Question:

The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:
V-I characteristic of a diode

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The forward I-V characteristic is when the diode is forward-biased, and the anode with respect to the cathode is positive.

Updated On: Sep 13, 2024
  • 10

  • 10-6

  • 106

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The Correct Option is B

Approach Solution - 1

The forward I-V characteristic is when the diode is forward-biased, and the anode with respect to the cathode is positive. On the other hand, the reverse I-V characteristics are observed when the cathode is positive with respect to the anode. 

Formula Used:

Forward bias resistance Rf is given as - \(R_f=\frac{ΔV_f}{ΔI_f}\)

  • where, \(ΔV_f\) is the difference between forward bias voltages,
  • \(ΔI_f\) is the difference between currents.

Reverse bias resistance \(R_r\) is given as - \(R_r=\frac{ΔV_r}{ΔI_r}\)

  • where, \(ΔV_r\) is the difference between reverse bias voltages,
  • \(ΔI_r\) is the difference between currents.

Complete step-by-step answer:

The forward resistance offered by a diode in forward bias is very small. On the other hand, a diode offers very high reverse resistance when it is in reverse bias. 

The formula for forward bias resistance \(R_f\) is given as \(R_f=\frac{ΔV_f}{ΔI_f}\)→................(1)

  • where, \(ΔV_f\) denotes the difference between forward bias voltages.
  • \(ΔI_f\) is the difference between currents approximating particular voltages.

From the given diagram, the current ranges from 10 to 20 milli-Amperes. Therefore, \(ΔI_f=20−10\)

\(⇒ΔI_f=10mA\)

\(⇒ΔI_f=10×10^{−3}A\)

and the voltage ranges from 0.7 to 0.8 V.

\(⇒ΔV_f=0.8−0.7\)

\(⇒ΔV_f=0.1V\)

Substituting the values in equation (1)

\(R_f=\frac{ΔV_f}{ΔI_f}\)

\(⇒R_f=\frac{0.1}{10×10^{−3}Ω}\)

\(⇒R_f=10Ω\)

The formula for reverse bias resistance \(R_r\) is given as:

\(R_r=\frac{ΔV_r}{ΔI_r}\)→................(2)

  • where, \(ΔV_r\) is the difference between reverse bias voltages 
  • \(ΔI_r\) is the difference between currents corresponding to particular voltages.

Also according to the diagram, the reverse bias current will be

\(R_r=\frac{ΔV_r}{ΔI_r}\)

\(⇒R_r=\frac{10}{10^{−6}}Ω\)

\(⇒R_r=10^7Ω\)

Then, the ratio of forward bias to reverse bias resistance can be written as:

\(Ratio=\frac{R_f}{R_r}\)

\(⇒Ratio=\frac{10}{10^7}Ω\)

\(∴Ratio=10^{−6}Ω\)

Hence, option B is the correct answer.

Note: The formula for reverse bias resistance and forward bias resistance can be written with the help of Ohm’s Law. Ohm’s law is given by V = IR.

  • where V is the voltage, 
  • I is the current 
  • R is the resistance.

Thus, resistance can be written as R = VI

Independently in terms of forward and reverse bias resistance, this equation becomes \(R_r=\frac{ΔV_r}{ΔI_r}\)

And \(R_r=\frac{ΔV_r}{ΔI_r}\) respectively.

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Approach Solution -2

Forward bias resistance \(= \frac{\Delta V}{\Delta I} = \frac{0.1}{10\times10^{-3}} = 10\,\Omega\) 

Reverse bias resistance \(= \frac{10}{10^{-6}} = 10^{7}\,\Omega\) 

Ratio of resistances \(= \frac{Forward \,bias\, resistance}{Reverse \,bias\, resistance} = 10^{-6}\)

As per the V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is 10-6.

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Approach Solution -3

The correct option is B)

V=IR

\(R_f=\frac{V_f}{I_f}=\frac{0.7}{10\times10^{-3}}=7\times10^{-5}\)

In a similar way, \(\frac{0.75}{15} = 5\times10^{-5}\)

and \(\frac{0.08}{20}=4\times10^{-5}\)

 ∴ Rf\(\frac{0.7-0.75}{(10-15)\times10^{-3}}\) = 10


and Rb =\(\frac{V_B}{I_B}\) = \(\frac{10}{10^{-6}}\)

  =107

 

\(\frac{R_f}{R_b}=\frac{10}{10^7}=10^{-6}\)

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Concepts Used:

Applications of diode

Diodes are used in many applications. By removing the signal component, several of them convert AC current to DC current also are known as rectifiers when used in this position. Diodes act as electric switches and can block voltage spikes, making them ideal for surgical protectors. They are hired to do digital logic. Some things are made from diodes such as power supply and voltage doubles. The sensors, as well as light on lighting devices and lasers, depend on LEDs. For electronic tuning, varactors are used, and for compressing transients in AC lines varistors are used. Zener diodes are used as voltage regulators where varactors are used for electronic tuning, and varistors are used for compressing transients in AC lines.

On the base of diodes, transistors and op-amps are built. The most common diode mode is the p-n junction. One (n) object with electrons as the charging conductor closes the second object (p) with holes (well-charged particles acted as depleted areas of electrons) as the charge carriers on this type of diode.  Where they connect, the narrowing space forms, where electrons are dispersed to fill holes in the p-side. This effectively stops the flow of electrons. When a positive voltage is given to the p-side of this junction, electrons may easily move from it and the current flows into the diode to fill the holes. The shrinkage area expands as the negative bias interacts (i.e., negative voltage is given to the p-side), making the electrons harder to move across.