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Student Manual FACET by Lab-Volt 57 Analog Communications Introduction to Analog Communications Exercise 3: Balanced Modulator EXERCISE OBJECTIVE When you have completed this exercise on the balanced modulator, you will be able to describe the circuit block, explain the output signal’s relationship to the two input signals, and describe the operation and applications. You will use an oscilloscope to make measurements. DISCUSSION On the ANALOG COMMUNICATIONS circuit board, balanced modulators function as the following devices: an AM modulator, a DSB suppressed modulator, mixers, an SSB product detector, and a phase detector. The balanced modulator is an IC that is designated as number 1496. There are six 1496 ICs on the circuit board.

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Student Manual

FACET by Lab-Volt 57

Analog Communications Introduction to Analog Communications

Exercise 3: Balanced Modulator

EXERCISE OBJECTIVE

When you have completed this exercise on the balanced modulator, you will be able to describe the

circuit block, explain the output signal’s relationship to the two input signals, and describe the operation

and applications. You will use an oscilloscope to make measurements.

DISCUSSION

On the ANALOG COMMUNICATIONS circuit board, balanced modulators function as the following

devices:

• an AM modulator,

• a DSB suppressed modulator,

• mixers,

• an SSB product detector,

• and a phase detector.

The balanced modulator is an IC that is designated as number 1496. There are six 1496 ICs on the circuit

board.

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The IC consists of eight NPN transistors (Q1 to Q8).

Q8).

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The bottom two transistors and associated bias circuitry form a constant current source for the lower

How many inputs are there in the balanced modulator?

a. 1

b. 2

c. 3

There are two outputs: the positive output at pin 6, and the negative output at pin 12.

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Usually, the pin 12 output connects to the circuits following the balanced modulator.

A null potentiometer may be provided in the message signal input to adjust the base voltages at Q5 and

Q6.

What is the function of the potentiometer?

a. It adjusts the amount of modulation at the output.

b. It adjusts the frequency of the output signal.

The potentiometer permits the same balanced modulator to be used as an AM modulator (with the carrier

frequency present), or as a DSB modulator (with the carrier frequency suppressed).

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Depending on the modulator’s function, the output may contain the carrier frequency, the sum and

difference frequencies (sidebands) of the carrier and modulating (message) signals, and sideband

harmonics.

In the output, the message signal frequency is highly attenuated and can be neglected.

saturation, so that they operate in a switching mode.

operates in a linear mode.

multiplication of the input signals.

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For sine wave signal inputs and suppressed carrier operation, the balanced modulator’s output voltage

(VOUT

VOUT = K x Vm x Vc x {cos[(wm + wc)t + P] + cos[(wm – wc)t + P]}

VOUT – the instantaneous dc output voltage

K – a constant that depends on the load and bias resistor values

Vm – the message signal peak value

Vc – the carrier signal peak value

(wm + wc) – the sum of the message and carrier frequencies

(wm – wc) – the difference between the message and carrier frequencies

t – time

P – the phase difference between the inputs

The expression shows that VOUT contains the

• sum frequency of the inputs.

VOUT = K x Vm x Vc x {cos[(wm + wc)t + P] + cos[(wm – wc)t + P]}

VOUT – the instantaneous dc output voltage

K – a constant that depends on the load and bias resistor values

Vm – the message signal peak value

Vc – the carrier signal peak value

(wm + wc) – the sum of the message and carrier frequencies

(wm – wc) – the difference between the message and carrier frequencies

t – time

P – the phase difference between the inputs

• difference frequency of the inputs.

VOUT = K x Vm x Vc x {cos[(wm + wc)t + P] + cos[(wm – wc)t + P]}

VOUT – the instantaneous dc output voltage

K – a constant that depends on the load and bias resistor values

Vm – the message signal peak value

Vc – the carrier signal peak value

(wm + wc) – the sum of the message and carrier frequencies

(wm – wc) – the difference of the message and carrier frequencies

t – time

P – the phase difference between the inputs

When input signals have equal frequencies but are out-of-phase, what is the output difference

component?

a. twice the input frequencies

b. a dc voltage that varies with the cosine of the phase difference

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When the balanced modulator functions as an amplitude modulator, you adjust the amount of modulation

with the null potentiometer in the message signal input circuit.

You can also adjust the null potentiometer to suppress the carrier frequency to output a DSB signal (only

the upper and lower sideband frequencies).

The balanced modulator is also used as a mixer that combines two frequencies to obtain a higher or

lower frequency.

output?

a. the sum and difference frequencies of the two inputs

b. the message and carrier signal frequencies

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The SSB product detector is a balanced modulator that combines a 455 kHz modulated single sideband

signal with a 452 kHz beat frequency oscillator (BFO) signal to output a message signal as the LSB.

The FM quadrature detector uses a balanced modulator as a phase detector; one of the outputs is a dc

voltage that varies with the phase difference, which is a function of the message signal.

Because the balanced modulator is used extensively in the circuits discussed in this course, a periodic

review of this exercise is recommended.

PROCEDURE

COMMUNICATIONS circuit board, and connect the circuit shown.

Set switches S1, S2, and S3 to OFF.

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Turn the NEGATIVE SUPPLY knob on the left side of the base unit fully CCW to adjust the

VCO-LO frequency to less than 452 kHz.

Connect the oscilloscope channel 1 probe to the message signal input M on the

MODULATOR.

While observing the signal on channel 1, adjust the SIGNAL GENERATOR for a 300 mVpk-pk,

3 kHz sine wave at M.

Connect the oscilloscope channel 2 probe to the carrier signal input C on the MODULATOR.

NOTE: When making oscilloscope measurements or observations, be sure that you connect the probe’s

ground clip to a ground terminal on the circuit board.

While observing the signal on channel 2, adjust the amplitude knob on VCO-LO for 100 mVpk-pk

at C.

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Connect the oscilloscope channel 2 probe to the MODULATOR’s output. Connect a two-

post jumper in the SSB position to connect the MODULATOR output to the input of the LSB

FILTER.

Set the oscilloscope vertical mode to channel 2, and trigger on channel 1 (the message

Turn the MODULATOR’s potentiometer knob fully CCW; and then slowly turn it CW until the

AM signal (channel 2) is less than 100% modulated, as shown.

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Continue to turn the knob slowly CW until the AM signal appears, as shown. What type of

AM modulated signal appears on channel 2?

a. SSB

b. DSB

c. 100% modulated

If the carrier signal frequency (fc) is 452 kHz and the message signal frequency (fm) is 3 kHz,

what frequencies would be present in the frequency spectrum of the DSB signal?

a. 3 kHz and 452 kHz

b. 449 kHz and 455 kHz

c. 449 kHz, 452 kHz, and 455 kHz

457 kHz.

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Connect the oscilloscope channel 2 probe to the output of the LSB FILTER.

Set the oscilloscope vertical mode to channel 2, trigger on channel 2, set the channel 2

NOTE: When making oscilloscope measurements or observations, be sure that you connect the probe’s

ground clip to a ground terminal on the circuit block.

Because the frequencies in the DSB signal from the MODULATOR are below 453 kHz, a

trace on the oscilloscope screen appears at the LSB FILTER output, as shown.

Increase the VCO-LO’s frequency to the MODULATOR by slowly turning the NEGATIVE

SUPPLY knob CW until the LSB FILTER’s output signal is maximum, as shown.

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What is the LSB FILTER’s output signal shown on channel 2?

a. a greater than 100% modulated AM signal

b. a DSB signal

c. an SSB signal

While observing the 455 kHz signal at the LSB FILTER’s output on channel 2, vary the

amplitude of the 3 kHz message signal to the MODULATOR by varying the AF LEVEL knob

on the SIGNAL GENERATOR.

Does the amplitude of the 455 kHz signal vary with the amplitude of the 3 kHz message signal?

a. yes

b. no

Connect the output of the VCO-HI circuit block to input C at the MIXER.

In the following PROCEDURE steps, you will observe how a balanced modulator functions as a

MIXER to increase the frequency of a signal.

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Turn the POSITIVE SUPPLY knob on the right side of the base unit fully CCW to set the

VCO-HI frequency above 1455 kHz.

Set the oscilloscope sweep to 0.5

input.

Adjust the VCO-HI oscillator signal to 100 mVpk-pk with the potentiometer knob on the VCO-HI

circuit block.

Connect channel 1 to the LSB FILTER’s output and trigger on channel 1.

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Adjust the MIXER’s potentiometer knob for a DSB signal at pin 6, as shown.

the MIXER, which is the FILTER’s output.

The signal should appear as a trace (DC level).

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Adjust the VCO-HI frequency to 1455 kHz by slowly turning the POSITIVE SUPPLY knob

CW until the SSB signal at pin 12 of the MIXER is maximum, as shown.

Measure the period (T) between peaks of the waveform. Each horizontal division is 0.5 s.

T = s (Recall Value 1)

From the period (T),calculate the frequency of the SSB at pin 12 of the MIXER.

T = s (Step 23, Recall Value 1)

f = kHz (Recall Value 2)

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While observing the 1000 kHz signal at the LC FILTER’s output (pin 12), vary the amplitude of

the 3 kHz message signal to the MODULATOR by varying the AF LEVEL knob on the SIGNAL

GENERATOR.

Does the amplitude of the 1000 kHz signal at pin 12 vary with the amplitude of the 3 kHz

message signal?

a. yes

b. no

CONCLUSION

• The balanced modulator on the circuit board is an IC (1496).

• The output of the balanced modulator is a full-wave multiplication of the two input signals.

• The balanced modulator may have a null potentiometer to adjust the amount of modulation or to

suppress the carrier frequency in the output.

• The balanced modulator’s output will contain the sum and difference frequencies of the inputs, plus

the carrier frequency, if the carrier frequency is not suppressed.

• The balanced modulator can be used as an AM modulator, an SSB modulator, a mixer, a product

detector, and a phase detector.

REVIEW QUESTIONS

1. On the circuit board, what is the function of a balanced modulator?

a. mixer

b. amplitude modulator

c. detector

d. All of the above

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2. On the circuit board, how is the balanced modulator constructed?

a.

b.

c.

d. The circuit consists of three transistors, a current source, and an LC network.

3. How is the carrier signal suppressed in the output of a balanced modulator?

a. It is suppressed by the NEGATIVE SUPPLY voltage.

b. It is suppressed by the POSITIVE SUPPLY voltage.

c. It is suppressed by the null potentiometer in the message signal input.

d. It is suppressed by adjusting the VCO-LO potentiometer.

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4. To a balanced modulator, the carrier signal is 1000 kHz and the message signal is 3 kHz.

What are the sum and difference frequencies in the output?

a. 997 kHz and 1003 kHz

b. 994 kHz and 997 kHz

c. 1003 kHz and 1006 kHz

d. 994 kHz and 1006 kHz

5. When only the sum frequency is desired from a balanced modulator, what device removes the lower

frequencies?

a. a mixer

b. a product detector

c.

d. an automatic gain control circuit