Mini Physics
Part of Mini EducationSwitch subject
Search

I–V Graph of a Semiconductor Diode

Key idea: Learn the diode I–V characteristic, forward vs reverse bias, how to sketch the graph, and how to interpret turn-on and near-zero reverse current (O Level).

  • SEC G3 Physics 2027
On this page

Learning objectives

  • State current as rate of charge flow measured in amperes
  • Distinguish conventional current from electron flow
  • Apply charge equals current multiplied by time
  • Define source e.m.f. as work done per unit charge around a circuit
  • Calculate total e.m.f. for sources in series
  • Define component potential difference as work done per unit charge
  • State resistance as potential difference divided by current
  • Apply resistance equals potential difference divided by current
  • Apply wire-resistance proportionalities for length and cross-sectional area
  • Describe the effect of temperature on metallic resistance
  • Sketch and interpret required current–voltage characteristics

1. Definition

A semiconductor diode allows current to flow easily in one direction (forward bias) but blocks current in the opposite direction (reverse bias). Its I–V graph is non-linear.

2. Key Ideas

  • Forward bias: diode conducts after a small “turn-on” voltage; current then increases rapidly.
  • Reverse bias: current is approximately zero (very small leakage).
  • The diode is non-ohmic (resistance is not constant).
  • Use a stated sign convention: V_D = V_A-V_K, and take positive conventional current from anode to cathode.
  • When sketching, show:
    • forward region with a steep rise after the turn-on
    • reverse region close to the I = 0 axis
What you need for this course

You should be able to sketch and interpret the I–V characteristic graph for a semiconductor diode.

3. Detailed Explanations

A. Forward bias vs reverse bias (in words)

  • Forward bias: the diode is connected so it conducts (current flows).
  • Reverse bias: the diode is connected the other way, so it blocks current (almost no current flows).

The anode is the p-side of the diode and the cathode is the n-side, marked by the bar on the circuit symbol. With V_D = V_A-V_K, positive V_D is forward bias and negative V_D is reverse bias.

Diode sign convention and qualitative I–V characteristicThe left panel labels the anode, cathode bar, positive diode voltage and conventional-current direction. The right panel is a qualitative current-against-voltage graph: the reverse-bias branch stays close to zero current, while the forward-bias branch bends upward and becomes steep. Reverse breakdown is outside the model shown.Define the signs firstanode, Acathode, Kbar marks the cathodepositive conventional current, I+−Vᴅ = Vₐ − VₖVᴅ > 0: forward biasVᴅ < 0: reverse biasA real diode is non-ohmicnot a perfect switch at one exact voltageQualitative I–V graphVᴅIforward biasreverse biasvery small reverse currentrapid risethrough turn-on regionSchematic shape — not measured dataturn-on depends on the diode and operating conditionsreverse breakdown is outside this displayed model
Scroll diagram horizontally to read all labels.
Positive diode voltage means the anode is at a higher potential than the cathode. Forward current then rises rapidly through a turn-on region; reverse current remains very small on this qualitative scale.

B. Interpreting the I–V graph

In forward bias, the current is very small at first. After the diode “turns on”, the current increases quickly with voltage.

In reverse bias, the current stays very close to zero on the same scale. The graph is qualitative: a real diode does not change from perfectly off to perfectly on at one exact voltage, and the turn-on region depends on the diode and operating conditions.

Typical value (not required)

For a silicon diode, the turn-on is often around 0.6 V, but exam questions usually focus on the shape of the graph.

4. Common Mistakes

  • Drawing a straight line through the origin (that is for an ohmic conductor).
  • Mixing up forward and reverse regions.
  • Saying reverse current is “exactly zero” (it is usually very small).
  • Treating the turn-on region as a perfect switch at one universal voltage.
  • Reading negative voltage as reverse bias without first checking how V_D is defined.

5. Exam Tips

  1. Forward bias: “almost no current at first, then steep rise”.
  2. Reverse bias: “current is approximately zero”.
  3. If asked for resistance at a point, use R = V/I at that operating point. This ratio is not the gradient of a curved I–V graph.

6. Worked Examples

Modelled example 1

Identify forward vs reverse bias

Core

Problem

A diode carries substantial current in one supply orientation, but almost zero current when the supply is reversed. Interpret both operating regions.
Study the worked solution

  1. Identify forward bias

    Method

    Call the conducting orientation forward bias.

    Reason

    After turn-on, a small additional p.d. produces a large current increase.

    Working

    Forward bias → steep positive-current branch.

  2. Identify reverse bias

    Method

    Call the reversed orientation reverse bias and treat current as approximately zero at this level.

    Reason

    The diode blocks ordinary reverse current apart from very small leakage.

    Working

    Reverse bias → near-zero current.

Guided practice 2

Explaining the steep rise

About 4 min

Problem

What does the steep forward-bias branch after turn-on tell you about how current responds to a small increase in p.d.?

Interpret the graph slope qualitatively

Current response

Hints

Hint 1: read the axes
Current is vertical and voltage horizontal.
Hint 2: translate steepness
Large Δ I occurs for small Δ V.
View solution step by step

  1. Interpret the steep branch

    Method

    State that current increases rapidly for a small additional forward p.d.

    Reason

    The diode has entered its strongly conducting, non-ohmic region.

    Working

    Small Δ V → large Δ I.

Common misconception 3

Sketching

Find and correct the mistake

Learner response

A learner sketches equal steep branches in positive and negative voltage, like a symmetric component. Diagnose and describe the correct diode characteristic.

Identify the defining graph property

Characteristic symmetry

View solution step by step

  1. Draw the forward region

    Method

    Keep current near zero initially, then curve into a steep positive-current rise after turn-on.

    Reason

    The diode conducts strongly only after sufficient forward bias.

    Working

    Positive V: near-zero then steep rise.

  2. Draw the reverse region

    Method

    Keep the negative-voltage branch close to the voltage axis with approximately zero current.

    Reason

    Ordinary reverse leakage is negligible on the course graph scale.

    Working

    Negative V: I ≈ 0.

Examiner practice 4

Identifying the region from voltage sign

3 marks

Examination question

On a graph where positive diode voltage is defined as forward bias, the operating point is V_D = -4.0 V. Identify the bias region and approximate current, with a reason. [3 marks]

Use the stated voltage convention

View solution step by step

  1. Identify the region

    1 mark

    Method

    Classify -4.0 V as reverse bias under the stated convention.

    Reason

    The voltage sign is opposite to the defined forward direction.

    Working

    V_D < 0: reverse bias.

  2. Infer the current

    2 marks

    Method

    State that current is approximately zero, with only negligible leakage on this scale.

    Reason

    The diode blocks current in ordinary reverse bias.

    Working

    I ≈ 0 A.

Challenge 5

Resistance at a point (forward bias)

Minimal support

Operating-point transfer

At a forward-bias point, V = 0.70 V and I = 0.20 A. Find V/I and explain what the result does and does not represent.

Calculate the ratio, then qualify it

Hints

Hint 1: use the point values
Compute V/I at the stated coordinates.
Hint 2: remember the curve
The diode is non-ohmic, so the ratio changes with operating point.
View solution step by step

  1. Calculate the operating-point ratio

    Method

    Divide voltage by current.

    Reason

    Both values describe the same forward-bias point.

    Working

    R = 0.70/0.20 = 3.5 Ω

  2. Qualify the result

    Method

    Call 3.5 Ω the V/I resistance at this point only.

    Reason

    The diode’s curved characteristic means resistance changes with voltage and current.

    Working

    Not a constant device resistance and not the curve gradient.

7. Mind Stretchers

Mind stretcher 1: Comparing three componentsExtension

Match each I–V shape to the component: metallic resistor (constant temperature), filament lamp, diode.

  1. straight line through origin
  2. curve that becomes less steep at higher voltage
  3. almost no current in reverse, steep rise in forward
Show Answer
  1. metallic resistor (ohmic)
  2. filament lamp
  3. diode

Mind stretcher 2: Reverse regionExtension

Why is the reverse current often treated as zero in simple sketches?

Show Answer

The reverse current is usually extremely small compared to forward current, so on the same scale it is negligible. For O Level sketches, you show it as “approximately zero”.

8. Practice and next step

Sketch the full forward- and reverse-bias graph from memory, then verify it in the I–V Characteristics Lab. Finish with the graph questions in Structured Current Electricity, then continue to D.C. Circuits.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027