What is a diode?

It is a semiconductor device in electronics applications where only a one-way current is required. It is like a one-way switch that allows current to flow in one direction only. A diode has two terminals, a cathode, and an anode. The cathode is the negative terminal that carries electrons, while the anode is the positive terminal that carries holes. Although no element has holes, it just means that there is a free space for electrons. Diodes are used to convert AC to DC. As we know, AC shifts between positive and negative cycles and DC stays positive constantly. As the diodes allow current only in one direction, they filter out the AC signal’s negative cycle and only keep the positive one.

Forward bias

When a diode is in working condition, then it is in forwarding bias. The negative terminal of the battery is connected to the cathode and the positive terminal of the battery is connected to the anode. As we know, like charges repel each other, this narrows the gap between the p-n junction, hence making the diode conduct current.

Reverse bias

If we reverse the polarity of the voltage supply, then the current flow stops. It is a reverse biased condition. Unlike charges that attract each other, this broadens the gap between the p-n junction. The current that exists under reverse biased conditions is reverse saturation current and is denoted by IS.

Symbol of Diode

CC BY-SA 3.0 | Image Credits: https://commons.wikimedia.org | Omegatron
I-V curve

CC BY-SA 3.0 | Image Credits: https://commons.wikimedia.org | Filip Dominec

We can analyze the graph and conclude that the first quadrant shows forward-biased current while the third quadrant shows reverse-biased current.

Characteristics of a diode

The general characteristics of a semiconductor diode can be expressed by the following equation, which is also known as Shockley’s equation.

ID=IS (eVDnVT-1)                             (i)

where IS is the reverse saturation current

            VD is the forward bias voltage of the diode

            n is ideality factor, which ranges from 1 to 2 for the operating conditions

In equation 1, VT is the thermal voltage.

VT=kTq                         (ii)

where k is Boltzmann’s constant, 1.38 x10-23 J/K

            T is the absolute temperature in kelvins.

            Kelvin temperature = Celsius temperature + 273

            q is the magnitude of electronic charge, 1.6 x 10-19 C

Ideal diode

An ideal diode’s characteristics look like the following diagram.
There is no breakdown voltage in the case of an ideal diode. Hence, it does not get destroyed when excess voltage is supplied in reverse polarity.
It is an ideal insulator. In the case of reverse polarity, it will not let any reverse current through it. Also, it can conduct unlimited forward current.

I-V characteristics of an ideal diode

CC BY 3.0 | Image Credits: https://commons.wikimedia.org | Jklmnop

Silicon diodes have an approximate voltage of 0.7 V to make them functional. It means if we supply less than 0.7 V, the diode will not switch on, and it will not let the current pass through it. This state is an open circuit. As soon as we exceed 0.7 V at the supply, the diode conducts current. Similarly, germanium diodes work at 0.3 V.

Diode circuit analysis

One way to analyze a diode is to assume the diode like a switch. In ON condition, it acts as a short circuit whereas, in OFF condition, it acts as an open circuit. Similarly, on testing voltage drop across the diode if we notice some voltage, then it is in working condition. Let us assume to analyze a circuit we tested the voltage across the diode, we should get voltage in one direction only, not in reverse polarity. If there is a voltage drop in reverse polarity, then the diode needs to be replaced. Suppose we assumed the diode to be conducting current, but on inspection, we came to know it is showing the flow of current in reverse polarity, this means the current we are seeing is reverse saturation current.

Load line analysis

Circuit diagram of a voltage source, a diode, and a resistor connected in series
CC BY-SA 2.5 | Image Credits: https://commons.wikimedia.org | Amr Bekhit

To solve electrical circuits, we first need to analyze them and find if there is a more efficient method to solve them. The above figure is a simple configuration of diode circuits. While doing circuit analysis, we see we need the value of ID, VD, VR, and IR so that these values satisfy both the diode characteristics and network parameters.

The load line meets the diode characteristics at point Q
Graph representing load line analysis

In the above figure, we can see the load line meeting the diode curve at one-point Q. It is called a load line because resistances are also known as load in practical examples. Since the line meets the current axis and is defined by load R, therefore this analysis is called load line analysis. It is a simplified way to approach diode circuits instead of solving mathematical equations.
As soon as we turn the voltage supply on, we see a clockwise current flowing through the diode circuit.Since it is a clockwise current, we can deduct from this that the diode is now in forwarding bias condition. Hence, the I-V characteristics of the diode can be seen in the figure in ideal diode section. If we apply Kirchhoff’s voltage law in the circuit, then we will get the following equation.


VS=VD+VR=VD+IDR                                (iii)

where  Vis the source voltage in volts.

                VD is the voltage drop across the diode.

            VR is the voltage drop across the resistor R.

                ID is the current through the diode in ampere.

                R is the resistor value in ohms.

VS=VD+IDRID=VSR-VDRID=IS(eVDnvT -1)                                             (iv)

Which is also known as Shockley’s equation. The sole purpose of using this analysis is because we are dealing with nonlinear components like diodes and transistors whose V-I curve is in exponential form rather than linear or a straight line. We use the resistors connected in series with the diode to form a load line on the graph and the point where it meets is where we get all our required values. Load line analysis solves the problem in a shorter time than the mathematical equation.

Instead of using the mathematical formula, we can use load line analysis to get the diode values graphically.

If we take ID=0,

VS=VD+IDRVS=VD+0VS=VD                             (v)

Let us take the second case when VD=0

VS=VD+IDRVS=0+IDRID=VSR                                                (vi)


By plotting both VS and ID obtained from equation 5 and 6 on the graph and connecting them, we will get the load line.

Point Q denotes the point of intersection between the device characteristics (known as a quiescent point). The operating point sometimes refers to Q. By drawing a horizontal line from the Q point to the current axis, we will get IDQ. Similarly, drawing a vertical line from the Q point of the voltage axis will give us VDQ.

Both these values are approximation values as compared to the mathematical solution.

Context and Applications

This concept is helpful for undergraduate and graduate courses such as:

  • Bachelors of Technology (Electrical Engineering)
  • Bachelors of Technology (Material Engineering)
  • Bachelors of Science (Physics)

Practice Problem

1. Which of the following is a possible voltage drop across a Germanium diode?

  1. 0.7 V
  2. 0.3 V
  3. 0.1 V
  4. 0.9 V

Answer- b

Explanation: Silicon diodes operate at approximately 0.7 V and Germanium diodes at 0.2~0.3 V.

2. Which among the following is also known by the name of barrier diode?

  1. PN junction diode
  2. Zener diode
  3. Schottky diode
  4. Avalanche diode

Answer- c

Explanation: Equipment like solar panels need reverse current protection while inactive at night. The solution to this problem is a Schottky diode.

3. Which among the following is not a linear component?

  1. Resistor
  2. Inductor
  3. Capacitor
  4. DiodeEngineering


Explanation: Apart from the diode, the rest of the electrical components mentioned in the options produce linear characteristics when plotted on a graph.

4. What is the thermal voltage at a temperature of 35°C?

  1. 26.56 mV
  2. 34.78 mV
  3. 21.68 mV
  4. 31.34 mV

Answer- a

Explanation: After changing the temperature to kelvin and then using equation 2, we will get the solution 26.56 mV.

5. Which of the following applies to ideal diodes?

  1. It has a breakdown voltage.
  2. It does not conduct unlimited forward current.
  3. It never performs reverse leakage current.
  4. It is an ideal conductor.

Answer- c

Explanation: In a diode, a reverse leakage current happens when the diode reaches breakdown voltage due to reverse polarity. In the case of an ideal diode, it never reaches breakdown voltage, hence no reverse leakage current. An ideal diode is practically impossible to achieve.

  • Schematics
  • Conduction and valence band
  • Rectifiers

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