malvino ep 04 cac mach diode

8/3/2019 Malvino EP 04 Cac Mach Diode

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Chng 4Các mch diode



T Vng (1) Bias = phân cc Capacitor-input filter = Mch lc ngõ vào

(dùng) t

Choke-input filter = Mch lc ngõ vào(dùng) cun dây Clamper = mch kp Clipper = mch xén

dc value of signal = giá tr DC ca tín hiu Filter = mch lc, b lc Half-wave signal = tín hiu bán k



T Vng (2) IC voltage regulator = Mch n áp IC Integrated circuit = IC = vi mch = mch

tích hp

Passive filter = mch lc th ng Peak detector = mch tách sóng nh Peak inverse voltage = in áp ngc

nh Polarized capacitor = t (in) hóa (hc) =

t có phân cc Power supply = ngun cp in



T Vng (3)

Rectifier = mch/b chnh lu Ripple = gn

Surge current = dòng in quá Surge resistor = in tr bo v quá Unidirectional local current = dòng in

cc b n hng

Volatge multiplier = mch nhân in áp Waveform = dng sóng



Ni dung chng 4

4-1 Mch chnh lu bán k4-2 Máy bin th4-3 Mch chnh lu toàn sóng4-4 Mch chnh lu cu4-5 Mch lc ngõ vào (dùng) cun dây4-6 Mch lc ngõ vào (dùng) t4-7 in áp ngc nh và dòng quá 4-8 Mt s vn khác v ngun cp in

4-9 Troubleshooting4-10 Mch xén và mch hn biên (limiter)4-11 Mch kp4-12 Mch nhân in áp



4-1 Mch chnh lu bán k

H. 4-1 (a) Mch chnh lu bán k lý tng;

(b) bán k dng (diode ON); (c) bán k âm (diode OFF)



4-1 Mch chnh lu bán k (tt)Các dng sóng lý tng

Hình 4-2



4-1 Mch chnh lu bán k (tt)

in áp ra nh bng in áp vào nh:

Giá tr DC ca tín hiu bán k Vdc

:

Tn s ra:

f out = f in Xp x bc 2:

Vp(out) = Vp(in) ± 0.7V (diode Si)



A simple Battery charger-Example of a

Rectifier Can be used to charge a car battery from the alternator



4-2 Máy bin th (Transformer)

Máy bin th là 1 cp cun dây có ghéph cm vi nhau ( truyn nng lng tcun này sang cun kia bng t trngbin thiên).

Vi s vòng dây khác nhau ta có máy binth tng áp (step up) hay gim áp (step

down).



4-2 Máy bin th (tt)




Ký hiu máy bin th The symbol for a transformer is a pair of thesame loopy lines used for inductors, but closetogether. If the inductor has a core of a

magnetic material, it is shown as a couple of lines between the coils.

The number of turns in the coils willBe written nearby.




Máy bin th lý tng in áp th cp (secondary) V2:

Vi:

± N1: s vòng dây s cp (primary) ± N2: s vòng dây th cp (primary)

± V1: in áp s cp

Chú ý ta ang nói n in áp AC

1

1

2

2 V N

N

V !




Mch chnh lu bán k vi máy bin th



4-3 Mch chnh lu toàn sóng



4-3 Mch chnh lu toàn sóng (tt)



4-3 Mch chnh lu toàn sóng (tt)

Các công thc Giá tr DC hay trung bình Vdc:

Vdc = 2Vp/T $ 0.636 Vp

Tn s ra:f out = 2f in

Xp x bc 2:

Vp(out) = Vp(in) ± 0.7V (diode Si)Chú ý: Vp(in) = 0.5 V2 (V2 là in áp th

cp vì ngõ ra có chu gia (center tap))



4-4 Mch chnh lu cu



4-4 Mch chnh lu cu (tt)



4-4 Mch chnh lu cu (tt)

Các công thc Giá tr DC hay trung bình Vdc:

Vdc = 2Vp/T $ 0.636 Vp

Tn s ra:f out = 2f in Xp x bc 2:

Vp(out) = Vp(in) ± 1.4V (diode Si)

Chú ý: Chnh lu cu có u im hn loi bán klà ta có th s dng toàn b in áp th cp



4-5 Mch lc ngõ vào (dùng)

cun dây

Hình 4-10 (a) Mch lc ngõ vào (dùng) cundây; (b) mch tng ng AC

Ý tng c bn: Nu XL >> XC thì

Vout $ XCVin/XL



Lc ngõ ra ca mch chnh lu



Lc ngõ ra ca mch chnh lu (tt)



4-6 Mch lc ngõ vào (dùng) t

Ý tng c bn (H.4-12, pp.108)



4-6 Mch lc ngõ vào (dùng) t (tt)

Hiu ng ca in tr ti (H.4-13)



4-6 Mch lc ngõ vào (dùng) t (tt)

Các công thc in áp gn nh-nh ngõ ra VR:

Vi: ± VR = in áp gn nh-nh

± f = tn s gn ( = f in vi bán k; = 2f in vi toàn sóng) ± C = in dung

fC

I V

R

!



4-7 in áp ngc nh PIV vàdòng quá (xét lc ngõ vào t)

H.4-18

(a) PIV = 2VP (b) PIV = VP (c) PIV = VP



Surge resistor

Hình 4-19 in tr bo v gii hn dòng quá



Half-wave Full-wave Full-wave

Bridge

# of diodes 1 2 4

DC output

voltage(**)

Vin (peak) Vin (peak) (*) Vin (peak)

DC diode

current

IL 0.5IL 0.5IL

PIV (***) 2Vin (peak) 2Vin (peak) Vin (peak)

Ripple

Frequency

f 2f 2f

Comparison of Rectifier Circuits

* With 1:2 transformer turns ratio; ** With Large C *** Peak Inverse Voltage (also with Large C)



4-10 Mch xén và mch hn biên

Có 3 loi mch xén: ± Mch xén trên (mch xén dng)

± Mch xén di (mch xén dng) ± Mch xén 2 mc c lp (mch xén kt hp)

Mch hn biên hay kp diode

Chú ý: Tng quát các mch xén có phâncc



Các Mch xén và hn biên(Clippers and Limiters)

- Small Signal Diode

The positive clipper

or

* 1N914 : IF = 10mA for 1V

RB =mA

V

10

7.01 = 30 Ohm

* Stiff Clipper : 100 RB

< RS

< 0.01 RL

c tuyn truyn t



The Negative clipper

or

The Limiter or Diode clamp

protect sensitive circuits

c tuyn truyn t

c tuyn truyn t



Biased Clippers



Combination Clipper (slicer)



Clipper Circuits using Zener Diodes

Zener diode provides a reference voltage of VZ



Clipper applications (1)

Transient-protection circuits



Clipper applications (2)

AM detector



4-11 Mch kp

Có 2 loi mch kp: ± Mch kp dng

± Mch kp âm

Mch tách sóng nh-nh: thng dùngcho diode tín hiu nh tn s cao



Clampers

Positive clamper

-

+ -

+ off

Vout = VP +

VP

Negative clamper



Biased clampers

Biased clampers.



4-12 Mch nhân in áp

Thí d mt s mch nhân áp: ± Nhân ôi in áp (voltage doubler)

± Nhân 3 in áp (voltage tripler)

± Nhân 4 in áp (voltage quadrupler)

± Nhân ôi in áp toàn sóng (full-wave voltagedoubler)



Voltage multipliers (1)

Half-wave voltage doublers.



Voltage multipliers (2)

Operation of half-wave voltage doublers.



Full-wave voltage doubler



Cockcroft-Walton Circuit



S khi ca mt ngun DC



DC Power Supply Requirement

Since this circuit has a significant dc component,

with no dc component at the input, it can be usedto produce a dc power supply. A dc power supplymust be a nonlinear circuit.



DC Power Supplies, PeakDetector

Power Supplies use a

rectifier with acapacitor at theoutput. We start witha half-wave rectifier,since it is simpler.Show Fig. 3.32 fromSedra and Smith.This is a half-waverectifier with a

capacitor, which holdsthe peak value of theinput source, v I (t).



DC Power Supplies, PeakDetector

This circuit will hold

the peak value onlybecause there isnothing connectedat the output.

When we connect aload, somethingdifferent happens.



DC Power Supplies

Here we

have connected aload, R, at theoutput. Lookcarefully at the

diagram for theoutput voltage,vO(t). Note that itno longer holdsthe peak value,but decreasesexponentially.



DC Power Supplies

There is a short

time period,which is referredto here as theConduction

interval (t , wherethe diode is on,and conducts.During this time,vO(t) = v I (t).



DC Power Supplies

There is a longer

time period, which isabout as long as theperiod of the inputwave, where thediode is off. Duringthis time, theresistor andcapacitor make anatural response

circuit, and

2

.t t

RC O p

v t V e

!



DC Power Supplies

The total waveform,

vO(t), is acomplicatedcombination of asinusoid, and adecayingexponential, evenfor a simple diodemodel. A moreaccurate diode

model makes for avery complicatedsolution.



DC Power Supplies, Approximations

The waveform,vO(t), is oftenapproximated asa simpler case,that is a dc

component equalto V p, and an accomponentwhose peak-to-peak value iscalled V r , theripple voltage.

C S




The dc

component of vO(t) is oftenapproximated bythe zero-to-peak

input sinusoid,perhapsdecreased by thethreshold voltageof the diode,and/or by half theripple voltage.

DC P S li A i i




The ac component

of vO(t) which wecalled the ripplevoltage, or V r , canalso beapproximated. The

following is aderivation of asimple estimate of the ripple voltage,Vr, which is defined

as the peak to peakvoltage on theoutput of the power supply.

Ri l V lt A i ti



Ripple Voltage Approximation

The ac component of vO(t) which we called the ripple

voltage, or V r , can also be approximated.1. Assume an ideal diode. (If the input amplitude

is much larger than V f , we make little error in

ignoring V f . If the input amplitude is not much

larger than V f , we can make a more accurate

estimate by using a better diode model.)

2. Assume that RC >> T , which we make possible

by picking C large.3. Since RC >> T , we can treat the exponentialdecay as a straight line.

Ri l V lt A i ti




The ripple voltage, or V r , can be approximated.

4. The charge gained by capacitor during charging,Qacq, is

Qacq = C V r .

5. The charge lost during discharge,

Qlost , isQlost = i L

(T-(t) } i L

T .

6. Here, we have assumed that(T-(t) } T ,

since it discharges for almost the entire period.

Ri l V lt A i ti





7. The voltage is a straight line, with a slope we willcall m, and a peak equal to V p. So, the current is

just that voltage divided by R,

. p

L

V mt i

R

!

Ri l V lt A i ti





8. Now, since RC >> T , we assume that m = 0. Then, withQacq = Qlost , and plugging in we get

where I L

is the load current, which is almost constantwith time, and so is expressed as a dc quantity.

.

Solving, we get

,

p

r L

p p L

r

V C V i T T

R

V T V I V

RC f RC fC

! !

! ! !

Ri l V lt A i ti





9. With a full wave rectifier, the frequency iseffectively doubled, so

which is a very handy little equation. It issurprisingly accurate, considering the number of approximations used to get it.

,2 2 2

p p Lr

V T V I V

RC f RC fC ! ! !



Các ng dng c bn ca diode

malvino ep 04 cac mach diode

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