/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A uniformly doped silicon npn bi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 9

A uniformly doped silicon npn bipolar transistor at \(T=300 \mathrm{~K}\) is biased in the forward-active mode. The doping concentrations are \(N_{E}=8 \times 10^{17} \mathrm{~cm}^{-3}\), \(N_{E}=2 \times 10^{16} \mathrm{~cm}^{-3}\), and \(N_{c}=10^{15} \mathrm{~cm}^{-3} \cdot(a)\) Determine the thermal- equilibrium values \(p_{E 0}, n_{B 0}\), and \(p_{c 0} \cdot(b)\) For \(V_{B E}=0.640 \mathrm{~V}\), calculate the values of \(n_{B}\) at \(x=0\) and \(p_{E}\) at \(x^{\prime}=0 .(c)\) Sketch the minority carrier concentrations through the device and label each curve.

Short Answer

Expert verified
Find minority carrier concentrations using doping levels; adjust for forward bias with the exponential term based on thermal voltage.

Step by step solution

01

Understand the Parameters

We have a silicon npn bipolar transistor with the following doping concentrations: \(N_{E} = 8 \times 10^{17} \text{ cm}^{-3}\) for the emitter, \(N_{B} = 2 \times 10^{16} \text{ cm}^{-3}\) for the base, and \(N_{C} = 10^{15} \text{ cm}^{-3}\) for the collector. The task involves finding various carrier concentrations by solving for thermal-equilibrium values and under given bias conditions.
02

Determine Thermal-Equilibrium Values

Thermal-equilibrium means no external bias is applied, and minority carrier concentrations can be found using the relationship between majority and minority carriers. The intrinsic carrier concentration \(n_i\) for silicon at 300 K is \(1.5 \times 10^{10} \text{ cm}^{-3}\).- **Emitter:** Since \(N_{E} = 8 \times 10^{17} \text{ cm}^{-3}\), the minority hole concentration is given by \(p_{E0} = \frac{n_i^2}{N_{E}} = \frac{(1.5 \times 10^{10})^2}{8 \times 10^{17}}\).- **Base:** Similarly, for \(N_{B} = 2 \times 10^{16} \text{ cm}^{-3}\), the minority electron concentration is \(n_{B0} = \frac{n_i^2}{N_{B}} = \frac{(1.5 \times 10^{10})^2}{2 \times 10^{16}}\).- **Collector:** For \(N_{C} = 10^{15} \text{ cm}^{-3}\), the minority hole concentration is \(p_{C0} = \frac{n_i^2}{N_{C}} = \frac{(1.5 \times 10^{10})^2}{10^{15}}\).
03

Calculate Values under Forward Bias

For a forward-biased base-emitter junction with \(V_{BE} = 0.640 \text{ V}\):- The electron concentration at the base-emitter junction boundary ( at \(x = 0\)) in the base is determined by \(n_{B} = n_{B0} \cdot e^{(V_{BE}/V_T)}\), where \(V_T = \frac{kT}{q} \approx 0.0259 \text{ V} \/ \text{at} \/ 300 \text{ K} \).- The hole concentration at the emitter-base junction boundary ( at \(x' = 0\)) in the emitter is \(p_{E} = p_{E0} \cdot e^{(V_{BE}/V_T)}\).
04

Sketch the Minority Carrier Concentrations

To sketch the minority carrier concentrations, plot each carrier concentration as a function of distance through the transistor.- In the base, start with minority carrier electrons \(n_B(x)\) from the base-emitter junction and show a gradient towards zero at the collector end.- In the emitter, plot the minority holes \(p_E(x')\), showing an exponential decay from the emitter-base junction into the emitter.- Finally, in the collector, minority holes \(p_C(x)\) remain close to \(p_{C0}\) as there is no significant forward bias affecting them.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Doping Concentrations
Doping concentrations refer to the number of impurities added to different regions of a semiconductor to control its electrical properties. In the context of an npn bipolar transistor, doping is conducted in the emitter, base, and collector. The concentration of doping affects how the transistor amplifies current.
  • **Emitter ({E})**: Typically has the highest doping concentration, like the given value of \(N_{E} = 8 \times 10^{17} \text{ cm}^{-3}\). This high concentration helps the emitter to efficiently inject a large number of charge carriers (electrons, in this case) into the base.
  • **Base ({B})**: The base is doped at a moderate level, such as \(N_{B} = 2 \times 10^{16} \text{ cm}^{-3}\). This ensures enough minority carriers are available to cross into the collector while keeping base current minimal.
  • **Collector ({C})**: This region is lightly doped, represented by \(N_{C} = 10^{15} \text{ cm}^{-3}\). The light doping allows the charge carriers to leave the base region easily and ensures that the collector can sustain a high voltage without breaking down.
In essence, understanding doping concentrations is crucial for analyzing how efficient a transistor is in controlling electrical current and its suitability based on required electronics specifications.
Thermal-Equilibrium Values
The thermal-equilibrium state describes a condition where the semiconductor device is not influenced by any external voltages or currents. In this state, the concentration of minority carriers becomes a vital aspect of analysis.

For semiconductors, thermal-equilibrium can be analyzed using the intrinsic carrier concentration \(n_i\). The intrinsic concentration for silicon at \(300 \text{ K}\) is \(1.5 \times 10^{10} \text{ cm}^{-3}\). Using this value, we calculate the minority carrier concentrations in each region of the transistor by understanding the equilibrium relationship:
  • **Emitter Minority Holes \(p_{E0}\)**: Given by \(p_{E0} = \frac{n_i^2}{N_E}\), resulting in a concentration due to the heavy doping in the emitter.
  • **Base Minority Electrons \(n_{B0}\)**: Calculated as \(n_{B0} = \frac{n_i^2}{N_B}\), this lets us understand the propensity of electrons to flow into the collector.
  • **Collector Minority Holes \(p_{C0}\)**: Determined by \(p_{C0} = \frac{n_i^2}{N_C}\), signifying the low concentration due to light doping.
Reaching these thermal-equilibrium values helps to predict how a transistor behaves when an external voltage is applied, essential for effective transistor design.
Minority Carrier Concentrations
In a bipolar transistor, minority carriers are critical to its operation. Essentially, minority carrier concentrations refer to the density of the minority charge carriers in any given region of the semiconductor. Identifying these carriers allows us to grasp how well the transistor can conduct when a forward bias is introduced.

**Understanding Key Concepts**:
  • **Balance in Forward Bias**: Upon applying a bias, carrier concentration changes exponentially at junctions. In this exercise, for a forward-biased junction \(V_{BE} = 0.640 \text{ V}\), the electron concentration at the base-emitter junction and the hole concentration at the emitter-base junction are recalculated.
  • **Calculation Using Bias**: For instance, \(n_{B}\) (electrons in the base) is influenced by \((V_{BE}/V_T)\), where \(V_T\) is the thermal voltage, calculated as about \(0.0259 \text{ V}\) at \(300 \text{ K}\).
Understanding minority carriers helps in predicting the transistor's response under various operating conditions, ensuring circuits are designed to meet desired performance criteria.
Forward-Active Mode
The forward-active mode is one of the primary operating modes of a bipolar junction transistor (BJT). In this mode, the base-emitter junction is forward-biased while the base-collector junction is reverse-biased. This setup allows the transistor to amplify a signal effectively.

**Significance and Behavior**:
  • **Signal Amplification**: Forward-active mode is crucial for amplification as it permits electrons to move easily from the emitter to the collector through the base, allowing the transistor to control larger currents with a smaller input current.
  • **Minority Carriers and Efficiency**: In this mode, the base's charge carrier distribution adapts to enhance the flow of electrons from emitter to collector, leading to high efficiency.
Operating a bipolar transistor in the forward-active mode focuses on maintaining low base current and high collector current, maximizing the current gain \(\beta\) of the device. This understanding is pivotal for designing circuits that require reliable amplification.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The parameters in the Ebers-Moll model are \(\alpha_{F}=0.9920, I_{E S}=5 \times 10^{-14} \mathrm{~A}\), and \(I_{C S}=10^{-13} \mathrm{~A} .\) Let \(T=300 \mathrm{~K} .\) Plot \(I_{C}\) versus \(V_{C B}\) for \(-0.5

Consider the Ebers-Moll model and let the base terminal be open so \(I_{B}=0 .\) Show that, when a collector-emitter voltage is applied, we have $$ I_{C}=I_{C E O}=I_{C S} \frac{\left(1-\alpha_{F} \alpha_{R}\right)}{\left(1-\alpha_{F}\right)} $$

Assume that an npn bipolar transistor has a common-emitter current gain of \(\beta=100 .(a)\) Sketch the ideal current-voltage characteristics \(\left(i_{C}\right.\) versus \(\left.v_{C R}\right)\), like those in Figure \(12.9\), as \(i_{B}\) varies from zero to \(0.1 \mathrm{~mA}\) in \(0.01-\mathrm{mA}\) increments. Let \(\nu_{C E}\) vary over the range \(0 \leq \nu_{C E} \leq 10 \mathrm{~V} .(b)\) Assuming \(V_{c c}=10 \mathrm{~V}\) and \(R_{c}=1 \mathrm{k} \Omega\) in the circuit in Figure \(12.8\), superimpose the load line on the transistor characteristics in part \((a) .(c)\) Plot, on the resulting graph, the value of \(i_{c}\) and \(\nu_{C E}\) corresponding to \(i_{6}=0.05 \mathrm{~mA}\)

A uniformly doped silicon npn bipolar transistor at \(T=300 \mathrm{~K}\) has parameters \(N_{E}=2 \times 10^{18} \mathrm{~cm}^{-3}, N_{B}=2 \times 10^{16} \mathrm{~cm}^{-3}, N_{C}=2 \times 10^{15} \mathrm{~cm}^{-3}, x_{10}=0.85 \mu \mathrm{m}\) and \(D_{B}=25 \mathrm{~cm}^{2} / \mathrm{s}\). Assume \(x_{B o} \ll L_{s}\) and let \(V_{B E}=0.650 \mathrm{~V}\). \((\) a \()\) Determine the electron diffusion current density in the base for (i) \(V_{C B}=4 \mathrm{~V},(i i) V_{C B}=8 \mathrm{~V}\), and (iii) \(V_{C B}=12 \mathrm{~V}_{.}(b)\) Estimate the Early voltage.

Consider \(\mathrm{a} \mathrm{p}^{++} \mathrm{n}^{+} \mathrm{p}\) bipolar transistor, uniformly doped in each region. Sketch the energy-band diagram for the case when the transistor is \((a)\) in thermal equilibrium, (b) biased in the forward-active mode, \((c)\) biased in the inverse-active region, and \((d)\) biased in cutoft with both the \(\mathrm{B}-\mathrm{E}\) and \(\mathrm{B}-\mathrm{C}\) junctions reverse biased.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Mechanics and Materials

Read Explanation

Electrostatics

Read Explanation

Translational Dynamics

Read Explanation

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.