Diod Rectifier 1v6o2g

DEE 2023 SEMICONDUCTOR DEVICES CHAPTER 2

DIODES

2.1 CHARACTERISTICS OF DIODE & DIODE AS A SEMICONDUCTOR DEVICES 2.1.1 The schematics diagram and physical structure of a diode a. Schematic diagram of a diode is same as P-N combination. b. A diode have 2 pins electronic component which is called anode and cathode. c. Anode is a P type material and Cathode is a N type material. d. The arrow from the diagram shows the current conventional direction.

2.1 Schematic diagram, physical and and schematic symbol of a diode DEE2023 SEMICONDUCTOR DEVICES

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2.1 CHARACTERISTICS OF DIODE & DIODE AS A SEMICONDUCTOR DEVICES 2.1.2 The schematics diagram of a diode in BIAS VOLTAGE The voltage across the diode call biased voltage. a. Forward bias If anode get voltage more positive from cathode, or cathode get voltage more negative from anode. b. Reverse bias If anode get voltage more negative than cathode or cathode get more positive than anode. (a)

(b)

2.1 Forward bias (a) and Reverse bias (b) circuit diagrams DEE2023 SEMICONDUCTOR DEVICES

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2.1 CHARACTERISTICS OF DIODE & DIODE AS A SEMICONDUCTOR DEVICES 2.1.3 The I-V Characteristics Curve for Diode When a diode in forward and reverse bias condition and each of the bias and current value are measured, then I-V Curve will appear like in diagram.

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2.1.3 The I-V Characteristics Curve for Diode – Cont. a. Forward current (Id)  The current ed through the diode when the diode in forward bias condition. Id measured in mA. b. Reverse current (Is)  The small current ed through the diode when the diode is in reverse bias condition. Is measured in  A(starting 0.1A). c. Knee voltage  Limit voltage when forward current is suddenly increased. Knee voltage is similar with barrier voltage. (Si = 0.7 V, Ge = 0.3 V) d. Breakdown Voltage  Limit voltage where reverse current suddenly happen. Huge current that exceed the limit can cause the malfunction of PN junction. e. Burning level  When Id,Vd exceeds Pmax DEE2023 SEMICONDUCTOR DEVICES

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2.2 DIODE IN RECTIFIERS CIRCUITS 2.2.1 Block Diagram Of Power Supply AC VOLTAGE

TRANSFORMER

RECTIFIER

FILTER

REGULATOR

VOLTAGE DEVIDER

DC VOLTAGE

2.2 : Block diagram for DC power supply

Summary :The a.c. voltage, typically 240V is connected to transformer, which steps that ac voltage down to the level for the desired dc output. A diode rectifier then provides a full-wave rectified voltage that it initially filtered by a simple capacitor filter to produce a dc voltage. This resulting dc voltage that not only has much less ripple voltage but also remains the same dc value even if the input dc voltage varies somewhat or the load connected to the output dc voltage changes. This voltage regulation is usually obtained using one of a number of popular voltage regulator IC units. DEE2023 SEMICONDUCTOR DEVICES

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2.2.1 Block Diagram Of Power Supply – Cont.

Function: To convert a.c.voltage to pulse d.c. voltage.

2.2.2 Type of rectifiers

a. Half Wave Rectifiers b. Full Wave Rectifiers c. Bridge Rectifiers

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2.2 DIODE IN RECTIFIERS CIRCUITS a. Half Wave Rectifiers

OPERATION:  During input signal positive cycle Diode D forward bias. D act as closed switch so current can go through. Voltage drop at RL is equal to positive cycle input signal positive magnitude  During input signal negative cycle D reverse bias and act as open switch so current cannot go through. Voltage drop at RL during negative cycle is equal to zero DEE2023 SEMICONDUCTOR DEVICES

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2.2 DIODE IN RECTIFIERS CIRCUITS a. Full Wave Rectifiers

OPERATION:  During input signal positive cycle M terminal will be positive and N negative. Diode D1 and D3 will be forward bias meanwhile diode D2 dan D4 get reverse bias. Current will be flow along M, E, A, B, C, F, N. One positive cycle will occur at RL load.  During input signal negative cycle M terminal will be negative and N will be positive. Diode D2 and D4 will be forward bias meanwhile diode D1 and D3 become reverse bias. Current will be flow along N, F, A, B, C, E, M. Due to current flow is same as positive cycle input signal, the result is also same as positive cycle. DEE2023 SEMICONDUCTOR DEVICES

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2.2 DIODE IN RECTIFIERS CIRCUITS a. Bridge Rectifiers

OPERATION:  During input signal positive cycle M terminal will be positive and N negative. Diode D1 and D3 will be forward bias meanwhile diode D2 dan D4 get reverse bias. Current will be flow along M, E, A, B, C, F, N. One positive cycle will occur at RL load.  During input signal negative cycle M terminal will be negative and N will be positive. Diode D2 and D4 will be forward bias meanwhile diode D1 and D3 become reverse bias. Current will be flow along N, F, A, B, C, E, M. Due to current flow is same as positive cycle input signal, the result is also same as positive cycle. 10

2.2 DIODE IN RECTIFIERS CIRCUITS 2.2.3 Equations of Average Current, Average Voltage and Roots Means Square Voltage

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2.2.3 Equations of Average Current, Average Voltage and Roots Means Square Voltage - Cont.

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EXAMPLE 1

A half wave rectifier build in 60 Vrms in the secondary of the transformer. Calculate:i.What is the output voltage. ii.Average voltage of the circuit

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Solution : i. Output voltage, Vo Known that, Vo = Vp sec – 0.7V From formula, Vp sec = √2 x Vrms = √2 x 60 = 84.85 V Therefore, Vo = Vp sec – 0.7V = 84.85 – 0.7 = 84.16V ii. Average voltage, Vavg Known that, Vavg = Vo

π = 84.16

π = 26.7 V

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EXAMPLE 2 Determine:i. Output voltage ii. Average voltage iii. Average current

100

2:1

Vp(sec)

Vp(pri)

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1k

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Solution : i. Output voltage, Vo Known that, Vo = Vp sec – 0.7V 2 From formula, Vp sec = (N sec) (Vp pri) N pri = (1) x (100) 2 = 50 V Therefore, Vo = 50 – 0.7V 2 = 24.3 V ii. Average voltage, Vavg Vavg = 2Vo

iii. Average voltage, Vavg Iavg = Vavg

π

= 15.5

= 2 (24.3)

1K

π = 15.5 V

RL

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= 15.5mA

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EXAMPLE 3 In a centre-tap full wave rectifier, load resistance is 2KΩ. The a.c. supply across the primary winding is 220V. Taking transformer turn ratio N1/N2 = ½ and Neglecting diode resistance. Determine:i. Output voltage ii. Average voltage iii. Average current

220V

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2k

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Solution : i. Output voltage, Vo Known that, Vo = Vp sec – 0.7V 2 From formula, Vp sec = (N sec) (Vp pri) N pri = (2) x (220) 1 = 440 V Therefore, Vo = 440 – 0V 2 = 220 V ii. Average voltage, Vavg iii. Average voltage, Vavg Known that, Vavg = 2Vo Iavg = Vavg

π = 2(220)

π

RL = 140 = 70mA 2K

= 140 V DEE2023 SEMICONDUCTOR DEVICES

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EXAMPLE 4: A full wave rectifier with a 120 Vrms sinusodial input had a load resistor of 1KΩ. If silicone diodes are employed, determine:120Vrms

i. Average voltage available at the load. ii. Average current

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2k

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Solution : i. Average voltage, Vavg Vavg = 2Vo

π Known that, Vo = Vp sec – 1.4V From formula, Vp sec = √2 x Vrms = √2 x 120 = 169.7 V

ii. Average Current Iavg = Vavg

RL = 107.15 = 107.15mA 1K

So, Vo = Vp sec – 1.4V = 169.7 – 1.4V =168.3 V Average voltage, Vavg Vavg = 2Vo π

= 2 x 168.3 = 107.15 V π

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2.2.5 D’Arsonval Meter Movement with Half-Wave Rectification.  In order to measure the alternating current with

the d’Arsonval meter movement, we must rectify the alternating current by use of diode rectifier .  Figure 2.2.5 is the DC voltmeter circuit modified to measure AC voltage.  The diode, assume to be ideal diode, has no effect on the operation of the circuit .  For example if the 10 V sine-wave input is fed as the source of the circuit, the voltage across the meter movement is just the positive half-cycle of the sine wave due to the rectifying effect of the diode. 2.2.5: DC Voltmeter Circuit Modified to Measure AC Voltage.

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2.2.6 D’Arsonval Meter Movement with Full-Wave Rectification. The full-wave rectifier provide higher sensitivity rating compare to the half-wave rectifier.  Bridge type rectifier is the most commonly used, Figure 2.2.6.

Figure2.2.6 : Full Wave Bridge Rectifier Used in AC Voltmeter Circuit. 22

Cont’d…

Operation;  (a) During the positive half cycle (red arrow), currents flows through diode D2, through the meter movement from positive to negative, and through diode D3. - The polarities in circles on the transformer secondary are for the positive half cycle. - Since current flows through the meter movement on both half cycles, we can expect the deflection of the pointer to be greater than with the half wave cycle. - If the deflection remains the same, the instrument using full wave rectification will have a greater sensitivity. (b) Vise-versa for the negative half cycle (blue arrow). 23

2.3 DIODE IN OTHER ELECTRONICS CIRCUIT APPLICATIONS 2.3.1 CLIPPER First clipper circuit a. Diode is forward biased during positive half cycle . This makes the positive potential 0.7V

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2.3.1 CLIPPER First clipper circuit – Cont. b. Output swings from 0.7V positive to 50V negative

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2.3.1 CLIPPER Second clipper circuit a. Diode is forward biased during positive half cycle . This makes the negative potential 0.7V

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2.3.1 CLIPPER Second clipper circuit – Cont. b. Output swings from 24V positive to 0.7V negative

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2.3.1 CLIPPER Third clipper circuit a. Since diodes are in series, each branch will drop 0.7V + 0.7V = 1.4V

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2.3.1 CLIPPER Third clipper circuit – Cont. b. Output swings from +1.4V positive to -1.4V negative

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2.3 DIODE IN OTHER ELECTRONICS CIRCUIT APPLICATIONS 2.3.1 CLAMPER First clamper circuit a. DC potential is pk – 0.7V = 14.3V  This becomes the zero reference line for the AC, so it will go 15V above and 15V below this line

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2.3.1 CLAMPER First clamper circuit – Cont. b. Output swings from +29.3V positive to -0.7V negative

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2.3 DIODE IN OTHER ELECTRONICS CIRCUIT APPLICATIONS 2.3.1 CLAMPER Second clamper circuit a. DC potential is pk – 0.7V = 29.3V  This becomes the zero reference line for the AC, so it will go 30V above and 30V below this line

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2.3.1 CLAMPER Second clamper circuit – Cont. b. Output swings from +0.7V positive to -59.3V negative

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2.3 DIODE IN OTHER ELECTRONICS CIRCUIT APPLICATIONS 2.3.1 CLAMPER Third clamper circuit a. This one charges the capacitor during the positive alternation, and this will then discharge during the negative alternation. b. This makes the output at maximum a straight line of 40V (2Vpk) c. However, since the diode is not perfect we need to use the first approximation (simplified diode)  This means we need to subtract 0.7V for the one diode that charges the capacitor (39.3V)  And lest we forget, subtract the potential for the second diode as well (38.6V)

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2.3.1 CLAMPER Third clamper circuit – Cont. b. Output is pulsating wave (ripple) that rises to a maximum potential of 39.3V and descends to a minimum potential of 38.6V

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