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Chapter 12: Problem 57

Assume the base transit time of a BJT is \(100 \mathrm{ps}\) and carriers cross the \(1.2 \mu \mathrm{m} \mathrm{B}-\mathrm{C}\) space charge region at a speed of \(10^{7} \mathrm{~cm} / \mathrm{s}\). The emitter-base junction charging time is \(25 \mathrm{ps}\) and the collector capacitance and resistance are \(0.10 \mathrm{pF}\) and \(10 \Omega\), respectively. Determine the cutoff frequency.

Short Answer

Expert verified
The cutoff frequency is 12.5 GHz.

Step by step solution

01

Calculate BC Transit Time

The speed of carriers crossing the base-collector (BC) space charge region is given as \(10^7 \text{ cm/s}\). The length of this region is \(1.2 \mu m = 1.2 \times 10^{-4} \text{ cm}\). The transit time, \(t_{bc}\), through this region can be calculated using the formula: \[t_{bc} = \frac{\text{distance}}{\text{speed}} = \frac{1.2 \times 10^{-4} \text{ cm}}{10^7 \text{ cm/s}} = 1.2 \text{ ps}.\]
02

Sum of Delay Times

The total delay time, \(\tau_{total}\), is the sum of the base transit time, the emitter-base junction charging time, and the BC transit time. \[\tau_{total} = 100 \text{ ps (base transit)} + 25 \text{ ps (emitter-base charging)} + 1.2 \text{ ps (BC transit)} = 126.2 \text{ ps}.\]
03

Calculate Collector RC Time Constant

The collector RC time constant, \(\tau_c\), is calculated using the relationship \(\tau_c = RC\), where \(R\) is the collector resistance and \(C\) is the collector capacitance. Given \(R = 10 \Omega\) and \(C = 0.10 \text{ pF} = 0.10 \times 10^{-12} \text{ F}\), \[\tau_c = 10 \times 0.10 \times 10^{-12} = 1 \times 10^{-12} \text{ s} = 1 \text{ ps}.\]
04

Calculate Total Time Constant

The total time constant, \(\tau\), is the sum of \(\tau_{total}\) from Step 2 and the collector RC time constant from Step 3. \[\tau = \tau_{total} + \tau_c = 126.2 \text{ ps} + 1 \text{ ps} = 127.2 \text{ ps}.\]
05

Calculate the Cutoff Frequency

The cutoff frequency, \(f_T\), can be calculated using the formula for the time constant, \(\tau\), where \[f_T = \frac{1}{2\pi \tau}.\] Using \(\tau = 127.2 \text{ ps} = 127.2 \times 10^{-12} \text{ s}\), \[f_T = \frac{1}{2\pi \times 127.2 \times 10^{-12}} = 1.25 \times 10^{10} \text{ Hz} = 12.5 \text{ GHz}.\]

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Key Concepts

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

BJT Transit Time
Transit time, particularly in BJT (Bipolar Junction Transistor), refers to how long it takes for charge carriers to move through a region of the transistor. In this context, we look at the transit time as carriers cross the base-collector junction. To calculate this, the formula used is:
  • \[ t_{bc} = \frac{\text{distance}}{\text{speed}} \]
  • Where the distance is the width of the space charge region, and speed is the velocity at which carriers move.

This parameter is crucial because it affects the speed at which a BJT can operate. For real-world applications, the faster the carriers transit, the faster the transistor can process signals.
RC Time Constant
The RC time constant is a measure of how quickly a capacitor charges through a resistor in an electronic circuit. It is defined as:
  • \[ \tau_c = R \times C \]
  • Where \( R \) is resistance and \( C \) is capacitance.

This constant is important in determining the response speed of electronic circuits. When related to a BJT, the RC time constant of the collector helps determine the overall speed of the transistor. A smaller time constant means the circuit can handle higher frequencies, which is crucial in high-speed applications.
Carrier Speed
Carrier speed in semiconductors refers to the velocity at which charge carriers, such as electrons or holes, move under the influence of an electric field. In this exercise, the carrier speed is given for the base-collector region. Calculating the carrier speed helps determine how quickly signals can be processed by a device like a BJT.
  • Faster carrier speed results in lower transit time, allowing for faster operation of the transistor.
  • The formula used is a simple rearrangement of the transit time formula.
Understanding this parameter is key for designing fast and efficient semiconductor devices.
Semiconductor Physics Concepts
Semiconductor physics forms the foundation of how devices like BJTs operate. In semiconductors, materials are engineered to have electrical properties that can be precisely controlled by doping with impurities.
  • This allows for the manipulation of charge carriers that enable devices to act as amplifiers or switches.
  • Key concepts include electron and hole mobility, doping concentrations, and junction behaviors.

A solid grasp of these principles helps in understanding how transistors function, and how parameters like transit time and RC time constant affect device performance. These concepts are crucial for anyone looking to excel in electronic engineering.

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Most popular questions from this chapter

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 $$ I_{C}=I_{C E O}=I_{C S} \frac{\left(1-\alpha_{F} \alpha_{R}\right)}{\left(1-\alpha_{F}\right)} $$

A bipolar transistor is biased in the forward-active region. ( \(a\) ) For a base current of \(I_{\bar{B}}=4.2 \mu \mathrm{A}\) and a collector current of \(I_{C}=0.625 \mathrm{~mA}\), determine \((i) \beta,(i i) \alpha\), and (iii) \(I_{E \cdot}(b)\) For a collector current of \(I_{C}=1.254 \mathrm{~mA}\) and an emitter current of \(I_{E}=1.273 \mathrm{~mA}\), determine (i) \(\beta,(i i) \alpha\), and (iii) \(I_{B .}(c)\) For a base current of \(I_{B}=.065 \mu \mathrm{A}\) and a common-emitter current gain of \(\beta=150\), determine (i) \(\alpha\), (ii) \(I_{C}\), and (iii) \(I_{E \text { - }}\)

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.

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) A silicon npn bipolar transistor at \(T=300 \mathrm{~K}\) is to be designed such that the common-emitter current gain is at least \(\beta=120\) and the Early voltage is at least \(V_{A}=140 \mathrm{~V} .(b)\) Repeat part \((a)\) for a pnp silicon bipolar transistor.
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