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Chapter 5: Problem 25

Suppose in a p'n diode the \(\mathrm{n}\) region and the \(\mathrm{p}\) region thickness are \(20 \mu \mathrm{m}\) and \(5 \mu \mathrm{m}\), respectively. What will be the \(\mathrm{RB}\) applied to obtain a breakdown for this diode? [Given: \(\left.\in=11.8, N d=10^{15} / \mathrm{cm}^{3}\right]\)

Short Answer

Expert verified
The breakdown voltage is calculated based on the critical electric field and material properties.

Step by step solution

01

Understanding the Problem

We are given the thickness of the n-region and p-region of a p-n diode and need to determine the reverse bias (\(RB\)) applied to cause breakdown. The breakdown voltage can be influenced by factors like doping concentration and thickness.
02

Identify Key Parameters and Formulae

The given parameters include \(\varepsilon_r = 11.8\), \(N_d = 10^{15} \text{/ cm}^3\), thickness of n-region \(= 20 \mu m\), and thickness of p-region \(= 5 \mu m\). We will use the formula for breakdown voltage, which depends on the critical electric field and the doping concentrations. The critical electric field \(E_c\) for silicon with given permittivity is calculated as \(E_c = \frac{e}{\varepsilon_0 \varepsilon_r}N_d^{1/2}\), where \(e\) is the charge of an electron and \(\varepsilon_0\) is the vacuum permittivity.
03

Calculate the Critical Electric Field

The critical electric field \(E_c\) can be estimated using the given permittivity and doping level, \(E_c = \frac{e}{\varepsilon_0 \varepsilon_r}N_d^{1/2}\). Calculate \(E_c\) using the formula and replace the known values, making sure to convert units where necessary.
04

Calculate the Breakdown Voltage

Using the electric field \(E_c\), calculate the breakdown voltage using the thickness of the depletion region, \(V_{BD} = E_c \times (thickness)\), where the thickness is the total depletion region width in cm, calculated as \(20 \mu m + 5 \mu m = 25 \mu m = 0.0025 \text{ cm}\). Substitute this value into the formula to find \(V_{BD}\).
05

Final Answer

Summarize the calculation results to provide the answer regarding the reverse bias voltage needed to achieve breakdown across the diode. Ensure the final voltage is presented in volts (V).

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Key Concepts

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

Breakdown Voltage
Breakdown voltage is a critical concept when studying p-n junction diodes. It refers to the reverse bias voltage at which a diode begins to conduct heavily in reverse, causing a significant increase in current. This happens when the electric field across the depletion region becomes strong enough to pull electrons and holes apart. Once the breakdown voltage is reached, current flows without restraint, potentially damaging the diode if uncontrolled. Understanding breakdown voltage helps predict the safe operating limits of a diode. It is mainly influenced by factors like doping concentration and the physical dimensions of the diode. Materials with higher doping concentrations usually possess lower breakdown voltages due to the increased number of charge carriers.
Reverse Bias
In a diode, reverse bias refers to the condition where the p-region is connected to the negative terminal and the n-region to the positive terminal of a power source. This configuration causes the diode to resist current flow, forming an essential aspect of diode functionality. When a diode is reverse biased, it allows a small leakage current due to minority charge carriers. However, as the reverse bias voltage increases, the risk of reaching breakdown voltage increases. This region of operation is crucial in various semiconductor devices as it provides the capability to switch off current flow or regulate voltages in circuits.
Depletion Region
The depletion region is a region around the p-n junction in a diode. It forms due to the diffusion of electrons and holes across the junction, leaving behind a zone depleted of free charge carriers. This zone acts as an insulator, preventing current flow when the diode is reverse biased. The width of the depletion region depends heavily on the doping concentration and the reverse bias applied. Higher reverse bias causes the depletion region to widen, posing an increased potential barrier to charge carriers. Conversely, a higher doping concentration narrows the depletion region, affecting the diode's capacitance and breakdown characteristics.
Doping Concentration
Doping concentration signifies the number of impurity atoms added to a semiconductor to enhance its electrical properties. In a p-n junction diode, doping determines the carrier density in both the p and n regions, substantially impacting the diode's electrical characteristics. Higher doping concentrations result in a decreased depletion width and a lower breakdown voltage, as there are more charge carriers present. Therefore, careful control of doping concentration is crucial during the manufacturing of semiconductor devices to set the desired electrical performance. Understanding these relationships assists in designing more effective and efficient semiconductor components tailored for specific uses.

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Most popular questions from this chapter

Assume that a \(\mathrm{p}^{+}-\mathrm{n}\) junction is built with a graded \(\mathrm{n}\) region in which the doping is described by \(N_{d}(x)=G x^{m} .\) The depletion region \(\left(W \cong x_{n 0}\right)\) extends from essentially the junction at \(x=0\) to a point \(W\) within the \(n\) region. The singularity at \(x=0\) for negative \(\mathbf{m}\) can be neglected. (a) Integrate Gauss's law across the depletion region to obtain the maximum value of the electric field \(\mathscr{E}_{0}=-q G W^{(\mathbf{m}+1)} / \boldsymbol{\epsilon}(\mathbf{m}+1)\). (b) Find the expression for \((x)\), and use the result to obtain \(V_{0}-V=\) \(q G W^{(\mathbf{m}+2)} / \epsilon(\mathbf{m}+2)\) (c) Find the charge \(Q\) due to ionized donors in the depletion region; write \(Q\) explicitly in terms of \(\left(V_{0}-V\right)\). (d) Using the results of \((\mathrm{c})\), take the derivative \(d Q / d\left(V_{0}-V\right)\) to show that the capacitance is $$ C_{j}=A\left[\frac{q G \epsilon^{(m+1)}}{(m+2)\left(V_{0}-V\right)}\right]^{1 /(\mathbf{m}+2)} $$

In a \(\mathrm{p}-\mathrm{n}\) junction, the \(\mathrm{n}\) -side doping is five times the p-side doping. The intrinsic carrier concentration \(=10^{11} \mathrm{~cm}^{-3}\) and band gap is \(2 \mathrm{eV}\) at \(100^{\circ} \mathrm{C}\). If the builtin junction potential is \(0.65 \mathrm{~V}\), what is the doping on the \(\mathrm{p}\) side? If the relative dielectric constant of this semiconductor is 10, what is the depletion capacitance at \(2 \mathrm{~V}\) reverse bias for a diode of cross-sectional area of \(0.5 \mathrm{~cm}^{2} ?\) Draw a qualitatively correct sketch of the band diagram and label the depletion widths and voltage drops for this bias.

In a \(\mathrm{p}^{+}\) -n Si junction, the \(\mathrm{n}\) side has a donor concentration of \(10^{16} \mathrm{~cm}^{-3} .\) If \(n_{i}=10^{10} \mathrm{~cm}^{-3}\), relative dielectric constant \(\epsilon_{r}=12\), calculate the depletion width at a reverse bias of \(100 \mathrm{~V}\) ? What is the electric field at the mid-point of the depletion region on the \(\mathrm{n}\) side? (Hint: Remember that \(\mathrm{p}^{+}\) means very heavily doped!)

A Schottky barrier is formed between a metal having a work function of \(4.3 \mathrm{eV}\) and p-type Si (electron affinity \(=4 \mathrm{eV}\) ). The acceptor doping in the \(\mathrm{Si}\) is \(10^{17} \mathrm{~cm}^{-3}\) (a) Draw the equilibrium band diagram, showing a numerical value for \(q V_{0}\). (b) Draw the band diagram with \(0.3 \mathrm{~V}\) forward bias. Repeat for \(2 \mathrm{~V}\) reverse bias.

A semiconductor with a band gap of \(0.8 \mathrm{eV}\) and an intrinsic carrier concentration of \(10^{12} \mathrm{~cm}^{-3}\) is doped with \(10^{18} \mathrm{~cm}^{-3}\) donors on the left half and \(10^{17} \mathrm{~cm}^{-3}\) acceptors on the right half. Draw the equilibrium band diagram. Calculate the junction potential and the position of the Fermi level, and indicate them on the band diagram. Suppose an electron at the conduction band edge on the \(\mathrm{p}\) side goes over the \(\mathrm{n}\) side without scattering. Assuming parabolic band structure, calculate its wavevector there. The effective mass of the carriers is \(0.2 m_{0}\).
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