Working of PN Junction Diode and VI Characteristics

Working of PN Junction Diode under biasing:  

Once depletion region is formed, the diode is ready to operate voltage and current through it. When we analyze the working of PN Junction diode, we have three conditions

1. Diode is unbiased (VD = 0 Volt or anode and cathode terminals are shorted.)

2. Diode is biased with positive voltage at P side (VD > 0 Volt, Forward Bias)

3. Diode is biased with negative voltage at P side (VD < 0 Volt, Reverse Bias)

Case 1: Diode is unbiased 

When diode is unbiased, no energy is provided to the electrons and holes, therefore any electrons or holes will not be able to move across the junction.

So net current across junction will be 0 mA.

Case 2: Diode is forward bias  

When Vi is positive (Vi > 0 Volt), Diode is said as forward bias. We know that the junction behaves as a potential barrier with Vk voltage (knee voltage). It means diode will not allow any electron or hole movement if Vi < Vk .

When Vi ≥ Vk , electrons and holes have sufficient energy to cross the junction. So majority holes from P side and majority electrons from N side cross the junction due to repulsion force from the battery to the majority charge carriers as shown in the diagram. 

The movement of charge carriers due to repulsion force of the battery is known as drift process. So current is found due to drift process in biasing condition.

Junction in this case, becomes narrow because of the large concentration of charge carriers at the edge of the junction due to movement of majority charge carriers towards the junction. This also helps electrons and holes to cross the junction easily.

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction.

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction. Finally both the direction becomes same to each other.

So, total current becomes the sum of electron current and hole current I = Ih + Ie  

Where Ih = Current due to holes and Ie = Current due to electrons

Total current follows an exponential equation of input voltage across the diode using following equation

           (Diode Current Equation)

I = Net current

I0 = Reverse Saturation Current

VD = Voltage across diode

η  = 1 (for Ge)

    = 2 (for Si)                                                                   (η  is called material constant)

(VT is called thermal constant) and 

Here T is temperature in Kelvin)

At room temperature i.e. T = 27 C = 273 + 27 K = 300 K

VT = 300/11600 = 0.02586 Volt = 0.026 Volt = 26 mV

Case 3: Diode is Reverse Bias:  

When Vi < 0 volt, minority charge carriers are energized to move towards the junction. Since minority charge carriers are small in numbers therefore a small current is found in the circuit. In this case, negative electrons from P side and positive holes from N side, cross the junction due to repulsion force from the battery to the minority charge carriers. Junction is wide in this case. We can see in the diagram. 

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction.

Net reverse saturation current in this case is Io = Ioe + Ioh   µA which is opposite to the forward diode current. This Io is used in diode current equation.

After understanding the working of PN diode, now we will try to understand its VI characteristics in detail.

Example: A germanium diode is used in a rectifier circuit and is operating at a temperature of 25℃ with a reverse saturation current of 1000 𝜇A. calculate the value of forward current at a forward voltage 0.2 Volts.

Solution:

Given: VD = 0.2 V, I0 = 1000µA = 1000 x 10-6 A, η=1 (Ge diode)

T = 250 C = 273 + 25 K = 298 K

VT = 298 / 11600 = 0.026 volt

putting all values in diode current equation

I = 1000 x 10-6 x (e(0.2/0.026) -1)

I = 2.19 A                                                                          (Answer: I = 2.19 A)

Example: A germanium diode has a reverse saturation current of 100  𝜇A. Calculate the voltage at which 10% of the rated current will flow through the diode, at room temperature if diode is rated for 1 A.

solution:

Given: I0 = 3 µA = 3 x 10-6 A, η=1 (Ge diode) and I = 10 % of rated 1 A = 0.1 A

T = 270 C = 273 + 27 K = 300 K

VT = 300 / 11600 = 0.026 volt,              To be calculated VD

putting all values in diode current equation

0.1 A = 100 x 10-6 x (e(VD/0.026) -1) A

e(VD/0.026) -1 = 0.1 x 10000  = 1000

e(VD/0.026)  = 1000 + 1 = 1001

VD/0.026 = ln(1001)         or              VD = 0.026 x ln(1001)

VD = 0.18 Volt                                                                        (Answer: VD = 0.18 Volt)

Symbol of Diode 

From the working of PN Junction Diode, we saw that a large current in diode is found from P side towards N side when diode is forward bias. However a small negligible current in diode is found from N side towards P side.

It means diode allows current in in forward direction however it restricts current in opposite direction. So the direction of arrow represents the allowed direction of current and a line like wall represents restriction of current from opposite side.

Terminal in P side is Anode and Terminal in N side is Cathode.

VI Characteristics of PN Junction Diode

Practical VI Characteristics

VI characteristics of PN junction diode is shown in the diagram. According to the working of diode

 In forward bias, For VD < Vk current is zero I = 0 mA

                           For VD > Vk current                                          

                   (Diode Current Equation)

In reverse bias, for VD < 0 current I = I0 (Opposite to I)

Here Vk = Knee Voltage: This is the minimum voltage required by the diode to make it ON or make it in forward bias.

VB = Breakdown Voltage: It is defined as the minimum reverse voltage at which junction gets broken down. After breakdown a large current is found in the circuit

Io = reverse saturation current in reverse bias diode. 

Ideal Characteristics

An ideal VI characteristic of a diode is shown in the diagram. This was the assumption about the correct usefulness of a diode. According to this concept

For VD > 0, diode becomes immediately ON and current tends towards infinite value.

For VD < 0, diode becomes immediately OFF and current becomes absolutely zero.

As diode voltage VD tends to 0+, current tends to its maximum value

So forward resistance Rf = VD / I = 0 / I = 0 Ω                 (Short Circuit)

As diode voltage becomes 0 or 0- , current also becomes 0.

So reverse resistance Rr = VD / I = VD / 0 = ꝏ Ω             (Open Circuit)

We always approach to fabricate a diode which has parameters very close to the Ideal diode. It is not possible to design an ideal diode.

Questions for Exercise

1. When PN Junction id forward biased:

(a) electrons from P side are injected into N side.

(b) holes from N side are injected into P side.

(c) Both of (a) and  (b)                                            (d) None of these.

2. When a reverse bias is applied to a junction diode, it

(a) rises in potential barrier,                                        (b) lower the potential barrier

(c) generally increases the majority carrier current    (d) None of these

3. The minimum voltage which is required to switch diode in ON state is known as

(a) Threshold voltage     (b) Knee voltage       (c) Cut in voltage      (d) All of these

4. Breakdown occurs when 

(a) Battery is not connected to the diode

(b) positive terminal of battery is connected to P side of diode and vice versa

(c) negative terminal of battery is connected to P side of diode and vice versa

(d) all of the above.

Answer: 1 – d,   2 - a ,     3 - d,      4 - c 

Numerical 1- A silicon diode is used in a rectifier circuit and is operating at a temperature of 36℃ with a reverse saturation current of 10 𝜇A. calculate the value of forward current at a forward voltage 0.0.18 Volts.

Numerical 2- A germanium diode has a reverse saturation current of 15  𝜇A. Calculate the voltage at which 1% of the rated current will flow through the diode, at room temperature if diode is rated for 100 mA.

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