A triangular input drives the circuit of Fig. 36-23a. The variable resistance has a maximum value of 10 kΩ. If the triangular input has a frequency of 1 kHz, what is the duty cycle when the wiper is at the middle of its range?

Question

A triangular input drives the circuit of Fig. 36-23a. The variable resistance has a maximum value of 10 k[latex]\Omega [/latex]. If the triangular input has a frequency of 1 kHz, what is the duty cycle when the wiper is at the middle of its range?

Accepted Answer

When the wiper is at the middle of its range, it has a resistance of 5 k[latex]\Omega[/latex]. This means that the reference voltage is

[latex]v_{ref}=\frac{5k\Omega }{15 k\Omega }15 V=5V[/latex]

The period of the signal is

[latex]T=\frac{1}{1 kHz}=1000 \mu s[/latex]

Figure 36-23b shows this value. It takes 500 ms for the input voltage to increase from-7.5 to +7.5 V because this is half of the cycle. The trip point of the comparator is +5 V. This means that the output pulse has a width of W, as shown in Fig. 36-23b.

Because of the geometry of Fig. 36-23b, we can set up a proportion between voltage and time as follows:

[latex]\frac{W/2}{500\mu s}=\frac{7.5 V-5V}{15 V}[/latex]

Solving for W gives

[latex]W = 167 \mu S[/latex]

The duty cycle is

[latex]D=\frac{167\mu s}{1000 \mu s}=0.167[/latex]

In Fig. 36-23a, moving the wiper down will increase the reference voltage and decrease the output duty cycle. Moving the wiper up will decrease the reference voltage and increase the output duty cycle. For all values given in Fig. 36-23a, the duty cycle can vary from 0 to 50%.

Question 36.7: A triangular input drives the circuit of Fig. 36-23a. The va...

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