Semiconductor Diode - Forward Bias And Reverse Bias

Consider the charger you use to power your smartphone. Inside that small device lies a crucial component known as a semiconductor diode. This diode ensures that electricity flows correctly, protecting your device from potential damage. A semiconductor diode can control the direction of current flow, which is essential for the functioning of electronic circuits.

This Story also Contains

1. Semiconductor Diode
2. P-N Junction Diode Under Forward Bias
3. P-N Junction Diode Under Reverse Bias
4. Formation of a Reverse Bias Diode
5. Critical Value of Reverse Bias Voltage
6. Experimental Study of the V-I characteristics of a Semiconductor Diode
7. Solved Examples Based on Semiconductor Diode - Forward Bias And Reverse Bias
8. Summary

To understand how a semiconductor diode operates, we need to look at its two main modes: forward bias and reverse bias. These modes dictate how the diode allows or restricts the flow of electrical current, playing a vital role in rectification, signal modulation, and protection in electronic circuits. Let's explore how forward bias and reverse bias with the help of a solved example.

Semiconductor Diode

If a p-n junction has metallic contacts at both the ends for application of external voltage. This is called a semiconductor diode.

The symbolical representation of a semiconductor diode is shown below -

In the figure given above, the arrow indicates the direction of current when the diode is under forward bias. One should note here that the equilibrium barrier potential can be altered. This can be done by applying an external voltage across the diode. Depending on how this voltage is applied, the diode is a forward-bias or a reverse-bias diode which we will study in the upcoming concept.

P-N Junction Diode Under Forward Bias

In the figure given above, we can see that an external voltage is applied across the semiconductor diode where the p-side of the diode is connected to the positive terminal and the n-side is connected to the negative terminal of the battery. This type of arrangement for the diode is forward-biased.

Formation of a Forward Bias Diode

As the depletion region has no charge so the resistance is very high there so the applied voltage drops primarily across this region. The drop in voltage across the p and n side of the junction is relatively negligible. And the direction of the applied voltage (V) being opposite to that of the built-in potential (V_o) due to which the depletion layer’s width decreases and the barrier height reduces.

If the applied voltage is small, then the barrier potential is reduced marginally only below the equilibrium value. Then only small number of carriers crossing the junction, so the current is small. Similarly for a significantly high value of voltage, more carriers have the energy to cross the junction so, the current will be high.

One should also note that when the voltage is applied, some electrons cross to the p-side and some holes cross to the n-side. Under forward bias, this process is the minority charge injection process. Hence, the minority charge concentration which is electrons on the p-side are a minority and holes on the n-side are a minority, is significantly higher at the junction boundary.

Due to this concentration gradient, the injected electron diffuse from the junction-end to the far-end of the p-side. Similarly, injected holes diffuse to the far end of the n-side. This gives rise to current too.

The total diode forward current = Hole diffusion current + Electron diffusion current (mA)

P-N Junction Diode Under Reverse Bias

In the figure given above, we can see that an external voltage is applied across the diode. We can see that the n-side of the diode connects to the positive terminal and the p-side connects to the negative terminal of the battery. This type of arrangement in the diode is a reverse-bias diode.

Formation of a Reverse Bias Diode

As the depletion region has no charges, the resistance will be very high, as a result, the applied voltage drops primarily across this region. Also, the drop in voltage across the p and n sides of the junction is relatively negligible. Now here the direction of the applied voltage (V) is the same as that of the built-in potential (V₀) (Opposite to the forward bias), because of this the depletion layer’s width widens and the barrier height also increases. This decreases the flow of electrons to the p-side and holes to the n-side. So, the diffusion current decreases to a great extent.

Because of the direction of the electric field, the electrons in the p-side and holes in the n-side are driven to their majority zones, if they come close to the junction. This will produce the drift current. The drift current is usually of a few microamperes. This current is very low even in the forward-biased diode as compared to the current due to the injected carriers.

Critical Value of Reverse Bias Voltage

A small amount of voltage applied to the diode is sufficient to sweep the minority charge carrier to the far side of the junction. This diode reverses current which is not dependent on the voltage but on the concentration of the minority charge carriers on both sides of the junction. However, the current is independent up to a critical value of reverse bias voltage which is the Breakdown Voltage (V_br). When the voltage applied crosses breakdown voltage i.e., V_br, even a small change in the bias voltage causes a huge change in current. There is an upper limit of current for every diode, beyond which the diode gets destroyed due to overheating. This is the rated value of the current.

Experimental Study of the V-I characteristics of a Semiconductor Diode

V-I Characteristics of Diode

The figure shows a diode connected in reverse bias. The battery connects to the diode through a potentiometer by which we can change the voltage for the sake of the experiment. A microammeter is also used (since the expected current is in milliAmpere) to measure the current.

Here is the result of the experiment

As we can see in the graph above, in the forward biased diode, initially when the current increases almost negligibly till a certain value is reached. After that, the current increases exponentially even for a small increase in diode bias voltage. This voltage is called as threshold voltage. (Its value is approximately ~0.7 V for silicon diode and ~ 0.2 V for germanium diode)

In the reverse biased diode, the current is very small and almost remains constant with a change in bias voltage. It is called as Reverse saturation current. It is observed that in some cases, beyond the breakdown voltage, the current increases suddenly.

Hence, from this experiment and the given graph, we can conclude that the p-n junction diode allows the flow of current only in one direction, i.e. forward-bias, which means that the forward bias resistance is lower than the reverse bias resistance.

Extra edge -

1. P -N junction as diode

R = 0 , Forward
R → ∞ Reverse

It is a one-way device. It offers a low resistance when forward-biased and high resistance when reverse-biased.

2. Dynamic Resistance

Since slope of potential vs current graph is non uniform hence resistance keep changing .

R d = d v d i

3. Knee voltage of P-N junction

Knee voltage for Ge is 0.3 V

Knee voltage for Si is 0.7 V

It is defined as the forward voltage at which the current through the junction starts rising rapidly with an increase in voltage.

4. Relation between current I & Voltage V

K = Boltzmann constant
I 0 = reverse saturation current
In forward bias

$

e^{\frac{\varepsilon v}{K T}}>>1

$

Then forward biasing current is

$

\begin{aligned}

& I=I_0 \cdot e^{\frac{\varepsilon v}{K T}} \\

& I=I_0\left(e^{\frac{c v}{K T}}-1\right)

\end{aligned}

$

Solved Examples Based on Semiconductor Diode - Forward Bias And Reverse Bias

Example 1: In an unbiased n-p junction electrons diffuse from n-region to p - region because :

1) holes in p - region attract them

2) electrons travel across the junction due to potential difference

3) electron concentration in n - region is higher as compared to that in p - region

4) only electrons move from n to p region and not the vice - versa

Solution:

In an unbiased p-n junction, electrons diffuse from the n-region to the p-region because of a concentration gradient.

Hence, the answer is the option (3)

Example 2: The circuit has two oppositely connected ideal diodes in parallel. What is the current (in ampere) following in the circuit?

1) 2.00

2) 1.71

3) 1.33

4) 2.31

Solution:

P -N junction as diode

It is a one-way device. It offers a low resistance when forward-biased and high resistance when reverse-biased.

wherein

R = 0, Forward

R→∞ Reverse

Diode D₁ is in reverse bias and hence this branch will offer infinite resistance and hence we can remove this.

Diode D₂ is forward bias and hence it offers no resistance.

The circuit can be reduced to

I=126=2A

The correct option is 1.

Example 3: If in a $p-n$ junction diode, a square input signal of 10 V is applied as shown

Then the output signal will be

1)

2)

3)

4)

Solution:

P -N junction as diode

It is a one-way device. It offers a low resistance when forward-biased and high resistance when reverse-biased.

wherein

R=0, Forward
R→∞ Reverse
For - ve applied voltage the diode is in reverse biased mode and hence there is no current output across RL=0

If + 5V is applied, the diode is in forward-biased mode and hence offers no resistance.

Output voltage = 5V

Example 4: The forward-biased diode connection is :

1)

2)

3)

4)

Solution:

To determine which diode connection is forward-biased, we need to identify the configuration where the anode (positive side of the diode) is at a higher potential than the cathode (negative side of the diode).

Let's analyze each option:

1. +2V on the anode and -2V on the cathode:

The voltage across the diode is +2V−(−2V)=+4V+2V - (-2V) = +4V+2V−(−2V)=+4V

This means the anode is at a higher potential than the cathode.

Forward-biased.

1. -3V on the anode and -3V on the cathode:

The voltage across the diode is −3V−(−3V)=0V-3V - (-3V) = 0V−3V−(−3V)=0V.

The anode and cathode are at the same potential.

Not forward-biased.

1. 2V on the anode and 4V on the cathode:

The voltage across the diode is 2V−4V=−2V2V - 4V = -2V2V−4V=−2V.

This means the anode is at a lower potential than the cathode.

Not forward-biased.

1. -2V on the anode and +2V on the cathode:

The voltage across the diode is −2V−2V=−4V-2V - 2V = -4V−2V−2V=−4V.

This means the anode is at a lower potential than the cathode.

Not forward-biased.

Therefore, the forward-biased diode connection is:

Option 1).

Example 5: A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery’s positive terminal is connected to A and in other case when the positive terminal of the battery is connected to B will respectively be :

1) 0.2 A and 0.1 A

2) 0.4 A and 0.2 A

3) 0.1 A and 0.2 A

4) 0.2 A and 0.4 A

Solution:

In one case
NO current will pass through D2
I=2/5 A=0.4 A

In the second case: no current will pass through D1
I=2/10=0.2 A

Summary

A junction of p-type and n-type semiconductors makes up a semiconductor. A p-side of the battery is positively linked to a p-type semiconductor while a pn junction connected to the n-side is negatively terminal. This means that the diode allows for the flow of electric current because the potential difference of the dopant atoms is decreased through frontward biasing. Nevertheless, by placing the p-side to the negative terminal and the n-side to the positive terminal refers to reverse biasing.