Transistor and MOSFET Switching Circuits
Switching High Currents from Low-Power Signals
A logic gate or comparator output can drive only a few milliamperes — not nearly enough to light a lamp, spin a motor or click a relay. A transistor or MOSFET acts as a current amplifier and electronic switch, letting a small control signal turn a much larger load current on or off. This is commonly misunderstood, so this lesson builds the skill in both directions: forward (given a signal, will the device switch?) and backward (given a required load current, what control signal do I need?).
The npn Bipolar Transistor as a Switch
An npn transistor has three terminals: base (B), collector (C) and emitter (E). In a switching circuit:
- The load (lamp, relay coil, etc.) is connected between the positive supply and the collector.
- The emitter connects to 0 V (ground).
- A base resistor connects the control signal to the base, limiting the base current to a safe value.
The transistor turns on (saturates) when a sufficient base current flows. The base-emitter voltage \(V_{BE}\) is approximately 0.7 V when the transistor is on. Any control voltage above about 0.7 V (after the base resistor voltage drop) will start to turn the transistor on. To ensure full saturation, the base current \(I_B\) must be at least \(I_C / h_{FE}\), where \(I_C\) is the required collector (load) current and \(h_{FE}\) is the transistor’s DC current gain from its data sheet.
$$I_B = \frac{I_C}{h_{FE}}$$
npn transistor switch: \(V_{BE} \approx 0.7 \text{ V}\) when on. Current gain \(h_{FE} = I_C / I_B\). To saturate the transistor: \(I_B \geq I_C / h_{FE}\).
Problem: A transistor with \(h_{FE}\) = 100 must switch a lamp drawing 200 mA. The control signal is 5 V. Find the minimum base current needed and a suitable base resistor value.
Step 1 — minimum base current: \(I_B = I_C / h_{FE} = 200 / 100 = 2 \text{ mA} = 0.002 \text{ A}\)
Step 2 — voltage across base resistor: \(V_{RB} = V_{IN} - V_{BE} = 5 - 0.7 = 4.3 \text{ V}\)
Step 3 — base resistor: \(R_B = V_{RB} / I_B = 4.3 / 0.002 = 2150 \text{ }\Omega\). Nearest E24 value below 2150 Ω is 2.0 kΩ. This gives \(I_B = 4.3/2000 = 2.15 \text{ mA}\), which meets the minimum.
The n-Channel Enhancement-Mode MOSFET
An n-channel MOSFET has three terminals: gate (G), drain (D) and source (S). Unlike the bipolar transistor, the MOSFET is voltage-controlled: almost no gate current flows, so the control circuit does not need to supply base current. This makes MOSFETs ideal for driving high-power loads.
The MOSFET turns on when the gate-to-source voltage \(V_{GS}\) exceeds the threshold voltage \(V_{GS(th)}\). For calculations, the threshold is taken as 3 V. When it is conducting, the drain current \(I_D\) is controlled by the transconductance:
$$I_D = g_m (V_{GS} - V_{GS(th)})$$
where \(g_m\) is the transconductance in siemens (S or A/V). In a typical exam question \(g_m\) is given; you substitute \(V_{GS}\) and \(V_{GS(th)} = 3 \text{ V}\) to find \(I_D\).
MOSFET: \(I_D = g_m(V_{GS} - V_{GS(th)})\). Threshold \(V_{GS(th)} = 3 \text{ V}\) (assumed). Gate current = 0. Turns on when \(V_{GS} > 3 \text{ V}\).
Problem: An n-channel MOSFET has \(g_m = 0.5 \text{ S}\) and \(V_{GS(th)} = 3 \text{ V}\). The gate voltage is 7 V. Calculate the drain current.
\(I_D = g_m(V_{GS} - V_{GS(th)}) = 0.5 \times (7 - 3) = 0.5 \times 4 = 2 \text{ A}\)
This 2 A drain current could drive a motor. A bipolar transistor base circuit would need \(I_B = 2/h_{FE}\) — at \(h_{FE}\) = 100, that’s 20 mA just to bias the base, which overloads a logic gate output. The MOSFET avoids this entirely because its gate draws negligible current.
Choosing between npn and MOSFET
Use an npn bipolar transistor for low-to-medium current loads (up to around 100–200 mA) where the control circuit can supply a few milliamps of base current. Use an n-channel MOSFET for high-current loads (motors, solenoids) or where the control source (such as a logic gate output) cannot supply even a small base current. Both are placed between the processing sub-system and the output device as a transducer driver.
Electromagnetic loads (relays, solenoids, motors) need a protection diode connected in reverse bias across the load. When the transistor or MOSFET switches off, the coil’s collapsing magnetic field generates a large voltage spike (back-EMF) that can destroy the switch device. The diode clamps this spike. Note: a protection diode is only needed for electromagnetic loads, not for resistive loads like lamps.