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Chapter 18: Problem 75

A transistor with \\[ \begin{array}{cl} h_{f e}=120, & h_{i e}=2 \mathrm{k} \Omega \\ h_{r e}=10^{-4}, & h_{o e}=20 \mu \mathrm{S} \end{array} \\] is used for a CE amplifier to provide an input resistance of \(1.5 \mathrm{k} \Omega\) (a) Determine the necessary load resistance \(R_{L}.\) (b) Calculate \(A_{v}, A_{i},\) and \(Z_{\text {out }}\) if the amplifier is driven by a \(4 \mathrm{mV}\) source having an internal resistance of \(600 \Omega.\) (c) Find the voltage across the load.

Short Answer

Expert verified
(a) \(R_L \approx 1.02\text{k} \Omega\), (b) \(A_v \approx -68\), \(A_i = 120\), \(Z_{out} = 50\text{k} \Omega\), (c) Voltage across load is \(-0.272\text{V}\).

Step by step solution

01

Given Parameters

We are given the following transistor parameters:\[ h_{f e} = 120, \ h_{i e} = 2 \Omega, \ h_{r e} = 10^{-4}, \ h_{o e} = 20 \mu S \]And it is desired to have an input resistance \( R_{in} = 1.5 \text{k} \Omega \).
02

Determine Load Resistance (R_L)

The input resistance for a CE amplifier is approximately given by:\[ R_{in} \approx \frac{h_{ie}(R_L || R_C)}{1 + h_{fe} \cdot h_{oe}(R_L || R_C)} \]Assuming \(R_C \approx R_L\) and replacing the input resistance, solve for \(R_L:\)\[ 1.5\text{k} \approx \frac{2 \text{k} \cdot R_L}{1 + 120 \cdot 20 \cdot 10^{-6} \cdot R_L} \]Solve the above equation to find \(R_L \approx 1.02\text{k} \Omega\).
03

Calculate Av (Voltage Gain)

The voltage gain is given by:\[ A_v = - h_{fe} \left(\frac{R_L}{r_{in} + R_s}\right) \]Where \(r_{in} = \frac{h_{ie}}{1 + h_{fe} \cdot h_{oe} \cdot (R_L || R_C)}\), and we approximate \(r_{in} \approx 1.5 \text{k} \Omega\).Plugging values:\[ A_v \approx -120 \cdot \left(\frac{1.02\text{k}}{1.5\text{k} + 600}\right) = -68 \]
04

Calculate Ai (Current Gain)

The current gain \(A_i\) can be calculated as:\[ A_i = h_{fe} \cdot \frac{R_L}{R_L + R_C} \]With \(R_L \approx R_C:\)\[ A_i = h_{fe} = 120 \]
05

Calculate Z_out (Output Impedance)

The output impedance \(Z_{out}\) is approximated as:\[ Z_{out} = \frac{1}{h_{oe}} = \frac{1}{20\cdot 10^{-6}} = 50\text{k} \Omega \]
06

Find Voltage Across the Load

The output voltage \(V_{out}\) is the product of the voltage gain and the input voltage.\[ V_{out} = A_v \cdot V_s = -68 \cdot 4 \times 10^{-3} = -0.272 \text{V} \]Thus, the voltage across the load is \(-0.272\text{V}\).

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Key Concepts

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

Common Emitter Amplifier
A Common Emitter (CE) Amplifier is a popular circuit configuration in transistor amplifiers, often used for amplifying voltage signals. It is known for its ability to provide a significant gain in a signal's amplitude. In this type of amplifier, the emitter leg of a transistor is common to both the input and output circuits. The CE amplifier essentially inverts the input signal, meaning that the output signal is 180 degrees out of phase with the input.
This phase inversion is a unique characteristic of the common emitter amplifier. Moreover, it typically offers a moderate input impedance and high output impedance, making it suitable for a broad range of applications in electronics. Understanding CE amplifiers is crucial as they form the backbone of many electronic circuits used in various devices.
  • High Voltage Gain: Often used to increase the amplitude of weak signals.
  • Inversion Capability: The output signal is inverted in relation to the input.
  • Wide Use: This design is prevalent in many audio and radio frequency applications.
Voltage Gain Calculation
Calculating the voltage gain (A_v) of a Common Emitter amplifier is essential to understand how much the amplifier can increase the amplitude of a signal. The voltage gain formula is given by:
\[ A_v = - h_{fe} \left(\frac{R_L}{r_{in} + R_s}\right) \]
where:
  • \( h_{fe} \): The current gain factor of the transistor.
  • \( R_L \): Load resistance connected to the collector.
  • \( r_{in} \): Effective input resistance of the amplifier.
  • \( R_s \): Source resistance of the signal.
The negative sign indicates the 180-degree phase shift provided by the amplifier. Proper selection of the load and source resistances, along with understanding the transistor characteristics, are fundamental to achieving the desired output.
Here's a simple walk-through: Calculate the effective resistance, use the given parameters for calculation, and substitute them into the formula to find the gain. For example, in our exercise, the gain calculated was approximately -68, indicating a significant amplification and phase inversion of the signal.
Load Resistance Determination
Choosing the correct load resistance (R_L) in a CE amplifier affects the performance of the amplifier significantly. The load resistance is crucial for setting the appropriate gain and stability. It can be determined using the relationship between the input resistance and the load resistance, as shown in the formula:
\[ R_{in} \approx \frac{h_{ie}(R_L || R_C)}{1 + h_{fe} \cdot h_{oe}(R_L || R_C)} \]
where \( R_C \) is the collector resistance. By solving this equation, you can determine the necessary \( R_L \) for your circuit requirements.
  • This formula shows that the input resistance is influenced by both the dynamic resistance parameters and the chosen load and collector resistances.
  • The load resistance directly impacts the voltage gain and the overall amplification factor of the circuit.
  • Correctly balancing these values ensures the desired input-output relationship, optimizing both signal strength and fidelity.
In practice, resistor networks, like potentiometers or fixed resistances, can be adjusted until the desired load resistance is achieved, ensuring the amplifier operates within its optimum range.
Output Impedance
Output impedance (Z_{out}) is a critical concept when analyzing and designing common emitter amplifiers. It represents the ability of the amplifier circuit to drive the load. The lower the output impedance, the more effectively the circuit can deliver power to the load. In a CE amplifier, the output impedance can be approximated using the formula:
\[ Z_{out} = \frac{1}{h_{oe}} \]
where \( h_{oe} \) is the output admittance of the transistor. This parameter is pivotal when it comes to matching the amplifier with other electrical devices or components in a network.
  • A proper match between the output impedance and the load ensures maximum power transfer.
  • If the output impedance is too high, it might fail to drive the load adequately, resulting in less efficient signal amplification.
  • For our case, the impedance was calculated as 50k惟, implying a good balance to drive most loads effectively, showcasing the design's efficiency.
Understanding and calculating output impedance helps in designing robust and high-performance amplifier systems, tailored for specific applications.

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Most popular questions from this chapter

Realize an \(L C\) ladder network such that \\[ y_{22}=\frac{s^{3}+5 s}{s^{4}+10 s^{2}+8} \\]

For a two-port network \\[ [\mathbf{z}]=\left[\begin{array}{ll} 12 & 4 \\ 4 & 6 \end{array}\right] \Omega \\] find \(V_{2} / V_{1}\) if the network is terminated with a \(2-\Omega\) resistor.

A transistor has the following parameters in a common-emitter circuit \\[ \begin{array}{c} h_{i e}=2640 \Omega, \quad h_{r e}=2.6 \times 10^{-4} \\ h_{f e}=72, \quad h_{o e}=16 \mu \mathrm{S}, \quad R_{L}=100 \mathrm{k} \Omega \end{array} \\] What is the voltage amplification of the transistor? How many decibels gain is this?

A two-port is described by \\[ \mathbf{V}_{1}=\mathbf{I}_{1}+2 \mathbf{V}_{2}, \quad \mathbf{I}_{2}=-2 \mathbf{I}_{1}+0.4 \mathbf{V}_{2} \\] Find: (a) the \(y\) parameters, (b) the transmission parameters.

Show that the transmission parameters of a two-port may be obtained from the \(y\) parameters as: \\[ \begin{array}{ll} \mathbf{A}=-\frac{\mathbf{y}_{22}}{\mathbf{y}_{21}}, & \mathbf{B}=-\frac{1}{\mathbf{y}_{21}} \\ \mathbf{C}=-\frac{\Delta_{y}}{\mathbf{y}_{21}}, & \mathbf{D}=-\frac{\mathbf{y}_{11}}{\mathbf{y}_{21}} \end{array} \\]
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