Analog Electronics I · 120 h · Topic 2 of 8

Bipolar transistors

The component that changed the 20th century: a small current in the base controls a large current between collector and emitter. With that you amplify, switch and regulate. Everything that comes later in the program builds on this.

Transistors Power electronics Q-point Biasing Switching

🎛️ Interactive course on bipolar transistors Eleven topics with characteristic curves that move, the Q-point shifting along the load line live, the transistor as a switch driving an LED, the drift of the operating point when β triples, and the common-emitter, common-collector and Darlington amplifiers. It is worth keeping it open alongside while reading this page. ›

01How it is built

A bipolar transistor (BJT, Bipolar Junction Transistor) is two PN junctions sharing a very thin central region. Depending on the order of the materials there are two types: NPN and PNP.

Figure 1. Structure and symbols. The three regions are not symmetrical: the emitter is heavily doped, the base is extremely thin and lightly doped, and the collector is large to dissipate heat. That is why the emitter and collector cannot be interchanged.

02Transistor action

For an NPN to work in its useful region, two conditions are needed at the same time:

The base-emitter junction forward biased (V_BE ≈ 0.7 V).
The base-collector junction reverse biased (the collector more positive than the base).

Once both are met, the following happens. The heavily doped emitter injects a flood of electrons into the base. Since the base is extremely thin and lightly doped, only 1% of those electrons find a hole to recombine with; the remaining 99% cross the base and are captured by the collector's field.

Result: a very small current flows through the base, and a current a hundred times larger flows through the collector — and the small one controls the large one.

I_{E} = I_{C} + I_{B}

Kirchhoff's current law applied to the transistor. The current entering through the base plus the current entering through the collector all leaves through the emitter (in an NPN).

The idea to hold on to

The bipolar transistor is a current-controlled current source. It does not amplify energy out of nothing: it takes energy from the power supply and lets more or less through depending on what the base tells it. It is a faucet, and the base is the handle.

03DC current gain

The ratio between the collector current and the base current is the gain β (beta), which appears in datasheets as h_FE.

β = \frac{I_{C}}{I_{B}}

α = \frac{I_{C}}{I_{E}} = \frac{β}{β + 1}

β is typically 100 to 300. α is always slightly less than 1 (0.99 for β = 100).

β is not a reliable figure

This is the most important point of the whole section. In the datasheet of a BC547, β is listed as 110 to 800: a range of 7 to 1. It also changes with temperature and with collector current. That is why no well-designed circuit depends on the value of β. The entire topic of biasing that follows exists precisely for this reason.

Worked example

A transistor with β = 150 has I_B = 40 µA. Calculate I_C, I_E and α.

I_C = β · I_B = 150 × 40 µA = 6 mA
I_E = I_C + I_B = 6 mA + 0.04 mA = 6.04 mA
α = 150 / 151 = 0.9934

As you can see, I_C ≈ I_E with an error of less than 1%. That approximation is used constantly and greatly simplifies the calculations.

04The three operating regions

Region	B-E junction	B-C junction	V_CE	Behavior	Use
Cutoff	Reverse (or < 0.6 V)	Reverse	≈ V_CC	I_C ≈ 0. Open switch.	Digital
Active	Forward (0.7 V)	Reverse	between 0.2 and V_CC	I_C = β·I_B. Amplifies.	Analog
Saturation	Forward	Forward	≈ 0.2 V	I_C is set by the load, not by β. Closed switch.	Digital

There is a fourth region, the reverse region (emitter and collector interchanged), which is hardly ever used: the gain falls below 5 because the structure is not symmetrical.

How to tell which region it is in, by measuring

With the transistor operating, measure V_CE:

V_CE ≈ V_CC → it is in cutoff.
V_CE ≈ 0.2 V → it is saturated.
V_CE intermediate (ideally ≈ V_CC/2) → it is in the active region.

05Characteristic curves

A transistor is not described by a single curve but by a family of them: I_C as a function of V_CE, one curve for each value of I_B.

Figure 2. Output curves with the load line, animated: the Q-point moves as the base current changes, from cutoff to saturation. Only in the middle does the transistor amplify.

Two regions can be distinguished:

Initial knee (V_CE < 0.2 V): the current rises very quickly. This is the saturation region.
Flat section: I_C hardly changes even as V_CE rises. This is the active region, where the transistor behaves as a current source.

The slight slope of the flat section is due to the Early effect (base-width modulation) and is the reason why the output resistance of the transistor is not infinite.

06Load line and Q-point

The curves describe the transistor; the load line describes the external circuit. Where they cross is the actual operating point, called the Q-point (from quiescent, at rest).

Applying Kirchhoff's law to the collector-emitter loop of a circuit with collector resistor R_C:

V_{C E} = V_{C C} - I_{C} \cdot R_{C}

DC load line equation.

This is the equation of a straight line, and to draw it its two end points are enough:

Cutoff end

I_C = 0 → V_CE = V_CC. The transistor does not conduct and the entire supply voltage drops across it.

Saturation end

V_CE = 0 → I_C(sat) = V_CC / R_C. The transistor conducts as much as the load allows.

Where to put the Q-point

It depends on what the transistor is used for:

Amplifier: in the middle of the line (V_CE ≈ V_CC/2). That way the signal can swing up and down equally without clipping. This is the maximum symmetrical swing.
Switch: jumping between the two end points (cutoff and saturation), without staying in the middle. At the end points it dissipates almost no power.

Power dissipation

The power the transistor dissipates is P = V_CE · I_C. In cutoff I_C = 0 and in saturation V_CE ≈ 0.2 V: in both cases the power is negligible. The midpoint of the line is precisely the point of maximum dissipation, and that is why a class A amplifier heats up even with no signal.

07The transistor as a switch

This is the most immediate application and the one that connects with Digital Electronics I: the output of a logic gate cannot drive a relay, a motor or a lamp, but it can drive the base of a transistor that can.

Figure 3. Transistor as a switch, animated: input at zero, input at 5 V, and the instant of turn-off, where the flyback diode gives the coil's spike a path.

Worked example · Calculating R_B

You want to saturate a BC547 (guaranteed minimum β = 110) to drive a 12 V relay that draws 30 mA. The control signal comes from a 5 V digital output.

1. Minimum base current to reach saturation:

I_{B \min} = \frac{I_{C}}{β_{\min}} = \frac{30 mA}{110} = 0.27 mA

2. Apply an overdrive factor of 3 to 5 to ensure saturation even with the worst-case β and at low temperature. With a factor of 4: I_B = 1.1 mA.

3. The base resistor:

R_{B} = \frac{V_{i n} - V_{B E}}{I_{B}} = \frac{5 - 0.7}{0.0011} = 3900 Ω

Adopt 3.9 kΩ (standard value) or 4.7 kΩ if a little less margin is acceptable.

4. Dissipation check: saturated, V_CE ≈ 0.2 V and I_C = 30 mA → P = 6 mW. A BC547 (500 mW) is more than enough.

Inductive loads

Relays, motors, solenoids and coils store energy in their magnetic field. When the current is cut off abruptly, that energy is released as a voltage spike of hundreds of volts (V = L·di/dt). The flyback diode gives that current a path and holds it at 0.7 V. Without it, the transistor breaks down on the first switching.

08Biasing

Biasing means setting the Q-point and, above all, making it stay there even when the temperature or the transistor changes. This is what separates a circuit that works from one that works only sometimes.

Fixed-bias (fixed-base) biasing

A single resistor between V_CC and the base. It is the simplest and the worst.

I_{B} = \frac{V_{C C} - 0.7}{R_{B}}, I_{C} = β \cdot I_{B}

The problem is obvious: I_C is directly proportional to β.

If the transistor is replaced by another of the same type but with β = 300 instead of 100, the collector current triples and the Q-point goes into saturation. The circuit stops amplifying. And since β also increases with temperature, the circuit drifts on its own while operating.

Voltage-divider biasing

This is the one used in practice. Two resistors set the base voltage and an emitter resistor introduces negative feedback.

Figure 4. Voltage-divider biasing. It is the standard configuration of every amplifier in Years 4 and 5.

The calculation has four steps, and β appears in none of them:

V_{B} = V_{C C} \cdot \frac{R 2}{R 1 + R 2}

V_{E} = V_{B} - 0.7 V

I_{E} = \frac{V_{E}}{R_{E}} \approx I_{C}

V_{C E} = V_{C C} - I_{C} \cdot (R_{C} + R_{E})

Calculation of the Q-point with a voltage divider. Valid if the divider current is at least 10 times I_B, a condition that is met by choosing R2 ≤ β·R_E/10.

Worked example · The Q-point and its stability

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

V_B = 12 × 10/(47+10) = 2.11 V
V_E = 2.11 − 0.7 = 1.41 V
I_E = 1.41 / 470 = 3.0 mA ≈ I_C
V_CE = 12 − 3.0 mA × (2200 + 470) = 12 − 8.0 = 4.0 V

Stability test: if the transistor is replaced by one with a β three times higher, the calculation does not change at all, because β does not appear in any formula. With fixed bias, the same substitution would triple I_C. That is the whole point of the voltage divider.

Why R_E stabilizes

It is a negative-feedback loop: if I_C tends to rise because of temperature, V_E = I_E·R_E rises. Since V_B is fixed by the divider, V_BE = V_B − V_E decreases, and that reduces I_C. The circuit corrects itself. The larger R_E is, the more stable — but the lower the AC voltage gain, which is why later it is bypassed with a capacitor.

09Regulated power supplies with transistors

The Zener regulator is only good for small currents. By adding a transistor in an emitter follower configuration, the available current is multiplied by β.

Figure 5. Regulated power supply with emitter follower, animated: the load changes and the output does not move. The output current is β times the base current, so the Zener hardly notices.

The output is V_out = V_Z − 0.7 V. The Zener only has to deliver I_C/β, that is, a few milliamperes, while the transistor handles the entire load current. With β = 100, a Zener that could handle 50 mA now allows 5 A — limited, of course, by the transistor's power dissipation:

P_{D} = (V_{i n} - V_{o u t}) \cdot I_{o u t}

With 18 V in, 12 V out and 1 A, the transistor dissipates 6 W: it definitely needs a heat sink.

🔋 Regulated power supply with the LM317 The integrated version of this same circuit, with built-in thermal protection and current limiting. It walks through the four stages with the real waveforms. ›

10Packages, identification and power dissipation

Transistor	Type	I_C max.	V_CEO	P_D	Typical β	Package
BC547	NPN	100 mA	45 V	500 mW	110-800	TO-92
BC557	PNP	100 mA	45 V	500 mW	110-800	TO-92 (complement of the 547)
2N2222	NPN	800 mA	40 V	500 mW	100-300	TO-92 / TO-18
BD139	NPN	1.5 A	80 V	12.5 W	40-160	TO-126, with heat sink
TIP31C	NPN	3 A	100 V	40 W	10-50	TO-220
TIP120	NPN Darlington	5 A	60 V	65 W	≥ 1000	TO-220

The official datasheets are at onsemi.com (Texas Instruments does not make generic BJTs). The pinout is not universal: a BC547 in a TO-92 package, viewed from the front with the flat face toward the viewer, has C-B-E from left to right; a 2N2222 in the same package has E-B-C. Mixing them up is the most common mistake in the shop.

Power dissipation and heat sinks

The P_D in the datasheet is valid at a case temperature of 25 °C, which almost never happens. In practice a safety factor is applied: do not exceed half of the rated power, and fit a heat sink with silicone thermal grease as soon as you go past 1 W. Watch out: in many TO-220 packages the metal tab is connected to the collector, so if two transistors are bolted to the same heat sink you must use an insulating mica pad and a plastic shoulder washer.

11In the lab

Lab 1 · Identifying the leads and type with a multimeter

A transistor is “two diodes back to back.” On the diode range:

Look for the lead that, with the red probe held fixed, reads ≈ 0.7 V against the other two: that is the base and the transistor is an NPN.
If that happens with the black probe held fixed, it is a PNP.
Between the two remaining leads the reading must be OL in both directions. If it conducts, the transistor is shorted.
Of those two, the emitter reads a few hundredths of a volt more than the collector against the base (0.72 vs 0.68 V, approximately). If the multimeter has an h_FE function, you can confirm by trying both orientations: the one that gives the higher gain is the correct one.

Lab 2 · Measuring β

Build the circuit with R_B = 470 kΩ from 12 V to the base and R_C = 1 kΩ from 12 V to the collector, emitter to ground. Measure the drop across R_B and across R_C, calculate I_B and I_C, and obtain β. Repeat with three transistors of the same type and compare: the spread among them is the lesson of the exercise.

Lab 3 · Cutoff and saturation

A transistor with an LED in the collector (with its resistor) and a pushbutton that takes the base to 5 V through 4.7 kΩ. Measure V_CE with the pushbutton released (it should read ≈ V_CC) and pressed (it should read ≈ 0.2 V). Calculate the power dissipated in both cases and check that it is minimal.

Lab 4 · Q-point stability

Build two circuits: one with fixed bias and one with a voltage divider, adjusted for the same initial Q-point. Replace the transistor with another of the same type (with a different β) in both and measure V_CE again. In the fixed-bias circuit the Q-point shifts noticeably; in the divider circuit it hardly moves. This is the exercise that justifies the whole topic.

12Common mistakes

Symptom	Usual cause
The transistor never conducts	Leads reversed (emitter and collector swapped), or V_BE below 0.6 V.
It always conducts, even with no signal	The base resistor is missing, or the divider is miscalculated and left it saturated.
It gets very hot with no load	Q-point in the middle of the line with high current. This is normal in class A, but it is worth checking the dissipation.
The transistor burns out when the relay is switched off	The flyback diode is missing.
It works with one transistor but not with another identical one	The circuit depends on β: fixed-bias biasing. You need to switch to a voltage divider.
The output signal is clipped at the top or bottom	Q-point off-center: the signal runs into saturation or cutoff before completing the cycle.
It works cold and fails when it warms up	Thermal drift. R_E is missing or too small.

13Self-assessment

Why does the base of a transistor have to be thin and lightly doped?

So that the carriers injected by the emitter find nothing to recombine with and cross over to the collector. If the base were thick or heavily doped, most of them would recombine, I_B would be large and β small: there would be no transistor action.

Can a transistor be used with the emitter and collector interchanged?

It works, but very poorly: the gain falls below 5 and the reverse voltage that the base-emitter junction can withstand is only 5 or 6 V. The three regions have different doping levels and sizes, so the transistor is not symmetrical.

A BC547 with β = 200 has I_B = 25 µA. What are I_C and I_E?

I_C = 200 × 25 µA = 5 mA. I_E = 5 + 0.025 = 5.025 mA. In practice, I_C ≈ I_E is assumed.

What are the two end points of the load line if V_CC = 15 V and R_C = 3 kΩ?

Cutoff: V_CE = 15 V with I_C = 0.
Saturation: I_C(sat) = 15 / 3000 = 5 mA with V_CE = 0.
For maximum swing, the Q-point goes at V_CE = 7.5 V with I_C = 2.5 mA.

Why is fixed-bias biasing bad?

Because I_C = β·I_B depends directly on β, which varies by up to 7 to 1 between units of the same model and also increases with temperature. The Q-point shifts when the transistor is replaced or when it heats up, and the circuit stops working.

Explain in one sentence how R_E stabilizes the circuit.

If I_C rises, V_E rises; since V_B is fixed by the divider, V_BE = V_B − V_E drops and the current falls back down. It is negative feedback.

Why is the base overdriven when the transistor is used as a switch?

To guarantee saturation with the worst possible β and at any temperature. 3 to 5 times the theoretical minimum I_B is used. If the transistor does not reach saturation, it stays in the active region, V_CE is high and it dissipates a lot of heat.

A power supply with an emitter follower delivers 12 V and 1.5 A, fed from 20 V. How much does the transistor dissipate?

P = (20 − 12) × 1.5 = 12 W. It absolutely needs a properly sized heat sink, and it is advisable to lower the input voltage to reduce that loss — which is the reason why switching power supplies replaced linear ones in power equipment.

What does a multimeter read between the base and emitter of a good NPN, on the diode range?

With the red probe on the base and the black one on the emitter: ≈ 0.7 V (junction forward biased). Reversed: OL. If it reads low in both directions, the junction is shorted and the transistor is useless.

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