The Follower Bench

The 5 volts that isn’t 5 volts

12 minutes · a circuit you trust, broken

Recall From chapter 2: what does the emitter of a transistor always carry more of than the collector? show answer

Two 10 kΩ resistors in series across 10 V. The midpoint sits at 5.00 V. You have drawn this a hundred times.

Now connect something to that midpoint.

Commit before you touch anything

You hang a 10 kΩ load on the midpoint of that divider. What does it read now?

Answer: C, 3.33 V. A third of the voltage, gone, and nothing inside the divider changed. Drag the load on the bench and watch it collapse further.

Why this matters more than it looks

Around 30–40% of students — including a third of second-year undergraduates — carry a model in their head where a source produces its voltage no matter what you connect to it. If you picked A or B, you are in good company, and you have just found the most useful thing in this chapter.

Every source has a hidden resistance in series with it. For our divider it is the two resistors in parallel: 10 k ∥ 10 k = 5 kΩ. That number is the divider’s output impedance, and the load forms a second divider with it.

Drag the load down to 1 kΩ on the bench. The output falls to 0.83 V. Your 5-volt reference is now a 0.83-volt reference, and you changed nothing except what you hung off it.

The rule this gives you: a voltage source is only as good as the current it can supply without sagging. Two circuits connected together always form a divider, whether you drew one or not. The one on the left has an output impedance; the one on the right has an input impedance. Keep the second much bigger than the first — ten times is a good habit — and almost nothing is lost.

Now press “insert follower.” One transistor and nothing else, dropped between the divider and the load. The output goes back up and, more importantly, it stays up when you drag the load. Swap to 1 kΩ: the bare divider gives 0.83 V, the follower gives about 4.2 V and barely flinches.

The follower’s voltage gain is one. It has just saved the circuit. Hold those two facts together and the rest of the chapter is easy.

One honest detail on the plot. The two curves cross at about 46 kΩ. To the right of that, the bare divider is better — the follower is costing you 0.65 V and giving nothing back, because a load that light was never a problem. The follower earns its place only when the load is heavy, which is exactly when you need it. Nothing in electronics is free; the skill is knowing what you are buying.

In the wild

Your multimeter is 10 MΩ on purpose. That number on the front panel is its input impedance, and it is enormous so that measuring a circuit does not change it. An old moving-coil meter had a much lower one, which is why measuring with the wrong meter used to give you the wrong answer.

Your oscilloscope probe is worse. A scope input is typically 1 MΩ. Probe a 1 MΩ source and you have just halved the signal you were trying to look at. This is why 10× probes exist.

The phone charger that dims your screen. A cheap charger has a high output impedance. Plug in something hungry and its 5 V sags to 4.6 V, and the phone charges slowly or refuses. A good one has milliohms of output impedance and does not move.

Before you move on

A sensor outputs 2 V through an internal 100 kΩ. You feed it to an amplifier whose input impedance is 100 kΩ. What does the amplifier actually see?

Two equal resistances in series across the source: half the voltage lands on each. You have thrown away 50% of your signal before doing anything with it. If the amplifier input were 1 MΩ instead, you would keep 91% of it.

Bench 01 · loading a divider NO BUFFER

● Locked until you commit a prediction above.

Load resistance 10.0 kΩ

100 Ω3.2 kΩ1 MΩ

Output—

Lost—

Load current—

Source impedance5.00 kΩ

A signal follower, not a DC follower

14 minutes · the distinction that trips up final-year students

Recall From lesson 1: what is the output impedance of two 10 kΩ resistors forming a divider? show answer

Here is the whole circuit. Collector straight to the supply. Signal in at the base. Output taken at the emitter, across a resistor to ground.

The collector has no resistor and no signal on it — it just sits at the supply rail. That is why the circuit’s formal name is common collector: the collector is the terminal common to input and output, doing nothing for either.

Commit before you touch anything

You sweep the input from −2 V up to +2 V. What does the output do?

Answer: C. In a study of 214 students who had already been taught this, 12–22% chose A and up to 12% chose B. Sweep it yourself and watch what happens below 0.6 V.

Why A is wrong, and why it matters

Answer A draws a straight line all the way down: at Vin = −2 V it claims Vout = −2.6 V. Look at what that would require. The emitter would have to sit below the base, and current would have to run backwards through the base–emitter junction — through a reverse-biased diode.

It cannot. Below about 0.6 V the transistor is simply cut off and the emitter resistor pulls the output to ground and holds it there. Drag the slider into the negative and watch the badge say CUT OFF.

The sentence worth memorising: the emitter follower is a signal follower, not a DC follower. Wiggle the input by 100 mV and the output wiggles by 100 mV — that is what “follower” means. But the output is not equal to the input; it sits about 0.65 V below it. Both statements are true at the same time, and confusing them is the single most common error on this circuit.

Watch the offset move

Look carefully at the readout as you sweep. The gap is not exactly 0.65 V — it is 0.59 V near the bottom and 0.68 V near the top. That is chapter 2, lesson 3, showing up again: more emitter current means more Vbe, at 60 mV per decade. The offset is a measurement, not a constant.

And that slight movement is exactly why the gain is not quite 1. Over a 1 V input swing the offset changes by a few millivolts, so the output moves 0.995 V rather than 1.000 V. We will put a number on that in the next lesson.

In the wild

Why an audio buffer needs two supplies, or a bias. Music swings both ways around zero. A follower on a single supply chops the negative half off completely, exactly as you are about to see on the bench. Real single-supply audio circuits lift the whole signal to the middle of the rail first — and that is what chapter 4 is about.

The 0.65 V you can cancel. Put a PNP follower after an NPN one and the second one’s +0.65 V offset cancels the first one’s −0.65 V. This is one of the reasons complementary pairs turn up everywhere, and it is the seed of the push-pull stage in chapter 6.

Before you move on

A follower’s input sits at 4.00 V DC with a 50 mV signal riding on it. What is at the output?

The DC level drops by one Vbe. The signal comes through essentially untouched. That is the whole meaning of “signal follower, not DC follower” in one line — and it is why you will see a capacitor used to strip that DC offset off again.

Bench 02 · input against output ACTIVE

● Locked until you commit a prediction above.

Input voltage +1.00 V

−2 V+4.5 V+11 V

Vin—

Vout—

The gap (Vbe)—

Emitter current—

Vcc = 12 V, R_E = 3.3 kΩ, source resistance 10 kΩ.

Impedance, as a number you dial in

15 minutes · the abstraction, made into a measurement

Recall From lesson 2: is the output of a follower equal to its input? show answer

“Input impedance” is usually introduced as a formula and stays an abstraction forever. So here it is as a procedure instead — something you do with a knob, on a bench, in two minutes.

Commit before you touch anything

To find a circuit’s output impedance you load it until the output voltage falls to exactly half its unloaded value. What is the answer then?

Answer: B. Two equal resistances in series split a voltage evenly — so if the load halves the output, the load equals the source. That is the whole method. Try it on the bench.

Do the measurement, twice

Output impedance. Note the unloaded output. Drag the test load down until the output reads exactly half. Read the load value. That is the output impedance. The bench tells you when you are there.
Input impedance. Same trick at the other end. Put a test resistor in series with the source and increase it until the voltage arriving at the base is half the source voltage. That resistor equals the input impedance.

This is how it is done on a real bench with a real signal generator. No model, no formula — just a divider and a knob.

Now the numbers, and where they come from

One new quantity first. A transistor has an internal resistance at its emitter, set entirely by how much current is flowing:

r_e = 25 mV / I_C(mA)

At 1 mA that is 25 Ω. At 0.1 mA it is 250 Ω. It is not a component you can see — it is the slope of the exponential from chapter 2, lesson 3, turned into ohms. Everything else in this chapter is built on it.

Input impedance

R_in = (β+1) × (r_e + R_E∥R_L)
Whatever hangs on the emitter looks β+1 times bigger when seen from the base.

Output impedance

R_out = r_e + R_source / (β+1)
And whatever drives the base looks β+1 times smaller when seen from the emitter.

Voltage gain

A_v = (R_E∥R_L) / (R_E∥R_L + r_e)
A divider between r_e and the load. Just under 1, and closer the more current you run.

The whole chapter in one line: a transistor is an impedance transformer with a ratio of β+1. Look through it one way and resistances get multiplied; look through it the other way and they get divided. That is what you buy with your unity gain, and it is worth far more than a voltage gain of 10.

Try it on the bench. With a 10 kΩ source, a 4.7 kΩ emitter resistor and a 1 kΩ load, you get an input impedance around 125 kΩ and an output impedance around 70 Ω. The circuit takes a source that could barely drive a 100 kΩ load and turns it into one that can drive 100 Ω. A ratio of a thousand, from one transistor.

In the wild

Why a guitar sounds dull through a cheap pedal. A magnetic pickup has an output impedance of hundreds of kilohms and hates being loaded. An input impedance of 1 MΩ is the accepted minimum for a guitar input, and pedals that get this wrong audibly lose treble. “Tone suck” is guitarists’ word for a loading error.

Why a long cable needs a driver. A few metres of coaxial cable is a substantial capacitance. Driving it from a 10 kΩ source rolls off anything fast; driving it from a follower’s 70 Ω does not. Every piece of test equipment buffers its output for exactly this reason.

The 10× scope probe. It deliberately throws away 90% of your signal to raise the impedance the probe presents from 1 MΩ to 10 MΩ. Engineers trade signal for impedance all the time, quite happily, because you can always amplify a small signal and you can never un-load a loaded source.

Before you move on

You need a follower’s input impedance to be higher. Which change helps most?

R_in = (β+1)(r_e + R_E∥R_L), so both factors are in your hands. In practice the load is fixed and you cannot pick β (chapter 2, lesson 5), which is why the real answer to “I need much more input impedance” is two followers in series — a Darlington — giving you β².

Bench 03 · measuring impedance —

● Locked until you commit a prediction above.

Test load on the output 1.00 kΩ

10 Ω3.2 kΩ1 MΩ

Unloaded—

With test resistor—

Fraction left—

r_e right now—

Why it follows: your first feedback loop

12 minutes · the idea the rest of electronics is built on

Recall From lesson 3: what is r_e at a collector current of 1 mA? show answer

Nothing in the circuit is telling the output to equal the input. So why does it?

Look at what the transistor actually responds to. It does not see the input voltage and it does not see the output voltage. It sees the difference between them — that is what Vbe is. And chapter 2, lesson 3 told you how violently it responds: 60 mV of extra difference and the current goes up tenfold.

Commit before you touch anything

The output is sitting at 5.00 V. You attach something that tries to drag it down to 4.00 V. What happens?

Answer: B. And the mechanism is worth watching happen. Drag the disturbance slider and keep your eye on Vbe.

The loop, step by step

Something pulls the output down by 10 mV.
The base has not moved, so Vbe just went up by 10 mV.
10 mV more Vbe is a 50% increase in emitter current — a lot of extra current, immediately.
That extra current flows into the disturbance and pushes the output back up.
It settles where the correction exactly cancels the disturbance, which is almost where it started.

That is negative feedback. The output is subtracted from the input, the difference drives the device, and the device acts to reduce the difference. You have just met the most important idea in analogue electronics, in its simplest possible form.

Two things you already know are now the same thing. Why is the gain almost exactly 1? Because feedback forces the output to track the input. Why is the output impedance only 70 Ω when there is a 4.7 kΩ resistor sitting right there? Because feedback fights any attempt to move the output. Low output impedance is negative feedback, seen from outside.

And the amount of correction available is set by the steepness of the exponential. That is why r_e — the same 25/I_C from the last lesson — turns up in both the gain and the output impedance. It is the loop’s strength, measured in ohms.

In the wild

An op-amp follower is this circuit, with a much stronger loop. Wire an op-amp’s output straight back to its inverting input and you get a buffer whose gain is 1.0000 and whose output impedance is milliohms. Same idea, same subtraction, gain of a hundred thousand inside the loop instead of a few hundred. When you meet op-amps in chapter 8 you will already understand why they behave the way they do.

Cruise control. Speed is subtracted from the setpoint, the difference works the throttle, and the car holds speed up a hill without you doing anything. The hill is the disturbance slider on this bench.

The cost, which is real. Feedback loops can oscillate. A follower driving a capacitive load — a long cable, say — can ring or sing at radio frequencies, which is why you will see a small resistor in series with the output of many buffers. Free lunches remain unavailable.

Before you move on

You increase the follower’s bias current from 1 mA to 10 mA. What happens to its output impedance?

r_e = 25/I_C(mA), so ten times the current is a tenth of the resistance and a stronger loop. This is the standard reason output stages are run at high quiescent current: you are buying stiffness. It costs power, which is why your amplifier gets warm doing nothing.

Bench 04 · the loop correcting itself HOLDING

● Locked until you commit a prediction above.

Disturbance — current pulled out of the output 2.50 mA

−3 mA in0+6 mA out

Output—

Moved by—

Vbe corrected to—

Transistor supplies—

Base held at 6.00 V. Without the transistor, this disturbance would move the output a long way.

It can push. It cannot pull.

14 minutes · where a DC number breaks an AC signal

Recall From lesson 4: what does the transistor actually respond to — the input, the output, or something else? show answer

Everything so far has been generous about the follower. Here is what it cannot do, and the failure is instructive rather than embarrassing.

To push the output up, the transistor turns on harder and delivers current from the supply. It can deliver a lot. To pull the output down, it turns off — and then the only thing left to drain current is the emitter resistor. The transistor is not pulling. It is getting out of the way and hoping the resistor is enough.

Commit before you touch anything

A follower is biased at 1 mA through a 4.7 kΩ emitter resistor, driving a 470 Ω load through a capacitor. You ask for a 2 V peak sine at the output. What comes out?

Answer: B. The bottom. On the negative peak the load wants 2 V / 470 Ω = 4.3 mA to flow back into the emitter — and there is only 1 mA of standing current available to give up. The transistor cuts off and the waveform flattens.

The number that decides it

The most the follower can sink is its own quiescent emitter current. Ask for more and it cuts off. So:

peak load current < standing emitter current

Now raise the bias slider from 1 mA to 10 mA and run the same signal. The clipping vanishes. Nothing else changed — not the transistor, not the load, not the signal. Only a DC current that has nothing to do with your signal at all.

This is the hinge into chapter 4. A quiescent current is a DC quantity. Clipping is an AC symptom. They are supposed to be separate analyses — and here is a case where a DC decision silently sets an AC limit. Separating DC from signal is how you analyse a circuit; remembering that the DC sets the signal’s parameters and its limits is how you design one. Chapter 4 is that argument in full.

The cost of fixing it this way

Raising the standing current to 10 mA solves the clipping and burns 10 mA × 12 V = 120 mW continuously, whether a signal is present or not. That is a Class A output stage: it works, it is beautifully linear, and it is enormously wasteful. A 50 W Class A amplifier draws 50 W of heat at idle.

The escape is to use a second transistor that is good at pulling down — a PNP, from chapter 2, lesson 6. One pushes, one pulls, and neither has to stand there burning current. That is the push-pull stage, and it is chapter 6.

In the wild

Why a headphone output has a minimum impedance. The spec sheet saying “16 Ω minimum” is this calculation. Below that, the peak current demanded exceeds what the output stage can deliver and the sound distorts on peaks — loud passages only, which is exactly what makes it hard to diagnose by ear.

Why audiophiles argue about Class A. A Class A amplifier never turns off, so it never has the small distortion that happens as one transistor hands over to another. It also runs hot enough to heat a room. Both facts come straight from this bench.

Why your Arduino can light an LED but not sink one. Many microcontroller pins can source and sink different amounts, and driving an LED “the other way round” sometimes works less well. It is the same asymmetry, in a different device.

Before you move on

Your follower clips on the negative peaks into a 100 Ω load. Which fix works?

The standing current is what you are allowed to give up on the negative half, and a smaller R_E raises it. β is irrelevant here — the limit is current available at the emitter, not current gain. (The better answer in a real design is a PNP doing the pulling, but that is chapter 6.)

Bench 05 · driving a real load CLEAN

● Locked until you commit a prediction above.

Emitter resistor — sets the standing current 4.7 kΩ

220 Ω1 kΩ6.8 kΩ

Load 470 Ω

47 Ω680 Ω10 kΩ

Signal amplitude 2.0 V pk

Standing current—

Load wants (peak)—

Output swing—

Idle power burnt—

The payoff: a power supply from three parts

12 minutes · everything so far, in one circuit

Recall From lesson 5: what limits how much current a follower can pull out of its output? show answer

A Zener diode — chapter 1, lesson 6 — holds a fixed voltage when you run current backwards through it. So a resistor and a Zener give you a voltage reference. Except they don’t, quite.

Commit before you touch anything

A 5.6 V Zener fed from 12 V through a 1 kΩ resistor. You draw 20 mA from it. What happens to the 5.6 V?

Answer: B. Lesson 1 again, in a different costume. The Zener sets a voltage but the resistor limits the current, so the reference has a high output impedance and sags under load.

Add one transistor

Put a follower between the Zener and the load and everything changes. Trace it:

The Zener holds the base at 5.6 V, and now only has to supply base current — microamps, not milliamps.
The output sits one Vbe below: about 4.95 V.
The load current comes from the collector, straight off the 12 V rail, and never touches the Zener.
The output impedance is r_e plus the Zener’s impedance divided by β+1 — a few ohms instead of a kilohm.

Drag the load on the bench and compare the two traces. The bare Zener collapses. The buffered one holds until the transistor runs out of current or the input rail sags.

Read the division of labour, because it recurs everywhere. One part decides what the voltage should be. A different part supplies the current to make it so. Almost every regulator, reference and control loop you will ever meet is organised this way, and the follower is the simplest possible version of the second half.

What it still cannot do

The output tracks the Zener, so it inherits the Zener’s temperature drift. It is a follower, not a regulator with real feedback — nothing measures the output and corrects it against a reference. And Vbe drifts −2 mV/°C on top. A 7805 does much better because it closes a loop around an amplifier, which is chapter 8.

In the wild

This is the ancestor of the 7805 in your parts drawer. Open one up and you find a bandgap reference (chapter 2, lesson 3), an error amplifier, and a big follower as the pass transistor. The follower is doing exactly this job, and it is the part that gets hot.

The heatsink tells you the whole story. A linear regulator dropping 12 V to 5 V at 1 A burns 7 W as heat in the pass transistor. It is not lost in the load — it is dissipated in that one component, which is why it needs a metal tab and why switching regulators exist.

Where you have already seen a follower without knowing. Every op-amp has one on its output. Every logic gate’s output stage is a version of the same idea. Every voltage reference chip. You have been using this circuit for years.

Before you move on

Your Zener-plus-follower supply gives 4.95 V at no load, and 4.90 V at 100 mA. What is its output impedance?

50 mV of sag divided by 100 mA of load is 0.5 Ω. Output impedance is just “how much does it droop per amp” — you never need the formula to measure it, only to predict it. Note this is a different measurement from lesson 3’s half-voltage trick and gives the same quantity.

Bench 06 · Zener reference, buffered and not —

● Locked until you commit a prediction above.

Load resistance 270 Ω

47 Ω680 Ω10 kΩ

Zener alone—

With follower—

R out, bare—

R out, buffered—

12 V in, 1 kΩ feed resistor, 5.6 V Zener.

✓

Chapter checkpoint

8 minutes · shuffled on purpose

Mixed up deliberately, and two of them reach back into chapters 1 and 2. Recognising which idea a question is about is the skill; solving it is the easy part.

Question 1

A follower’s input is at −1 V. Where is its output?

A needs current running backwards through a reverse-biased junction. The emitter resistor pulls the output to ground and holds it there. This is the error 12–22% of students make after being taught the circuit.

Question 2

Why is a follower’s output impedance so much lower than its emitter resistor?

Low output impedance and negative feedback are the same fact described from two sides. This is also why the gain is almost exactly 1 — one mechanism, two consequences.

Question 3

Same follower, biased at 2 mA. What is r_e?

25/I_C(mA) = 25/2 = 12.5 Ω. Worth being able to do in your head: this one number sets the gain, the output impedance and the strength of the feedback loop.

Question 4

A 12 V supply feeds a divider. You need the output to stay within 1% when a 220 Ω load is connected. What is required?

Losing 1% into 220 Ω needs a source impedance around 1/100 of the load. Resistor tolerance is a different problem entirely — it changes the unloaded value, not the sag. A divider that stiff would burn a lot of current, which is exactly the argument for a buffer.

Question 5

A follower biased at 5 mA drives a 100 Ω AC-coupled load. What is the largest clean output peak?

On the negative peak the load must push 5 mA back at most, and 5 mA through 100 Ω is 0.5 V. Beyond that the transistor cuts off. The supply is nowhere near the limit — the standing current is.

Question 6

Reaching back to chapter 2: a follower’s emitter current doubles. What happens to its base current?

The whole current relationship still holds. It is why the input impedance falls when you load the output harder — more emitter current means more base current means the source is loaded more. The follower buffers well, not perfectly.

Checkpoint score

0 / 6

Answer all six. Four or more and the chapter has landed.

Every word this chapter introduced

Loading L1

What happens when you connect two circuits: they form a divider whether you drew one or not.

Output impedance L1

How much a source droops per amp you draw. Low is stiff.

Input impedance L1

How much current a circuit steals from whatever drives it. High is polite.

Emitter follower L2

Collector to the rail, output at the emitter. Voltage gain 1, current gain β.

Common collector L2

The formal name. The collector is common to input and output and carries no signal.

Signal follower, not DC follower L2

δV_out = δV_in, but V_out = V_in − 0.65. Both true at once.

r_e L3

25 mV / I_C(mA). The transistor’s own emitter resistance — the exponential expressed in ohms.

Half-voltage method L3

Load a circuit until its output halves; that load equals its impedance. Works at either end.

Impedance transformation L3

×(β+1) looking from base to emitter, ÷(β+1) looking the other way.

Negative feedback L4

Output subtracted from input, difference drives the device, device reduces the difference.

Sourcing and sinking L5

Pushing current out versus letting it flow in. An NPN follower does the first actively and the second only through R_E.

Quiescent current L5

What flows with no signal. Sets how much signal current you are allowed.

Class A L5

An output stage that conducts all the time. Linear, and wasteful.

Pass transistor L6

The follower in a linear regulator. Carries the load current and dissipates the difference as heat.

Where this goes next

CH 04

Bias, and DC versus signal

The single most documented confusion in the whole subject, and the one intervention shown to halve it. Lesson 5 here was the trailer.

CH 05

Gain, and the wall

−R_C/R_E, then shrink R_E to nothing and meet r_e again as the thing that stops you.

CH 06

The output stage

The fix for lesson 5’s one-sided clipping: a PNP that does the pulling, and the crossover distortion it brings with it.

CH 07–12

FETs, op-amps, integrated circuits

The other kind of transistor, the loop that makes all of this automatic, and the circuits nobody can solder.

About the simulations. Every operating point is solved with the same damped Newton–Raphson iteration on the Ebers–Moll transport equations used in chapter 2, with the Early effect and temperature-dependent saturation current included — so the offset really does move with current, and the gain really is 0.995 rather than 1. Small-signal quantities (r_e, R_in, R_out) are computed from the solved operating point rather than assumed. The waveform in bench 5 is solved point by point rather than drawn from a formula, which is why the clipping appears on its own instead of being painted on. Two simplifications: β is held flat with current, and the coupling capacitor in bench 5 is treated as an ideal DC block, so you are seeing the mid-band behaviour with no low-frequency roll-off. Formulas follow Horowitz & Hill’s r_e = 25/I_C(mA) convention; Sedra & Smith write the same results as r_π = (β+1)r_e and g_m = I_C/V_T, which are the same equations in different clothes.