experiment no. 7
DESCRIPTION
I. 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 - PowerPoint PPT PresentationTRANSCRIPT
I
V
R dVdI
dynamic
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
noiseMeasurement 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
ddt
i Rddt
d
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
H IrIrA
V ddt
ArdIdt
V ArdIdt
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 mI I t tac sin( ) sin( ) 100 2 60
100 amp. peak 60 HZ
VAr
dIdt
V t
V t
2 10
2 1012
100 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 V ArdIdt
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 rdexp
rdn=1
mA V mV mV A
0.2 ~10 47k 200k 6.1m 8.93 4.5 136 1250.51.02.05.0102040
Fall 2000 Data Table For Forward rd
ID VDD R1 R2 Vdpp VR2pp Id rdexp
rdn=1
mA V mV V A
0.2 ~10 47k 200k 6.1m 8.93 4.5 136 1250.51.02.05.0102040
mVVqkT
nkTqI
nkTqeI
dVdI
eII
dVdI
r
DnkTqV
SD
D
nkTqV
SD
D
D
d
D
D
25 025.0
1
DDd
ImVxn
IqnkTr 25 1
n=1 to 2
rtheoretical ?
ln( ) ln( )
(ln )
I I qVnkT
d IdV
qnkT
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 rdexp
mA V mV V mA
0.20.51.02.05.010 16 470 10k 4.8 8.6 0.86 5.62040
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.