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I
dV
Rdynamic 
dI
V
Experiment No. 7
EE 312
Basic 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 ?
V
R=
I
rd =
dV
dI
resistance
dynamic resistance
V-I Characteristics
diode I
R I
V
I
V
V
transistor
I
Tube
I
Operating Point
slope
dV
dI
V
V
R= V
I
rd
iD
slope di /dv
D
D
iD
Id
VD
Vd
vD
rd =
dvD
diD
=
 VD = Vd
 iD Id
IEEE Standard Notation
+
-
ID
id
dc circuit
ac circuit
Id
~
 sin t
iD
Id
ID
iD total diode current
id
iD
X d or D
ID
dc diode current
id ac diode current
Id ac amplitude
VD
vD
+
vd
dc circuit
ac circuit
-
Vd  sin t
Vd
vd
vD total diode voltage
VD
VD
~
vD
dc diode voltage
vd ac diode voltage
Vd ac amplitude
Small-Signal Condition
Id, Vd
ID, VD
Id  ID
and
Vd VD
Dynamic Resistance Measurement
iD
Vd
rd =
Id
ID
2Id
id  Id  sin(t )
2Vd
VD
vD
vd  Vd  sin(t  )
Input Signal Too Large
iD
ID
VD
vD
vd  Vd  sin(t  )
Input Signal Too Small
iD
ID
VD
noise
vD
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:
R
id
!
Oscilloscope
A
d
v
dt
d
id  R 
dt

B
Stray
magnetic
flux
Questions: Where does  come from ?
How large is it ?
Questions: Where does  come from ?
Answer:
1. Current iac in
power lines on
bench & drops
from ceiling
2. fluorescent
lights
3. AC machines
iac  Iac  sin(t )

r
Question: How large is  ?
  B A    H  A
 H  dl  I
I
H
2r
I
 
A
2r
d A dI
V 


dt 2r dt
A dI
7
V  2  10  
r dt
I
, B, H
r
Assume our experiment is about 2 meters from the power lines: r = 2 m
I  Iac  sin(t )  100  sin(2  60  t )
100 amp. peak
60 HZ
A dI
V  2  10  
r dt
7 1
V  2  10   100  cos(2    60  t )
2
V  0.00377  cos(120 )t
7
Peak value is 3.77 mV and
this may be comparable to
signal amplitudes being
measured!
R
Oscilloscope
2
Area=1 m
A B
Must be concerned about  in all parts of circuit.

+
-

~

How is this problem avoided?
remember
V  2  10
7
A dI
 
r dt
We have control over A.
We can‘t do much about r or I.
So, we must minimize A.
Step 1: Make the area small
OSC.
Step 2: Twist wires together
OSC.
Twisting wires does two things,
1- Holds wires together
2- voltages induced in adjacent sections cancel
V1
V2
1
2
V1~ -V2
So induced signals cancel
Keep track of grounded leads
R
Oscilloscope
A B

Single Point Grounding
Use 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 = jCV 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: CRO
Z = 1 Meg
VCRO = 377x1pFx120Vx1Meg
= 45 mV(rms) = 130 mVpp
Procedures:
• I-
• II-
Measure dynamic resistance of
a Zener diode in the forward
bias region.
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
R1
CH. 2
R2
C
A
0-20V
+
-
ID
id
dc circuit
ac circuit
vD, iD
~
Vg  sin t
~10 VDC
~10.4V
to
~10.8V
+
-
R1
A
ID
~0.4 to 0.8V
~
dc circuit
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
R1
~10.7V
A
+
-
R2
id
ac circuit
C
~
~20Vpp
1 kHz
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 .
A
~10.7V
+
-
R2
id
ac circuit
C
~
~20Vpp
1 kHz
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
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
0.5
1.0
2.0
5.0
10
20
40
Vdpp VR2pp Id
mV mV


~10 47k 200k 6.1m 8.93
mA V
0.2
R2
rd
exp
A

4.5 136
rd
n=1

125
Fall 2000 Data Table For Forward rd
ID
VDD R1
0.5
1.0
2.0
5.0
10
20
40
Vdpp VR2pp Id
mV V


~10 47k 200k 6.1m 8.93
mA V
0.2
R2
rd
exp
A

4.5 136
rd
n=1

125
rtheoretical
?
1 dID

rd dVD
ID  IS  e
qVD
nkT
qVD
dID
q
q
nkT
 IS  e 
 ID 
dVD
nkT
nkT
kT
 0.025 V  25 mV
q
nkT 1 n x 25mV
rd 
 
q
ID
ID
n=1 to 2
n?
qVD
ln( ID)  ln( Is ) 
nkT
d (ln ID)
q

dVD
nkT
d(lnID)
dVD
1/T
Examples:
ID = 0.2 mA
n=1
rd = 1X25mV/0.2mA = 125 
n=2
rd = 2X25mV/0.2mA = 250 
2- Simulation
1
0-20V
+
-
R1
2
R2
D1
3
4
C
~
0
a- Simulate Part 1 of experiment
b- Plot I(D1) and V(2) on separate graphs
c- Calculate dynamic impedance of the diode
1
0-20V
+
-
R1
2
R2
D1
3
4
C
~
0
[mA]
DYNAMIC IMPEDANCE
I1 0 1 PWL(0 .5M .00249 .5M .0025 1M .00499 1M .005 5M .00749 5M .0075
10M)
R1 1 2 1.5K
time [s]
D1 2 0 DIODE
.MODEL DIODE D((RS=2 IS=2E-9 N=1.8)
R2 3 2 15K
C1 4 3 .22U
V1 4 0 SIN(0 5 1KHZ)
.TRAN .05M 10M 0 .05M
.PRINT TRAN V(2) i(D1)
.END
3- Dynamic Resistance of Zener
in the Breakdown Region
CH. 1
R1
CH. 2
R2
C
A
0-20V
+
-
ID
id
dc circuit
ac circuit
vD, iD
~
Vg  sin t
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
mA V

R2
Vdpp VR2pp Id
rd
exp
mA 

mV V
0.2
0.5
1.0
2.0
5.0
10
20
40
16
470 10k 4.8
8.6
0.86 5.6
Assume Zener Diode Breakdown Voltage
VZ = 12V
The 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.