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

Background: What is dynamic impedance ? R= V I r d = dV d I dynamic resistance resistance

V-I Characteristics V I R V I V I V I V I slope rdrd dV dI R= VIVI diode transistor Tube Operating Point

iDiD vDvD slope di D /dv D di D r d = dvDdvD = VDVD iDiD = VdVd IdId IdId VdVd

I D dc diode current I d ac amplitude i d ac diode current i D total diode current ~ + - dc circuitac circuit IDID idid iDiD idid iDiD IDID IdId X d or D IEEE Standard Notation

V D dc diode voltage V d ac amplitude v d ac diode voltage v D total diode voltage vdvd vDvD VDVD VdVd ~ + - dc circuitac circuit VDVD vdvd vDvD

I D, V D I d, V d Small-Signal Condition

Dynamic Resistance Measurement iDiD vDvD IDID VDVD 2I d 2V d r d = V d I d

iDiD vDvD IDID VDVD Input Signal Too Large

iDiD vDvD IDID VDVD Input Signal Too Small noise

Measurement of r d looks simple. The problem is that v d in the millivolt range for forward bias. Thus, noise and stray pickup may cause trouble if you are not careful.

Example: Questions: Where does come from ? ! AB Oscilloscope R idid Stray magnetic flux How large is it ?

Questions: Where does come from ? Answer: 1. Current i ac in power lines on bench & drops from ceiling 2. fluorescent lights 3. AC machines r

Question: How large is ? rr I

AB Oscilloscope R 1 meter Area=1 m 2 Assume our experiment is about 2 meters from the power lines: r = 2 m 100 amp. peak60 HZ 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 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 smallStep 2: Twist wires together OSC.

Twisting wires does two things, 1- Holds wires together 2- voltages induced in adjacent sections cancel V1V2 1 2 V 1~ -V 2 So induced signals cancel

AB Oscilloscope R Keep track of grounded leads

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 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  V CRO = 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 1N VDC-0.5 W 2 Heathkit Resistance Substitution Boxes 1-kohm & 10 kohm Resistors Decade Capacitor Box

1- Dynamic Resistance in Forward Region ~ CH. 1CH A 0-20V R1R2 dc circuitac circuit IDID idid C v D, i D

~ + - A ~10.4V to ~10.8V R1 dc circuit IDID The values of R1 and the voltage source are selected to control the dc bias current I D. Suppose we want I D = 10 mA. Make the dc voltage across R1 = ~10 VDC. Assume V D = 0.7 V. V=10.7 volts & I D =10 mAR1=1000 Ohms ~0.4 to 0.8V ~10 VDC

~ + - A R2 ac circuit idid C R2 is selected so that ac current peak is ~10% of dc current. R1=1000 OhmsR2=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. I D =10 mA & i d =1 mA. To obtain other values of I D & i d 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 I D & i d. ~10.7V ~20Vpp 1 kHz R1

~ + - A R2 ac circuit idid C R2 is selected so that ac current peak is ~10% of dc current. R1=1000 OhmsR2=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 ~20Vpp 1 kHz

Selection of R2 The values of R2 and the function generator voltage amplitude V gen should be chosen to make the ac current amplitude i d 10% to 20 % of I D. The corresponding diode peak ac voltage V d will be 10% V to 20 % of nV T where V T = 25 mV at T = 290 K. (~20 C). Thus V d 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 r d

n=1 to 2 rtheoretical ?

1/T d(lnI D ) dV D slope gives n n?

Examples: I D = 0.2 mA n = 1 r d = 1X25mV /0.2mA = 125  n = 2 r d = 2X25mV/0.2mA = 250 

2- Simulation a- Simulate Part 1 of experiment b- Plot I(D1) and V(2) on separate graphs c- Calculate dynamic impedance of the diode ~ V R1R2 C D1

DYNAMIC IMPEDANCE I1 0 1 PWL(0.5M M M M.005 5M M M) R K D1 2 0 DIODE.MODEL DIODE D((RS=2 IS=2E-9 N=1.8) R K C U V1 4 0 SIN(0 5 1KHZ).TRAN.05M 10M 0.05M.PRINT TRAN V(2) i(D1).END ~ V R1R2 C D1 time [s] [mA]

3- Dynamic Resistance of Zener in the Breakdown Region ~ CH. 1CH A 0-20V R1R2 dc circuitac circuit IDID idid C v D, i D 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).

Assume Zener Diode Breakdown Voltage V Z = 12V The values of R1 and the dc voltage source are selected to control the dc bias current I D. Suppose we want I D = 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 .