Repeat Example 2.1 if the input voltage is
The peak diode current, the maximum reverse-biased voltage of the diode and the fraction of the cycle over which diode conducts for a half-wave battery charging circuit.
Answer to Problem 2.1EP
The peak diode current is
Maximum reversed biased voltage of the diode is
Fraction of the cycle diode is conducting is
Explanation of Solution
Given Information:
The given values are
Calculation:
Apply Kirchhoff’s voltage law,
Substituting the given values,
Maximum reversed biased voltage of the diode,
Consider the time at which diode start conducting as
By symmetry,
Fraction of the cycle diode is conducting can be calculated as,
Conclusion:
Therefore, the peak diode current is
Want to see more full solutions like this?
Chapter 2 Solutions
Microelectronics: Circuit Analysis and Design
- Calibri (Body) -11 by AaBbCcDx AaBbCcD AaBbC AaBbCc AaB AaBbCc. mat Painter BIU 1Normal 1No Spaci. Heading 1 Heading 2 Title Subtitle Chang Styles d Font Paragraph Styles IMPEDANCE QUIZ # 2 1. What capacitance when connected in series with a 500Q resistor will limit the current drawn from a 48-mV 465-kHz source to 20µA? * a. 144pF b. 145pF c. 146pF d. 147pF MY ANSWER: c.146pF 2. What is the total impedance at 20kHz of a series circuit consisting of a 1.5mH inductance, a 100Q resistance, and a 0.08uF capacitance? * a. 1340 41.7 b. 1930 L-37.2" C. 920 290 d. 530 2-12.6°arrow_forwarda student is doing work on circuit in which he employs a voltage af 60 sin(314t+60) by observing the peak voltage determine following below calculate the form factor of the circuit a=1.87 b=1.59 c=1.11 d=1.48arrow_forwardFor the given circuit for a 5.63 Vpeak sinusoidal input vi, what is the value of vo (in V) when vi is a negative half cycle? ... Vin Vout Round your answer to 2 decimal places. + Rarrow_forward
- Multicolour LED Mono LED Dimensional LED The capacitor in a clamper circuit is in parallel with the voltage source * None True False For the following circuit, the vo waveform is wwwwwarrow_forward6V Ⓒ 3 кД +0 300 HF з ка I(+) In the transient circuit for +0. 4MA Solve for I (+)arrow_forwardCalculate 11, 12, and 13. The frequency is fixed at 500 Hz, and its output voltage magnitude to 5 V (rms). Rs=4792 w R₁=472 13 R3-472 5 V AC R2=4802 88 Ω 0.22 μ 100 mHarrow_forward
- 29. Write the analytical expression for the wayeforms with the phase angle in degrees: 4 (mV) 4i (mA) f= 60 Hz f= 2 kHz 6 30 20 (a) (b)arrow_forwarda student is doing work on circuit in which he employs a voltage af 60 sin(314t+60) by observing the peak voltage determine following below calculate the value of rms voltage a=25.25v b=45v c=42.42v d=30.5varrow_forwardB- Calculate the input and output power for the circuit of Figure shown. The input signal results in a base current of 6 mA rms. +Vcc (14V) RB 1.2 ΚΩ C₁ 100 µF Rc-1692 V₂ B=50arrow_forward
- is In a BJT, Ico = Icso = 2 μ4. Given a = 0.99 , the value of Iceo A 2 μ.Α Β 99 μ.Α C c 198 μ.Α D 200 μ.Αarrow_forwarda) Sketch the o/p waveform of the circuit shown in Figure (1), when vi-2sinwt. Vec 2ΚΩ vb 3ΚΩ Va 1kΩ Vb ViWW www + ΣΑΚΩ Vc Figure (1) Vg 5ΚΩ Vf www Ve ΣΚΩ Vd www 10 ΚΩ www www 6ΚΩ Voarrow_forwardThe sum of the following two e.m.fs will be e1 = 10 sin wt and e2 = 10 cos wt O 10 O 20 sin wt O 14.14 cos wt O 14.14 sin (wt + TT/4)arrow_forward
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning