Problem 2: BJT Small-Signal Voltage Amplifiers amplifier that utilizes a In the
ID: 2082803 • Letter: P
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Problem 2: BJT Small-Signal Voltage Amplifiers amplifier that utilizes a In the lectures, we covered in detail the analysis of an npn BJT additional amplifier single-supply DC biasing. In this problem, you will meet two a DC current source, architectures one that is based on dual DC supplies combined with and the other that utilizes a pnp BJT with single-supply biasing. Q-point (that is, the In both problems (below) you need to manually find the amplifier's and ro), drawing of DC values of lo and VCE), the BJT small-signal parameters (re, r, gm Rin and the the amplifier's small-signal diagram, the amplifier's input resistance amplifier's voltage gain Av vo/vs. Av (using both In the PSPICE simulation verify the DC conditions and measure R and simulation modes Time Domain and AC Sweep). The supply voltages all involve 10V voltages (that is Voc VEE 10V). In both problems the signal source resistance is and the load resistance is RL 15K2. The large capacitors are assumed infinite (for hand-calculations) and in PSPICE take each to be equal to C 1mF. In both problems let Rc 3.5KS2. 2.1. Let I 1mA (and it's beyond the scope of this course to discuss how such DC current sources are implemented using transistor circuits) and let RB 100KQExplanation / Answer
The BJT is biased in the active region by dc voltage source VBE. e.g., Q-point is set at (IC, VCE) = (1 mA, 10 V) with IB = 15 µA (F = 100) Total base-emitter voltage is: vBE = VBE + vbe Collector-emitter voltage is: vCE = VCC – iCRC This is the load line equation.
If changes in operating currents and voltages are small enough, then iC and vCE waveforms are undistorted replicas of the input signal. A small voltage change at the base causes a large voltage change at collector. Voltage gain is given by: Minus sign indicates 180o phase shift between the input and output signals. Av = Vce Vbe = 1.65180o 0.0080o = 206180o = 20Minus sign indicates 180o phase shift between the input and output signals.
8 mV peak change in vBE gives 1mA change in iB and 0.5 mA change in iC. 0.5 mA change in iC produces a 1.65 V change in vCE
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