I V Experiment No. 7 EE 312 Basic Electronics Instrumentation Laboratory Wednesday, October 11, 2000

I

V

RdV

dIdynamic

Experiment No. 7

EE 312Basic Electronics Instrumentation Laboratory

Wednesday, October 11, 2000

Objectives:• Measure dynamic impedance of a

forward-biased diode & Zener diode

• Learn about small-signal techniques

• Learn about interference reduction through the use of proper grounding and twisted-pair techniques

Background:

What is dynamic impedance ?

R=V I

rd =dVd I

dynamic resistance

resistance

V-I Characteristics

V

IR

V

I

V

I

V

I

V

I

slope rd

dVdI

R= VI

diode

transistor

TubeOperating Point

iD

vD

slope diD /dvD

iD

VD

diD rd =dvD

=VD

iD

=Vd

Id

Id

Vd

ID dc diode current

Id ac amplitude

id ac diode current

iD total diode current

I td sin~

+

-dc circuit ac circuit

ID id

iD

id

iDID

Id

X d or D

IEEE Standard Notation

VD dc diode voltage

Vd ac amplitude

vd ac diode voltage

vD total diode voltagevd

vDVD

Vd

V td sin~

+

-dc circuit ac circuit

VD vdvD

ID, VD

Id, Vd

Small-Signal Condition

I I

V V

d D

and

d D

i I td d sin( )

Dynamic Resistance MeasurementiD

vD

ID

VD

2Id

2Vd

v V td d sin( )

rd = Vd

I d

iD

vD

ID

VD

v V td d sin( )

Input Signal Too Large

iD

vD

ID

VD

Input Signal Too Small

noise

noise

Measurement of rd looks simple.

The problem is that vd in the millivolt range for forward bias.

Thus, noise and stray pickup may cause trouble if you are not careful.

Example:

v

d

dt

i Rd

dtd

Questions: Where does come from ?

!A B

OscilloscopeR id

Stray magnetic flux

How large is it ?

Questions: Where does come from ?

i I tac ac sin( )Answer: 1. Current iac in power lines on bench & drops from ceiling 2. fluorescent lights3. AC machines

r

Question: How large is ?

r

B A H A

H dl I

HI

rI

rA

Vd

dt

A

r

dI

dt

VA

r

dI

dt

2

2

2

2 10 7

r

, ,B H

I

A B

OscilloscopeR

1 meter

Area=1 m2

Assume our experiment is about 2 meters from the power lines: r = 2 m

I I t tac sin( ) sin( ) 100 2 60100 amp. peak 60 HZ

VA

r

dI

dt

V t

V t

2 10

2 101

2100 2 60

0 00377 120

7

7 cos( )

. cos( )

Peak value is 3.77 mV and this may be comparable to signal amplitudes being measured!

~+

-

Must be concerned about in all parts of circuit.

How is this problem avoided?

remember VA

r

dI

dt 2 10 7

We have control over A. We can‘t do much about r or I.So, we must minimize A.

OSC.

Step 1: Make the area small Step 2: Twist wires together

OSC.

Twisting wires does two things,

1- Holds wires together2- voltages induced in adjacent sections cancel

V1 V2

1 2

V1~ -V2So induced signals cancel

A B

OscilloscopeR

Keep track of grounded leads

Single Point GroundingUse Only One Ground Connection Such As

CRO ground

Can only one ground connection be realized? e. g. CRO ground. Not with BNC’s because the each outer connector is another ground.

Capacitive Coupling

1. Assume 1 pF between your circuit and 120 VAC power lines.

2. 60-Hz current I = jCV where = 377 rad/s at f = 60 Hz, C = 1 pF, and V = 120 VAC(rms)

3. The voltage produced by I = ZxI where Z is the impedance I flows through.

4. Example: CROZ = 1 MegVCRO = 377x1pFx120Vx1Meg

= 45 mV(rms) = 130 mVpp

Procedures:

• I- Measure dynamic resistance of a Zener diode in the forward

bias region.

• II- Simulation for Part I. • (In Bell 242)

• III- Measure dynamic resistance in the Zener breakdown region.

Components:

• Zener Diode 1N4742-12VDC-0.5 W

• 2 Heathkit Resistance Substitution Boxes

• 1-kohm & 10 kohm Resistors

• Decade Capacitor Box

1- Dynamic Resistance in Forward Region

~

CH. 1 CH. 2

+

-

A

0-20V

R1 R2

dc circuit ac circuit

ID id

C

vD, iD V tgsin

~+-

A~10.4V to ~10.8V

R1

dc circuit

ID

The values of R1 and the voltage source are selected to control the dc bias current ID. Suppose we want ID = 10 mA. Make the dc voltage across R1 = ~10 VDC. Assume VD = 0.7 V.

V=10.7 volts & ID =10 mA R1=1000 Ohms

~0.4 to 0.8V

~10 VDC

~+-

AR2

ac circuitid

C

R2 is selected so that ac current peak is ~10% of dc current.

R1=1000 Ohms R2=10,000

By setting the dc power supply voltage to ~10.7 VDC & the FG amplitude to ~20 Vpp and R2 to ~10R1, the ac current peak is ~10% of dc current. I. E. ID =10 mA & id =1 mA .

To obtain other values of ID & id change both R1 & R2 with R2/R1 = ~10. The dc & ac voltage levels in the circuit change very little as R1 & R2 are changed to change the currents ID & id .

~10.7V ~20Vpp1 kHz

R1

~+-

AR2

ac circuitid

C

R2 is selected so that ac current peak is ~10% of dc current.

R1=1000 Ohms R2=10,000

C blocks dc current in the ac circuit & C should be large enough so that capacitance reactance is small compared with R2

Note that R1 must be >> diode dynamic resistance so that most of the ac current goes through the diode & not the dc circuit

~10.7V ~20Vpp1 kHz

Selection of R2• The values of R2 and the function generator

voltage amplitude Vgen should be chosen to make the ac current amplitude id 10% to 20 % of ID. The corresponding diode peak ac voltage Vd will be 10% V to 20 % of nVT where VT = 25 mV at T = 290 K. (~20 C). Thus Vd will be 2.5 to 5 mV for n = 1 and the peak-to-peak diode ac voltage will be 5 to 10 mV.

Fall 2000 Data Table For Forward rd

ID VDD R1 R2 Vdpp VR2pp Id rd

exprd

n=1mA V mV mV A

0.2 ~10 47k 200k 6.1m 8.93 4.5 136 125

0.5

1.0

2.0

5.0

10

20

40

Fall 2000 Data Table For Forward rd

ID VDD R1 R2 Vdpp VR2pp Id rd

exprd

n=1mA V mV V A

0.2 ~10 47k 200k 6.1m 8.93 4.5 136 125

0.5

1.0

2.0

5.0

10

20

40

mVVq

kT

nkT

qI

nkT

qeI

dV

dI

eII

dV

dI

r

DnkT

qV

SD

D

nkT

qV

SD

D

D

d

D

D

25 025.0

1

DDd

I

mVxn

Iq

nkTr

25 1 n=1 to 2

rtheoretical ?

ln( ) ln( )

(ln )

I IqV

nkTd I

dV

q

nkT

D sD

D

D

1/T

d(lnID)dVD

slopegives n

n?

Examples: ID = 0.2 mA

n = 1rd = 1X25mV/0.2mA = 125

n = 2rd = 2X25mV/0.2mA = 250

2- Simulation

a- Simulate Part 1 of experiment b- Plot I(D1) and V(2) on separate graphsc- Calculate dynamic impedance of the diode

~+-

0-20V

R1 R2 C

0

1 2 3 4

D1

DYNAMIC IMPEDANCEI1 0 1 PWL(0 .5M .00249 .5M .0025 1M .00499 1M .005 5M .00749 5M .0075 10M)R1 1 2 1.5KD1 2 0 DIODE.MODEL DIODE D((RS=2 IS=2E-9 N=1.8)R2 3 2 15KC1 4 3 .22UV1 4 0 SIN(0 5 1KHZ).TRAN .05M 10M 0 .05M.PRINT TRAN V(2) i(D1).END

~+-

0-20V

R1 R2 C

0

1 2 3 4

D1

time [s]

[mA]

3- Dynamic Resistance of Zener in the Breakdown Region

~

CH. 1 CH. 2

+

-

A

0-20V

R1 R2

dc circuit ac circuit

ID id

C

vD, iD V tgsin

Choose values of dc bias current so that the dc power dissipation in the diode is less than 1/2 of its max rated power dissipation (1/2 Watt).

ID VDD R1 R2 Vdpp VR2pp Id rd

expmA V mV V mA

0.2

0.5

1.0

2.0

5.0

10 16 470 10k 4.8 8.6 0.86 5.6

20

40

Assume Zener Diode Breakdown Voltage VZ = 12VThe values of R1 and the dc voltage source are selected to control the dc bias current ID. Suppose we want ID = 10 mA. Make the dc voltage across R1 = ~5 VDC. Then R1 = ~5 VDC/10mA = 0.5 k. Use the closest value which is 470 . The FG peak voltage is set at 10 V. The value of R2 is selected so that the peak ac current = 10% of the dc current = 0.1 X 10 mA. Thus R2 = ~10V/1mA= 10 k.