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Chapter 5: Problem 7

For the common-base configuration of Fig. 18, an ac signal of \(10 \mathrm{mV}\) is applied, resulting in an ac emitter current of \(0.5 \mathrm{~mA}\). If \(\alpha=0.980\), determine: a. \(Z_{i}\) b. \(V_{o}\) if \(R_{L}=1.2 \mathrm{k} \Omega\). c. \(A_{v}=V_{d} / V_{i}\). d. \(Z_{o}\) with \(r_{o}=\infty \Omega\). e. \(A_{i}=I_{o} / I_{i}\) f. \(I_{b}\)

Short Answer

Expert verified
The solved results are as follows: \(Z_i = 20 \, k\Omega\), \(V_o = 0.588 \, V\), \(A_v = 0.059\), \(Z_o = \infty \, \Omega\), \(A_i = 0.980\) and \(I_b = 0.01 \, mA\).

Step by step solution

01

Calculation of \(Z_i\)

The input impedance \(Z_i\) is given by the formula \(Z_i = V_i / I_i\). Given that \(V_i = 10 mV\) and \(I_i = 0.5 mA\), just plug the values into the formula and solve for \(Z_i\).
02

Calculation of \(V_o\)

Since the output voltage \(V_o\) is the result of the emitter current \(I_e\) running through the load \(R_L = 1.2 k \Omega\), it can calculated by \(V_o = I_e * R_L\). As given, \(I_e = \alpha * I_i\), so first calculate \(I_e\), then substitute \(I_e\) in the previously mentioned equation to solve for \(V_o\).
03

Calculation of \(A_v = V_d / V_i\)

Here, \(A_v\) is the Voltage gain and \(V_d\) is written as \(V_o\). So simply substitute \(V_o\) and \(V_i\) from the previous steps to find out the Voltage gain \(A_v\).
04

Calculation of \(Z_o\) with \(r_o = \infty \Omega\)

The output impedance \(Z_o\) is equal to \(r_o\) when it is not zero. Since \(r_o = \infty \Omega\), \(Z_o\) will also be \( \infty \Omega \).
05

Calculation of \(A_i = I_o / I_i\)

To calculate the current gain \(A_i\), use the given formula. The current \(I_o\) is same as \(I_e = \alpha * I_i\) and \(I_i\) is given in the problem. Substitute these values in the equation to solve for \(A_i\).
06

Calculation of \(I_b\)

The base current \(I_b\) can be determined from the relation \(I_i = I_b + I_e\). Formulating this in terms of \(I_b\) gives us \(I_b = I_i - I_e\). Substituting the given values will yield \(I_b\).

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Key Concepts

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

Input Impedance Zi
Understanding the input impedance (\(Z_i\)) of a common-base amplifier is crucial for analyzing its compatibility with the signal source. The input impedance is essentially the resistance that the input signal encounters when it tries to enter the amplifier. It's calculated by the formula \(Z_i = V_i / I_i\), where \(V_i\) is the input voltage and \(I_i\) is the input current. A lower \(Z_i\) means more current from the signal source is drawn, which might be problematic for sources with high internal resistance. For the common-base configuration, it's common to find a low \(Z_i\) owing to its direct connection to the emitter.
Output Voltage Vo
The output voltage (\(V_o\)) of a common-base configuration can be thought of as the 'final product' that comes out of the amplifier when an AC signal is applied. It is found by multiplying the emitter current \(I_e\) with the load resistance \(R_L\), as shown by the equation \(V_o = I_e * R_L\). Since the emitter current is directly proportional to the input current (\(I_i\)), modified by the transistor's current gain (\(\alpha\)), it is crucial to consider internal device characteristics and the external load to accurately predict the output voltage.
Voltage Gain Av
Voltage gain (\(A_v\)) is a measure of how much an amplifier increases the strength of a signal. In the common-base amplifier, this is expressed as the ratio of the output voltage (\(V_o\)) over the input voltage (\(V_i\)), given by the equation \(A_v = V_o / V_i\). High \(A_v\) values mean greater amplification of the original signal, which is often sought in amplifier design. However, the gain can vary with frequency, so it's important to consider this factor for AC signals.
Output Impedance Zo
Output impedance (\(Z_o\)) is an indicator of how the amplifier affects the connected load and is the electrical resistance 'felt' by the load connected to the amplifier's output. Ideally, in a common-base configuration with an infinite \(r_o\), like what is given in the exercise, the \(Z_o\) is also considered infinite. An infinite output impedance in theory may not always be desirable as it prevents the circuit from efficiently delivering power to the load, underscoring the need to balance impedance in actual amplifier design.
Current Gain Ai
Current gain (\(A_i\)), in the context of a common-base circuit, is defined as the ratio of the output current to the input current, which mathematically can be expressed as \(A_i = I_o / I_i\). However, because we're dealing with common-base, we remember the output current is essentially the emitter current (\(I_e\)) and input current is base current (\(I_i\)). Since \(I_e\) is approximately equal to \(\alpha * I_i\), where \(\alpha\) is the common-base current gain of the transistor, \(A_i\) is typically slightly less than 1 for this configuration.
Base Current Ib
The base current (\(I_b\)) is a relatively small current that controls the larger emitter and collector currents within a transistor. For the common-base configuration, the emitter current (\(I_e\)) is produced when the base current and the collector current come together. Thus, knowing \(I_e\) and the input current \(I_i\), which in this case is the emitter current, you can find \(I_b\) by subtracting \(I_e\) from \(I_i\), because in this configuration, \(I_i\) is equivalent to \(I_b + I_e\). A higher base current would lead to a higher collector current, which affects the overall amplification properties of the transistor.

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Most popular questions from this chapter

a. What is the expected amplification of a BJT transistor amplifier if the de supply is set to zero volts? b. What will happen to the output ac signal if the dc level is insufficient? Sketch the effect on the waveform. c. What is the conversion efficiency of an amplifier in which the effective value of the current through a \(2.2-\mathrm{k} \Omega\) load is \(5 \mathrm{~mA}\) and the drain on the \(18-\mathrm{V}\) de supply is \(3.8 \mathrm{~mA}\) ?

Given the typical values of \(R_{L}=2.2 \mathrm{k} \Omega\) and \(h_{o e}=20 \mu \mathrm{S}\), is it a good approximation to ignore the effects of \(1 / h_{o e}\) on the total load impedance? What is the percentage difference in total loading on the transistor using the following equation? $$ \% \text { difference in total load }=\frac{R_{L}-R_{L} \|\left(1 / h_{o e}\right)}{R_{L}} \times 100 \% $$

a. Given an Early voltage of \(V_{A}=100 \mathrm{~V}\), determine \(r_{o}\) if \(V_{C E_{e}}=8 \mathrm{~V}\) and \(I_{C_{Q}}=4 \mathrm{~mA}\). b. Using the results of part (a), find the change in \(I_{C}\) for a change in \(V_{C E}\) of \(6 \mathrm{~V}\) at the same \(Q\) -point as part (a).

The input impedance to a common-emitter transistor amplifier is \(1.2 \mathrm{k} \Omega\) with \(\beta=140\), \(r_{o}=50 \mathrm{k} \Omega\), and \(R_{L}=2.7 \mathrm{k} \Omega\). Determine: a. \(r_{e}\) b. \(I_{b}\) if \(V_{i}=30 \mathrm{mV}\). c. \(I_{c}\). d. \(A_{i}=I_{o} / I_{i}=I_{L} / I_{b}\). e. \(A_{v}=V_{o} / V_{i}\)

What is the reactance of a \(10-\mu \mathrm{F}\) capacitor at a frequency of \(1 \mathrm{kHz}\) ? For networks in which the resistor levels are typically in the kilohm range, is it a good assumption to use the short-circuit equivalence for the conditions just described? How about at \(100 \mathrm{kHz}\) ?
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