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Published byDominique Steward Modified over 3 years ago

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**Chap. 3 Diodes Simplest semiconductor device Nonlinear**

Used in power supplies Voltage limiting circuits

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

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**I-V characteristics of an ideal diode**

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Ideal diode operation on off

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Ideal diode operation diode on diode off

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**Ideal diode operation sinwt = 12/24 Vin = 24 sinwt 24 12 Vout**

on off on off 30 Diode conducts when 24 sinwt = 12 sinwt = 12/24 wt = 30

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**Exercise 3.4(a) I + 2.5KW V 5V - I 2.5KW 5V Find I and V**

Assume diode is on. V = 0, I = 5V/ 2.5KW I = 2mA, implies diode is on. Correct assumption I 2.5KW 5V

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**Exercise 3.4(b) I + 2.5KW V 5V - 2.5KW 5V Find I and V**

Assume diode is off. VD = - 5, ID = 0 implies diode is off. Correct assumption V = 5, ID = 0 2.5KW 5V

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**Exercise 3.4(e) Find I and V +3 (Start with largest voltage)**

Assume D1 on, then D2 will be off, and D3 will be off V = 3V, and I = 3V/1KW = 3mA. Check assumption, VD1 = 0, on VD2 = -1, off VD3 = -2, off Correct assumption (old-style OR gate) +2 +1 + V - I

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**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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**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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**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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**Prob. 3.9(b) Are the diodes on or off? Assume both diodes are on.**

10V = (10K)I1 I1 = 10V/10K = I1 = 1mA 0 = (5K)I2 - 10V, I2 = 2mA Current in D2 = I2 = 2mA, on Current in D1 = I1 - I2 = -1mA, off Does not match assumption; start over. I1 I2

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**Prob. 3.9(b) Are the diodes on or off? I 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.

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**I-V characteristics of an ideal diode**

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**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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**Prob. 3.10(b) Is the diode on or off? Assume diode on.**

15V = (10K)I1 + (10K)(I1- I2) 15 = (20K)I1 - (10K)I 0 = (10K)(I2- I1) + (10K)(I2- I3) 0 = -(10K)I1 + (20K)I2 - (10K)I3 2 0 = (10K)( I3- I2) + (10K)I3 + 10 -10 = -(10K)I2 + (20K)I I3 I1 I2 Put 3 into = -(10K)I1 + (15K)I2, Put 1 into this equation, solve for I2. I2 = 0.875mA, Current through diode is negative! Diode can’t be on.

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**Prob. 3.10(b) Assume diode off. 15V = (10K)I1 + (10K)I1 I1 = 0.75mA**

Find V1. V1 = (10K)I1 = 7.5V Find V2. V2 = -(10K)I3 = 5V Voltage across diode is V2 - V1 = -2.5V, diode is off

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**3.2 Real diodes Characteristics of a real diode breakdown Reverse bias**

Forward bias

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**Reverse bias region A small current flows when the diode is**

reversed bias, IS IS is called the saturation or leakage current IS 1nA -VZ is the reverse voltage at which the diode breaks down. VZ is the Zener voltage in a Zener diode (controlled breakdown). Otherwise, VZ is the peak inverse voltage (PIV) IS

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**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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**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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**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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**Constant-voltage drop model**

I-V characteristics and equivalent circuit + 0.7V - 0.7V

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**Constant-voltage drop with resistor model**

I-V characteristics A straight line with a slope is used to represent the fast-rising characteristics. Resistance of diode is 1/slope.

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**Constant-voltage drop with resistor model**

I-V characteristics and equivalent circuit + 0.7V 50W - 0.7V

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**Prob. 3.9(b) (using constant voltage-drop model)**

Are the diodes on or off? Assume both diodes are on. 10V = (10K)I I1 = 9.3V/10K = I1 = 0.93mA 0 = (5K)I2 - 10V, I2 = 2mA Current in D2 = I2 = 2mA, on Current in D1 = I1 - I2 = -1.07mA, off Does not match assumption; start over. I1 I2

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**Prob. 3.9(b) (using constant voltage-drop model)**

Are the diodes on or off? I Assume D1 off and D2 on. 10V = (10K)I (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.

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**Prob. 3.10(b) (using constant voltage-drop model)**

Is the diode on or off? Assume diode on. 15V = (10K)I1 + (10K)(I1- I2) 15 = (20K)I1 - (10K)I 0 = (10K)(I2- I1) (10K)(I2- I3) 0.7 = -(10K)I1 + (20K)I2 - (10K)I3 2 0 = (10K)( I3- I2) + (10K)I3 + 10 -10 = -(10K)I2 + (20K)I I3 I1 I2 Put 3 into = -(10K)I1 + (15K)I2, Put 1 into this equation, solve for I2. I2 = 0.91mA, Current through diode is negative! Diode can’t be on.

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**Prob. 3.10(b) (using constant voltage-drop model)**

Assume diode off. 15V = (10K)I1 + (10K)I1 I1 = 0.75mA I2 = 0 0 = (10K)I3 + (10K)I3 + 10 I3 = -0.5mA V1 V2 I3 I1 I2 Find V1. V1 = (10K)I1 = 7.5V Find V2. V2 = -(10K)I3 = 5V Voltage across diode is V2 - V1 = -2.5V, diode is off

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3.7 Rectifier circuits Block diagram of a dc power supply

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Half-wave rectifier Simple Wastes half the input

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**Full-wave rectifier VS > 0 VS < 0**

Current goes through load in same direction for + VS. VO is positive for + VS. Requires center-tap transformer

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Full-wave rectifier Entire input waveform is used

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**Bridge rectifier VS > 0 D1, D2 on; D3, D4 off**

A type of full-wave rectifier Center-tap not needed Most popular rectifier

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Bridge rectifier VO is 2VD less than VS

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**Filter Capacitor acts as a filter.**

Vi charges capacitor as Vi increases. As Vi decreases, capacitor supplies current to load.

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**Filter Diode off Diode on**

When the diode is off, the capacitor discharges. Vo = Vpexp(-t/RC) Assuming t T, and T=1/f VP - Vr = Vpexp(-1/fRC) half-wave rectifier (t T) VP - Vr = Vpexp(-1/2fRC) full-wave rectifier (t T/2)

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