such as the TIP120 (NPN) and the TIP125 (PNP), with the bleeder resistors included.

β_{T} = β_{1} \cdot β_{2}

With β1 = β2 = 100, β_T = 10,000: 10 µA of base current controls 100 mA of collector current.

Advantages

Enormous gain: a large load is handled with a minimal base current.
Extremely high input impedance (β_T·r_e).
It can drive relays and motors directly from a microcontroller.

Disadvantages

1.4 V drop between base and emitter (two junctions).
V_CE(sat) of 0.9 to 1.2 V: it dissipates considerably more than a single transistor.
Slower to turn off (the stored charge takes time to clear).
The leakage current of the first transistor is amplified by the second: worse thermal drift.

08Coupling between stages

A single stage is rarely enough. Several are chained together, and the total gain is the product of the individual gains (or the sum in decibels).

A_{T} = A_{1} \cdot A_{2} \cdot A_{3} \dots

A_{d B} = 20 \cdot \log (A_{v})

A gain of 200 is 46 dB. Two identical stages give 92 dB, that is, 40,000 times.

There is a detail that textbook calculations tend to ignore: the Z_in of the second stage loads the first. That is, to calculate the gain of stage 1 you have to use r_c = R_C1 ∥ Z_in2, not R_C1 alone. If you forget, the calculation gives a gain much higher than the measured one.

Coupling	How	Advantage	Disadvantage
Capacitive (RC)	Capacitor between stages	Simple, cheap, isolates the bias points	Does not pass DC or very low frequencies
Direct	Collector of one to the base of the next	Responds down to 0 Hz	DC drifts are amplified along the cascade
Transformer	Winding between stages	Matches impedances, provides galvanic isolation	Expensive, heavy, limited bandwidth

09In the lab

Lab 1 · Common emitter, gain measurement

Build the circuit from the example (12 V, 47 k / 10 k / 2.2 k / 470 Ω, C1 = C2 = 10 µF, C_E = 100 µF). Steps:

With no signal, measure V_B, V_E and V_C. Compare with what was calculated. If they do not match, the problem is in DC and there is no point going on.
Inject 10 mV peak-to-peak at 1 kHz with the function generator.
With the oscilloscope on two channels, view input and output at the same time: you should see the phase inversion.
Measure the two amplitudes and calculate A_v = v_out/v_in.
Raise the input until the output clips. Note at what amplitude it happens and on which side it clips first: that shows which way the Q point is off-center.

Lab 2 · The role of C_E

On the same circuit, disconnect C_E and measure the gain again. It should fall from about 200 to about 4. Also check that the output signal looks cleaner (less distortion) without the capacitor. It is the practical demonstration of the trade-off between gain and linearity.

Lab 3 · Emitter follower as a matching stage

Connect the function generator through a 100 kΩ series resistor (it simulates a high-impedance source) directly to the common emitter: the signal almost disappears. Insert an emitter follower between the two and measure again: the signal recovers. The lab shows that a circuit with a gain below 1 can increase the overall gain of the system.

Lab 4 · Frequency response

With a constant input amplitude, sweep the frequency from 10 Hz to 1 MHz and note the output amplitude. Plot it on a logarithmic scale. The two cutoff frequencies appear: the lower one is set by the coupling and bypass capacitors, and the upper one by the parasitic capacitances of the transistor. The bandwidth is the distance between the two.

10Common mistakes

Symptom	Usual cause
No signal at the output, but the DC levels are fine	Coupling capacitor open, reversed, or too small a value for that frequency.
The output clips only at the top (or only at the bottom)	Q point off-center. It clips first on the side with less margin.
The measured gain is much lower than the calculated one	The load (or the Z_in of the next stage) was left out of the calculation of r_c. Or C_E is wired wrong.
The gain is far lower (about 4)	C_E open, or missing. R_E is acting on the signal.
The amplifier oscillates or “squeals”	Feedback through the power supply. Missing decoupling (100 nF + 100 µF next to the stage) or the input and output wires run together.
50 Hz hum	Poor grounding (ground loops) or unshielded input. The high-impedance input picks up the mains.
Audible distortion even though the output does not look clipped	Input signal too large: it is leaving the small-signal range (v_in must be much smaller than r_e·I_E).

11Self-assessment

Why can DC and AC be analyzed separately?

Because of the superposition theorem: around the Q point the circuit behaves linearly, so the total response is the sum of the response to DC (with capacitors open) and the response to AC (with capacitors shorted and DC sources at ground).

An amplifier has I_E = 2 mA. What is r_e?

r_e = 26 mV / 2 mA = 13 Ω.

What exactly is capacitor C_E for?

So that R_E exists in DC (stabilizing the Q point) but disappears in AC (recovering the full gain). Without it the gain falls to the order of R_C/R_E.

Why does the common emitter invert the phase?

Because V_C = V_CC − I_C·R_C. If the input signal makes I_C rise, the drop across R_C increases and the collector voltage falls. Input up, output down: a 180° phase shift.

An emitter follower has a gain of 0.98. What is it good for, then?

For impedance matching: it presents a high-impedance input (it does not load the signal source) and a low-impedance output (it can drive demanding loads). Its current gain is β, so it does amplify power.

With R_C = 4.7 kΩ, R_L = 4.7 kΩ and I_E = 1 mA, what is the gain?

r_e = 26 Ω. r_c = 4.7 k ∥ 4.7 k = 2.35 kΩ. A_v = −2350 / 26 = −90. Note that the load cut the gain in half compared with having no load.

What is the price of the Darlington's enormous gain?

A base-emitter drop of 1.4 V (two junctions), a V_CE(sat) of almost 1 V that makes it dissipate more heat when switching, slower turn-off and worse thermal behavior, because the leakage of the first transistor is amplified by the second.

Three stages with a gain of 20 each. What is the total gain as a ratio and in dB?

20 × 20 × 20 = 8000 times. In decibels: 20·log(20) = 26 dB per stage, 78 dB in total. Check: 20·log(8000) = 78 dB.

What limits the lower working frequency of an RC-coupled amplifier?

The capacitors: as the frequency drops X_C rises and they stop behaving as short circuits. The one that most often limits is C_E, because it works against a very small resistance (r_e) and needs a large value.