A simple fixed-biased small signal amplifier made of the following electronic componen-ts produce an output signal of 4 V (peak to peak) across the collector load resistancewhen the input signal source provides a base current variation of 20 µA (peak topeak).A 12V DC supplyAn input signal sourceA 2.2 µF coupling capacitorBC547 BJT having a DC current gain (βDC ) of 325 and Ri of 1 kΩA Collector load having 820 Ω resistanceA variable resistor with variation ranging from 200 kΩ to 2 MΩ(a) Sketch the circuit diagram of the amplifier. Properly label all circuit components,voltages and currents. [3 marks](b) Find the required bias resistance to set the DC base bias current to 22.6 mA.[2 marks](c) Find the quiescent collector current and voltage. [2 marks](d) Derive the equation of the DC load line. [2 marks](e) Sketch the DC load line. Mark the quiescent point on it and comment on itslocation on the DC load line. [2 marks](f) Using the equation of the DC load line, find the peak to peak variation of thecollector current. [2 marks](g) Find the AC signal power delivered to the load. [2 marks](h) Find the DC power dissipation in the load. [2 marks](i) Mark all relevant points on the DC load line and graphically illustrate theoperation of the circuit using appropriate waveforms on the graph. Clearlylabel the axes and the waveforms. [2 marks](j) Find the voltage gain, current gain and the power gain of the circuit. [3 marks](k) In what scenario you should not use the DC load line for gain calculations of anamplifier. Briefly explain the method used for gain calculations in such cases
Question
A simple fixed-biased small signal amplifier made of the following electronic componen-ts produce an output signal of 4 V (peak to peak) across the collector load resistancewhen the input signal source provides a base current variation of 20 µA (peak topeak).A 12V DC supplyAn input signal sourceA 2.2 µF coupling capacitorBC547 BJT having a DC current gain (βDC ) of 325 and Ri of 1 kΩA Collector load having 820 Ω resistanceA variable resistor with variation ranging from 200 kΩ to 2 MΩ(a) Sketch the circuit diagram of the amplifier. Properly label all circuit components,voltages and currents. 3 marks Find the required bias resistance to set the DC base bias current to 22.6 mA.2 marks Find the quiescent collector current and voltage. 2 marks Derive the equation of the DC load line. 2 marks Sketch the DC load line. Mark the quiescent point on it and comment on itslocation on the DC load line. 2 marks Using the equation of the DC load line, find the peak to peak variation of thecollector current. 2 marks Find the AC signal power delivered to the load. 2 marks Find the DC power dissipation in the load. 2 marks Mark all relevant points on the DC load line and graphically illustrate theoperation of the circuit using appropriate waveforms on the graph. Clearlylabel the axes and the waveforms. 2 marks Find the voltage gain, current gain and the power gain of the circuit. 3 marks In what scenario you should not use the DC load line for gain calculations of anamplifier. Briefly explain the method used for gain calculations in such cases
Solution
This is a complex question that involves multiple steps. Here's how you can approach it:
(a) The circuit diagram of the amplifier would include the DC supply, input signal source, coupling capacitor, BJT, collector load, and variable resistor. The BJT would be connected to the DC supply through the variable resistor (which sets the base current), and the input signal source would be connected to the base of the BJT through the coupling capacitor. The collector load would be connected to the collector of the BJT and the other end to the DC supply.
(b) The required bias resistance can be found using Ohm's law. The base current is given as 22.6 mA and the supply voltage is 12V. So, R = V/I = 12V / 22.6mA = approximately 530.97 kΩ.
(c) The quiescent collector current can be found using the formula Ic = βIb, where β is the DC current gain and Ib is the base current. The quiescent collector voltage can be found using the formula Vc = Vcc - IcRc, where Vcc is the supply voltage, Ic is the collector current, and Rc is the collector load resistance.
(d) The equation of the DC load line can be derived from the formula Vc = Vcc - Ic*Rc. This equation represents the relationship between the collector current and the collector-emitter voltage.
(e) The DC load line can be sketched on a graph with Vc on the y-axis and Ic on the x-axis. The quiescent point can be marked on the graph at the coordinates (Icq, Vcq), where Icq is the quiescent collector current and Vcq is the quiescent collector voltage.
(f) The peak to peak variation of the collector current can be found by substituting the peak to peak base current variation into the equation of the DC load line.
(g) The AC signal power delivered to the load can be found using the formula P = V^2 / R, where V is the peak to peak output voltage and R is the load resistance.
(h) The DC power dissipation in the load can be found using the formula P = Ic*Vc, where Ic is the quiescent collector current and Vc is the quiescent collector voltage.
(i) The operation of the circuit can be illustrated graphically by plotting the DC load line and the AC signal waveform on the same graph.
(j) The voltage gain, current gain, and power gain of the circuit can be found using the formulas Av = Vo / Vi, Ai = Io / Ii, and Ap = Av * Ai, respectively, where Vo is the output voltage, Vi is the input voltage, Io is the output current, and Ii is the input current.
(k) The DC load line should not be used for gain calculations of an amplifier when the amplifier is operating in the non-linear region. In such cases, the gain can be calculated using the small signal model of the amplifier.
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