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I am trying to solve a numerical where the Speed of a DC motor operating at 15V was measured to be 2000 rad/sec. Now a PWM signal is applied to the same motor to get a speed of 400 rad/sec. If the voltage amplitude of the PWM signal is 5V, then what should be the value of duty cycle?

I know the formula of duty cycle is

D=Vout/Vin

D=duty cycle

Vin=15V

Vout=5V

therefore the duty cycle is =5/15

i.e 1/3

therefore duty cycle in percentage is 33.33%

is my calculation right or whether I have to consider the speed of motor at 15V(2000 rad/sec) and at 5V(400 rad/sec)? If yes then how?

this the exact question

enter image description here

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  • \$\begingroup\$ Your way of thinking is simple. And correct. It simply IS a share of the voltage, nothing more. It's exactly like calculating average. If you have 15V for 1/3 of time and 0V for 2/3 of time, then the average is indeed 5V (15V*1/3+0V*2/3)/(1/3+2/3)=(5/1)=5V. So yes, you're right. \$\endgroup\$
    – Ilya
    Commented Oct 26, 2020 at 11:33
  • \$\begingroup\$ the options that i have for above question is a.60% b.50% c. 80% d. 100% i am getting answer as 33.33%@llya \$\endgroup\$
    – power machines
    Commented Oct 26, 2020 at 11:43
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    \$\begingroup\$ No you have to factor in that both the driving voltage ( 5 vs 15) and duty cycle (?? vs 100%) have changed to get the reduced speed. One of the multiple choices is correct. (In the real world you'd have to factor in motor losses so the real answer would be different. But this is ideal homework-land) \$\endgroup\$
    – user16324
    Commented Oct 26, 2020 at 15:14
  • \$\begingroup\$ Another hint then: When you go from 15V to 5V, you will get 1/3 of the original rotation speed. You want 1/5 of the original rotation speed. \$\endgroup\$
    – ocrdu
    Commented Oct 26, 2020 at 15:56
  • \$\begingroup\$ 400 rad/sec is 20% of 2000rad/sec so can the duty cycle be 20%? @ocrdu \$\endgroup\$
    – power machines
    Commented Oct 26, 2020 at 17:36

3 Answers 3

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When you go from 15 V to 5 V, you get \${5 \over 15} = {1 \over 3}\$ of the original speed.

What you actually want is \${400 \over 2000} = {1 \over 5}\$ of the original speed, so you will somehow have to reduce the speed further by "PWM-ing your way" from \$1 \over 3\$ to \$1 \over 5\$ of the original speed. So:

\${1 \over 3} \times dutycycle = {1 \over 5} \implies dutycycle = {3 \over 5} = 60\%\$

Going from 15 V to 5 V has reduced the speed to \${2000 \over 3} \mathrm{\, rad/s}\$, reducing the speed further to 60% of that gives \$\small 60\% \normalsize \times {2000 \over 3} \small = 400 \mathrm{\, rad/s}.\$

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    \$\begingroup\$ I was getting the OP to work it out for her/himself. It's generally better for homework questions. \$\endgroup\$
    – Transistor
    Commented Oct 26, 2020 at 18:31
  • \$\begingroup\$ @Transistor: I know, and tried also, but gave up. Maybe the answer will make the penny drop. \$\endgroup\$
    – ocrdu
    Commented Oct 26, 2020 at 18:46
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Duty cycle is just the ratio of on-time to the period time.

enter image description here

Figure 1. PWM duty cycles. Source: Robotic Controls.

I know the formula of duty cycle is *D = Vout / Vin ...

Nope. $$ D = \frac {t_{on}} {t_{period}} $$.

If you work 6 hours every day then your duty cycle = 6 / 24 = 25%. It doesn't matter how hard you work, it's still 25% duty.

If the voltage amplitude of the PWM signal is 5V, then what should be the value of duty cycle?

It sounds as though your motor supply voltage is 15 V and your control signal is 5 V logic. In that case you use the 5 V PWM signal to switch the 15 V motor power at the same rate to achieve the lower average voltage. If the 15 V supply is on 30% of the time then you'll get 0.3 x 15 V = 4.5 V equivalent.

enter image description here

Figure 2. The actual question.

From the question it appears that two things are changing. The supply voltage has dropped from 15 V to 5 V and PWM is being applied. The motor speed has dropped from 2000 rad/s to 400 rad/s.

Questions for you:

  1. What is the new speed as a percentage of 2000 rad/s?
  2. What would the speed be at 5 V DC?
  3. What PWM do you need to get 400 rad/s on this new 5 V supply?
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  • \$\begingroup\$ the options that i have for above question is a.60% b.50% c. 80% d. 100% i am getting answer as 33.33%@Transistor \$\endgroup\$
    – power machines
    Commented Oct 26, 2020 at 11:42
  • \$\begingroup\$ Show the exact question then. \$\endgroup\$
    – Transistor
    Commented Oct 26, 2020 at 11:46
  • \$\begingroup\$ I suspect the 5V PWM in the question is used to drive the motor directly. \$\endgroup\$
    – ocrdu
    Commented Oct 26, 2020 at 11:46
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    \$\begingroup\$ @ocrdu: I suspect that it's there to mislead the student and the misdirection has worked. That's why I added my 6 hours/day example. \$\endgroup\$
    – Transistor
    Commented Oct 26, 2020 at 11:48
  • \$\begingroup\$ i have added the exact question @Transistor \$\endgroup\$
    – power machines
    Commented Oct 26, 2020 at 11:51
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I think you are not getting the scenario.

At 15V with D=1 the speed is 2000 rad/s.

At 5V with D=? the speed is 400 rad/s.

Speed is proportional to V * D. I don't want to solve it for you. But it seems like you were not understanding the givens in the problem. You just have to solve for the question mark.

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