Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- d) Parasitic capacitances usually exist in a practical amplifier. Comment on the possible sources of the parasitic capacitances. How they are likely to affect the frequency response of the amplifier.arrow_forwardDescribe the method for calculating gain in a cascade analysis.arrow_forwardExplain the following tradeoffs: Error performance vs bandwidtharrow_forward
- I need the answer as soon as possiblearrow_forward7 The function of a loudspeaker crossover network is to channel frequencies higher than a given crossover frequency, f-, into the high-frequency speaker (tweeter) and frequencies below f. into the low-frequency speaker (woofer). Figure P6.37 shows an approximate equivalent circuit where the amplifier is represented as a voltage source with zero internal resistance and each speaker acts as an 8-2 resistance. If the crossover frequency is chosen to be 1,200 Hz, evaluate C and L. Hint: The break frequency would be a reasonable value to set as the crossover frequency. 10 Vms R1 R2 R1 = R2 = 8 2 wwarrow_forwardFigure 6 shows a circuit where feedback is provided between the drain and gate of the transistor M1 Q6 Draw the small signal equivalent circuit of the open loop circuit obtained by breaking the loop at the gate of transistor M1 (suitable for the analysis in the mid band frequency range). Consider Cg as short circuit in the model and state any assumption. (a) (b) State the necessary conditions for the oscillation to occur in the circuit. Using the conditions from (b), derive an expression for the amplifier oscillation frequency fo. (c) Derive the necessary condition for the oscillation to occur in the circuit and state if you can conclude that the circuit will oscillate when considering the values of the inductances L2 = 8 LI = 2 mH; capacitors C = 2 nF; transconductance of the transistor gml = 1.4 mS; and resistors R1 = R2 = Rs : 10 k2. (d) %3D L2 L, R, Cg. C Vout Cp M1 R2 CG Rs Cs Vss = 0 V Figure 6arrow_forward
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