Diode UWO二极管西安大略大学

1、0chap. 3 diodessimplest semiconductor devicenonlinearused in power suppliesvoltage limiting circuits13.1 ideal diodesforward bias(on)reverse bias(off)2i-v characteristics of an ideal diode3ideal diode operationononoffoff4ideal diode operationdiode ondiode off52412on off on offvin = 24 sinwt voutidea

2、l diode operationdiode conducts when 24 sinwt = 12 sinwt = 12/24 wt = 30306exercise 3.4(a)5v2.5kwfind i and vassume diode is on.v = 0, i = 5v/ 2.5kwi = 2ma, implies diode is on.correct assumption5v2.5kwi+v-i7exercise 3.4(b)5v2.5kwfind i and vassume diode is off.vd = - 5, id = 0implies diode is off.c

3、orrect assumptionv = 5, id = 05v2.5kwi+v-8exercise 3.4(e)(start with largest voltage)assume d1 on,then d2 will be off, and d3 will be offv = 3v, and i = 3v/1kw = 3ma.check assumption,vd1 = 0, onvd2 = -1, offvd3 = -2, offcorrect assumption (old-style or gate)+v-i+3+2+1find i and v93.6 zener diodesdes

4、igned to break down at a specific voltageused in power supplies and voltage regulatorswhen a large reverse voltage is reached, the diode conducts. vz is called the breakdown, or zener voltage.10typical use of zener diodethe zener diode will not usually conduct, it needs vs 12.5v to break downassume

5、vs fluctuates or is noisyif vs exceeds 12.5v, the diode will conduct, protecting the load11solving 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 unti

6、l assumption is true.12prob. 3.9(b)assume both diodes are on.10v = (10k)i1i1 = 10v/10k = i1 = 1ma 0 = (5k)i2 - 10v, i2 = 2macurrent in d2 = i2 = 2ma, oncurrent in d1 = i1 - i2 = -1ma, offdoes not match assumption; start over.are the diodes on or off?i1i213prob. 3.9(b)assume d1 off and d2 on.10v = (1

7、0k)i + (5k)i -10v 20v = (15k)ii = 20v/15k = 1.33macurrent in d2 = i = 1.33ma, onvoltage across d1 10v - 10k(1.33ma) = -3.33v, offmatches assumption; done.are the diodes on or off?i14i-v characteristics of an ideal diode15solving ideal diode problems(determining if the diode is on or off)assume diode

8、s 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.16prob. 3.10(b)assume diode on.15v = (10k)i1 + (10k)(i1- i2)15 = (20k)i1 - (10k)i2 10 = (10k)(i2- i1) + (10k)(i2- i3)0 = -(10k)i1 + (20k)i2 - (10k)i3

9、 2 0 = (10k)( i3- i2) + (10k)i3 + 10-10 = -(10k)i2 + (20k)i3 3is the diode on or off?i1i3i2put 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 cant be on.17prob. 3.10(b)assume diode off.15v = (10k)i1 + (10k)i1i1 = 0.75m

10、ai2 = 0 0 = (10k)i3 + (10k)i3 + 10i3 = -0.5mai1i3i2find v1. v1 = (10k)i1 = 7.5vfind v2. v2 = -(10k)i3 = 5vvoltage across diode is v2 - v1 = -2.5v, diode is off v1v2183.2 real diodescharacteristics of a real diodeforward biasreverse biasbreakdown19reverse bias regiona small current flows when the dio

11、de is reversed bias, isis is called the saturation or leakage currentis 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)is20forward bias regionfor silicon diodes, very little

12、 current flows until v 0.5vat v 0.7v, the diode characteristics arenearly verticalin the vicinity of v 0.7v, a wide range ofcurrent may flow.the forward voltage drop of a diode is oftenassumed to be v = 0.7vdiodes made of different materials have different voltage drops v 0.2v - 2.4valmost all diode

13、s are made of silicon, leds are not and have v 1.4v - 2.4v213.4 analysis of diode circuits(simplified diode models) p. 159-162ideal diodeconstant-voltage drop modelconstant-voltage drop model with resistorall use assumptions because actual diode characteristics are too difficult to use in circuit an

14、alysis22constant-voltage drop modeli-v characteristicsa straight line is used to represent the fast-rising characteristics.resistance of diode when slope is vertical is zero.23constant-voltage drop modeli-v characteristics and equivalent circuit0.7v0.7v+-24constant-voltage drop with resistor modela

15、straight line with a slope is used to represent the fast-rising characteristics.resistance of diode is 1/slope.i-v characteristics25constant-voltage drop with resistor modeli-v characteristics and equivalent circuit0.7v0.7v 50w+-26prob. 3.9(b) (using constant voltage-drop model)assume both diodes ar

16、e on.10v = (10k)i1 + 0.7i1 = 9.3v/10k = i1 = 0.93ma 0 = -0.7 + 0.7 + (5k)i2 - 10v, i2 = 2macurrent in d2 = i2 = 2ma, oncurrent in d1 = i1 - i2 = -1.07ma, offdoes not match assumption; start over.are the diodes on or off?i1i227prob. 3.9(b) (using constant voltage-drop model)assume d1 off and d2 on.10

17、v = (10k)i + 0.7 + (5k)i -10v 19.3v = (15k)ii = 19.3v/15k = 1.29macurrent in d2 = i = 1.29ma, onvoltage across d1 10v - 10k(1.29ma) = -2.9v, offmatches assumption; done.are the diodes on or off?i28prob. 3.10(b) (using constant voltage-drop model)assume diode on.15v = (10k)i1 + (10k)(i1- i2)15 = (20k

18、)i1 - (10k)i2 10 = (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 3is the diode on or off?i1i3i2put 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 cant be on.29prob. 3.10(b) (using constant voltage-drop model)assume diode off.15v = (10k)i1 + (10k)i1i1 = 0.75mai2 = 0 0 = (10k)i3 + (10k)i3 + 10i3 = -0.5mai1i3i2find v1. v1 = (10k)i1 = 7.5vfind v2. v2 = -(10k)i3 =