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What is the mechanism that causes more inductive spikes with a higher duty cycle, even if the frequency of the pwm is held constant? Doesn't the motor create voltage kickback only when it is turned off? If that is the case increasing duty cycle doesn't increase the frequency of the motor turning off(since PWM frequency is constant) and so should not increase motor spikes.

I am observing the PWM signal on a scope, and the spikes are more frequent and of larger magnitude. This is my setup: enter image description here

The left portion of the circuit controls the duty cycle of the pwm, which is generated on the out pin.

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    \$\begingroup\$ How do you know you are seeing "more" spikes? And what does "more" mean? Bigger" Greater number at a voltage which triggers a scope? Describe your setup and exactly what you're seeing. \$\endgroup\$ Commented Jul 15, 2018 at 23:09
  • \$\begingroup\$ Added some more info \$\endgroup\$ Commented Jul 15, 2018 at 23:33
  • \$\begingroup\$ Where are you measuring these "spikes"? Is the motor current continuous? \$\endgroup\$ Commented Jul 15, 2018 at 23:41
  • \$\begingroup\$ @Sphero Pefhany Measuring at the 'out' pin and the 'in' pin. How could I check if the motor current is continuous? \$\endgroup\$ Commented Jul 16, 2018 at 0:18
  • \$\begingroup\$ Could be a resonance issue at some pulse-width. \$\endgroup\$ Commented Jul 16, 2018 at 1:50

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With more duty-cycle, the windings have more time to ramp up the current further, thus storing more energy.

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Why am I seeing more inductive spikes from my motor at a higher duty cycle?

Just to enhance analogsystemsrf's answer....

The formula for an inductor is: -

$$V = L\dfrac{di}{dt}$$

So if you apply a voltage across an inductor, the current rises in a ramp fashion at a rate of V/L amps per second. So, if you hold the voltage for a longer period of time, the current attains a peak value that is naturally higher than if you held the voltage for a shorter period.

The inductive spike happens when you switch off the PWM transistor. The current attained during conduction of the transistor stores energy in the magnetic field and thus, more energy is stored for a higher duty cycle and a bigger spike happens when you open circuit the transistor.

Additionally, if the current doesn't fall to zero when spiking but (say) 100 mA, the subsequent ramp up in current when the transistor begins conducting again begins at 100 mA and rises to a value that is 100 mA higher than if it had began at zero amps. This means that above a certain duty cycle the current can start to attain higher peak values than those suggested by the simple V = L di/dt formula.

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