A curve you are allowed to call a straight line
Every gain formula in this chapter is a lie, and it is worth knowing exactly how big a lie before you start trusting them.
A transistor obeys IC = IS eVBE/VT. That is an exponential, and a sine wave pushed into an exponential does not come out as a sine wave. So where does a single number called “gain” come from at all?
Commit before you touch anything
A stage biased at 1 mA with RC = 2.2 kΩ. You feed the base a sine and slowly raise it from 0.5 mV to 30 mV peak. What happens?
The trick is calculus, not electronics
Zoom far enough into any smooth curve and it becomes indistinguishable from its tangent. That is the entire justification for small-signal analysis. We do not claim the transistor is linear. We claim that over a small enough piece of its curve the error is beneath our notice, and then we are careful about how small “small enough” is.
Expand the exponential about the operating point and the size of the lie falls out directly:
ex = 1 + x + x²/2 + … so HD2 ≈ v̂be / 4VT
The second term relative to the first is x/2, and working it through for a sine gives a second-harmonic distortion of the peak signal divided by four thermal voltages. Put numbers in and the rule of thumb writes itself.
| Signal at the base–emitter junction | Second-harmonic distortion | Verdict |
|---|---|---|
| 1 mV peak | about 1% | a straight line for any purpose |
| 2.6 mV peak | about 2.5% | fine for most things |
| 10 mV peak | about 10% | audible, and visible on a scope |
| 26 mV peak | about 25% | this is a fuzz box, not an amplifier |
Notice the number that sets the window is not a property of the transistor you bought. It is VT, which is temperature. Every bipolar transistor ever made has the same linear window, and it is about ten millivolts wide.
And this is a different failure from chapter 4’s. Clipping is the output hitting a wall. This is the output being the wrong shape while sitting comfortably in the middle of its range. On the bench the trace never goes near a rail, and it is still visibly crooked.
In the wild
Why a microphone preamp is the easy stage and a power amp is the hard one. A microphone gives you a few millivolts, which lands inside the linear window by luck. A power output stage has to swing volts, which is a long way outside it, and every bit of its linearity has to be bought back with feedback.
Why every amplifier data sheet quotes THD. Total harmonic distortion is exactly the quantity on this bench: how much of the output is not the frequency you put in. 0.01% is hi-fi, 1% is a guitar amp being polite, 10% is a effect pedal.
Fuzz pedals are this lesson, sold. Drive a single transistor stage well past its linear window and the harmonics it manufactures are the product. Nothing is broken; the promise was simply never made.
Before you move on
What is the small-signal condition for a bipolar transistor?
The slope has a name. Three names, and one fact.
Lesson 1 said the tangent is what we use. This lesson works out its slope, and then gives that one slope three different names, because you will meet all three and they are not three ideas.
Commit before you touch anything
Two transistors, one with β = 100 and one with β = 400. Both biased at exactly 1 mA. Which has the larger gm?
Differentiate the exponential. That is the whole derivation.
Take the transport equation from chapter 1 and ask how much the collector current changes for a small change in base–emitter voltage:
gm = dIC/dVBE = (IS/VT) eVBE/VT = IC / VT
The exponential differentiates into itself, so IS vanishes and what is left is the current you biased it at, divided by a temperature. No β. No device geometry. No part number. At 1 mA and room temperature, gm = 1 mA / 25.9 mV = 38.7 mA/V, and that is true of every silicon bipolar transistor on the planet biased at 1 mA.
The same slope, seen from three terminals
Boylestad tends to write re; Sedra & Smith tend to write gm and rπ; Horowitz & Hill write the shorthand re = 25/IC(mA). Three notations, one number, and the only reason to know all three is that you will read all three.
Where the 26 millivolts comes from
VT = kT/q — Boltzmann’s constant times absolute temperature, divided by the electron charge. It is not a property of silicon. It is the average thermal energy of a charge carrier expressed in volts, and it is the same number in germanium, in a vacuum tube’s space charge, and in the chemistry of a battery.
| Temperature | VT | re at 1 mA | Gain, RC = 4.7 kΩ |
|---|---|---|---|
| −20 °C | 21.8 mV | 21.8 Ω | −216 |
| 25 °C | 25.7 mV | 25.7 Ω | −183 |
| 85 °C | 30.9 mV | 30.9 Ω | −152 |
So a fully bypassed stage loses about a third of its gain between a cold morning and a hot enclosure — and the gain moves with absolute temperature, which is to say about 0.33% per degree. That is the price of the “wonderful” gain of chapter 4, lesson 5, and it is why real designs give most of it away.
In the wild
Every chip’s temperature sensor is this equation, run backwards. Bias one junction at I and another at 10I, subtract their VBE, and you get VT ln 10 — a voltage exactly proportional to absolute temperature, with the device parameters cancelled out. Your CPU reads its own temperature this way.
Why the bandgap reference works. VBE falls with temperature and that ΔVBE rises with it. Add them in the right proportion and you get a voltage that does neither — about 1.2 V, the bandgap of silicon. Every voltage reference in your parts drawer is that trick.
Why data sheets specify hFE at a particular current and still spread 3:1. β is a manufacturing accident. gm is physics. Design so your answer depends on the second one and the 3:1 spread stops mattering — which is what chapter 2, lesson 5 was really about.
Before you move on
A stage is biased at 2 mA. Roughly what is re?
−RC/re, and what a load does to it
Now assemble it. A signal vbe at the base produces a collector current ic = gmvbe. That current has nowhere to go except through whatever is hanging on the collector. Voltage is current times resistance, and the sign is negative because more current means more drop and a collector that falls:
Av = −gm(RC ∥ RL) = −(RC ∥ RL) / re
Two lines of algebra, and the whole common-emitter amplifier is in them.
Commit before you touch anything
A stage with RC = 4.7 kΩ measures a gain of about −180 with nothing connected to its output. You connect the next stage, whose input impedance is 4.7 kΩ. Now what?
The inversion is a resistor, not a convention
People memorise “common emitter inverts” as a rule. It is not a rule, it is a consequence, and it takes one sentence: the base goes up, so the collector current goes up, so the drop across RC goes up, so the collector — which is VCC minus that drop — goes down. Remove RC and there is no inversion because there is no output.
The second load line, which catches people out
Chapter 4 drew a load line: everything the collector circuit permits, slope −1/RC. That line is a DC object — it is drawing one.
Now put a coupling capacitor and a load on the collector. In drawing two the capacitor is a wire, so the collector sees RC in parallel with RL, which is smaller. The signal therefore moves along a different, steeper line — the AC load line — which passes through the same Q point because with no signal nothing has changed.
The DC load line tells you where the circuit sits and what the transistor is dissipating. Slope −1/(RC+RE).
The AC load line tells you how far the signal can travel before it clips. Slope −1/(RC∥RL), through the Q point.
Your maximum clean output is twice the smaller of ICQ×Rac and VCEQ − Vsat — measured along the steeper line, not the gentle one.
This is why an amplifier that looked comfortably centred on paper can clip on one side as soon as it is loaded. Drag the load resistor on the bench and watch the second line pivot about the Q point while the first stays put.
In the wild
Why the volume drops when you plug headphones into a line output. The headphones are 32 Ω and the output was expecting 10 kΩ. RC∥RL collapses to almost nothing, and so does the gain. It is the same arithmetic as the prediction above, with a more dramatic ratio.
Why a scope probe is ×10. A 1 MΩ probe across a high-impedance node would load it badly; the ×10 divider makes it 10 MΩ at the cost of a tenth of the signal. Measuring a circuit is connecting a load to it, and this lesson is why that matters.
Why data sheets quote output impedance. It is precisely the number that lets somebody else predict how much of your gain their load will take. Publishing it is publishing the parallel-resistor calculation in advance.
Before you move on
Why does a common-emitter stage invert?
Where the gain stops
If Av = −RC/re then the way to more gain is obvious: use a bigger RC. And it works, for a while, and then it stops working in a way that turns out to be the most important fact about analogue integrated circuits.
The first objection is chapter 4’s. A bigger RC at the same current drops more voltage, so the collector falls towards saturation and there is no room left to swing. Fine — cheat. Replace RC with a current source, which by definition has a huge resistance to the signal while dropping only whatever DC voltage you allow it. That is a real technique and you will build it in chapter 9.
Commit before you touch anything
So: perfect current source in the collector, infinite resistance, no wasted volts. What gain do you get?
The transistor has its own resistance across the collector
Chapter 2 mentioned that the collector characteristics are not quite flat — raise VCE at fixed base drive and IC creeps up. The mechanism is the Early effect: more collector voltage widens the collector depletion region, which eats into the base, which makes the base thinner, which lets more carriers across. Extend those sloping characteristics backwards and they all meet at a single negative voltage, −VA, the Early voltage — typically 50 to 150 V for a small-signal transistor.
Written as a resistance in parallel with the collector:
ro = VA / IC
So the honest gain equation has one more term in the parallel combination, and it was always there:
Av = −gm(RC ∥ RL ∥ ro)
Now take RC to infinity and watch the current cancel
With nothing but ro left in the bracket:
Av,max = −gmro = −(IC/VT)(VA/IC) = −VA/VT
IC divides out. Every bias current gives the same answer. With VA = 75 V and VT = 25.9 mV that is about 2900, or 69 dB, and it is called the transistor’s intrinsic gain.
You cannot buy your way past it. More current does not help — gm rises and ro falls in exactly compensating proportion. A bigger transistor does not help. A more expensive transistor helps only if it has a larger VA, and VA is proportional to base width, so a high-gain device is a slow device. That is a physics trade, not an engineering one.
In practice you rarely see even 2900, because the next stage’s input impedance loads you long before ro does. That is lesson 5, and it is the difference between the ceiling and the room you are actually standing in.
In the wild
Why every op-amp has two or three gain stages. Not for elegance. One transistor cannot reach 120 dB, so the topology is forced by VA/VT.
The cascode, which is the standard dodge. Stack a common-base transistor on top of a common-emitter one and the effective output resistance becomes roughly β ro instead of ro. It does not break the ceiling so much as build a second storey. Chapter 9.
Why MOSFET analogue design is harder than bipolar. A MOSFET’s gm is smaller for the same current and its intrinsic gain is often only 20 to 50. Everything you can do casually with a BJT has to be fought for in CMOS — which is chapter 11’s problem, and part of why chapter 7 exists.
Before you move on
Two stages with ideal current-source loads, one biased at 0.1 mA and one at 10 mA. Which has more voltage gain?
Three stages of a hundred is not a million
One stage gives you a hundred and something. You need ten thousand. Obvious: build two and multiply.
Commit before you touch anything
Two identical common-emitter stages, each measuring −170 on its own with nothing on the output. Wire them in series. What do you measure?
The input impedance of a common-emitter stage is disappointing
Look into the base of a bypassed stage and you see rπ — about 3.9 kΩ at 1 mA. In parallel with that sit the two bias divider resistors, which chapter 4 said should be reasonably stiff, so call it 2.5 kΩ altogether:
Rin = R1 ∥ R2 ∥ rπ
And the stage before it is trying to drive that through its own RC of 4.7 kΩ. The parallel combination is about 1.6 kΩ, so the first stage’s gain falls from 170 to about 62. Multiply out and you have roughly ten thousand rather than twenty-nine thousand — a factor of three lost to nothing but connecting two things together.
Notice which stage suffers. The last stage in a chain often drives a light load and keeps most of its gain; the ones in the middle are the ones being squeezed. That asymmetry is worth remembering when you go looking for missing gain in a real circuit.
Three ways out, and you have met two of them
Raise the loading stage’s input impedance. Leave some emitter resistance unbypassed and Rin becomes (β+1)(re+RE′) — much larger. You pay in gain at that stage, so you are moving the problem rather than solving it, but sometimes the total improves.
Put a follower between them. Chapter 3’s circuit, doing exactly the job it was built for: high input impedance, low output impedance, gain of one. It costs a transistor and some current and buys back most of the loss. Try the toggle on the bench.
Use a device whose input takes no current at all. A MOSFET’s gate draws nothing, so its input impedance is essentially infinite and this whole problem evaporates. That is one of the two reasons chapter 7 exists.
In the wild
Why a buffer is worth its current. A follower burning 2 mA that recovers a factor of three in overall gain is a much better deal than another gain stage, because it adds no extra phase shift to worry about later. Chapter 12 will make that sentence mean something.
Why radio systems standardise on 50 Ω. If every block is specified to present 50 Ω and to drive 50 Ω, loading stops being a surprise and becomes part of the specification. You give up the freedom to have high impedances and you get composability in return.
Why op-amp datasheets quote open-loop gain into a stated load. “100 dB into 2 kΩ” and “110 dB into 10 kΩ” are the same amplifier. Without the load, the number is not a fact.
Before you move on
You cascade two identical stages and measure far less than the product of their individual gains. The most likely reason?
Which terminal you drive, and which you listen to
A transistor has three terminals. You feed a signal into one and take it out of another, which leaves the third one common to both — and that third terminal is what the configuration is named after. There are only three, and you have already built two of them without being told they had names.
Commit before you touch anything
One transistor, one bias point. Which configuration presents an input impedance of only a few tens of ohms?
The whole family, at one bias point
Read down the gain column and notice that common emitter and common base have the same magnitude. Of course they do — both convert a base–emitter voltage into a collector current through the same gm, and both push that current through the same collector resistance. What differs is where you injected the signal, and therefore the sign and the input impedance.
Why anybody uses common base
A 26 Ω input is a feature at radio frequencies. If the world has agreed on 50 Ω source impedances, an amplifier that naturally wants a low input impedance is easy to match and needs no transformer.
It has no Miller effect. In a common-emitter stage the collector–base capacitance sits between the input and an inverted, amplified output, so it looks (1+|Av|) times larger from the input — and that is usually what limits the bandwidth. In common base the base is grounded and that path does not exist. Chapter 12 makes this precise.
Put those two facts together and you get the cascode: a common-emitter stage with a common-base stage stacked on its collector. The common-emitter transistor now sees a load of about 26 Ω, so its gain is roughly one and there is no Miller multiplication; the common-base transistor passes that current up to a high-impedance node where the voltage gain actually happens. Full gm, full output resistance, and none of the bandwidth penalty. It is in every op-amp and every RF amplifier, and it is chapter 9.
In the wild
The emitter follower is probably the most-built circuit in history. Output stage of every op-amp, every voltage regulator’s pass transistor, the buffer before every long cable. Chapter 3 gave you it before you had a name for the family.
Common gate in RF front ends. The FET version of common base, chosen in low-noise amplifiers for exactly the two reasons above — easy matching and no Miller.
Why the “three configurations” table is worth memorising once. Handed an unfamiliar schematic, the fastest orientation is: where does the signal go in, where does it come out, which terminal is left over. That tells you the sign, the rough gain and both impedances before you write a single equation.
Before you move on
Which configuration gives substantial voltage gain without inverting?
Chapter checkpoint
Question 1
A transistor is biased at 2 mA at room temperature. Roughly what is gm?
Question 2
A fully bypassed common-emitter stage: RC = 3.9 kΩ, IC = 1 mA, nothing on the output. Roughly what is the gain?
Question 3
You double the bias current of a bypassed stage and halve RC, so the collector sits in the same place as before. What happens to the gain?
Question 4
What ultimately limits the voltage gain of a single bipolar stage?
Question 5
Reaching back to chapter 4: a stage with RC = 4.7 kΩ at 1 mA leaves 100 Ω of its emitter resistor unbypassed. Gain?
Question 6
Reaching back to chapter 3: why does an emitter follower placed between two common-emitter stages recover most of the lost gain?
Every word this chapter introduced
Where this goes next
About the simulations. Every operating point on these benches comes from the same damped Newton–Raphson solve of the Ebers–Moll transport equations used since chapter 2, with the Early effect and the temperature-dependent saturation current included. Bench 1’s waveform is solved point by point and its distortion figure is measured from those samples by correlating against the fundamental — it is not the textbook formula drawn as a curve, which is why it agrees with v̂/4VT at small signals and departs from it at large ones. Bench 2’s gm, re, rπ and ro are read off the solved point at the temperature you set, so VT = kT/q really is doing the work. Benches 3 to 6 combine solved bias points with the standard hybrid-π small-signal expressions. Three simplifications worth knowing: β is held constant with current, all capacitors are ideal so nothing here has a frequency response, and the Early effect is modelled with a single VA = 75 V rather than a bias-dependent one. Notation follows Sedra & Smith; re = 25/IC(mA) is Horowitz & Hill’s shorthand for the same quantity.