--- title: "Q-point of diode; avalanche vs Zener breakdown" subject: "Electronics Devices and Circuit" chapter: 1 key: "semester_3/electronics_devices_and_circuit:diode-q-point-avalanche-zener" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-q-point-avalanche-zener site: BCT Engineering --- # Q-point of diode; avalanche vs Zener breakdown Electronics Devices and Circuit · Chapter 1: Diodes - Asked in: 2076 Chaitra, 3+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-q-point-avalanche-zener - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1 ## Question Define Q-point in pn junction diode operation. Show it graphically with necessary derivations. Differentiate between avalanche and zener break down. ## Short answer The Q-point is the dc operating point (VDQ,IDQ) at the intersection of diode curve and load line. Avalanche breakdown (>~6 V) is impact ionization, positive temp co; Zener (<~5 V) is field emission, negative temp co. ## Answer #### Q-point The **Q-point (quiescent or operating point)** is the dc voltage and current (VDQ, IDQ) of the diode when no signal is applied. It is the point where the diode curve and the dc load line intersect. #### Graphical derivation For a series circuit of VDD, R and the diode, KVL gives ```text VDD = ID R + VD → ID = VDD/R - VD/R ``` This is the load line, with intercepts ID = VDD/R and VD = VDD, slope -1/R. The diode equation ID = IS[exp(VD/(η VT)) - 1] is the second relation. Their intersection is the Q-point. ```text ID VDD/R |\ | \ diode curve IDQ | \ _.-Q | \| +----+\------- VD VDQ VDD ``` Steps: draw the characteristic, draw the load line with the two intercepts, read the intersection. For VDD = 5 V and R = 1 kΩ, the line runs from 5 mA to 5 V, and the Q-point is about 0.75 V, 4.25 mA. #### Avalanche vs Zener breakdown | Point | Avalanche breakdown | Zener breakdown | |---|---|---| | Cause | Impact ionisation: carriers accelerate, knock out more carriers (multiplication) | Strong electric field pulls electrons directly from covalent bonds (field emission) | | Junction doping | Lightly doped, wide depletion region | Heavily doped, very narrow depletion region | | Voltage | Above about 6 V (typically 6 to hundreds of V) | Below about 5 V | | Temperature coefficient | Positive (VB rises with temperature) | Negative | | Characteristic knee | Less sharp | Very sharp | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Q-point of a diode? A. The cut-off point B. The breakdown voltage C. The peak ac current D. The dc voltage and current at the intersection of the diode curve and dc load line, with no signal Answer: D. It is the quiescent operating point. (Key point: Q-point) 2. What is the slope of the load line for a VDD, R series circuit? A. R B. -1/R C. -R D. VDD/R Answer: B. ID = VDD/R - VD/R gives slope -1/R. (Key point: Load line) 3. Which statement correctly matches mechanism to breakdown? A. Avalanche: impact ionisation and carrier multiplication; Zener: field emission from covalent bonds B. Avalanche: field emission; Zener: impact ionisation C. Both are due to impact ionisation D. Both are due to thermal runaway only Answer: A. Avalanche is carrier multiplication, Zener is direct field-assisted bond breaking. (Key point: Cause of breakdown) 4. Which is true of Zener breakdown compared with avalanche? A. Lightly doped, above 6 V, positive temperature coefficient B. Heavily doped, below about 5 V, negative temperature coefficient C. Wide depletion, sharp knee, positive tempco D. Lightly doped, below 5 V, positive Answer: B. Zener breakdown occurs in heavily doped, narrow-depletion junctions at low voltage with negative tempco. 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