## [answered] Lab 6: Time-Proportioning DC/AC Control Circuit Objectives

Here is another lab. this week is a little easier, being that itske only one lab.

Lab 6: Time-Proportioning DC/AC Control Circuit

Objectives

1. To simulate a DC/AC time-proportioning circuit using a schematic diagram.

2. To alter the duty cycle of a DC time-proportioning amplifier circuit to vary the

current applied to a DC/AC actuator.

Introduction

The DC time-proportioning circuit can be used to control an actuator device that operates

on DC/AC voltages. An electromagnetic relay, solid-state relay or an opto-triac can be

used to interface to high power circuit to produce a time-proportioning signal at the

actuator output. These interface devices are to separate circuits that function at two

different voltages.

Time proportioning is a method in which the amplifier output is switched alternately to

fully on and fully off. Changing the ratio of signal-on to signal-off varies the average

voltage produced. Figure 42-1 illustrates various time-proportioning output signals that

are in the form of a square wave. When the on-time is shorter than the off -time, as shown

in Figure 42-1(a), the average output voltage is low. When the on-time and off-time are

the same, as shown in Figure 42-1(b), the average voltage is at the mid-range between

low and high. When the on-time is greater than the off-time, as shown in Figure 42-1(c),

the average voltage is high. The ratio of time at which the square wave is on to the total

time period of one cycle is called the duty cycle. The average DC voltage produced can

be determined by multiplying the duty cycle times the on-state DC voltage of the square

wave.

Suppose the square wave produces +10V when it is on, and 0V when it is off . If the duty

cycle is 25 percent, as in Figure 42-1(a), the average voltage is 2.5V (10V x 25%=2.5V).

When the duty cycle is 50 percent, as in Figure 42-1(b), the average voltage is +5V. If the

duty cycle is 0%, the average voltage is 0V. Figure 42-1 Waveforms at various duty cycles A voltage level?detector op amp, as shown in Figure 42-2, can be used as a timeproportioning circuit. The voltage level?detector output is 0V when the non-inverting

input is less than the inverting input. When the voltage at the non-inverting input is

greater than the voltage at the inverting input, the op amp produces a positive saturation

voltage at its output. Figure 42-2 Voltage level detector op amp

In Figure 42-3, a 12V Peak sawtooth is applied to the inverting input of the op amp. The

wiper arm of a potentiometer is connected to the non-inverting input. When the wiper

arm voltage is higher than the sawtooth voltage the output of the op amp will be high.

When the wiper arm varies between its maximum and minimum the output pulse width

will change and so does the average DC output voltage. The result is that the op amp

produces a square wave with a variable percent duty cycle. The output of the op amp has

been amplified using a transistor to drive a heater. If the transistor drives an

electromagnetic relay, or opto-triac it can easily drive a time-proportion high power AC

heater. Figure 42-3 Time-proportion control of a Heater

Figure 42-4 shows how time-proportion will be used for AC signals and the way it will

appear on the AC load. If the actuator is a heater, it will become hotter as the duty cycle

increases. The types of actuators driven by AC time-proportioning amplifiers are

elements for heaters and for lighting systems that require variable-intensity control.

Motors do not use them because motors require an applied AC signal that is not

interrupted. Figure 42-4 Time-proportion AC control circuit Procedure

Step 1

Draw the time-proportioning circuit in Figure 42-3 in Multisim.

Step 2

Adjust the sawtooth waveform generator to 12V voltage amplitude and 10ms period.

Step 3

Draw the waveform at inverting, non-inverting and output voltage of the op amp when

duty cycle is 50% in the space provided. (+) Input waveform (-) Input waveform Op amp output Step 4

Fill out the table 42-1 by changing the potentiometer value and record op amp output

duty cycle and the DC voltage across the heater.

Table 42-1

Potentiometer Potentiometer Potentiometer Potentiometer Potentiometer

0%

25%

50%

75%

100% Op amp

output

Duty Cycle

%

Voltage

Across

Heater Experiment Questions

1. The term duty cycle refers to the amount of time a signal is ?..... compared to the

period of one complete cycle.

A. off

B. on

2. When the voltage applied to the ?........ input is greater than the ?.......... input, the

voltage level?detector op amp produces a +V saturation voltage at its output.

A. inverting

B. non-inverting

3. A square wave that is 20V at its high state and 0V when it is off will produce an

average DC voltage of ?......... when its duty cycle is 75 percent.

A. 7.5V

B. 10V

C. 15V

4. AC time-proportioning signals are not applied to ?.........

A. light bulbs

B. heater elements

C. AC motors

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