Farzana R. ZakiCSE 177/ EEE 1771 Lecture 6 & 7 Diode Models (to be continued) Zener diode Block diagram of DC power supply

Farzana R. Zaki CSE 177/ EEE 177 1

Lecture 6 & 7

•Diode Models (to be continued)•Zener diode•Block diagram of DC power supply


Small-Signal Model • The diode is biased to

operate (in this case) at 0.7V.

• The AC response can be modeled as a resistance equal to the inverse slope of the tangent IF it is small enough (small-signal model)

• This concept of restricting an AC signal to the short, linear region around some DC bias point is used throughout this course.


Graphical Representation of Small Signal model


Small-Signal Model• Small-Signal Approximation

– Valid for signals whose amplitudes are smaller than about 10mV for n=2 and 5mV for n=1

/

/

/

//

/

2 3 4

1 ..., 1! 2! 3! 4!

if 1,

1

D T

D T

D d T

d TD T

d T

V nVD S

D D d

v nVD S

V v nVD S

v nVV nVD S

v nVD D

x

d

T

dD D

T

I I e

v t V v t

i t I e

i t I e

i t I e e

i t I e

x x x xe

v

nV

vi t I

nV

DD D d

T

D D d

Dd d

T

Ii t I v

nV

i t I i

Ii v

nV

is dependent

on the Bias Cur

1Note that

r

is

entd

Td

D

dD

nVr

r

I

rI

Small Signal Approximation

1 dD D

T

vi t I

nV


Problem 1• Consider the same circuit for the case in which R=10k. The power

supply has a DC value of 10V on which is superimposed a 60-Hz sinusoid of 1V peak amplitude. Calculate both the dc voltage of the diode and the amplitude of the sine wave that appears across it. Assume the diode to have a 0.7V drop at 1mA and an n=2.

( )

10 0.70.93mA

102 25

53.80.93

0.05381 5.35mV

10 0.0538

D

Td

D

dsd peak

d

I

nVr

I

rv V

R r


Problem 2• Diode Regulator

– Design the following circuit to provide an output voltage of 2.4V. Assume the diodes have a current of 1 mA at a voltage of .7 V and that its voltage drop changes by .1 V for every decade of change in current.

2.4V. Each diode must therefore drop .8V

The current must be 1 decade above 1mA in order

for the diode to change from .7 to .8V drops.

Thus, the current is 10mA, and the resistance must be

10 2.410m

OV

R

A, 760R


Problem 3• Voltage Regulation

– Consider the following circuit. What is the percentage change in the regulated voltage caused by (a) a 10% change in the power-supply voltage and (b) connection of a 1k load resistance?

For a 10% power supply change,

10 2.17.9mA

12 25

6.37.9

3 18.9

0.01892 2 37.1mV

0.0189 1so the change in the diode is 18.5 mV

or 6.2 mV per diode. (Remember

that we said that the

Td

D

d

O

I

nVr

I

r r

rv

r R

small-signal is

valid for Amplitudes less than 10mV

for n=2, and 5mV for n=1



Operation in the reverse bias region-Zener diodes• The very steep i-v curve that the diode exhibits in the breakdown

region and the almost constant voltage drop that indicates suggest that diodes operating in the breakdown region can be used in the design of voltage regulators.

• Normal Si diode cant operate in breakdown region. So, special diodes are manufactured to operate specially in the breakdown region. Such diodes are called breakdown diodes or zener diodes.

i-v curve for normal Si diode


i-v characteristic curve of zener diode


Zener diode• For currents greater than knee current Izk , the i-v curve

is almost constant.• The manufacturer usually specifies the voltage drop

across the zener diode VZ at a specifies test current IZT. This point is labelled as Q.

• As the current through zener diode deviates from IZT, the voltage across it will change slightly.

For ΔI current change, voltage change, ΔV= ΔI×rZ

Where rz = incremental resistance of zener diode at operating point Q = dynamic resistance of the zener• rz is in the range of few ohms to a few tens of ohms.• Vz is in the range of few volts to a few hundred volts.


• In addition to specifying Vz (at a particular IzT), rz,Vzk, the manufactures also satisfies the maximum power that the device can safely dissipate.

• Say, a 0.5W, 6.8V zener diode can operate safely at currents up to a maximum of about 70mA.

• VZ denotes the point at which the straight line of slope 1/rz intersects the voltage axis. VZ = VZ0 + IZrZ


Problem 4(a)

a) Find with no load and

with at its nominal value.

OV

V

0

0 0

0

0

6.8 20 5 , 6.7

10 6.76.346

0.5 0.02

6.7 6.346 0.02 6.827

Z Z z Z

Z Z

Zz

z

O Z Z z

V V r I

V mA V V

V VI mA

R r

V V I r V

• A 6.8-V Zener diode in the circuit below is specified to have Vz=6.8V at Iz=5mA, rz=20 ohms, and Izk=0.2mA. The supply voltage is nominally 10V but can vary by +/- 1V.


Problem 4b

201 38.5

500 2

Line Regulation=38.5mV/

0

V

zO

z

rV V mV

R r

b) Find the change in resulting from the 1V change in .

Note that / usually expressed in mV/V, is known

line regulation as .

O

V

V V


Problem 4c

c) Find the change in resulting from connecting a load

resistance that draws a current of 1 mA, and hence

load regulation find the in mV/mA.

O

L L

O L

V

R I

V I

The load draws a current of 1mA

from the diodes .

Lo

..

ad

20 1 20

Regulation=-20mV/mAO z ZV r I mV


Problem 4dd) Find the change in when 2O LV R k


Problem 4e

of .5k would draw a load current of

6.8 / 0.5 13.6mA. This is not possible as

the current though R is only 6.4mA. Therefore,

the Zener must be cut off.

0.510 5V

0.5 0.5

Therefore, the zener i

L

LO

L

R

RV V

R R

s not in breakdown.

e) Find the change in when 0.5O LV R k


Problem 4f

L

To be at breakdown, 0.2mA

and 6.7V. In this case, the

worst-case (lowest) current through R is

9 6.74.6mA. The load current is therefore

0.54.6 0.2 4.4mA. R is therefore

6.71.523k

4.4

Z ZK

Z ZK

L

I I

V V

R

f) What is the minimum value of for which the diode

still operates in the breakdown region?LR


Diode Rectifiers power

Block diagram of a DC power supply


Components of DC power supply

• The power supply is fed from 120V rms 60Hz ac line, and it delivers a dc voltage (usually in the range of 5-20V) to an electronic circuit represented by the load block.

• Power Transformer: consists of 2 separate coils wound around an iron core that magnetically couples the 2 windings.

• Primary windings have N1 turns and is connected to 120V ac power supply & secondary windings have N2 turns and is connected to the circuit of dc power supply.


* ac voltage, vs = 120×(N2/N1) V(r.m.s.) develops between two terminals of secondary windings.

* By choosing appropriate turns ratio ( N2/N1) for the transformer, particular dc voltage output can be supplied.

• [For 8V r.m.s. in secondary winding may be appropriate for a dc output of 5V. For this, 15:1 turns ratio is required.


Components of DC power supply (cont.)

• Diode Rectifier & filter: Diode rectifier converts input sine wave (vs) to a unipolar pulsating output waveform. Pulsating nature makes it unsuitable as a dc source for electronic circuits, hence filter is required.

• Voltage regulator: The output of filter contains some ripple. To reduce ripple and to stabilize the magnitude of dc output supply, voltage regulator is employed. (Zener shunt regulator)

Documents

Farzana R. ZakiCSE 177/ EEE 1771 Lecture 6 & 7 Diode Models (to be continued) Zener diode Block diagram of DC power supply