Q₁: Refer to the circuit shown in Fig. 1, the input voltages V and V are an AC signal and a Constant positive DC signalRespectively:a) Derive expressions for the Upper and Lower threshold voltage limits of the Output voltage Voutb) Sketch the hysteresis curve. Assume R₁ = 2R, and the DC signal V₂ = 17.32% of the Effective value (r.m.s value)of the AC signal Va (See Fig. 2).c) Sketch the output voltage Vour as a function of time for the same Conditions in (b)V₂VawwwR₂wwwR₁Fig. 1-10 V+10 VM0 1 2 3VoutFig. 2D 1012

Answered: Image /qna-images/answer/8fa8cddd-0ef1-4190-adbf-fca4763f3a82.jpg

Answered: Q₁: Refer to the circuit shown in Fig.…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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RI Rz Vo %3D 2R.R.CG. REZ 1.+ Redrawthe circuit-equivalentwhenoperatingat-a-DC-voltage(0Hz).1 VerifyusingKCL-that-the circuit-operation will-be-that-of a-non-invertingamplifier. 2.→ Redrawthe circuit-equivalentat-veryhigh-frequencies.-VerifythatVoutisshorted.-(Hint.-Note- thereactance at-thenon-invertinginputat-extremelyhighfrequencies.--What-will-bethe- voltage at-that-node?-What-doesthis-meanfor-the voltage at-Rf2?)1 3.+ Let-C1=C2..-Calculate thevaluesofC1-and-C2-for-a-cut-offfrequency of500-Hz-if-R1-and-R2. are-33kohm-and-choosereasonablevaluesof Rf1 and-RF2•50-thatthe-gain-at-low-frequenciesis- 4-dB.--1 4.→ Simulate the circuit-on-Multisim-andrepeatsteps4-and5-of-Part-1-again.--Produce its-Bode Plot- using-the AC-Sweepfunction.-Identify-the type of responsecurve andtheroll-offrate. +,
7. When the automatic gain control voltage of a radio receiver is measured with a vacuum tube voltmeter whose input resistance is 10 M-ohm, it reads 4.5 V. When an ordinary voltmeter whose input resistance is 10 k-ohm is also connected across the AGC voltage source, both meters read 0.40. What is the internal resistance of the AGC voltage source? 8. A phonograph pick up develops an open-circuit emf of 50 mV and has an internal resistance of 1200 ohm. A. What voltage will it develop across a 4 ohms load resistance? B. What value of load resistance must be used to obtain a 75% eff? C. What is the maximum power output of the pickup? 9. What bleeder resistor is required to complete a voltage divider to give 300 v at 40 mA from a 500 V source of emf if the series dropping resistor is 4 k-ohm.
A controlled rectifier whose firing angle at 60 supplies a load RLE. The voltage is 120 V, R 30 ohms, E-80 V, and w = 2760 rad/s. a) Estimate the value of the inductance in series with R and E such that continuous conduction is ensured. (Consider that the peak-to-peak variation of the load current is determined solely by the first AC term of the current series Fourier.) ..
1. Given a circuit in figure 1.1 where an amplifier model is proposed. The input signal is a courant source with values provided in the figure. 1.1 Calculate the expression T(jw) = Vo(jw) is(jw) * 1.2 If the input frequency win = 10 rad/s, evaluate the value of T(jw) and use the obtained results to determine the equation for the output voltage Vo(t) when Is = 2mA · sin [ 10 t] 20Ω + Ro + 10µF RLS Vo i 5000 is (1 R;S Vi 100 v 1kQ Is = 2mA · sin [Wint ] Figure 1.1 Amplifier circuit
Consider the following circuit with Vs(t) = 2.0 + 0.02 sin(2 pi t) V,R1 = R2 = 25ohms Idc = 1mA, Vdc = 5V. Assume Vy = 0.7V and VT = 25mV.a) Draw the equivalent circuit under DCb) Determine the DC voltage and current through the diode.c) Draw the equivalent circuit under acd) Determine the total current through the diode
Show the downward shift of the output signal with the series capacitor and diode.Clampers or DC restorers are clamp a peak of a waveform to a specific DC level comparedwith a capacitively coupled signal which swings about its average DC level (usually 0V).If the diode is removed from the clamper, it defaults to a simple coupling capacitor– noWhat is the clamp voltage? And, which peak gets clamped? Why is this so? Voltage values: Capasitor value: 1000 pFDC offset: 0 VAmplitude: 5 VFreq: 1 KHz
20. A biased clamper is connected to a rectangular AC input with 20V on the positive half cycle and -10V on the negative half cycle. The orientation of the Si diode is the same with Vo while the 5V biasing voltage is in opposite connection. The value of R is 100K ohms while the capacitor is 1 microFarad.
A crystalline silicon photovoltaic cell has an open circuit voltage of 0.5V and a short circuit current of 1.8 A at Standard Test Conditions. What is the maximum output power at Standard Test Conditions and at a cell temperature of 50oC? What is the percentage decrease in power per oC? Assume a Fill Factor of 0.70 at 25oC. (As a rule of thumb for crystalline silicon solar cells, the increase in short circuit current is about 0.058% per oC while the decrease on open circuit voltage is 0.30% per oC and decrease in the fill factor is by 0.15% per oC).
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A 1000-VA, 230/115-V transformer has been tested to determine its equivalent cir- cuit. The results of the tests are shown below. Open-circuit test (on secondary side) Short-circuit test (on primary side) Voc = 115 V Vsc = 17.1 V %3D %3D Isc 8.7 A Psc loc =0.11 A Poc = 3.9 W = 38.1 W (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer. (b) Find the transformer's voltage regulation at rated conditions and (1) 0.8 PF lag- ging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer's efficiency at rated conditions and 0.8 PF lagging.
EMG devices are equipped with circuits to rectify and next integrate the EMG signals to assessthe muscle force.Design an improved precision half-wave rectifier circuit with a gain of -2 ready to simulate to process signals having amplitudes ranging from -4V to +4V. draw the signal at the output of the designed rectifier circuit for the input signal1×sin(2πt) V
The waveform below is from a 70-Vac source. - 50 ms 180 V Determine: a. Vpk (ans: 98.9949 V) b. Vpk-pk (ans: 197.9899 V) c. Vrms (ans: 70 V) d. Period(ans: 100 ms) e. Frequency(ans: 10 Hz) f. Angular velocity(ans:20n rad/sec) g. Phase angle(ans: 125.0876°) h. equation of the waveform(in time domain) i. the instantaenous voltage when t = 120ms(ans: -29.0880 V) j. the angle (1st occurence)after t = 0 when the voltage is + 80 V(ans: 0.99959 °) k. the time (2nd occurence) after t = 0 when the voltage is – 10 V(ans: 63.643 ms) (ans: v(t) = 98.9949sin(20nt – 125.0876°))

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