BJT and MOSFET as amplifiers and switches

BJT and MOSFET operating regions, biasing, transconductance and small-signal gain, and how to drive each as a switch, with a switch-design, a divider-bias CE amplifier and a MOSFET region-check example.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Every mechatronic system has a microcontroller pin that must switch a relay, solenoid, LED strip or motor that needs far more current than the pin can supply, and sensor signals that must be amplified before they can be read. A BJT or MOSFET does both jobs. Knowing which region the device sits in, and how to bias or drive it, decides whether the circuit works, wastes power or burns out.

Key ideas

BJT basics. A bipolar junction transistor (NPN or PNP) has emitter, base and collector. A small base current controls a much larger collector current, so it is a current-controlled device. The base-emitter junction behaves like a diode: V_BE ≈ 0.7 V for a conducting silicon transistor.

BJT operating regions (NPN).

  • Cutoff: base-emitter junction not forward biased (V_BE below about 0.5 V). I_C ≈ 0; the transistor is an open switch.
  • Active: base-emitter forward biased, base-collector reverse biased. I_C = β·I_B, nearly independent of V_CE. This is the region for amplification.
  • Saturation: both junctions forward biased. V_CE falls to V_CE(sat) ≈ 0.1–0.3 V and I_C is set by the external circuit, so I_C < β·I_B. This is the "on" state of a switch.

MOSFET basics. An enhancement n-channel MOSFET has gate, source and drain. The gate is insulated from the channel by a thin oxide, so the DC gate current is practically zero and the device is voltage controlled. A channel forms only when V_GS exceeds the threshold voltage V_TH.

MOSFET operating regions (NMOS).

  • Cutoff: V_GS < V_TH; I_D ≈ 0.
  • Triode (linear, ohmic): V_GS > V_TH and V_DS < V_GS − V_TH. The channel behaves like a resistor whose value falls as V_GS rises. A switch that is fully on sits here, characterised by R_DS(on).
  • Saturation (active): V_GS > V_TH and V_DS ≥ V_GS − V_TH. I_D depends mainly on V_GS (square law), so the device is a voltage-controlled current source. This is the amplifier region.

Note that "saturation" means opposite things for the two devices: BJT saturation is the fully-on switch state, MOSFET saturation is the amplifier state.

As amplifiers. The transistor is biased at a quiescent point (Q-point) in the active/saturation region, and a small signal is superimposed. The key small-signal parameter is transconductance g_m = ∂I_out/∂V_in. The common-emitter (BJT) and common-source (MOSFET) stages give large inverting voltage gain, about −g_m·R_C or −g_m·R_D. The emitter follower and source follower (common collector/common drain) give voltage gain just under 1 with low output impedance, used as buffers. Voltage-divider bias with an emitter resistor R_E stabilises the BJT Q-point against changes in β and temperature; bypassing R_E with a capacitor restores the AC gain.

As switches. The aim is to sit in one of two states with little power loss. A BJT switch must be driven hard into saturation: choose I_B larger than I_C/β_min by an overdrive factor of 2–10. A MOSFET switch must be driven with V_GS well above V_TH (logic-level MOSFETs are specified at 4.5 V or 2.5 V) so that R_DS(on) is small; conduction loss is I_D²·R_DS(on). MOSFETs need almost no steady gate current, switch fast and are easy to parallel, so they dominate in PWM motor drives; BJTs remain common for small, cheap loads. Inductive loads (relays, solenoids, motors) need a flyback diode across the load, otherwise the turn-off voltage spike L·di/dt can destroy the transistor.

Limits. Respect maximum collector/drain current, breakdown voltage (V_CEO, V_DSS) and power dissipation. A BJT has positive temperature dependence of I_C (V_BE falls about 2 mV/°C), which can lead to thermal runaway without emitter degeneration.

Formulas

I_C = β · I_B; I_E = I_B + I_C = (β + 1) · I_B; α = β / (β + 1)

  • I_C, I_B, I_E: collector, base, emitter currents (A); β: common-emitter current gain (dimensionless); α: common-base gain. Valid in the active region only.

I_D = ½ · μ_n · C_ox · (W/L) · (V_GS − V_TH)² (saturation)

  • I_D: drain current (A); μ_n: electron mobility (m²/(V·s)); C_ox: gate-oxide capacitance per unit area (F/m²); W/L: channel width-to-length ratio; V_GS: gate-source voltage (V); V_TH: threshold voltage (V). Often written I_D = (k/2)·(V_GS − V_TH)² with k = μ_n·C_ox·W/L in A/V². Channel-length modulation is neglected.

I_D = k · [(V_GS − V_TH)·V_DS − V_DS²/2] (triode); R_DS(on) ≈ 1 / [k·(V_GS − V_TH)] for small V_DS

g_m = I_C / V_T (BJT), with V_T = kT/q ≈ 25 mV at room temperature; r_π = β / g_m g_m = k·(V_GS − V_TH) = 2·I_D / (V_GS − V_TH) (MOSFET, saturation)

  • g_m in siemens (A/V).

A_v ≈ −g_m · R_C (CE, R_E bypassed) ; A_v ≈ −g_m · R_D (CS) — ignoring output resistance and load.

I_B ≥ k_od · I_C(sat) / β_min, I_C(sat) = (V_CC − V_CE(sat)) / R_C (BJT switch, overdrive factor k_od ≈ 2–10)

Worked examples

Example 1 (standard: BJT as a switch). A 5 V microcontroller pin drives an NPN transistor that switches a 240 Ω load from a 12 V supply. β_min = 100, V_BE = 0.7 V, V_CE(sat) = 0.2 V. Choose R_B for an overdrive factor of 2.

  1. I_C(sat) = (V_CC − V_CE(sat))/R_C = (12 − 0.2)/240 = 49.2 mA.
  2. I_B = 2 × I_C(sat)/β_min = 2 × 49.2/100 = 0.984 mA.
  3. R_B = (V_pin − V_BE)/I_B = (5 − 0.7)/0.984 mA = 4.37 kΩ.
  4. Pick the next lower standard value, 3.9 kΩ: I_B = 4.3/3.9 kΩ = 1.10 mA, β_min·I_B = 110 mA > 49.2 mA, so the transistor is saturated.

R_B ≈ 4.37 kΩ maximum; use 3.9 kΩ (I_B = 1.10 mA).

Example 2 (GATE level: CE amplifier with voltage-divider bias). V_CC = 12 V, R1 = 40 kΩ (top), R2 = 10 kΩ, R_C = 4 kΩ, R_E = 1 kΩ (fully bypassed), β = 100, V_BE = 0.7 V, V_T = 25 mV. Find the Q-point and the mid-band voltage gain (no load).

  1. Thevenin equivalent of the divider: V_TH = V_CC·R2/(R1 + R2) = 12 × 10/50 = 2.4 V; R_TH = R1‖R2 = 8 kΩ.
  2. Base loop: I_B = (V_TH − V_BE)/(R_TH + (β + 1)·R_E) = 1.7/(8 + 101) kΩ = 15.6 µA.
  3. I_C = β·I_B = 1.56 mA; I_E = 1.575 mA.
  4. V_CE = V_CC − I_C·R_C − I_E·R_E = 12 − 6.24 − 1.58 = 4.19 V. Since V_CE > 0.2 V, the transistor is in the active region.
  5. g_m = I_C/V_T = 1.56 mA/25 mV = 62.4 mS; r_π = β/g_m = 1.60 kΩ.
  6. A_v = −g_m·R_C = −62.4 mS × 4 kΩ = −250.

Q-point: I_C = 1.56 mA, V_CE = 4.19 V; A_v ≈ −250.

Example 3 (MOSFET region check). An NMOS with k = μ_n·C_ox·W/L = 2 mA/V², V_TH = 1 V, V_GS = 2 V, R_D = 2 kΩ, V_DD = 10 V.

  1. Assume saturation: I_D = (k/2)(V_GS − V_TH)² = 1 × 1² = 1 mA.
  2. V_DS = V_DD − I_D·R_D = 10 − 2 = 8 V ≥ V_GS − V_TH = 1 V, so saturation is confirmed.
  3. g_m = k(V_GS − V_TH) = 2 mS; A_v = −g_m·R_D = −4.

I_D = 1 mA, saturation, A_v = −4.

Common mistakes

  • Using I_C = β·I_B in saturation; there I_C is fixed by the load and is less than β·I_B.
  • Mixing up "saturation" for BJT (switch on) and MOSFET (amplifier region).
  • Forgetting to check the assumed region after solving (V_CE > V_CE(sat); V_DS ≥ V_GS − V_TH).
  • Ignoring the (β + 1) factor when R_E is reflected into the base circuit.
  • Driving a standard MOSFET from a 3.3 V pin: V_GS barely above V_TH leaves R_DS(on) high and the device overheats.
  • Omitting the flyback diode on a relay or motor.
  • Writing g_m in mA/V and R in kΩ, then forgetting that their product is dimensionless only when the units match.

For GATE ME

Expect short numericals: I_C or β from given currents, Q-point of a fixed-bias or divider-bias BJT stage, MOSFET drain current and region check from the square law, transconductance, and the base resistor needed to saturate a switching transistor. Conceptual questions test which region is used for switching versus amplification. Practise "assume a region, solve, then verify" until it is automatic.

Quick check

  1. In which BJT region is I_C = β·I_B valid?
  2. An NMOS has V_TH = 1 V, V_GS = 3 V and V_DS = 1.5 V. Which region is it in?
  3. A BJT has I_C = 2 mA at room temperature. What is g_m?
  4. A MOSFET in saturation carries 4 mA at V_GS − V_TH = 2 V. What is I_D at V_GS − V_TH = 1 V?
  5. Why is a diode placed across a relay coil driven by a transistor?

Answers: 1. Active region. 2. Triode (V_DS < V_GS − V_TH = 2 V). 3. 80 mS. 4. 1 mA. 5. To clamp the inductive voltage spike at turn-off and protect the transistor.

Interview questions

All Electrical Circuits and Electronics interview questions

Try answering each one aloud before you open it.

  1. 1.What is a BJT and how does it function as an amplifier?Concept

    A bipolar junction transistor is a three-layer NPN or PNP device with emitter, base and collector. Biased in the active region (base-emitter junction forward biased, base-collector reverse biased), its collector current is β times the base current and nearly independent of V_CE, so it acts as a controlled current source. A small signal added to the base bias changes I_C by g_m·v_be, where g_m = I_C/V_T, and passing that current through a collector resistor gives a voltage gain of about −g_m·R_C in the common-emitter stage. The Q-point must be set (for example by voltage-divider bias with an emitter resistor) so the signal swing does not drive it into cutoff or saturation.

  2. 2.Explain how a MOSFET operates as a switch.Concept

    In an enhancement n-channel MOSFET, a gate-source voltage above the threshold V_TH induces a conducting channel between drain and source; below V_TH the device is off and only leakage flows. As a switch it is driven between cutoff and the triode (ohmic) region, where it looks like a small resistance R_DS(on), so conduction loss is I_D²·R_DS(on). The gate is insulated, so no steady gate current is needed, but the gate capacitance must be charged and discharged quickly by the driver to keep switching losses low. V_GS must be well above V_TH (for example 10 V for a standard device, 4.5 V for a logic-level one) to get the rated R_DS(on).

  3. 3.Why is a MOSFET preferred over a BJT in certain applications?Application

    A MOSFET is voltage driven with an insulated gate, so it needs almost no steady drive power, whereas a BJT needs a continuous base current of at least I_C/β. MOSFETs are majority-carrier devices with no stored minority charge, so they switch faster, which suits high-frequency PWM drives and SMPS. Their R_DS(on) has a positive temperature coefficient, so parallel devices share current naturally. At high voltages, however, R_DS(on) rises steeply, which is why IGBTs or BJTs can be better for high-voltage, high-current conduction.

  4. 4.What happens if the base current of a BJT is increased beyond its maximum rating?Application

    Increasing base current first drives the transistor from the active region into saturation: V_CE drops to about 0.2 V and I_C stops rising, being limited by the external circuit, so I_C becomes less than β·I_B. Any extra base current beyond what is needed for saturation only adds base-emitter dissipation and stored charge, which slows turn-off. If the absolute maximum base current or the junction power rating is exceeded, the base-emitter junction overheats and the device can be permanently damaged. In design, the base current is chosen with a modest overdrive factor (about 2 to 10) above I_C/β_min.

  5. 5.Describe the role of the gate oxide layer in a MOSFET.Concept

    The gate oxide layer in a MOSFET is a thin insulating layer that separates the gate terminal from the underlying semiconductor material. It plays a crucial role in controlling the channel conductivity by allowing the gate voltage to create an electric field without direct current flow between the gate and the channel. This enables the MOSFET to operate efficiently as a switch or amplifier.

  6. 6.How does temperature affect the operation of a BJT?Application

    As temperature rises, V_BE for a given collector current falls by about 2 mV/°C, β increases and the collector leakage current I_CBO roughly doubles every 10 °C. All three increase the collector current for a fixed bias, which raises dissipation and temperature further and can cause thermal runaway. Designers counter this with emitter degeneration (an emitter resistor), voltage-divider bias that is insensitive to β, and adequate heat sinking.

  7. 7.Calculate the collector current of a BJT in the active region with a base current of 50 µA and a current gain (β) of 100.Numerical

    In the active region I_C = β·I_B = 100 × 50 µA = 5000 µA = 5 mA. This holds only if the external circuit allows the transistor to stay in the active region; if the collector resistor and supply cannot support 5 mA, the transistor saturates and I_C is set by the circuit instead.

  8. 8.What is the threshold voltage in a MOSFET and why is it important?Concept

    The threshold voltage in a MOSFET is the minimum gate-to-source voltage required to create a conductive channel between the source and drain terminals. It is important because it determines the MOSFET's switching characteristics and influences the device's on-state and off-state behavior. Proper threshold voltage ensures efficient operation in digital and analog circuits.

  9. 9.Explain the concept of saturation in a BJT.Concept

    A BJT is saturated when both the base-emitter and base-collector junctions are forward biased. V_CE falls to V_CE(sat), typically 0.1 to 0.3 V, and the collector current is fixed by the external circuit at about (V_CC − V_CE(sat))/R_C, which is less than β·I_B. The transistor then behaves like a closed switch with low power loss, which is why switching circuits drive it into saturation. Amplifiers avoid saturation because the output no longer follows the input, causing clipping.

  10. 10.A MOSFET in saturation has a threshold voltage of 1 V and carries 10 mA at V_GS = 3 V. What is the drain current at V_GS = 4 V?Application

    In saturation the square law gives I_D ∝ (V_GS − V_TH)², ignoring channel-length modulation. The overdrive rises from 2 V to 3 V, so I_D = 10 mA × (3/2)² = 22.5 mA. This assumes V_DS stays at least V_GS − V_TH = 3 V so the device remains in saturation.

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