manf360 chapter 2 diode applications

3/5/2012 MANF360 Ch.2 Diode Applications ١ of 38

MANF360

Electronics and Instrumentation

CHAPTER 2 Diode Applications

Prof. Ismail Mohammed Hafez

Electronics and Communication Department College of Engineering

Ain Shams University (ASU)

3/5/2012 MANF360 Ch.2 Diode Applications

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Chapter Contents

2.1 Load Line Analysis of Diode Circuits 2.2 Half Wave Rectifier 2.3 Full Wave Rectifier 2.4 Power Supplies 2.5 Diode Clippers 2.6 Diode Clampers 2.7 Zener Diodes 2.8 Digital Logic Circuits


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2.1 Load Line Analysis of Diode Circuits

• The analysis is performed in a graphical manner, a line can be drawn on the characteristics of the device that represent the applied load. The intersection of the load line with the characteristics will determine operating point or Quiescent Point (Q-Point) of the system.

• Consider the diode circuit shown in Fig. 2.1(a), employing a diode having characteristic shown in Fig. 2.1 (b). Apply KVL to this circuit, we have:

E – VD –VR = 0 or E= VD + IDR Load line equation When VD = 0 gives ID = E / R When ID = 0 gives VD = E • As shown in Fig 2.2, a straight line drawn between the two points

will define the load line. • The point of intersection the load line with the characteristic

determines operating point or Quiescent Point (VDQ and IDQ)


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2.1 Load Line Analysis (Continued)

Fig. 2.2 Drawing the load line and finding the point of operation

Fig. 2.1 Series diode configuration: (a) circuit; (b) characteristics


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2.1 Load Line Analysis (Continued) Example 2.1 For the diode circuit shown in Fig. 2.3, if E = 10 V and R = 0.5 kΩ

determine the diode voltage VDQ current IDQ and voltage VR using: a) The load line method. b) The diode piecewise linear equivalent circuit with VD=0.7V and rav=10 Ω. Solution

a) The load line equation is given by: E= VD + IDR When VD = 0 gives ID = E / R = 10/0.5 kΩ = 20 mA When ID = 0 gives VD = E = 10 V • The resulting load line appear in Fig 2.4. The point of intersection

the load line with the characteristic determines operating point as: VDQ = 0.78 V IDQ= 18.5 mA VR = 18.5 x 0.5 = 9. 25 V

b) Replace the diode by its piecewise linear equivalent circuit as shown in Fig. 2.5, we have:

E = VD + ID (rav + R) 10 = 0.7 + ID (10 +500) ID = 9.3/510 = 18.23 mA VR = 9.21 mA x 0.5 k Ω = 9.12 V


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Example 1.1 Solution (Continue)

Fig. 2.5 The diode circuit including the diode model

2.1 Load Line Analysis (Continued)

Fig. 2.3 (a) A diode circuit (b) characteristics

Fig. 2.4 Load line of Example 2.1


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2.2 Half-Wave Rectification • The half wave rectifier circuit is shown in Fig. 2.6, Vo has an average value of

particular use in ac to dc conversion process. • Assuming an ideal diode, during the interval t = 0 to T/2, the diode is on or short

circuit, Vo = Vi as shown in Fig. 2.6 (a). • For the interval t = T/2 to T, the diode is off or open circuit, Vo = 0 as shown in

Fig. 2.6 (b).

Fig. 2.6 Half wave rectifier (a) Conduction region (0:T/2); (a) Nonconduction region (T/2:T)


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2.2 Half-Wave Rectification (Continued)

• The process of removing one half of the input is called half wave rectification. The output is Vo has a net positive area above the axis over a full period and an average value determined by:

Vdc = 0.318 Vm shown in Fig. 2.7 • The effect of using a silicon diode with VT = 0.7 is demonstrated in Fig.

2.8, and the average value of Vo is given by: Vdc = 0.318 (Vm – VT)

Fig. 2.7 Half wave rectified signal Fig. 2.8 Effect of VT on half wave rectified signal


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2.2 Half-Wave Rectification (Continued)

Example 2.2: For the half wave rectifier circuit shown in Fig. 2.9, if Vm = 20 V, assuming an ideal diode do the following:

a) Sketch the output voltage VO. b) Calculate the dc level of the output. c) Repeat Part (a) and Part (b) when using a silicon diode. Solution (a) Vo is shown in Fig. 2.10 (a), note that the diode will conduct during the

negative part of the input. (b) Vdc = -0.318 Vm = -0.318 x 20 = - 6.36 V (c) Using silicon diode, Vo is shown in Fig. 2.10 (b), and Vdc = -0.318 (Vm – 0.7) = -0.318 x 19.3 = - 6.14 V

Fig. 2.10 (a) Resulting vo; (b) Effect of VK on output Fig. 2.9 Half wave rectifier circuit


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2.3 Full-Wave Rectification • The full wave rectifier circuit is shown in Fig. 2.11. Assuming ideal diodes, during

the interval t = 0 < t ≤ T/2, D2 and D3 are conducting “on” while D1 and D4 are “off”, the output Vo = Vi as shown in Fig. 2.12 (a).

• For the negative input T/2 < t ≤ T, D1 and D4 are “on”, while D2 and D3 are “off” the polarity across the load resistance is the same as shown in Fig. 2.12 (b).

• The area above the axis for full cycle is twice that for half wave, therefore the dc level is given by Vdc = 2 (0.318) Vdc = 0.636 Vm as shown in Fig. 2.13.

• If silicon rather than ideal diodes are employed the Vomax = Vm – 2VT, and Vdc = 0.636 (Vm – 2VT)

Fig. 2.11 Full wave rectifier circuit


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2.3 Full-Wave Rectification (Continue)

Fig. 2.12 Conduction path for (a) positive vi and (b) negative vi.

(b)

(a)

Fig. 2.13 Input and output waveform for a full wave rectifier


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It requires: • Two diodes • Center-tapped transformer

Vdc = 0.636(Vm)

2.3 Full-Wave Rectification (Continue)

Fig. 2.14 Center-Tapped full wave Transformer Rectifier

Fig. 2.15 Conduction path for (a) positive vi and (b) negative vi.

(a) (b)


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Summary of Rectifier Circuits Rectifier Ideal Vdc Realistic Vdc

Half Wave Rectifier Vdc = 0.318 Vm Vdc = 0.318 (Vm – 0.7)

Bridge Rectifier Vdc = 0.636 Vm Vdc = 0.636 (Vm – 2(0.7))

Center-Tapped Transformer Rectifier

Vdc = 0.636 Vm Vdc = 0.636 (Vm – 0.7)

• PIV (PRV) • Because the diode is only forward biased for one-half of the AC

cycle, it is also reverse biased for one-half cycle. • It is important that the reverse breakdown voltage rating of the

diode be high enough to withstand the peak, reverse-biasing AC voltage. PIV (or PRV) > Vm

• PIV = Peak inverse voltage PRV = Peak reverse voltage • Vm = Peak AC voltage


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2.4 The Power Supply Diagram • The block diagram of the power supply is shown in Fig. 2.16. It

consists of of the following modules: 1. Step down transformer which transforms the supply voltage to the

desired low level voltage [e.g. vi = 220 sin(2πx50) to vtrans = 12 sin(2πx50) or from AC 220 V to AC 12 V and the frequency is 50 Hz].

2. Full wave rectifier which rectifies the output of the transformer with voltage amplitude vm in our example vm = 12V. The output from the rectifier section is a pulsating DC with Vdc = 0.636(Vm)

Fig. 2.16 A Power supply block diagram


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2.4 Power Supply Diagram (Continued) 3. Capacitor Filter as shown in Fig. 2.17 is connected to the rectifier output

and a dc voltage is obtained across the capacitor. The filter circuit reduces the peak-to-peak pulses to a small ripple voltage.

4. IC Voltage Regulator as shown in Fig. 2.18 is the basic connection of three terminal voltage regulator IC to a load.

Fig. 2.17 Capacitor Filter and its output waveform

Fig. 2.18 Three terminal voltage regulator block

diagram


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2.4 Power Supply Diagram (Continued) • The series 78 regulators provide fixed regulated positive voltages from +5

to +24V as given by the manufactureur‘s specification sheets listed in Table 2.1.

• Fig. 2.19 shows on IC, a 7812, is connected to provide voltage regulation of 12 V.

• An unregulated voltage input Vi is filtered by capacitor C1 and connected to the IC’s IN terminal. The IC’s OUT terminal provides a regulated +12 V.

• C2 filters the DC output for any high frequency noise.

5. The Load which could be a mobile phone, computer radio …..etc.

Fig. 2.19 Connection of 7812 voltage regulator


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2.4 Power Supply Diagram (Continued) • The series 79 regulators provide fixed regulated negative voltages from -5

to -24V as given by the manufactureur‘s specification sheets listed in Table 2.2.

Table 2.1 Positive regulators in 7900 series

Minimum Voltage [V]

Output Voltage [V]

IC Part #

- 7.3 - 5 7905

- 8.4 - 6 7906

- 10.5 - 8 7908

- 11.5 - 9 7909

- 14.6 - 12 7912

- 17.7 - 15 7915

- 20.8 - 18 7918

- 27.1 - 24 7924

Minimum Voltage [V]

Output Voltage [V]

IC Part #

7.3 + 5 7805

8.3 + 6 7806

10.5 + 8 7808

12.5 + 10 7810

14.6 + 12 7812

17.7 + 15 7815

21.0 + 18 7818

27.1 + 24 7824

Table 2.1 Positive regulators in 7800 series


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2.4 Power Supply Diagram (Continued) Example 2.3: Consider a voltage power supply using full wave rectifier and IC

regulator, if the desired output is +5V, do the following: a) Draw the circuit diagram. b) Calculate the voltage amplitude at each node. b) Draw the waveform at each node. Solution a) The circuit diagram is shown in Fig. 2.20 b) Since Vo = 5V, from table 2.1 Vi at node (1) = 7.3 V The rectifier Vm = 7.3 / 0.636 = 11.48. Therefore the transformer used is step down one from 220V to12V

Fig. 2.20 Circuit diagram of +5 V power supply


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2.5 Diode Clippers

The diode in a series clipper circuit “clips” any voltage that does not forward bias it. The unbiased series clipper circuit is shown in Fig. 2.21 while the biased series circuit clipper is shown in Fig. 2.22.

2.5.1 Series Clipper

Fig. 2.21 Series unbiased clipper (Ideal diodes)

(a)

(b)


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2.5 Diode Clippers (Continued) 2.5.1 Series Clipper (Continued)

Fig. 2.22 (a) Simple biased series clippers (Ideal diodes)


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2.5 Diode Clippers (Continued) 2.5.1 Series Clipper (Continued)

Fig. 2.22 (b) Simple biased series clippers (Ideal diodes)


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2.5 Diode Clippers (Continued)

The diode in a parallel clipper circuit “clips” any voltage that forward bias it. The unbiased parallel clipper circuit is shown in Fig. 2.23 while the biased parallel circuit clipper is shown in Fig. 2.24.

2.5.2 Parallel Clipper

Fig. 2.23 Simple unbiased parallel clippers (Ideal diodes)

(a)

(b)


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2.5 Diode Clippers (Continued) 2.5.2 Parallel Clipper (Continued)

Fig. 2.24 Simple biased parallel clippers (Ideal diodes)


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Example 2.4: determine the output waveform for the network shown in Fig.2.25, assuming an ideal diode for the following inputs:

a) sine wave. b ) square wave.

Fig. 2.25 Series clipper

Fig. 2.26 Input / output waveform for part (a).


Solution (a) VO = Vi + 5V For V0 = 0 V => Vi = -5V For Vi more negative than -5V Diode is open circuit For more positive Vi than -5V the diode is short circuit the output waveform shown in Fig.2.26,


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Example 2.4 Solution (Continued) (b) VO = Vi + 5V, for Vi = 20 V => V0 = 25V the diode is short circuit as in Fig. 2.27 (a) For Vi = -10V => V0 = 0V the diode is open

(a) vo at vi = +20 V

(b) vo at vi = -10 V


Fig. 2.27 Input / output waveform for part (a).


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Example 2.5: determine the output waveform for the network shown in Fig.2.28, assuming an ideal diode for the following input triangular wave.

Fig. 2.28 Parallel clipper



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Example 2.5 Solution For negative cycle: VO = V = 4V The transition state at id =0A and Vd =0V this would be at Vi = V = 4V For the open circuit state Vo = Vi

Input / output waveform is shown in Fig. 2.29


Fig. 2.29 Input / output waveform for Example 2.5.


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2.6 Diode Clampers A diode and capacitor can be combined to “clamp” an AC signal to a specific DC level as shown in Fig. 2.30.

Fig. 2.30 Diode clamper circuit.


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Example 2.6: Determine the output waveform for the network shown in Fig. 2.31, assuming an ideal diode for the following input indicated.

2.6 Diode Clampers (Continued)

Fig. 2.31 Diode clamper circuit with indicated input.


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Example 2.6: Solution The diode equivalent circuits and the input / output

waveforms are shown in Fig. 2.32.

2.6 Diode Clampers (Continued)

Fig. 2.32 Input / output waveform for Example 2.6.


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2.8 Zener Diodes • The analysis on networks employing Zener diode is similar to that applied to the

analysis of the semiconductor diodes. • First determine of the diode then substitute the appropriate model to calculate

the unknown quantities of the network. • The Zener is a diode operated in reverse bias at the Zener Voltage (Vz). Its

equivalent circuit is shown in Fig. 2.33.

Fig. 2.33 Zener diode equivalent circuit (a) “on’ state and (b) is “off” state

(a) (b)


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2.8 Zener Diodes (Continued) Zener Diode Analysis • The analysis on networks employing Zener diode can be broken down into to

steps as follows: Step #1 Determine the state of the Zener diode by removing it from the network

and calculating the voltage across the resulting open circuit. • Applying step #1 to the network of Fig. 2.34 (a) will result in the network of Fig.

2.34 (b). V is calculated by the application of the voltage divider we get:

Fig. 2.34 (a) Basic Zener diode regulator (b) Determining its state

RR

RV iVV

L

LL +

==

(a) (b)

• If V ≥ VZ, the diode is “on” and the equivalent model of Fig. 2.33 (a) is substituted. • If V < VZ, the diode is “off” and the open circuit equivalent model of Fig. 2.33 (b) is

substituted


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2.8 Zener Diodes (Continued) Step #2 Substitute the appropriate equivalent circuit and solve for the desired

unknowns. • When the Zener diode is “on” state it is replaced by a voltage source as shown

in Fig. 2.35 and the circuit parameters are as follows: VL = VZ

Fig. 2.35 (a) Basic Zener diode regulator (b) Its equivalent circuit

RVV

RV

IandR

VI

LiRR

L

LL

−===

(a) (b)

IR = IZ + IL

IZ = IR - IL

• The power dissipated by the Zener diode is determined by: PZ = VZ * IZ

(b)


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Example 2.7: For the Zener diode network shown in Fig. 2.36, Determine VL, VR, IZ, and PZ for the following cases:

(a) RL = 1.2kΩ (b) RL = 3kΩ

Fig. 2.36 Zener diode network

2.8 Zener Diodes (Continued)


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solution (a) RL = 1.2kΩΩΩΩ Step #1 Determine the state of the Zener diode by removing it from the

network and calculating the voltage across the resulting open circuit as shown in Fig. 3.37.

Vkk

Vk

RR

RVV

L

Li 73.82.11

)16(2.1 =Ω+Ω

Ω=+

= VZ = 10V, V < VZ, Zener diode is “off”

VL= V = 8.73V VR = Vi – VL= 16 - 8.73 = 7.27 V IZ = 0 A and PZ = 0 W

Step #2 Substitute the appropriate equivalent circuit (Zener diode is off ) as shown in Fig. 3.37 (a) and solve for the desired unknowns as follows:

2.8 Zener Diodes (Continued)

Fig. 2.37 (a) Zener diode equivalent circuit


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Vkk

Vk

RR

RVV

L

Li 1231

)16(3 =Ω+Ω

Ω=+

= VZ = 10V, V > VZ, Zener diode is “on”

VL= VZ = 10V

VR = Vi – VL=16 – 10 = 6V

IL = VL/RL = 10V/3kΩ = 3.33 mA

IR = VR /R = 6V / 1 kΩ = 6 mA

IZ = IR – IL = 6 – 3.33 = 2.67 mA PZ = VZ I Z = (10V)(2.67 mA)

= 26.7 mW

2.8 Zener Diodes (Continued) solution (b) RL = 3 k ΩΩΩΩ Step #1 Determine the state of the Zener diode by removing it from the

network and calculating the voltage across the resulting open circuit as:

Step #2 Substitute the appropriate equivalent circuit (Zener diode is voltage source with VZ = 10 V) as shown in Fig. 3.37 (b) and solve for the desired unknowns as follows:

Fig. 2.37(b) Zener diode equivalent circuit


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2.8 Diode Logic Circuit

Fig. 2.38 Logic OR gate circuit

2.8.1 OR Gate as shown in Fig. 2.38

OR gate symbol


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2.8 Diode Logic Circuit (Continued)

Fig. 2.39 Logic AND gate circuit

2.8.2 AND Gate as shown in Fig. 2.39

AND gate symbol

manf360 chapter 2 diode applications

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