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Chapte r 2 Diode applications Ir. Dr. Rosemizi Abd Rahim 1 Ref: Electronic Devices and Circuit Theory, 10/e, Robert L. Boylestad and Louis Nashelsky.

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Presentation on theme: "Chapte r 2 Diode applications Ir. Dr. Rosemizi Abd Rahim 1 Ref: Electronic Devices and Circuit Theory, 10/e, Robert L. Boylestad and Louis Nashelsky."— Presentation transcript:

1 Chapte r 2 Diode applications Ir. Dr. Rosemizi Abd Rahim 1 Ref: Electronic Devices and Circuit Theory, 10/e, Robert L. Boylestad and Louis Nashelsky

2 Objectives ● Explain and analyze the operation of both half and full wave rectifiers ● Explain and analyze filters and regulators and their characteristics ● Explain and analyze the operation of diode limiting and clamping circuits ● Explain and analyze the operation of diode voltage multipliers ● Interpret and use a diode data sheet ● Troubleshoot simple diode circuits 2

3 Load-Line Analysis The load line plots all possible combinations of diode current (I D ) and voltage (V D ) for a given circuit. The maximum I D equals E/R, and the maximum V D equals E. The point where the load line and the characteristic curve intersect is the Q-point, which identifies I D and V D for a particular diode in a given circuit. 3

4 Load-Line Analysis Example For the series diode configuration of Fig. 2.3a, employing the diode characteristics of Fig. 2.3b, determine: a. VDQ and IDQ. b. VR 4

5 Series Diode Configurations Forward Bias Constants Silicon Diode: V D = 0.7 V Germanium Diode: V D = 0.3 V Analysis (for silicon) V D = 0.7 V (or V D = E if E < 0.7 V) V R = E – V D I D = I R = I T = V R / R 5

6 Series Diode Configurations Reverse Bias Diodes ideally behave as open circuits Analysis V D = E V R = 0 V I D = 0 A 6

7 Series Diode Configurations Example For the series diode configuration of Figure below, determine VD, VR, and ID. 7

8 Series Diode Configurations Example For the series diode configuration of Figure below, determine VD, VR, and ID. 8

9 Series Diode Configurations Example For the series diode configuration of Figure below, determine I, V1, V2 and V0. 9

10 Series Diode Configurations Example For the series diode configuration of Figure below, determine I, V1, V2 and V0. 10

11 Parallel Configurations Example 11 Determine Vo, I1, ID1, and ID2 for the parallel diode configuration of Figure below

12 Parallel Configurations Example 12 Determine I1, I2, and ID2 for the parallel diode configuration of Figure below

13 Parallel Configurations Example 13 Determine I1, I2, and ID2 for the parallel diode configuration of Figure below

14 Half-Wave Rectification The diode only conducts when it is forward biased, therefore only half of the AC cycle passes through the diode to the output. The DC output voltage (VDC) is 0.318V m, where V m = the peak AC voltage. 14

15 Half-Wave Rectification VK =0.7 V and vo = vi – VK When Vm » VK 15

16 Half-Wave Rectification Example a.determine the dc level of the output for the network of Figure belowSketch the output vo and b.Repeat part (a) if the ideal diode is replaced by a silicon diode. 16

17 Half-Wave Rectification Example a.determine the dc level of the output for the network of Figure belowSketch the output vo and b.Repeat part (a) if the ideal diode is replaced by a silicon diode. 17

18 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) > V m PIV = Peak inverse voltage PRV = Peak reverse voltage V m = Peak AC voltage 18

19 Full-Wave Rectification The rectification process can be improved by using a full-wave rectifier circuit. Full-wave rectification produces a greater DC output: Half-wave: V dc = 0.318V m Full-wave: V dc = 0.636V m 19

20 Full-Wave Rectification Bridge Rectifier Four diodes are connected in a bridge configuration V DC = 0.636V m 20

21 Full-Wave Rectification Center-Tapped Transformer Rectifier Requires Two diodes Center-tapped transformer V DC = 0.636V m 21

22 Summary of Rectifier Circuits V m = peak of the AC voltage. In the center tapped transformer rectifier circuit, the peak AC voltage is the transformer secondary voltage to the tap. RectifierIdeal V DC Realistic V DC Half Wave Rectifier V DC = 0.318V m V DC = 0.318V m – 0.7 Bridge RectifierV DC = 0.636V m V DC = 0.636V m – 2(0.7 V) Center-Tapped Transformer RectifierV DC = 0.636V m V DC = 0.636V m – 0.7 V 22

23 Diode Clippers The diode in a series clipper “clips” any voltage that does not forward bias it: A reverse-biasing polarity A forward-biasing polarity less than 0.7 V (for a silicon diode) 23

24 Diode Clippers Series clipper with dc supply 24

25 Diode Clippers 25 Adding a DC source in series with the clipping diode changes the effective forward bias of the diode.

26 Biased Clippers Example 26 Determine the output waveform for the sinusoidal input of Figure below.

27 Parallel Clippers The diode in a parallel clipper circuit “clips” any voltage that forward bias it. DC biasing can be added in series with the diode to change the clipping level. 27

28 Parallel Clippers Example Determine vo for the network of Figure below. 28

29 Parallel Clippers Example Determine vo for the network of Figure below. 29

30 Summary of Clipper Circuits Series 30

31 Summary of Clipper Circuits Parallel 31

32 Clampers A diode and capacitor can be combined to “clamp” an AC signal to a specific DC level. 32

33 Biased Clamper Circuits Example 1 The input signal can be any type of waveform such as sine, square, and triangle waves. The DC source lets you adjust the DC camping level. 33

34 Summary of Clamper Circuits 34

35 Voltage-Multiplier Circuits Voltage multiplier circuits use a combination of diodes and capacitors to step up the output voltage of rectifier circuits. Voltage Doubler Voltage Tripler Voltage Quadrupler 35

36 Voltage Doubler This half-wave voltage doubler’s output can be calculated by: V out = V C2 = 2V m where V m = peak secondary voltage of the transformer 36

37 Voltage Doubler Positive Half-Cycle oD 1 conducts oD 2 is switched off oCapacitor C 1 charges to V m Negative Half-Cycle oD 1 is switched off oD 2 conducts oCapacitor C 2 charges to V m V out = V C2 = 2V m 37

38 Voltage Tripler and Quadrupler 38

39 Practical Applications Rectifier Circuits –Conversions of AC to DC for DC operated circuits –Battery Charging Circuits Simple Diode Circuits –Protective Circuits against –Overcurrent –Polarity Reversal –Currents caused by an inductive kick in a relay circuit Zener Circuits –Overvoltage Protection –Setting Reference Voltages 39


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