HW1_Solution.pdf

8/11/2019 HW1_Solution.pdf

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V 1V:=I 1 mA=

IVDD 1volt( )

R:=

Assume D1ON and D2OFF

b) Since we have two diodes sharing the same anode so either both willbe OFF or the one with lower cathode voltage will be ON.

For D2-> ID2= I = +3 mA > 0 -> Our assumption of D2ON is correct

For D1-> VD1= -2 V < 0 -> Our assumption of D1OFF is correctCheck :

V 3volt:=I 3 mA=

I3volt VSS( )

R:=

Assume D1OFF and D2ON

a) Since we have two diodes sharing the same cathode so either both willbe OFF or the one with higher anode voltage will be ON.

Solution

VSS 3 volt:=R 2 k:=VDD 3 volt:=

Assume that the diodes are ideal and:

Given

Determine the currentIand the voltage Vfor the circuit shown below.

3.3

Diodes

Homework # 1

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The diode voltages of the diodes under consideration are measuredexperimentally at a current I and at a current I/10

Given

For the following diodes determine Is, n and the diode voltage at 10I

3.22

PIV 169.706 V=PIV VP:=

R RminRmin 3.394 10

3 =Rmin

VP

IDmax

:=

IDmax 50mA:=

VP 169.706 V=VP 2 Vrms:=

Solution

Vrms 120volt:=

Assume that the diode is ideal and the input is sinusoidal with:

Given

Determine the value of R to limit the diode current to 50mA for the

rectifier circuit shown below. Calculate the PIV?

3.11

For D2-> VD2= -2 V < 0 -> Our assumption of D2OFF is correct

For D1-> ID1= I =1 mA > 0 -> Our assumption of D1ON is correct

Check :

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To find the equation for the diode, we first must find ISand n.

VT 25mV:= nVD1 VD2( )

2.3 VT

log 10( ):= IS

I

e

VD1

n VT1

:=

IS 1.018 10 10

A=n 1.739=

The diode voltage at 10I for the diode is then

VD3 VD1 2.3 n VT log 10( )+:= VD3 0.8V=

(c) I=10.0 A, VD@I=VD1=800 mV, VD@I/10=VD2=700mV

I 10.0A:= VD1 800mV:= VD2 700mV:=



2.3 VT log 10( ):= IS

I

e

VD1

n VT1

:=

Solution

(a) I=10.0 mA, VD@I=VD1=700 mV, VD@I/10=VD2=600mV

I 10.0mA:= VD1 700mV:= VD2 600mV:=



2.3 VT log 10( ):= IS

I

e

VD1

n VT1

:=

IS 1.018 10 9

A=n 1.739=


VD3 VD1 2.3 n VT log 10( )+:= VD3 0.8V=

(b) I=1.0 mA, VD@I=VD1=700 mV, VD@I/10=VD2=600mV

I 1.0mA:= VD1 700mV:= VD2 600mV:=

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VD3 0.76 V=VD3 VD1 2.3 n VT log 10( )+:=


n 1.043= IS 22.204 10

12 A=

ISI

e

VD1

n VT1

:=nVD1 VD2( )

2.3 VT log 10( ):=VT 25mV:=


VD2 640mV:=VD1 700mV:=I 10.0A:=

(e) I=10.0 A, VD@I=VD1=700 mV, VD@I/10=VD2=640mV

VD3 0.82 V=VD3 VD1 2.3 n VT log 10( )+:=


n 2.087= IS 1.49 10

9 A=

IS

I

e

VD1

n VT1

:=nVD1 VD2( )

2.3 VT log 10( ):=V

T 25mV:=


VD2 580mV:=VD1 700mV:=I 1.0mA:=

(d) I=1.0 mA, VD@I=VD1=700 mV, VD@I/10=VD2=580mV

VD3 0.9V=VD3 VD1 2.3 n VT log 10( )+:=


IS 1.021 10 7

A=n 1.739=

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ID1

I

D2

Find ID1' ID2',( ):=

ID2'

ID1'

e

V

n VT=

ID1' ID2'+ 10mA=

Given

ID2' 5 mA:=ID1' 5mA:=Guessvalues:

From KVL -> V=VD2-VD1From KCL -> ID1+ID2=10mA

IS 1.018 10 9

A=n 1.739=

IS

I1

e

V1

n VT1

:=nV2 V1( )

2.3 VT logI2

I1

:=VT 25mV:=


V 80mV:=

V2 800mV:=I2 100.0mA:=V1 700mV:=I1 10.0mA:=

Solution

The two diodes are identical conducting 10mA at 0.7V and 100mA at 0.8V

Given

Find the value of R to have V=50 mV for the circuit shown below

3.26

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IS

ID

e

VD

n VT

1

:=n 1:=VT 25mV:=

To find the equation for the diode, we first must find IS

rD 11.616 = VD0 0.641 V=

VD0 VD2 rDID2:=rDVD2 VD1( )ID2 ID1( )

:=

VD2 0.758 V=ID2 10mA=

VD1 0.643 V=ID1 0.1mA=

VD2 VD 2.3 n VT logID2

ID

+:=VD1 VD 2.3 n VT log

ID1

ID

+:=

ID2 ID10:=ID1ID

10:=

Assume the two points of interest 1 and 2

n 1:=VD 700mV:=ID 1.0mA:=

(a) ID=1.0 mA, VD=700 mV, and n=1

Solution

The diode voltage of the diode under consideration is measuredexperimentally at a current IDand n is given

Given

For the following diodes find rDand VD0for which the straight line of thebattery plus resistor model intersect the real characteristics at 0.1IDand10IDwhere IDis the given current in each case.

3.40

R 58.372 =R

V

ID1:=

ID2 8.629 mA=ID1 1.371 mA=So the solution is

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ID

+:=VD1 VD 2.3 n VT log

ID1

ID

+:=

ID2 ID10:=ID1ID

10:=


n 1:=VD 700mV:=ID 1.0A:=

(b) ID

=1.0 A, VD

=700 mV, and n=1

0 0.2 0.4 0.6 0.8 10

0.002

0.004

0.006

0.008

0.01

(Volts)

(A

mps)

IDVD( )

ImodelVD( )

VD

VD 0 volt .01 volt, 1volt..:=

Both equations are then plotted.

ImodelVD( )VD VD0( ) VD VD0( )

rD

:=

The equation for the diode model is then (is the unit step function)

IDVD( ) IS e

VD

n VT1

:=

The equation for the diode is then

IS 6.914 10 13

mA=

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Both equations are then p lotted.


rD

:=


IDVD( ) IS e

VD

n VT1

:=


IS 6.914 10 10

mA=

IS

ID

e

VD

n VT1

:=n 1:=VT 25mV:=


rD 0.012= VD0 0.641 V=


:=

VD2 0.758 V=ID2 10 A=

VD1 0.643 V=ID1 100 mA=

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rD 1.162 103

= VD0 0.641 V=


:=

VD2

0.758 V=ID2

0.1mA=

VD1 0.643 V=ID1 1 A=


ID

+:=


ID

+:=

I

D2

I

D

10:=ID1

ID

10

:=


n 1:=VD 700mV:=ID 10.0A:=

(c) ID=10.0 A, VD=700 mV, and n=1

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

(Volts)

(Amps)IDVD( )

ImodelVD( )

VD

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VT 25mV:= n 1:= ISID

e

VD

n VT1

:=

IS 6.914 10 15

mA=


IDVD( ) IS e

VD

n VT1

:=



rD

:=

Both equations are then p lotted.


0 0.2 0.4 0.6 0.8 10

2 .10 5

4 .10 5

6 .10 5

8 .10 5

1 .10 4

(Volts)

(Amps)

IDVD( )

ImodelVD( )

VD

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I

VS VS0

+ VZ0

R rz+:=

VO VZ0 rzI+:= VO 7.554 V=

(d) For the zener to be at the edge of the breakdown region while VSis 10% lower,

IZK 0.5mA= VZK VZ0 IZKrz+:= VZK 7.155 V=

At this point, the lowest current supplied through Ris

IR

VS VS0

VZK

R:= IR 7.38 mA=

Thus the load current is

ILmin IR IZK:= ILmin 6.88 mA=

and the corresponding value of the load resistance is

RLmin

VZK

ILmin

:= RLmin 1.04 k=

(a) With the load connected, the value of R may be calculated as follows:

VO 7.5volt:= ILVO

RL

:=IL 6.25 mA=

Assume we will design for I 10mA:=IZ I IL:= IZ 3.75 mA= > IZK which will guarantee that our

diode in zener region

RVS VO( )

I:= R 250 =

Please note that you can design for IZ=IZKwhich will give the maximum resistance(better for line regulation). However, if VSis lowered your zener will be out of zenerregion and it will not act as a regulator anymore.

(b) For a 10% ( 1V) change in VS, the change in output voltage can be found from

rz1=rz||RL rz1

rzRL( )rz RL+

:=

VO VSrz1

R rz1+:= VO

104.803

104.803

mV=

(c) When the load is removed, and VSis 10% high

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The equation is plotted.below

VOVI( ) VI VD0+( ) 1 VI VD0+( )( ):=

The equation for output voltage is then (is the unit step function)

At first you have to find the regions at which the diode is on or off:If vs is less than -0.7V the diode will be off -> vo= 0

If vs is more than -0.7V the diode will be on -> vo= vs + 0.7V

(a) Transfer characteristics:

The rectifier circuit withthe diode replaced by itsmodel

Solution

R 1.5k :=VP 15volt:=VD0 0.7volt:=

Use the CVD model for the diode with VDO=0.7V and assume the inputis sinusoidal with maximum VP

Given

For the rectifier circuit shown below:

(a) Sketch the transfer characteristics

(b) Sketch the waveform of VO

(c) Find the average voltage of VO

(d) Find the peak current in the diode

(e) Find the PIV of the diode

3.72

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3 2 1 0 13

2

1

0

(Volts)

(Volts)

VOVI( )

VI

(b) Output Waveform

VIt( ) VPsint( ):= VOt( ) VIt( ) VD0+( ) 1 VIt( ) VD0+( )( ):=

10 5 0 5 10

10

0

10

(rad)

(Volts)VO t( )

VIt( )

t

To see the effect of VD0on the output waveform compared to the ideal diodecase let us consider the waveform if the peak input voltage is 2V only.

VI1t( ) 2 sint( ):= VO1t( ) VI1t( ) VD0+( ) 1 VI1t( ) VD0+( )( ):=

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10 5 0 5 102

1

0

1

2

(rad)

(Volts)

VO1 t( )

VI1 t( )

t

(c) Average value of VO

VOav0

2

tVOt( )

d

2:= VOav 4.43 V=

Please note to calculate the average voltage by hand do the following:(a) Assume a period from 0 to 2

(b) Calculate the angle at which the diode conduct =sin-1(VDO/VP)

(c) The diode will conduct from t= +to t= 2-(d) Integrate VI(t) from t= +to t= 2- and divide be 2

(d) Peak diode current (IDP)

IDP

VP VD0( )R

:=IDP 9.533 mA=

(d) Peak inverse voltage (PIV)

This happens when VO= 0V (off diode) and VI(voltage at cathode) ismaximum

PIV VP:= PIV 15V=

3.95

Sketch the transfer characteristics if the inputs of the following two circuitsare tied together and the outputs are tied together

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vOvI( )2.7 1 2 vI( )( )

1 2.7

vI

m

+

1

m

:=m 10:=

vO1vI( ) 2.7 2.7 vI+( ) 2.7 vI+( )+ vI 2.7+( ) vI 2.7( )+:=

We can redraw the circuit as shown,Now we can define three major areaswhen the input voltage is +ve.(a) If 0 D2is off and D1isoff -> vO=vI

(b) If 3.2 D2is off and D1is fullyconduct -> vO=2.7V

(c) If 2.5 D2is off and D1is on -> vO changes smoothly from2.5V to 2.7VSame arguments are true for the -veinput voltage except that the signchange and the role of D1and D2

Solution

IDfull 1mA:=atVDfull 0.7volt:=VDon 0.5volt:=

R 1 k:=VDD 3 volt:= VSS 3 volt:=

Assume that the diodes conducting at 0.5V and fully conduct at 0.7Vwith I=1mA

Given

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4 2 0 2 44

2

0

2

4

(Volts)

(Volts)

vO1vI( )

vOvI( )

vI

The solid red curve is representing the transfer characteristics if we assume CVDwith VD=0.7V while the dotted blue one represent the smoothing characteristics asin our problem.

3.104

Calculate the average (DC) value of the output from the shown circuit

Given

Assume that the diode is ideal

VIrms 10volt:=

Solution

VIP VIrms 2:= VOav VIP:=

VOav 14.142 V=

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Documents

HW1_Solution.pdf