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Chapter 12: Problem 6

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) \(I_{E \text { - }}\)

Short Answer

Expert verified
(a) 尾 鈮� 148.81, 伪 鈮� 0.9933, I_E = 0.6292 mA. (b) 尾 鈮� 66, 伪 鈮� 0.9852, I_B = 0.019 mA. (c) 伪 鈮� 0.9934, I_C = 9.75 渭A, I_E = 9.815 渭A.

Step by step solution

01

Calculate 尾 for Part (a)

To calculate the current gain \( \beta \) for a bipolar junction transistor (BJT), use the relationship between the collector current \( I_C \) and base current \( I_B \): \[ \beta = \frac{I_C}{I_B} \] Substitute the given values: \[ \beta = \frac{0.625 \text{ mA}}{4.2 \mu \text{A}} \approx 148.81 \] So, \( \beta \approx 148.81 \).
02

Calculate 伪 for Part (a)

The \( \alpha \) parameter is related to \( \beta \) through the equation: \[ \alpha = \frac{\beta}{\beta + 1} \] Substitute the calculated \( \beta \): \[ \alpha = \frac{148.81}{148.81 + 1} \approx 0.9933 \] Thus, \( \alpha \approx 0.9933 \).
03

Calculate I_E for Part (a)

The emitter current \( I_E \) is the sum of the base current \( I_B \) and the collector current \( I_C \): \[ I_E = I_B + I_C \] Convert \( I_B \) and \( I_C \) to milliamperes and substitute: \[ I_E = 0.0042 \text{ mA} + 0.625 \text{ mA} = 0.6292 \text{ mA} \] Therefore, \( I_E = 0.6292 \text{ mA} \).
04

Calculate 尾 for Part (b)

Given \( I_C = 1.254 \text{ mA} \) and \( I_E = 1.273 \text{ mA} \), first calculate: \[ I_B = I_E - I_C = 1.273 \text{ mA} - 1.254 \text{ mA} = 0.019 \text{ mA} \] Now use \( \beta = \frac{I_C}{I_B} \): \[ \beta = \frac{1.254}{0.019} \approx 66 \] So, \( \beta \approx 66 \).
05

Calculate 伪 for Part (b)

Use the \( \alpha \) equation from \( \beta \): \[ \alpha = \frac{\beta}{\beta + 1} = \frac{66}{66 + 1} \approx 0.9852 \] Therefore, \( \alpha \approx 0.9852 \).
06

Calculate I_B for Part (b)

The base current \( I_B \) is calculated as: \[ I_B = I_E - I_C = 1.273 \text{ mA} - 1.254 \text{ mA} = 0.019 \text{ mA} \] So \( I_B = 0.019 \text{ mA} \).
07

Calculate 伪 for Part (c)

Use the relationship between \( \alpha \) and \( \beta \): \[ \alpha = \frac{\beta}{\beta + 1} = \frac{150}{151} \approx 0.9934 \] Thus, \( \alpha \approx 0.9934 \).
08

Calculate I_C for Part (c)

Given \( I_B = 0.065 \mu \text{A} \) and \( \beta = 150 \), use: \[ I_C = \beta \cdot I_B = 150 \times 0.065 \mu \text{A} = 9.75 \mu \text{A} \] Therefore, \( I_C = 9.75 \mu \text{A} \).
09

Calculate I_E for Part (c)

Calculate the emitter current \( I_E \) as: \[ I_E = I_B + I_C = 0.065 \mu \text{A} + 9.75 \mu \text{A} = 9.815 \mu \text{A} \] Thus, \( I_E = 9.815 \mu \text{A} \).

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Key Concepts

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

Current Gain (Beta and Alpha)
In a bipolar junction transistor (BJT), two important parameters representing current gain are beta (\( \beta \)) and alpha (\( \alpha \)). Beta, also known as the current gain or amplification factor in the common-emitter configuration, measures the transistor's ability to amplify a small base current into a larger collector current. It's calculated using the formula:
  • \( \beta = \frac{I_C}{I_B} \)
where \( I_C \) is the collector current and \( I_B \) is the base current.

Alpha, on the other hand, pertains to the common-base configuration and shows the relation between the collector current \( I_C \) and the emitter current \( I_E \). It can be deduced from beta using:
  • \( \alpha = \frac{\beta}{\beta + 1} \)
This relationship is crucial because it highlights that as \( \beta \) increases, \( \alpha \) approaches 1, indicating that almost all the current flowing through the emitter reaches the collector.

Understanding \( \beta \) and \( \alpha \) helps in analyzing and designing circuits utilizing BJTs for amplification purposes.
Base, Collector, and Emitter Current Relationships
The currents flowing through the BJT鈥攂ase current \( I_B \), collector current \( I_C \), and emitter current \( I_E \)鈥攁re deeply interlinked. In the simplest terms, the current entering the transistor through the emitter is nearly equal to the sum of the currents exiting through the base and collector.
  • \( I_E = I_B + I_C \)

The base current is typically very small compared to the collector and emitter currents, given the amplifying nature of BJTs. Often, engineers and students calculate one current if the others are known, such as determining \( I_B \) when given \( I_C \) and \( I_E \):
  • \( I_B = I_E - I_C \)

This simple set of relationships is fundamental in analyzing and designing circuits, ensuring currents are within expected ranges. Also, these calculations assist in verifying the operation of the transistor in desired modes, such as in the forward-active region.
Forward-Active Region of BJT
The forward-active region is the most common mode of operation for a BJT and critical for amplification purposes. In this state, the base-emitter junction is forward-biased, allowing current to flow through easily, while the base-collector junction is reverse-biased, controlling the flow from collector to emitter.

In this region, the collector current \( I_C \) is nearly proportional to the base current \( I_B \) multiplied by \( \beta \), effectively creating amplification. Despite changes in \( I_B \), \( I_C \) remains predictable and stable due to the transistor's characteristics.

Operating within the forward-active region allows the transistor to function efficiently as an amplifier. It's crucial that the biasing conditions are correctly set for the BJT to maintain this region without transitioning to saturation (where both junctions are forward-biased) or cutoff (where both junctions are reverse-biased), as those states would limit or stop the transistor's ability to amplify signals.

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Most popular questions from this chapter

An npn silicon bipolar transistor at \(T=300 \mathrm{~K}\) has uniform dopings of \(N_{E}=\) \(10^{19} \mathrm{~cm}^{-3}, N_{B}=10^{17} \mathrm{~cm}^{-3}\), and \(N_{C}=7 \times 10^{15} \mathrm{~cm}^{-3}\). The transistor is operating in the inverse-active mode with \(V_{B E}=-2 \mathrm{~V}\) and \(V_{\mathrm{AC}}=0.565 \mathrm{~V} .(a)\) Sketch the minority carrier distribution through the device. (b) Determine the minority carrier concentrations at \(x=x_{\bar{B}}\) and \(x^{\prime \prime}=0 .(c)\) If the metallurgical base width is \(1.2 \mu \mathrm{m}\), determine the neutral base width.

In a particular bipolar transistor, the base transit time is 20 percent of the total delay time. The base width is \(0.5 \mu \mathrm{m}\) and the base diffusion coefficient is \(D_{B}=20 \mathrm{~cm}^{2} / \mathrm{s}\). Determine the cutoff frequency.

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.

The output resistance of a pnp bipolar transistor is \(r_{a}=180 \mathrm{k} \Omega\). The Early voltage is \(V_{A}=80 \mathrm{~V}\). Determine the change in collector current if \(V_{E C}\) increases from 2 to \(5 \mathrm{~V}\).

A silicon pnp bipolar transistor at \(T=300 \mathrm{~K}\) has a \(\mathrm{B}-\mathrm{E}\) cross-sectional area of \(A_{\overline{B E}}-5 \times 10^{-4} \mathrm{~cm}^{2}\), neutral base width of \(x_{B}-0.70 \mu \mathrm{m}\), a neutral emitter width of \(x_{E}=0.50 \mu \mathrm{m}\), and uniform doping concentrations of \(N_{E}=5 \times 10^{17} \mathrm{~cm}^{-3}\), \(N_{B}=10^{16} \mathrm{~cm}^{-3}\), and \(N_{C}=10^{15} \mathrm{~cm}^{-3} .\) Other transistor parameters are \(D_{B}=10 \mathrm{~cm}^{2} / \mathrm{s}\) \(D_{E}=15 \mathrm{~cm}^{2} / \mathrm{s}, \tau_{E 0}=\tau_{B 0}=5 \times 10^{-7} \mathrm{~s}\), and \(\tau_{c 0}=2 \times 10^{-6} \mathrm{~s}\). The transistor is biased in the forward-active mode and the recombination factor is \(\delta=0.995 .\) Determine the collector current for \((a) V_{E B}=0.550 \mathrm{~V},(b) I_{B}=0.80 \mu \mathrm{A}\), and (c) \(I_{E}=125 \mu \mathrm{A}\).
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