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Dr T
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Dr T
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Dr T
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Dr T
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المملكة العربية السعوديةوزارة التعليم العالي - جامعة
أم القرىكلية الهندسة و العمارة
اإلسالميةقسم الهندسة الكهربائية
802311-4 ELECTRONIC DEVICES
KINGDOM OF SAUDI ARABIAMinistry of Higher Education
Umm Al-Qura UniversityCollege of Engineering and Islamic Architecture
Electrical Engineering Department
Dr T
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Dr T
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Dr T
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Dr T
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Lecture 6
By: Dr Tarek Abdolkader
17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
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OUTLINE
•Diode circuit analysis
•Rectifiers
•Clipping circuits
•Clamping circuits
•Zener diode
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17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
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Dr T
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der Diode circuit analysis
• To analyze a diode dc circuit, you must determine first: Is the diode ON or OFF?
• To determine the status of the diode:
1. Assume that the diode is OFF.2. Determine the voltage on the diode in this
case (which is equal to thevinin voltage Vth).3. If Vth < Vcut, then the diode is really OFF and it
is replaced by open circuit.4. If Vth ≥ Vcut, then the diode is ON and it is
replaced by a battery of Vcut
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Diode dc circuit analysis:
Vcut is 0.7 V for silicon and 0.3 for Ge. (assuming practical model)
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For the series diode configuration shown, determine VD, VR, and ID.
See Example 2.6 page 60 in Boylestad
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Diode circuit analysis
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For the series diode configuration shown, determine VD, VR, and ID.
See Example 2.8 page 61 in Boylestad
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Diode circuit analysis
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For the series circuit shown, determine Vo and ID.
See Example 2.9 page 62 in Boylestad
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Diode circuit analysis
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For the circuit shown, determine ID , VD2 and Vo.
See Example 2.10 page 62 in Boylestad
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Diode circuit analysis
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For the series diode configuration shown, determine I , V1, V2, Vo.
See Example 2.11 page 63 in Boylestad
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Diode circuit analysis
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For the network shown, determine I .
See Example 2.13 page 65 in Boylestad
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Diode circuit analysis
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der Diode circuit analysis
• To analyze a diode ac circuit, you must determine at what value of the input the diode transfers from OFF to ON or vise versa?
• To determine the transition value at which the diode switches:
1. Assume that ID = 0 and VD = 0.7 V for silicon and of 0.3 for Ge. (assuming practical model)
2. Determine Vi in this case. It is the transition value of input (Vit).
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Diode ac circuit analysis:
• Calculate Vo for Vi > Vit, and for Vi > Vit. In one of these cases, the diode will be ON and in the other it will be OFF.
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der Rectifiers
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Half-wave rectifier:It transforms sinusoidal input to half-wave rectified signal
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(a) Sketch the output vo assuming ideal diode.
(b) Repeat part (a) assuming practical diode model.
(c) Repeat parts (a) and (b) if Vm is increased to 200 V and compare solutions
See Example 2.18 page 71 in Boylestad
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Half-wave rectifier
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der Rectifiers
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Full-wave rectifier:• It transforms sinusoidal input to full-wave rectified signal
• There are two types of full-wave rectifiers:1. Using center-taped transformer.2. Using bridge network.
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center-taped transformer full-wave rectifier :
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der Rectifiers
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Bridge full-wave rectifier:
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der Rectifiers
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Peak Inverse Voltage(PIV):• The peak inverse voltage (PIV) is the maximum negative bias
applied on the diode. If the diode is biased in the reverse mode with a voltage more than PIV, it will go into breakdown
Bridge full-wave rectifier:
center-taped transformer full-wave rectifier :
Half-wave rectifier:
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Half-wave rectifier Center-tap full-wave rectifier Bridge full-wave rectifier
Vdc = 0.318 Vm Vdc = 0.636 Vm Vdc = 0.636 Vm
PIV ≥ Vm PIV ≥ 2Vm PIV ≥ Vm
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• To reach the required dc voltage, a filter and regulator should be added to a rectifier circuit. The overall system is called power supply unit (PSU).
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• While the input is increasing, the diode D is ON and the capacitor is charging with a small time constant.
• After reaching the peak, the voltage at the anode will be less than the cathode and D will be OFF. The capacitor will discharge through the load with a large time constant.
• When the input exceeds the capacitor voltage, D is again ON.
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Full-wave rectifier is much better than Half-wave rectifier because it gives less ripple.
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• Diode networks that have the ability to “clip” off a portion of the input signal without distorting the remaining part of the alternating waveform are called clippers.
• The half-wave rectifier is an example of the simplest form of diode clipper—one resistor and diode.
• There are two general categories of clippers:series clipper: where the diode is in series with the load,parallel clipper: the diode is in parallel to the load.
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Series clipper Parallel clipper
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• To determine the output voltage:
1. Assume that ID = 0 and VD = 0.7 V for silicon and of 0.3 for Ge. (assuming practical model)
2. Determine Vi in this case. It is the transition value of input (Vit).
3. Calculate Vo for Vi > Vit, and for Vi > Vit. In one of these cases, the diode will be ON and in the other it will be OFF.
4. Sketch Vi and Vo.
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Determine the output waveform of the network shown assuming ideal diode model.
See Example 2.20 page 78 in Boylestad
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Clippers
Vit = ‒ 5 V For vi > ‒ 5 V, D is ON vo = vi + 5 For vi < ‒ 5 V, D is OFF vo = 0
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Repeat the previous example if the input waveform is the square wave shown.
See Example 2.21 page 79 in Boylestad
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Clippers
Vit = ‒ 5 V For vi > ‒ 5 V, D is ON vo = vi + 5 For vi < ‒ 5 V, D is OFF vo = 0
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Determine the output waveform of the network shown assuming ideal diode model.
See Example 2.22 page 80 in Boylestad
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Clippers
Vit = 4 V For vi > 4 V, D is OFF vo = vi For vi < 4 V, D is ON vo = 4 V
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Repeat the previous example assuming silicon diode practical model.
See Example 2.23 page 81 in Boylestad
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Clippers
Vit = 4 ‒ 0.7 = 3.3 V For vi > 3.3 V, D is OFF vo = vi For vi < 3.3 V, D is ON vo = 3.3 V
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• The clamping network is one that will “clamp” a signal to a different dc level.
• The network must have a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an additional shift.
• The total swing of the output is equal to the total swing of the input signal.
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clampers
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• The time constant with the diode is OFF is very large ( ).• The time constant with the diode is ON is very small ( 0 ).
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Assumptions:
1. Start by considering that part of the input signal that will make the diode ON.
2. In the ON state, assume that the capacitor will charge up instantaneously ( 0).
3. Find the voltage on the capacitor after charging (with the proper polarity) and the output voltage in the ON state.
4. Move to the OFF state and assume that the capacitor will hold on to its established voltage level in the ON state( ).
5. Find the output voltage in the OFF state.6. Keep in mind the general rule that the total swing of the total
output must match the swing of the input signal.
Procedure:
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Find the output voltage of the shown circuit.
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Vit = 0 V For vi > 0, D is ON
For vi < 0, D is OFF
Clamping circuits
D is ON
D is OFF
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Determine vo for the network shown.
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Vit = 5 V For vi < 5, D is ON
For vi > 5, D is OFF
Clamping circuits
D is ON
D is OFF
Note that:T = 1/f = 1 ms(OFF) = RC = 100 ms(ON) ~ rDC ~ few seconds
See Example 2.24 page 84 in Boylestad
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Repeat the previous example assuming Si diode.
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Clamping circuits
D is ON
D is OFF
Note that:T = 1/f = 1 ms(OFF) = RC = 100 ms(ON) ~ rDC ~ few seconds
See Example 2.25 page 85 in Boylestad
Vit = 4.3 V For vi < 4.3 V, D is ON
For vi > 4.3 V, D is OFF
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Clamping circuits
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Clamping circuits
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-IoI ~- Io
V
I
Vz
When a sufficiently large reverse voltage is applied to a p-n junction, the junction breaks down and conducts a very large current.
The reverse voltage at which the current increases sharply is called the BREAKDOWN VOLTAGE, or the zener voltage Vz
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1. Determine the state of the Zener diode by removing it from the network.
2. Substitute the appropriate equivalent circuit and solve for the desired unknowns.
Procedure for dc inputs:
Procedure for ac inputs:1. Determine the transition input at which the Zener diode will
switch its state by assuming that ID = 0 and VD = ‒ Vz .2. Determine Vi in this case. It is the transition value of input
(Vit).3. Calculate Vo for Vi > Vit, and for Vi > Vit. In one of these
cases, the zener diode will be ON and in the other it will be OFF.
4. Sketch Vi and Vo.
17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
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(a) For the Zener diode network shown, determine VL, VR, Iz, and Pz.(b) Repeat part (a) with RL = 3 kΩ.
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D is OFF
See Example 2.26 page 88 in Boylestad
Zener diode
17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
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Abd
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Dr T
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Dr T
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Dr T
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Dr T
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Dr T
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Dr T
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(a) For the Zener diode network shown, determine VL, VR, Iz, and Pz.(b) Repeat part (a) with RL = 3 kΩ.
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D is ON
See Example 2.26 page 88 in Boylestad
Zener diode
17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
arek
Abd
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der
Dr T
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Dr T
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Dr T
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(a) For the network shown, determine the range of RL and IL that will result in VL being maintained at 10 V.(b) Determine the maximum wattage rating of the diode.
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D must be ON:
See Example 2.27 page 91 in Boylestad
Zener diode
17/4/1433 Electronic devices (802311) Lecture 6 Dr Tarek Abdolkader Dr T
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Abd
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der
Dr T
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Abd
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Dr T
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Dr T
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Determine the range of values of Vi that will maintain the load voltage in the Zener diode circuit shown at 20 V.
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D must be ON
See Example 2.28 page 92 in Boylestad
Zener diode
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