Concept explainers
Videos
The two possible situations that will happen for the given conditions.
The diode current and terminal voltage across diode for two situations.
Answer to Problem 4.1P
The value of diode current and diode terminal voltage is:
For situation 1:
VD = 1.5 V
iD = 0 A
For situation 2:
Explanation of Solution
Given:
Thevenin equivalent of AA Flashlight cell with:
Thevenin voltage (Vth) = 1.5 V
Thevenin resistance (Rth) = 1 Ω
Positive side of battery is connected to the cathode of the ideal diode.
Calculation:
The two possible situations can be:
Situation 1: Cathode terminal of the diode is connected to the positive terminal of the battery.
Situation 2: Anode terminal of the diode is connected to the positive terminal of the battery.
Case 1:
When the cathode is connected to the positive terminal of battery.
The Thevenin equivalent circuit with diode connected as given can be drawn as:
The diode is operating in a reversed biased region as the positive side of the battery is connected to the cathode terminal of diode.
Hence, the ideal diode will behave as an open circuit.
So,
Terminal voltage can be calculated by applying mesh law in the circuit:
By putting the given values and from equation (1):
Case 2:
When anode terminal of the diode is connected to the positive terminal of battery.
The Thevenin equivalent circuit with diode connected as given can be drawn as:
The diode is operating in forward biased region as the positive side of the battery is connected to the anode terminal of diode.
As it is ideal diode, the cut-in voltage across it will be 0.
Hence,
Current can be calculated by applying mesh law in the circuit:
By putting the given values and from equation (2):
Want to see more full solutions like this?
Chapter 4 Solutions
Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
- Determine (a) the average and (b) rms values of the periodiccurrent waveform shown in Fig. P8.9arrow_forwardRepairs have to be carried out on HV cir- cuit breaker No. 6 shown in Fig. 26. If the three 220 kV lines must be kept in service, which disconnecting switches must be kept open?arrow_forwardFind the voltage v(t) for t>=0 show all steps and redraw circuit as necessary, the switch closes at t=0 and v(t) is the voltage over the 4ohm resistor as shown in the circuit.arrow_forward
- Find the voltage v(t) for t>=0 please redraw circuit as necessary and show all steps.arrow_forwardDetermine (a) the average and (b) rms values of the periodiccurrent waveform shown in Fig. P8.9arrow_forwardFind Eigenvalues and Eigenvectors for the following matrices: [5 -6 1 A = 1 1 0 3 0 1arrow_forward
- Use Gauss-Jordan Elimination method to solve the following system: 4x1+5x2 + x3 = 2 x1-2x2-3x3 = 7 3x1 x2 2x3 = 1. -arrow_forward3. As the audio frequency of Fig. 11-7 goes down, what components of Fig. 12-4 must be modified for normal operation? OD C₂ 100 HF R₁ 300 Re 300 ww 100A R 8 Voc Rz 10k reset output 3 R7 8 Voc 3 reset output Z discharge VR₁ 5k 2 trigger 2 trigger 7 discharge R 3 1k 5 control voltage threshold 6 5 control voltage 6 threshold GND Rs 2k C. C. 100 GND Uz LM555 1 Ce 0.01 U, LM555 0.01 8.01.4 PRO Fig. 11-7 Audio lutput Pulse width modulator R4 1k ww C7 Re 1k ww R7 100 VR 50k 10μ Ra R10 C₁. R1 3.9k 3.9k 0.14 100k TO w Rs 51 82 3 H 10 Carrier U₁ Ca Input A741 2.2 Us MC1496 PWM signal input R2 0.1100k Uz A741 41 Cs 1 Re 10k VR2 50k VR3 100k 14 12 C3. 3% + Ce 0.1 10μ 5 1A HH C +12V 0.1 O PWM Output C 0.02- R 100k +12 V Demodulated output 6 Ca 0.33 w R 10k R12 100k ww 31 о + 4A741 -12 V Fig. 12-4 PWM demodulator C 1500parrow_forwardDUC 1. In Fig. 12-4, what are the functions of the VR1 and VR2? 2. In Fig. 12-4, what is the function of the VR3? VR₁ 50k C₁ R1 0.1 100k Carrier Input U₁ A741 PWM signal input R41k www Re 1k w C7 ± 10μT R7 100 ww =L H C4 2.2 H W82 Rs 51 3 10 U3 MC1496 C2 R2 U2 A741 22 0.1 100k VR2 50k VR3 100kr 14 C3 10μ 1k 0.1 4 5 6 12 m Re 10k R9 R102 3.9k 3.9k HHI C10 0.1 -0 +12V C11 R 0.02 100k +12 V Demodulated output C R11 R12 A741 0.33 10k 100k -12 V Ca 1μ C12 1500p PRODUC Fig. 12-4 PWM demodulator PRODUCTSarrow_forward
- 10.37 Use mesh analysis to find currents I₁, I2, and I3 in the circuit of Fig. 10.82. ML 120-90° V 120 -30° V Figure 10.82 For Prob. 10.37. N N Z=80-135arrow_forward3. Find the phasor current I。 in the circuit shown below. Be aware of the direction markings. (15 pts) 1052 I 5057 ①520 Amps 2012 j5052arrow_forward10.93 Figure 10.135 shows a Colpitts oscillator. Show that the ed oscillation frequency is 1 fo= 2π √√LCT where CTC₁C2/(C₁ + C₂). Assume R; >>> R₁ + Rf ww Vo L m C₂ C₁ 5 Xci Figure 10.135 A Colpitts oscillator; for Prob. 10.93. (Hint: Set the imaginary part of the impedance in the feedback circuit equal to zero.)arrow_forward
