Bipolar Transistors Notes FE Sem 2 Semiconductor Physics | MU

The transistor, invented at Bell Labs in 1947 (point-contact) and 1948 (junction transistor, proposed by William Shockley and first made in 1950 to 1951), replaced vacuum tubes and started the electronics age. The bipolar junction transistor (BJT) is still used in amplifiers, switches, sensors and power circuits. It uses a small current to control a much larger one. This module of the Semiconductor Physics elective covers the BJT structure and its operation, its modes of operation, the input and output I-V characteristics of the common base (CB) and common emitter (CE) configurations, current gain, and the BJT as an amplifier and as a switch. It maps to course outcome CO4 (calculate current gain and plot I-V characteristics for CB and CE configurations).

Summary

BJT structure: A BJT is a three-layer, three-terminal semiconductor device with two p-n junctions. In an npn transistor a thin p-type base is sandwiched between an n-type emitter and an n-type collector; in a pnp transistor a thin n-type base lies between p-type emitter and collector. The junction between emitter and base is the emitter-base junction (EBJ) and the one between collector and base is the collector-base junction (CBJ). The three regions differ: the emitter is heavily doped and of moderate size, so that it can inject a large number of carriers into the base; the base is very thin (a few micrometres or less) and lightly doped, so that most injected carriers cross it without recombining; the collector is moderately doped and the largest region, because it collects the carriers and must dissipate most of the heat. The emitter and collector are therefore not interchangeable. It is called bipolar because both electrons and holes take part in conduction. In the symbol the arrow on the emitter shows the direction of conventional current: pointing out for npn and in for pnp.

BJT operation (npn, active mode): The emitter-base junction is forward biased (V_BE about 0.7 V for Si) and the collector-base junction is reverse biased. Forward bias lowers the EB barrier, so a large number of electrons are injected from the emitter into the base, forming the emitter current I_E. Since the base is thin and lightly doped, only a small fraction (typically 1 to 5 percent) of these electrons recombine with holes in the base; this recombination, together with holes injected from base into emitter, forms the small base current I_B. The remaining electrons diffuse across the base to the edge of the collector-base depletion region, where the strong field of the reverse-biased CBJ sweeps them into the collector, forming the collector current I_C. By Kirchhoff's current law, \(I_E = I_B + I_C\). In addition a small leakage current I_CBO flows across the reverse-biased CBJ due to minority carriers. Because I_C is controlled by the small I_B (or V_BE), and nearly independent of the collector voltage, the transistor acts as a current-controlled current source. In a pnp transistor the same happens with holes and with all polarities reversed.

Modes of operation: Depending on the bias of the two junctions, a BJT works in four modes. (i) Active (forward active): EBJ forward, CBJ reverse; I_C = βI_B; used for amplification. (ii) Saturation: both junctions forward biased; the collector cannot collect all the carriers and I_C is limited by the external circuit; V_CE is small, about 0.1 to 0.3 V (V_CE(sat)); the transistor behaves as a closed switch (ON). (iii) Cut-off: both junctions reverse biased; only tiny leakage currents flow; the transistor behaves as an open switch (OFF). (iv) Reverse active (inverse): EBJ reverse, CBJ forward; emitter and collector roles swap but the gain is very poor because of the doping asymmetry; rarely used.

Transistor configurations: Since a transistor has three terminals, one terminal is made common to the input and output circuits. Common base (CB): input between emitter and base, output between collector and base. Common emitter (CE): input between base and emitter, output between collector and emitter; the most widely used configuration. Common collector (CC, emitter follower): input between base and collector, output between emitter and collector; used as a buffer.

CB I-V characteristics: Input characteristics are plots of I_E against V_EB (or V_BE) at constant V_CB. They look like the forward characteristic of a diode: I_E is negligible until V_EB reaches the cut-in voltage (about 0.5 to 0.7 V for Si), then rises steeply. Increasing V_CB shifts the curve slightly left (more current for the same V_EB) because the base effectively becomes narrower (Early effect or base-width modulation). The input resistance \(r_i = \Delta V_{EB}/\Delta I_E\) is very low (tens of ohms). Output characteristics are plots of I_C against V_CB for different constant values of I_E. In the active region (V_CB reverse) the curves are almost horizontal lines with \(I_C \approx \alpha I_E\), nearly independent of V_CB; I_C continues to flow even at V_CB = 0 and becomes zero only when the CBJ is slightly forward biased (saturation region, to the left of the vertical axis). Below the I_E = 0 curve lies the cut-off region, where only I_CBO flows. Output resistance \(r_o = \Delta V_{CB}/\Delta I_C\) is very high (around a megohm). Current gain \(\alpha = I_C/I_E\) is slightly less than 1 (0.95 to 0.995), so the CB configuration gives no current gain but can give voltage and power gain; it is used in high-frequency amplifiers.

CE I-V characteristics: Input characteristics are plots of I_B against V_BE at constant V_CE. They resemble a forward-biased diode curve with currents in μA. Increasing V_CE shifts the curve slightly to the right (less base current for the same V_BE), because the wider collector depletion region narrows the base and reduces recombination. Input resistance \(r_i = \Delta V_{BE}/\Delta I_B\) is moderate (around 1 kΩ). Output characteristics are plots of I_C against V_CE for constant values of I_B. Each curve rises steeply for very small V_CE (below about 0.2 to 0.3 V, the saturation region), then becomes almost flat (active region) with a slight upward slope due to the Early effect; the spacing between curves represents β. The region below the I_B = 0 curve is cut-off, where a small leakage current I_CEO flows. At high V_CE the curves rise sharply (breakdown). Output resistance \(r_o = \Delta V_{CE}/\Delta I_C\) is a few tens of kilohms. Current gain \(\beta = I_C/I_B\) is large (typically 20 to 500), so CE gives high current, voltage and power gain, and its output is 180° out of phase with the input.

Current gain and relations: \(\alpha_{dc} = I_C/I_E\) (CB gain), \(\beta_{dc} = I_C/I_B\) (CE gain), \(\gamma = I_E/I_B = 1 + \beta\) (CC gain). Using \(I_E = I_B + I_C\): \(\beta = \dfrac{\alpha}{1 - \alpha}\) and \(\alpha = \dfrac{\beta}{1 + \beta}\). AC (small signal) gains are defined with changes: \(\alpha_{ac} = \Delta I_C/\Delta I_E\), \(\beta_{ac} = \Delta I_C/\Delta I_B\). Including leakage: \(I_C = \alpha I_E + I_{CBO}\) and \(I_C = \beta I_B + (1 + \beta)I_{CBO}\), so \(I_{CEO} = (1 + \beta)I_{CBO}\); since I_CBO doubles every 10 °C, I_CEO is sensitive to temperature, which is why bias stabilisation is needed.

BJT amplification: In the CE amplifier the base-emitter junction is forward biased by a DC bias network so that the transistor sits at a quiescent (Q) point in the middle of the active region. A small AC input voltage added to V_BE causes a small change ΔI_B, which causes a β times larger change \(\Delta I_C = \beta\Delta I_B\). Flowing through the collector load resistor R_C, this produces a large change in output voltage \(\Delta V_{CE} = -\Delta I_C R_C\). Voltage gain \(A_v = \Delta V_{out}/\Delta V_{in} \approx -\beta R_C/r_i\), which can be tens to hundreds; the minus sign shows phase reversal. The energy comes from the DC supply; the transistor only controls it. The Q point is chosen on the DC load line (\(V_{CE} = V_{CC} - I_CR_C\)) so that the signal swings without entering saturation or cut-off, which would clip and distort the output.

BJT as a switch: A transistor operated only between cut-off and saturation acts as an electronic switch. With the input at 0 V, V_BE is below cut-in, I_B = 0, I_C ≈ 0 and V_CE ≈ V_CC: the switch is OFF (open) and no current flows through the load. With a high input, enough base current is supplied to drive the transistor into saturation: I_C = I_C(sat) ≈ (V_CC − V_CE(sat))/R_C, V_CE ≈ 0.2 V: the switch is ON (closed) and the load gets nearly the full supply. The condition for saturation is \(I_B \geq I_{C(sat)}/\beta\); designers supply 2 to 10 times this value (overdrive) to make sure of saturation despite variations in β. In both states the power dissipated in the transistor is small (I_C ≈ 0 when OFF, V_CE ≈ 0.2 V when ON), so it runs cool. Because input high gives output low, the switch also works as a NOT gate (inverter). Applications: driving relays, LEDs, lamps, buzzers and motors from microcontroller pins, logic gates, and switching regulators. When driving an inductive load like a relay, a freewheeling diode is placed across the coil to protect the transistor from voltage spikes. Switching speed is limited by the charge stored in the base during saturation.

Mind map

Carrier flow in an npn transistor (active mode)

Important definitions

Formulas

Key points to remember

• Doping: emitter > collector > base. Size: collector > emitter > base.
• Active mode needs EBJ forward and CBJ reverse biased.
• α is slightly less than 1; β is large (20 to 500).
• CB: low input, very high output resistance, current gain α < 1; used at high frequency.
• CE: moderate input, high output resistance, high current and voltage gain, 180° phase shift; most used.
• Output curves are flat in the active region because IC depends on IE or IB, not on the collector voltage.
• ICEO is (1 + β) times ICBO, so CE is more temperature sensitive.
• Switch: OFF in cut-off, ON in saturation; transistor dissipates little power in both states.
• Use base overdrive (IB several times IC(sat)/β) to guarantee saturation.

Solved examples

Example 1. A transistor has α = 0.98. Find β. If IE = 5 mA, find IC and IB.

\(\beta = \alpha/(1 - \alpha) = 0.98/0.02 = 49\). \(I_C = \alpha I_E = 0.98\times5 = 4.9\) mA. \(I_B = I_E - I_C = 0.1\) mA \(= 100\) μA. Check: \(I_C/I_B = 4.9/0.1 = 49\).

Example 2. A transistor has β = 100 and IB = 20 μA. Find IC, IE and α.

\(I_C = \beta I_B = 100\times20\) μA \(= 2\) mA. \(I_E = I_B + I_C = 2.02\) mA. \(\alpha = \beta/(1 + \beta) = 100/101 = 0.990\).

Example 3. For a transistor with β = 100 and ICBO = 10 nA, find ICEO. What is IC when IB = 20 μA?

\(I_{CEO} = (1 + \beta)I_{CBO} = 101\times10\) nA \(= 1.01\) μA. \(I_C = \beta I_B + I_{CEO} = 2\) mA \(+ 0.00101\) mA \(\approx 2.001\) mA; the leakage is negligible at room temperature for Si.

Example 4. In a CE characteristic, IC changes from 2 mA to 3 mA when IB changes from 20 μA to 30 μA at constant VCE, and VBE changes by 20 mV for this change in IB. Find βac and the input resistance.

\(\beta_{ac} = \Delta I_C/\Delta I_B = 1\ \text{mA}/10\ \mu\text{A} = 100\). \(r_i = \Delta V_{BE}/\Delta I_B = 20\ \text{mV}/10\ \mu\text{A} = 2\) kΩ.

Example 5. A BJT switch drives a load RC = 1 kΩ from VCC = 5 V. β = 100, VCE(sat) = 0.2 V, VBE = 0.7 V, input high = 5 V. Find IC(sat), minimum IB, the maximum RB, and RB for an overdrive factor of 5.

\(I_{C(sat)} = (5 - 0.2)/1000 = 4.8\) mA. \(I_{B(min)} = 4.8\ \text{mA}/100 = 48\) μA. \(R_{B(max)} = (5 - 0.7)/48\times10^{-6} = 89.6\) kΩ. With overdrive 5: \(I_B = 240\) μA, \(R_B = 4.3/240\times10^{-6} = 17.9\) kΩ, so a standard 15 kΩ resistor (giving slightly more base current) is used.

Example 6. Find the power dissipated in the transistor of Example 5 in the ON and OFF states.

ON: \(P = V_{CE(sat)}I_C = 0.2\times4.8\times10^{-3} = 0.96\) mW (plus a negligible base power \(0.7\times0.24\) mA \(= 0.17\) mW). OFF: \(I_C \approx 0\), so \(P \approx 0\). The switch wastes very little power, while the load gets \(4.8\times4.8 \approx 23\) mW.

Important questions and answers

Q1. Describe the structure of a BJT and explain why the three regions are doped differently. (5 marks)

Diagrams: npn and pnp layer structures with E, B, C and symbols with emitter arrows. The emitter is heavily doped to inject plenty of majority carriers into the base and give high emitter efficiency. The base is very thin and lightly doped so that most injected carriers cross it without recombining, keeping I_B small and the gain high. The collector is moderately doped and largest because it must collect the carriers, withstand a large reverse voltage across the CBJ (wider depletion region in a lighter-doped region) and dissipate most of the heat. Hence the transistor is not symmetrical and emitter and collector cannot be interchanged without severe loss of gain.

Q2. Explain the operation of an npn transistor in the active mode. (5 marks)

See the flow above: EBJ forward, CBJ reverse; electrons injected from emitter into base; small fraction recombine (I_B); most diffuse to the CBJ and are swept into the collector (I_C); I_E = I_B + I_C; I_C = αI_E + I_CBO. Include a diagram of the npn transistor with V_EE and V_CC batteries and arrows for electron flow and conventional currents. Mention that I_C is almost independent of V_CB in this mode.

Q3. Explain the modes of operation of a BJT. (3 marks)

Active: EBJ forward, CBJ reverse; I_C = βI_B; amplifier. Saturation: both forward; V_CE ≈ 0.2 V; I_C set by external circuit; ON switch. Cut-off: both reverse; I_C ≈ 0; OFF switch. Reverse active: EBJ reverse, CBJ forward; low gain; rarely used. Show the regions on the CE output characteristics.

Q4. Draw and explain the input and output characteristics of a transistor in CB configuration. (6 marks)

Circuit: pnp or npn transistor with base grounded, V_EE with rheostat, milliammeter and voltmeter in the emitter circuit, V_CC with rheostat, milliammeter and voltmeter in the collector circuit. Input: I_E versus V_EB at V_CB = 0, 5, 10 V; diode-like curves; higher V_CB shifts curves slightly left (Early effect); input resistance low. Output: I_C versus V_CB for I_E = 1, 2, 3 mA; flat lines at I_C ≈ αI_E in the active region; I_C falls to zero only for a small forward V_CB (saturation region); cut-off below the I_E = 0 line where I_C = I_CBO; output resistance very high. Current gain α = ΔI_C/ΔI_E < 1.

Q5. Draw and explain the input and output characteristics of a transistor in CE configuration. (6 marks)

Circuit: emitter grounded, V_BB with rheostat, microammeter and voltmeter in the base circuit, V_CC with rheostat, milliammeter and voltmeter in the collector circuit. Input: I_B (μA) versus V_BE at V_CE = 1 V, 5 V; diode-like; higher V_CE shifts the curve right; input resistance about 1 kΩ. Output: I_C versus V_CE for I_B = 0, 20, 40, 60 μA; steep rise below about 0.3 V (saturation), nearly flat with slight slope in the active region (Early effect), cut-off below I_B = 0 with I_CEO; draw the load line. β = ΔI_C/ΔI_B is large. Compare with CB: higher current gain, moderate input resistance, lower output resistance.

Q6. Derive the relation between α and β. (2 marks)

\(I_E = I_B + I_C\). Divide by I_C: \(1/\alpha = 1/\beta + 1\), so \(1/\beta = (1 - \alpha)/\alpha\), giving \(\beta = \alpha/(1 - \alpha)\). Similarly \(\alpha = \beta/(1 + \beta)\). Since α is close to 1, a small change in α causes a big change in β (α = 0.99 gives 99; α = 0.995 gives 199).

Q7. Explain how a BJT acts as an amplifier. (5 marks)

Circuit: CE amplifier with bias resistors, collector resistor R_C, coupling capacitors and the input signal at the base. The DC bias sets the Q point in the active region. The input signal varies V_BE slightly, which varies I_B by ΔI_B; I_C varies by βΔI_B; the drop across R_C varies by βΔI_BR_C, so V_CE varies by a much larger amount than the input. Voltage gain \(A_v \approx -\beta R_C/r_i\); output is inverted. Power comes from V_CC. Correct biasing prevents clipping by saturation or cut-off. Show input and output waveforms with phase reversal.

Q8. Explain the BJT as a switch. (5 marks)

Circuit: npn transistor with load R_C to V_CC, base resistor R_B from the control input, emitter grounded. Input low (0 V): I_B = 0, cut-off, I_C ≈ 0, V_out = V_CC: switch open. Input high: I_B ≥ I_C(sat)/β drives the transistor into saturation, V_CE ≈ 0.2 V, I_C ≈ V_CC/R_C: switch closed. Mark the two operating points at the ends of the load line. Power loss is small in both states. Uses: relay and LED drivers, logic inverter, motor and lamp control from microcontrollers, switching power supplies. Mention the freewheeling diode for inductive loads and the switching times (delay, rise, storage, fall).

Exam tips

• Always draw the circuit for characteristics with correct meter placement and battery polarities.
• Label regions (active, saturation, cut-off) on output characteristics; examiners give marks for this.
• Remember the shift directions: CB input shifts left with V_CB; CE input shifts right with V_CE.
• Write β = α/(1 − α) before substituting; check I_E = I_B + I_C at the end.
• In switch problems find I_C(sat) first, then I_B(min), then R_B with overdrive.

FAQs

Why is the base made thin?

A thin base lets almost all injected carriers reach the collector before they recombine, so the base current is small and the current gain is high.

Can two diodes connected back to back act as a transistor?

No. Two separate diodes have no thin shared base, so carriers injected by one junction recombine before reaching the other. Transistor action needs both junctions in one crystal with a very thin base.

Why is CE configuration most popular?

It gives high current gain, high voltage gain and the highest power gain, with convenient input and output resistances for cascading stages.

What is the difference between β and hFE?

hFE is the datasheet symbol for the DC current gain I_C/I_B in CE, the same as β_dc; h_fe (lower case) is the small-signal AC current gain.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Semiconductor Physics, Semester II) and were prepared with AI help and checked.