Diodes

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Diodes: From Theory to Hands-On Practice

In this post, we’ll introduce diodes, explain their basic theory, and walk through practical exercises that you can do with just a calculator, a breadboard, and a few simple components.

What is a Diode?

A diode is a two-terminal semiconductor device that allows current to flow in one direction only. Its symbol looks like an arrow pointing towards a bar:

The left hand side with the white bar is the Cathode, the right hand side without the white bar is the Anode

  • Forward bias: When the anode is more positive than the cathode, the diode conducts (after reaching a certain threshold voltage, usually ~0.7V for silicon).

  • Reverse bias: When the cathode is more positive, the diode blocks current (almost no conduction until breakdown).

This directional property makes diodes essential in rectifiers, protection circuits, voltage clamps, and logic circuits.

Key Theory: Ideal vs. Real Diode

  • Ideal diode: Conducts perfectly in one direction, blocks completely in the other.

  • Real diode: Requires a small forward voltage (e.g., ~0.7V for silicon, ~0.3V for germanium, ~0.2–0.4V for Schottky). In reverse bias, a tiny leakage current flows until breakdown voltage is reached.

The diode’s I–V curve helps visualize this: almost no current in reverse, then an exponential rise in forward conduction after threshold.

Practical Exercise 1: Testing a Diode with a DC Supply

Materials

  • 1 × DC power source (5V supply or battery)

  • 1 × Silicon diode (e.g., 1N4001)

  • 1 × Resistor (1 kΩ)

  • Breadboard & jumper wires

  • Multimeter

Circuit Setup

Step 1: Forward Bias

  • Connect the diode anode to the positive terminal of the DC supply through the resistor.

  • Connect the cathode to the negative terminal.

+Vdc → Resistor → Diode → Ground

Step 2: Reverse Bias

  • Flip the diode so the cathode is connected to +Vdc and the anode to ground (through the resistor).

Forward Bias

  • Supply voltage: Vsupply=5 VV_{supply} = 5 \, \text{V}

  • Resistor: R=1 kΩ=1000 ΩR = 1 \, k\Omega = 1000 \, \Omega

  • Diode drop (approx): Vdiode≈0.7 

1. Voltage across resistor:

VR=Vsupply−Vdiode=5.0−0.7=4.3 VV_R = V_{supply} - V_{diode} = 5.0 - 0.7 = 4.3 \, \text{V}

2. Current through circuit:

I=VRR=4.31000=0.0043 A=4.3 mAI = \frac{V_R}{R} = \frac{4.3}{1000} = 0.0043 \, \text{A} = 4.3 \, \text{mA}

Reverse Bias

  • In reverse bias, the diode ideally does not conduct, so:

    I≈0 mAI \approx 0 \, \text{mA}

  • The voltage across the diode will be essentially the full supply voltage (since almost no current flows through the resistor):

    Vdiode≈Vsupply=5 V





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