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

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

I-V characteristics of an ideal diode

Ideal diode operation on off

Ideal diode operation diode on diode off

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

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

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

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

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.

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

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.

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

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.

I-V characteristics of an ideal diode

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.

Prob. 3.10(b) Is the diode on or off? Assume diode on. 15V = (10K)I1 + (10K)(I1- I2) 15 = (20K)I1 - (10K)I2 1 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)I3 3 I3 I1 I2 Put 3 into 2. -5 = -(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.

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

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

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

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

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

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.

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

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.

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

Prob. 3.9(b) (using constant voltage-drop model) Are the diodes on or off? Assume both diodes are on. 10V = (10K)I1 + 0.7 I1 = 9.3V/10K = I1 = 0.93mA 0 = -0.7 + 0.7 + (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

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 + 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.

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)I2 1 0 = (10K)(I2- I1) - 0.7 + (10K)(I2- I3) 0.7 = -(10K)I1 + (20K)I2 - (10K)I3 2 0 = (10K)( I3- I2) + (10K)I3 + 10 -10 = -(10K)I2 + (20K)I3 3 I3 I1 I2 Put 3 into 2. -4.3 = -(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.

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

3.7 Rectifier circuits Block diagram of a dc power supply

Half-wave rectifier Simple Wastes half the input

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

Full-wave rectifier Entire input waveform is used

Bridge rectifier VS > 0 D1, D2 on; D3, D4 off A type of full-wave rectifier Center-tap not needed Most popular rectifier

Bridge rectifier VO is 2VD less than VS

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

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)