---
title: "Short note: Graphical analysis of diode circuit"
subject: "Basic Electronics Engineering"
chapter: 2
key: "semester_2/basic_electronics_engineering:diode-graphical-analysis"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/diode-graphical-analysis
site: BCT Engineering
---

# Short note: Graphical analysis of diode circuit

Basic Electronics Engineering · Chapter 2: Diodes

- Asked in: 2067 Chaitra, 4 marks
- Video: [Load Line Analysis of Diode](https://www.youtube.com/watch?v=TaTGnbxIMdY) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/diode-graphical-analysis
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2

## Question

Write a short note on graphical analysis of diode circuit.

## Short answer

Graphical analysis (load line method) finds a diode circuit's operating point by drawing the load line ID = (VS - VD)/R on the diode's I-V curve. The intersection gives the quiescent current and voltage, making it easy to see how changes in source or load resistance shift the Q point.

## Answer

**Graphical analysis (load line method)** finds the operating point of a diode circuit when the diode's I-V curve is non-linear and an exact equation is hard to solve.

Consider a series circuit: supply VS, resistor R and diode D.

KVL gives: VS = ID R + VD, so ID = (VS − VD)/R. This is a straight line called the **load line**.

#### Procedure
1. Draw the diode characteristic (ID versus VD) from the data sheet or from measurement.
2. Plot the load line:
   - VD = 0 → ID = VS/R (point on the current axis)
   - ID = 0 → VD = VS (point on the voltage axis)
3. Join the two points. The intersection with the diode curve is the **operating (quiescent) point Q**, giving the actual ID and VD.

```text
 ID
 VS/R |\
      | \       diode curve
      |  \    /
      |   \  /
      |    Q*
      |   / \
      +--/----\------ VD
       0.7      VS
```

**Example:** VS = 5 V, R = 1 kΩ. Load line: 5 mA on the ID axis and 5 V on the VD axis. If the curve crosses at VD ≈ 0.7 V, then ID = (5 − 0.7)/1 k = 4.3 mA.

The method also shows how Q shifts when VS or R changes, and it can be extended to circuits with ac signals to find the output waveform.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a series circuit of VS, R and a diode, what is the load line equation?
   A. ID = (VS - VD)/R
   B. ID = VS/R + VD
   C. ID = VD/R
   D. ID = (VS + VD)/R
   Answer: A. KVL gives VS = ID R + VD, so ID = (VS - VD)/R. (Key point: Load line equation)

2. For VS = 5 V and R = 1 kΩ, where does the load line cut the axes?
   A. 5 V on the ID axis and 5 mA on the VD axis
   B. 0.7 V and 4.3 mA
   C. 5 mA on the ID axis and 5 V on the VD axis
   D. 1 mA and 1 V
   Answer: C. At VD = 0, ID = VS/R = 5 mA; at ID = 0, VD = VS = 5 V. (Key point: Load line end points)

3. What does the intersection of the load line and the diode curve give?
   A. The breakdown voltage
   B. The operating (Q) point with the actual ID and VD
   C. The maximum power rating
   D. The cut-in voltage only
   Answer: B. The intersection is the Q point, giving the actual diode current and voltage. (Key point: Operating point)

4. Why is the graphical method used for diode circuits?
   A. The diode I-V curve is non-linear, so an exact equation is hard to solve
   B. Diodes obey Ohm's law exactly
   C. It avoids needing the supply voltage
   D. It works only for ac circuits
   Answer: A. Because the curve is non-linear, the intersection with a straight load line is an easy way to find the operating point. (Key point: Use of the method)

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