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Chapter 1: Problem 54

The forward-bias currents in a pn junction diode and a Schottky diode are \(0.72 \mathrm{~mA}\). The reverse-saturation currents are \(I_{s}=5 \times 10^{-13} \mathrm{~A}\) and \(I_{S}=5 \times 10^{-8} \mathrm{~A}\), respectively. Determine the forward-bias voltage across each diode

Short Answer

Expert verified
Approximate forward-bias voltages are 0.71 V for the pn junction diode and 0.36 V for the Schottky diode.

Step by step solution

01

Understand the parameters

We have two diodes, a pn junction diode and a Schottky diode, each carrying a forward-bias current of \(0.72 \text{ mA}\). The reverse-saturation current for the pn junction diode is \(I_S = 5 \times 10^{-13} \text{ A}\), and for the Schottky diode, it is \(I_S = 5 \times 10^{-8} \text{ A}\). We need to find the forward-bias voltage \(V_f\) for each diode.
02

Use diode equation for pn junction diode

For the pn junction diode, we use the diode equation: \[ I = I_S \left( e^{\frac{qV_f}{kT}} - 1 \right) \]Given the forward-bias current \(I = 0.72 \times 10^{-3} \text{ A}\), reverse-saturation current \(I_S = 5 \times 10^{-13} \text{ A}\), and assuming room temperature \(T = 300 \text{ K}\), the thermal voltage \(V_T = \frac{kT}{q} \approx 25.86 \text{ mV}\), we can solve for \(V_f\).
03

Calculate forward-bias voltage for pn junction diode

Substitute \(I = 0.72 \times 10^{-3} \text{ A}\), \(I_S = 5 \times 10^{-13} \text{ A}\), and \(V_T = 25.86 \text{ mV}\) into the diode equation:\[ 0.72 \times 10^{-3} = 5 \times 10^{-13} \left( e^{\frac{V_f}{25.86 \times 10^{-3}}} - 1 \right) \]Solving for \(V_f\), we find that:\[ e^{\frac{V_f}{25.86 \times 10^{-3}}} \approx 1.44 \times 10^{9} \]Thus, \( V_f \approx 0.71 \text{ V}\).
04

Use diode equation for Schottky diode

For the Schottky diode, we use the same diode equation: \[ I = I_S \left( e^{\frac{qV_f}{kT}} - 1 \right) \]With \(I = 0.72 \times 10^{-3} \text{ A}\) and \(I_S = 5 \times 10^{-8} \text{ A}\), we solve for \(V_f\) using the same thermal voltage \(V_T = 25.86 \text{ mV}\).
05

Calculate forward-bias voltage for Schottky diode

Substitute \(I = 0.72 \times 10^{-3} \text{ A}\), \(I_S = 5 \times 10^{-8} \text{ A}\), and \(V_T = 25.86 \text{ mV}\) into the diode equation:\[ 0.72 \times 10^{-3} = 5 \times 10^{-8} \left( e^{\frac{V_f}{25.86 \times 10^{-3}}} - 1 \right) \]Solving for \(V_f\), we find that:\[ e^{\frac{V_f}{25.86 \times 10^{-3}}} \approx 1.44 \times 10^{4} \]Therefore, \( V_f \approx 0.36 \text{ V}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pn Junction Diode
A pn junction diode is a semiconductor device that allows current to flow in one direction. It consists of two types of semiconductor material: p-type, which has an abundance of holes, and n-type, which has an abundance of electrons. When these two materials are joined together, a p-n junction is formed. At this junction, a depletion region is created where no charge carriers exist. This region acts as a barrier to current flow.

In the forward-bias condition, the external voltage applied reduces this barrier, allowing charge carriers to move across the junction:
  • The holes in the p-type region move toward the junction.
  • The electrons in the n-type region migrate toward the junction.
As a result, current flows across the diode. The forward-bias voltage needed for significant current conduction in a typical silicon pn junction diode is about 0.7 volts.
Schottky Diode
The Schottky diode is known for its low forward voltage drop and fast switching speed. It is made by forming a junction between a metal and a semiconductor, usually n-type silicon. Unlike the pn junction diode, it does not have a significant depletion region, making it more efficient for certain applications.

The key advantages of Schottky diodes include:
  • Lower forward voltage drop (typically around 0.2 to 0.4 volts).
  • Reduced power loss, which makes them ideal for power-sensitive applications.
  • Faster switching speeds due to the absence of charge carrier storage in a depletion region.
These characteristics make them suitable for applications like power rectification and RF applications.
Diode Equation
The diode equation is a critical tool for determining the behavior of diodes in circuits. It describes the relationship between the current flowing through a diode and the voltage across it. The equation is given by:

\[ I = I_S \left( e^{\frac{qV_f}{kT}} - 1 \right) \]

In this equation:
  • \( I \) is the diode current.
  • \( I_S \) is the reverse-saturation current.
  • \( V_f \) is the forward-bias voltage.
  • \( q \) is the charge of an electron.
  • \( k \) is the Boltzmann constant.
  • \( T \) is the absolute temperature in Kelvin.
Understanding this equation is essential when calculating parameters like forward-bias voltage, as it helps predict how a diode will behave under different conditions.
Reverse-Saturation Current
Reverse-saturation current, denoted as \( I_S \), is a measure of the small amount of current that flows through a diode even when it is reverse-biased. In this state, a diode ideally blocks current, but a tiny leakage current still exists due to minority carriers in the semiconductor.

The value of the reverse-saturation current is important because:
  • It determines how quickly a diode can reach its forward-conduction state.
  • It affects the diode's voltage drop in forward-bias condition.
For example, in the original exercise, the reverse-saturation current for a pn junction diode is much smaller than that of a Schottky diode, explaining why the Schottky diode reaches conduction with a lower forward voltage.

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Most popular questions from this chapter

Consider a silicon pn junction. The n-region is doped to a value of \(N_{d}=10^{16} \mathrm{~cm}^{-3}\). The built-in potential barrier is to be \(V_{b i}=0.712 \mathrm{~V}\). Determine the required p-type doping concentration.

(a) The reverse-saturation current of a pn junction diode is \(I_{S}=10^{-11} \mathrm{~A}\). Determine the diode current for diode voltages of \(0.3,0.5,0.7,-0.02,-0.2\), and \(-2 \mathrm{~V}\). (b) Repeat part (a) for \(I_{S}=10^{-13} \mathrm{~A}\).

The output current of a pn junction diode used as a solar cell can be given by $$ I_{D}=0.2-5 \times 10^{-14}\left[\exp \left(\frac{V_{D}}{V_{T}}\right)-1\right] \quad \mathrm{A} $$ The short-circuit current is defined as \(I_{S C}=I_{D}\) when \(V_{D}=0\) and the opencircuit voltage is defined as \(V_{O C}=V_{D}\) when \(I_{D}=0\). Find the values of \(I_{S C}\) and \(V_{O C}\).

A pn junction diode is in series with a \(1 \mathrm{M} \Omega\) resistor and a \(2.8 \mathrm{~V}\) power supply. The reverse-saturation current of the diode is \(I_{S}=5 \times 10^{-11} \mathrm{~A}\). (a) Determine the diode current and voltage if the diode is forward biased. (b) Repeat part (a) if the diode is reverse biased.

The zero-bias capacitance of a silicon pn junction diode is \(C_{j o}=0.02 \mathrm{pF}\) and the built-in potential is \(V_{B}=0.80 \mathrm{~V}\). The diode is reverse biased through a \(47-\mathrm{k} \Omega\) resistor and a voltage source. (a) For \(t<0\), the applied voltage is \(5 \mathrm{~V}\) and, at \(t=0\), the applied voltage drops to zero volts. Estimate the time it takes for the diode voltage to change from \(5 \mathrm{~V}\) to \(1.5 \mathrm{~V}\). (As an approximation, use the average diode capacitance between the two voltage levels.) (b) Repeat part (a) for an input voltage change from \(0 \mathrm{~V}\) to \(5 \mathrm{~V}\) and a diode voltage change from \(0 \mathrm{~V}\) to \(3.5 \mathrm{~V}\). (Use the average diode capacitance between these two voltage levels.)
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