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Drexel UniversityElectrical and Computer Engr. Dept.Electrical Engineering Laboratory III, ECEL 303E.L. Gerber

WAVEFORM GENERATORS (VCO)

Object

The object of this experiment is to learn some of the basics of electronicwaveform generation. Standard IC chips will be used to produce various waveformsas well as a frequency modulation circuit.

Introduction

One of the old standard integrated circuit voltage-controlled oscillator (VCO)devices is the LM566. It is capable of generating a fixed frequency square wave andtriangle wave. The frequency of these waves can also be varied via an external inputvoltage. This device has a maximum operating frequency of 1 MHz with a 10-to-1range of frequency variation with a change in modulating input voltage. It is a verybasic chip and only requires two external circuit elements to operate. Figure 1illustrates a basic block diagram of the LM566 VCO chip and the external timingcircuit.

Theory

Voltage-Controlled Oscillator Operation

Fig. 1. Voltage Controlled Oscillator Block Diagram

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The chip has a basic oscillator circuit built in but requires external timingelements RT and CT. The chip provides a constant DC current source which isreversed under internal chip command. The magnitude of the current, however, canbe controlled externally via the timing resistor RT. The fundamental principle of thesystem is the current flow in a capacitor, i(t) = Cdvdt. Since the current in thecapacitor is constant, hence, the voltage across it (pins 7-1) will be a positive ornegative ramp as seen in Fig. 2. Referring to Figure 1, the current source/sink circuitprovides a constant charging or discharging current to the external timing capacitorCT. The amount of current is controlled by the timing resistor, RT. Increasing thevalue of RT decreases the capacitor current. Control of this current is also possible bychanging the voltage across the resistor via the modulating input. The voltage at pin 6is normally maintained at the same voltage as pin 5. Thus, if the modulating voltageat pin 5 is increased, the voltage at pin 6 increases, resulting in less voltage across RTand, therefore, less charging current. All voltages must be positive.

Refer to Fig. 1 again, the voltage developed on capacitor CT is applied to theSchmitt trigger circuit U2 via the buffer amplifier U1. The output voltage swing on theSchmitt trigger goes from VCC to 0.5 VCC. Resistors Ra and Rb form a positive feedbackloop from the output of U2 to its non-inverting input. With equal dividing resistors Raand Rb, the non-inverting input swing is from 0.5VCC to 0.25VCC. If the voltage on thetiming capacitor CT exceeds 0.5VCC during charging, it will cause the Schmitt triggeroutput to go low (0.5 VCC). A low level on the output of U2 causes the current source tochange to a sink (discharging CT). When CT discharges to 0.25VCC, the output of the U2will swing high (VCC), causing the current sink to return to a source (charging CT).Since the source and sink currents are equal, it takes the same amount of time to chargeCT as it does to discharge this capacitor. This results in a triangular voltage waveform(Fig.2) on CT which is available as a buffered output at pin 4. A square wave appears atthe output of the Schmitt trigger and is inverted by inverter U3 for a second output atpin 3. If the current from the source/sink is increased, the charge/discharge time forthe capacitor is reduced and the output frequency is increased (shorter period). See Fig.2.

Determining VCO Output Frequency

The output frequency can then be changed by three methods:

1. Changing the value of CT.2. Changing the value of RT.3. Changing the voltage at the modulating input terminal.

We can determine the actual frequency of oscillation from the time it takes tocharge and discharge the capacitor. The basic equation for a capacitor is:

i(t) = C dVdt

or v(t) = 1C

i(t)dt (1)

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where dV is the voltage change on the capacitor during the time change dt. The totalvoltage on the capacitor changes from 0.25 VCC to 0.5 VCC because of the limits of thecontrolled current source/sink circuit.

Thus V = 0.5VCC - 0.25VCC = 0.25VCC (2)

From Eq.1, for a constant current,

Dt = 0.25 VCCCT

I (3)

The triangular waveform on the capacitor has a period T = 2t (equal chargingand discharging time). The frequency of oscillation is:

f =

1T

=1

2Dt(4)

Fig. 2. Voltage Waveform Across CT.

Substituting t from Eq. 3 and Eq. 4, the frequency of oscillation is:

f = I

0.5 VCC CT(5)

Ohms law gives,

I = VCC - V5

RT(6)

Where V5 is the modulating input voltage at pin 5, then:

f = 2(VCC - V5 )

CTRTVCC(7)

For best operation, the resistance of RT should be between 2 to 20 kW.

In normal operation, RT and CT are selected for the desired center operatingfrequency. With no modulation signal the output frequency is fixed by Eq. 5. Themodulating input voltage can be varied to give a variation in the output frequency i.e.,

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frequency modulation. The range of allowable variation of the modulating inputsignal is from 0.75 VCC to VCC, which yields an output frequency variation of about 10to 1. With no modulating input signal, the voltage at pin 5 should be biased at 7/8VCC. This allows us to simplify Eq. 7 to give the unmodulated frequency fo,

f0 =

2(VCC - 7 8 VCC)CTRTVCC

=1

4 CTRT(8)

If we wish to determine what input modulation voltage (V) is required to produce agiven output frequency deviation (f) we can calculate the original frequency is fo andthe new frequency f1 from Eq.7,

f = f1 - f0

=2(VCC - V5 + DV)

CTRTVCC-

2(VCC - V5 )CTRTVCC

=2DV

CTRTVCC(9)

Solving for V: DV = Df CTRTVCC

2

Substituting RTCT from Eq. 8:

DV = Df VCC8 f0

=Df VCC RTCT

2(10)

Circuit Applications (8038):

1-Oscillator.

The ICL 8038 chip operates on the same principle as the older LM 566. It has areversible constant current source and it requires external R-C elements to set theoscillation frequency, fo. Unlike the LM 566 the duty-cycle of the output signal can bechanged by varying one of the timing resistors, RA or RB. See Fig. 3. The period of theoutput signal, To (= 1/fo ) is given in equation 11, where CT is the timing capacitor. Fora symmetrical output signal, 50 % duty-cycle, RA = RB , then equation 11 becomesequation 12.

To =32

RACT 1 +RB

2RA - RB

(11)

fo = 1/3 RT CT (12)

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2-DC Sweep Input.

The output frequency can also be controlled directly by an external DC voltageconnected to pin 8, see Fig. 4. VDC applied across pins 8 and 6 will change the voltageacross the timing resistor, see equations 6 and 7, and therefore change the outputfrequency. With RA, RB and CT fixed then VDC will control the output frequency of theVCO. The modulation rate of the system is f/V.

3-AC Modulation.

The output frequency of the VCO can be varied as a function of time by an ACinput signal, Vin = V sin(2pifmt), see Fig. 5. This is known as frequency modulation orFM. This voltage must be connected to pin 8 through a coupling capacitor, CC. Theamplitude of the input signal V will control the frequency change f of the VCO.Whereas the frequency of the input signal, fm, controls the rate of change of the outputsignal. When fm is very low you can see the changing output signal on an scope.

PreLab

Design a symmetrical square and triangle wave generator, as shown in Fig. 3,to generate a signal of approximately 620 Hz. Design means calculate the values forCT and RA, with RB = 10 k.

Repeat the design for 1.2 kHz.

Laboratory

1 Oscillation:a) Build the 620-Hz symmetrical square and triangle generator (Fig. 3)

designed in the PreLab with RB = 10 k, and VCC = 22-V DC. Use one or twocapacitors to obtain the correct value of CT. Adjust RA (the pot) for 50%duty-cycle which can be measured directly on the scope using the "TimeDisplay". Measure and calculate the VCOs output frequency. Capture bothoutputs on the 'scope include VPP, Freq, and duty-cycle of the square waveonly. Also measure the pot resistance.Compare frequency measurement with your design.

b) Now vary the pot (RA) until the duty-cycle of the square wave only is 25%.Measure and record the resistance values and calculate the total period, TO ,and frequency from Eq. 11. Capture the square wave output only on the'scope include VPP, Freq, and duty-cycle.

c) Repeat b for 66% duty-cycle.

d) Repeat b for 75% duty-cycle.

e) Tabulate these results for a, b, c, and d as: RA, RB, %D.C., fo, and To.

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f) Reset RA for 50% duty-cycle. Replace the capacitor with a substitution boxand vary CT over five orders of magnitude starting at 0.1 nF. Measure thefrequency and the duty-cycle. Plot these results on a log-log graph.

2 DC Sweep:Connect the circuit shown in Fig. 4. Apply a second DC power supply between pins

8 and VCC with the polarity as shown. Start with VDC set to zero. Increase the DC sweepvoltage from 0V, in one-volt steps, until the output signal shuts down. Measure and recordthe sweep voltage, VDC, and the frequency. Plot these results. Determine the VCO'smodulation rate, k, from this plot.

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3 AC Modulation:Connect the circuit in Fig. 5. Let RC = 10 k and CC = 200 F. Apply a 1-VPP, 100 Hz

sine wave from the HP function generator to the AC input to ground. Measure the ACinput at pin 8 not the HP generator. The capacitor will drop a large voltage at lowfrequencies so you will need to increase the amplitude of the input as you lower thefrequency. Observe the VCOs square wave output on the scope. Reduce the inputfrequency to 10 Hz and observe the VCOs output on the scope. Reduce again to 1 Hz andobserve the VCOs output on the scope. Reduce again to 0.1 Hz, and increase the inputeach time. Has the output frequency range changed?

Repeat the last part with a ramp function at 10 Hz and 0.1 Hz.Repeat the last part with a square function at 10 Hz and 0.1 Hz.Now increase the input AC to 2 VPP at low frequency, observe and capture the

output. Has the output frequency range changed?

Parts List:

1 ICL 8038 circuit boardResistors; 10 k, 25 k potCapacitors; 200 F, substitution box.2 Twisted DC leads2 BNC/clip

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ICL8038