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0 Chap. 3 Diodes Simplest semiconductor device Nonlinear Used in power supplies Voltage limiting circuits

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1 3.1 Ideal Diodes Forward bias (on) Reverse bias (off)

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2 I-V characteristics of an ideal diode

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3 Ideal diode operation on off

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4 Ideal diode operation diode on diode off

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5 24 12 on off V in = 24 sin t V out Ideal diode operation Diode conducts when 24 sin t = 12 sin t = 12/24 t = 30 30

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6 Exercise 3.4(a) 5V 2.5K Find I and V Assume diode is on. V = 0, I = 5V/ 2.5K I = 2mA, implies diode is on. Correct assumption 5V 2.5K I +V-+V- I

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7 Exercise 3.4(b) 5V 2.5K Find I and V Assume diode is off. V D = - 5, I D = 0 implies diode is off. Correct assumption V = 5, I D = 0 5V 2.5K I +V-+V-

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8 Exercise 3.4(e) (Start with largest voltage) Assume D1 on, then D2 will be off, and D3 will be off V = 3V, and I = 3V/1K mA. Check assumption, V D1 = 0, on V D2 = -1, off V D3 = -2, off Correct assumption (old-style OR gate) +V-+V- I +3 +2 +1 Find I and V

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9 3.6 Zener diodes Designed to break down at a specific voltage Used in power supplies and voltage regulators When a large reverse voltage is reached, the diode conducts. Vz is called the breakdown, or Zener voltage.

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10 Typical use of Zener diode The Zener diode will not usually conduct, it needs Vs > 12.5V to break down Assume Vs fluctuates or is noisy If Vs exceeds 12.5V, the diode will conduct, protecting the load

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11 Solving ideal diode problems (determining if the diode is on or off) Assume diodes are on or off. Perform circuit analysis, find I & V of each diode. Compare I & V of each diode with assumption. Repeat until assumption is true.

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12 Prob. 3.9(b) Assume both diodes are on. 10V = (10K)I 1 I 1 = 10V/10K = I 1 = 1mA 0 = (5K)I 2 - 10V, I 2 = 2mA Current in D2 = I 2 = 2mA, on Current in D1 = I 1 - I 2 = -1mA, off Does not match assumption; start over. Are the diodes on or off? I1I1 I2I2

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13 Prob. 3.9(b) Assume D1 off and D2 on. 10V = (10K)I + (5K)I -10V 20V = (15K)I I = 20V/15K = 1.33mA Current in D2 = I = 1.33mA, on Voltage across D1 10V - 10K(1.33mA) = -3.33V, off Matches assumption; done. Are the diodes on or off? I

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14 I-V characteristics of an ideal diode

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15 Solving ideal diode problems (determining if the diode is on or off) Assume diodes are on or off. Perform circuit analysis, find I & V of each diode. Compare I & V of each diode with assumption. Repeat until assumption is true.

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16 Prob. 3.10(b) Assume diode on. 15V = (10K)I 1 + (10K)(I 1 - I 2 ) 15 = (20K)I 1 - (10K)I 2 1 0 = (10K)(I 2 - I 1 ) + (10K)(I 2 - I 3 ) 0 = -(10K)I 1 + (20K)I 2 - (10K)I 3 2 0 = (10K)( I 3 - I 2 ) + (10K)I 3 + 10 -10 = -(10K)I 2 + (20K)I 3 3 Is the diode on or off? I1I1 I3I3 I2I2 Put 3 into 2. -5 = -(10K)I 1 + (15K)I 2, Put 1 into this equation, solve for I 2. I 2 = 0.875mA, Current through diode is negative! Diode can’t be on.

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17 Prob. 3.10(b) Assume diode off. 15V = (10K)I 1 + (10K)I 1 I 1 = 0.75mA I 2 = 0 0 = (10K)I 3 + (10K)I 3 + 10 I 3 = -0.5mA I1I1 I3I3 I2I2 Find V 1. V 1 = (10K)I 1 = 7.5V Find V 2. V 2 = -(10K)I 3 = 5V Voltage across diode is V 2 - V 1 = -2.5V, diode is off V1V1 V2V2

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18 3.2 Real diodes Characteristics of a real diode Forward biasReverse bias breakdown

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19 Reverse bias region A small current flows when the diode is reversed bias, I S I S is called the saturation or leakage current I S 1nA -V Z is the reverse voltage at which the diode breaks down. V Z is the Zener voltage in a Zener diode (controlled breakdown). Otherwise, V Z is the peak inverse voltage (PIV) ISIS

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20 Forward bias region For Silicon diodes, very little current flows until V 0.5V At V 0.7V, the diode characteristics are nearly vertical In the vicinity of V 0.7V, a wide range of current may flow. The forward voltage drop of a diode is often assumed to be V = 0.7V Diodes made of different materials have different voltage drops V 0.2V - 2.4V Almost all diodes are made of Silicon, LEDs are not and have V 1.4V - 2.4V

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21 3.4 Analysis of diode circuits (Simplified diode models ) p. 159-162 Ideal diode Constant-voltage drop model Constant-voltage drop model with resistor All use assumptions because actual diode characteristics are too difficult to use in circuit analysis

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22 Constant-voltage drop model I-V characteristics A straight line is used to represent the fast-rising characteristics. Resistance of diode when slope is vertical is zero.

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23 Constant-voltage drop model I-V characteristics and equivalent circuit 0.7V + -

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24 Constant-voltage drop with resistor model A straight line with a slope is used to represent the fast-rising characteristics. Resistance of diode is 1/slope. I-V characteristics

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25 Constant-voltage drop with resistor model I-V characteristics and equivalent circuit 0.7V 50 + -

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26 Prob. 3.9(b) (using constant voltage-drop model) Assume both diodes are on. 10V = (10K)I 1 + 0.7 I 1 = 9.3V/10K = I 1 = 0.93mA 0 = -0.7 + 0.7 + (5K)I 2 - 10V, I 2 = 2mA Current in D2 = I 2 = 2mA, on Current in D1 = I 1 - I 2 = -1.07mA, off Does not match assumption; start over. Are the diodes on or off? I1I1 I2I2

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27 Prob. 3.9(b) (using constant voltage-drop model) Assume D1 off and D2 on. 10V = (10K)I + 0.7 + (5K)I -10V 19.3V = (15K)I I = 19.3V/15K = 1.29mA Current in D2 = I = 1.29mA, on Voltage across D1 10V - 10K(1.29mA) = -2.9V, off Matches assumption; done. Are the diodes on or off? I

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28 Prob. 3.10(b) (using constant voltage-drop model) Assume diode on. 15V = (10K)I 1 + (10K)(I 1 - I 2 ) 15 = (20K)I 1 - (10K)I 2 1 0 = (10K)(I 2 - I 1 ) - 0.7 + (10K)(I 2 - I 3 ) 0.7 = -(10K)I 1 + (20K)I 2 - (10K)I 3 2 0 = (10K)( I 3 - I 2 ) + (10K)I 3 + 10 -10 = -(10K)I 2 + (20K)I 3 3 Is the diode on or off? I1I1 I3I3 I2I2 Put 3 into 2. -4.3 = -(10K)I 1 + (15K)I 2, Put 1 into this equation, solve for I 2. I 2 = 0.91mA, Current through diode is negative! Diode can’t be on.

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29 Prob. 3.10(b) (using constant voltage-drop model) Assume diode off. 15V = (10K)I 1 + (10K)I 1 I 1 = 0.75mA I 2 = 0 0 = (10K)I 3 + (10K)I 3 + 10 I 3 = -0.5mA I1I1 I3I3 I2I2 Find V 1. V 1 = (10K)I 1 = 7.5V Find V 2. V 2 = -(10K)I 3 = 5V Voltage across diode is V 2 - V 1 = -2.5V, diode is off V1V1 V2V2

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30 3.7 Rectifier circuits Block diagram of a dc power supply

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31 Half-wave rectifier Simple Wastes half the input

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32 Full-wave rectifier V S > 0 V S < 0 Current goes through load in same direction for + V S. V O is positive for + V S. Requires center-tap transformer

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33 Full-wave rectifier Entire input waveform is used

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34 Bridge rectifier A type of full-wave rectifier Center-tap not needed Most popular rectifier V S > 0 D 1, D 2 on; D 3, D 4 off V S < 0 D 3, D 4 on; D 1, D 2 off

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35 Bridge rectifier V O is 2V D less than V S

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36 Filter Capacitor acts as a filter. Vi charges capacitor as Vi increases. As Vi decreases, capacitor supplies current to load.

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37 Filter Diode on Diode off When the diode is off, the capacitor discharges. V o = V p exp(-t/RC) Assuming t T, and T=1/f V P - V r = V p exp(-1/fRC) half-wave rectifier (t T) V P - V r = V p exp(-1/2fRC) full-wave rectifier (t T/2)

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