3.4 Electronics & Devices

Diodes and rectification

A diode conducts current in one direction. In the common constant-voltage-drop model, a silicon diode is ON (a fixed 0.7 V forward drop) or OFF (open, reverse-biased). To analyze, assume a state, solve, and confirm it is consistent (a forward-ON diode must carry positive current; a reverse-OFF diode must have negative terminal voltage). If the check fails, flip the assumption and re-solve.

Key applications:

• Half-wave rectifier: passes one polarity of an AC waveform.
• Full-wave bridge rectifier: four diodes convert both half-cycles to one polarity; output ripple is smoothed by a filter capacitor.
• Zener diode: operated in reverse breakdown at a fixed V_Z to regulate voltage.

For the ideal diode model the forward drop is neglected (0 V). FE problems usually specify which model to use, so read carefully — using 0.7 V when the problem says "ideal" is a needless error.

Worked diode example: a 10 V source drives a series diode and a 1 kohm resistor. Assume the diode is ON with a 0.7 V drop; then the resistor sees 10 - 0.7 = 9.3 V and the current is 9.3/1000 = 9.3 mA, which is positive, so the ON assumption is consistent. If the source were reversed (-10 V), the assumed-ON current would come out negative, contradicting the assumption, so the diode is actually OFF and no current flows.

The full diode equation I = I_s*(e^(V/(nV_T)) - 1) describes the exponential I-V curve, where V_T = kT/q is about 26 mV at room temperature, but FE problems almost always use the simpler constant-drop or ideal models.

Bipolar junction transistor (BJT)

A bipolar junction transistor (BJT) has three regions of operation:

• Cutoff: both junctions reverse-biased, transistor off (I_C = 0); acts as an open switch.
• Active (forward-active): base-emitter forward, base-collector reverse; acts as a linear amplifier with I_C = beta * I_B and I_E = I_C + I_B = (beta + 1) * I_B.
• Saturation: both junctions forward-biased; acts as a closed switch with a small V_CE(sat) around 0.2 V.

The current gain beta (also written h_FE) typically ranges from about 50 to 250. The base-emitter junction drop V_BE is about 0.7 V when conducting. Worked bias check: if a base resistor sets I_B = 20 microamps and beta = 100, then in the active region I_C = 100 * 20e-6 = 2 mA — but only if the collector supply and load resistor allow that current without driving V_CE below ~0.2 V; otherwise the device saturates and I_C is limited by the external circuit, not by beta*I_B.

MOSFET regions

A metal-oxide-semiconductor field-effect transistor (MOSFET) is voltage-controlled — the insulated gate draws essentially no DC current. For an n-channel enhancement MOSFET with threshold voltage V_th:

• Cutoff: V_GS < V_th, no drain current (I_D = 0).
• Triode (ohmic): V_DS < V_GS - V_th; behaves as a voltage-controlled resistor.
• Saturation (active): V_DS >= V_GS - V_th; I_D = (k/2)(V_GS - V_th)^2, the square-law region used for amplification, where k is the transconductance parameter (k = muC_ox*W/L).

The quantity (V_GS - V_th) is the overdrive voltage. Because the gate is insulated, MOSFETs have very high input impedance, which is why CMOS dominates digital logic. Note the naming inversion versus the BJT: a BJT amplifies in active, and a MOSFET amplifies in saturation — "saturation" means opposite things in the two devices, a frequent FE trap.

Worked MOSFET example: an n-channel device with V_th = 1 V and k = 0.5 mA/V^2 is biased at V_GS = 3 V with V_DS = 5 V. First confirm the region: V_GS - V_th = 2 V overdrive, and V_DS (5 V) >= 2 V, so it is in saturation. Then I_D = (k/2)(V_GS - V_th)^2 = (0.5e-3/2)(2)^2 = 0.25e-3*4 = 1 mA. Doubling the overdrive to 4 V would quadruple I_D to 4 mA because of the square law. For amplifier biasing the device must stay in saturation across the signal swing; for switching (logic) the device toggles between cutoff (open) and deep triode (low-resistance on).

Ideal op-amp golden rules and worked gain

An operational amplifier (op-amp) has very high open-loop gain and high input impedance. With negative feedback, two golden rules apply:

1. No current flows into either input (infinite input impedance).
2. The inputs are forced to the same voltage, V+ = V- (the virtual short; a virtual ground when V+ is grounded).

Apply these with KCL at the inverting node to derive every standard circuit.

Worked op-amp example: an inverting amplifier with R_in = 2 kohm and R_f = 20 kohm has gain -R_f/R_in = -10. The inverting node is a virtual ground (0 V) because V+ is grounded; the input current V_in/R_in flows entirely through R_f (golden rule 1), so V_out = -(R_f/R_in)*V_in. For V_in = +0.5 V, V_out = -10 * 0.5 = -5 V. The non-inverting gain can never be less than 1; the inverting gain is negative (polarity inverts).