Math Bridge: Why PWM Becomes Smooth Motion

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Math BridgeActuatorsStruggle-friendly runway

Why does full-voltage switching look like a smaller, smooth push?

One thread from period and duty cycle to current ripple and motor inertia.

Max, the actuators guideMax guides
The one targetCompute what duty and frequency control.
The chapter case5 kHz at 50%; 20 kHz at 35%.
What it buys youPredict average drive and ripple separately.

A technician must decide whether switching period is safe before changing pwm frequency on the real device. The result is unresolved until the rule and units are checked. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is pwm frequency. The middle card applies this page's rule. The green card is switching period. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only pwm frequency, so the numeric fixture does not switch without explanation.

PWM frequency changes switching period An input card leads through the rule period = 1,000,000 us/s / frequency to the switching period result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Higher switching frequency makes each PWM period shorter.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 20000 Hz.

  2. 2

    Name the relationship. period = 1,000,000 us/s / frequency

  3. 3

    Substitute with units. 1,000,000 / 20,000 = 50.0 us

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change pwm frequency

Try Predict the direction of period = 1,000,000 us/s / frequency. Test another pwm frequency, then compare switching period.

20000 Hz
Chapter baseline
Switching period

Observe Higher switching frequency makes each PWM period shorter. Reset pwm frequency to 20000 and compare switching period.

Explain Higher switching frequency makes each PWM period shorter.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only pwm frequency moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only pwm frequency. It is not a field prediction.

1. PWM is a timed switch

The output is fully on or fully off. Duty D is the fraction of each period spent on. The load's electrical and mechanical storage smooths the pulses.

Max: Duty sets the mean push. Frequency sets how long current can rise during each pulse.

2. Find period, on-time, and mean

1

Invert frequencyT=1/fPWM.

2

Take the duty fractionton=DT.

3

Average the pulseVavg=DVs.

3. Add the winding and rotor

di/dt≈(Vs−Eb)/L; Δion≈(Vs−Eb)DT/L; τe=kti; Jdω/dt=τe−τload−bω

Inductance slows current change. Inertia slows speed change. A short period gives both less pulse-to-pulse movement.

4. Try the PWM frequency

T=1/f; ton=DT; Vavg=DVs; Δi≈(Vs−Eb)ton/L

TryMove between 2 and 20 kHz while the greenhouse fan stays at 35% duty.

Period
On-time at 35%
Average terminal voltage
Rough average current
Input power
On-pulse current rise
Cycles per 50 ms

ObserveAt 20 kHz, T=50.0 µs, ton=17.5 µs, Vavg=4.20 V, average current is 0.175 A, input power is 2.10 W, and the short-pulse current-rise estimate is 0.140 A. At 2 kHz it becomes 1.40 A.

ExplainChanging frequency leaves the ideal average voltage fixed because duty is fixed. It changes pulse duration, so current and torque ripple change.

Technical boundaries.

The short-pulse estimate freezes winding resistance and back-EMF inside one pulse.

exact RL steady-state ripple
Needs separate evidence
driver drops
Needs separate evidence
dead time
Needs separate evidence
core loss
Needs separate evidence
current limiting
Needs separate evidence
speed-dependent back-EMF
Needs separate evidence
load torque
Needs separate evidence
switching loss
Needs separate evidence
acoustics
Needs separate evidence
thermal limits
Needs separate evidence

Use field evidence or a deeper model before release.

5. Reproduce the 5 kHz fan example

At 5 kHz, T=200 µs. A 50% command gives ton=100 µs and 2.50 V from 5 V. A 50 ms mechanical time constant spans 250 cycles, so the shaft cannot follow each pulse.

6. Verify smooth operation

Measure current ripple, shaft speed, audible noise, driver temperature, supply sag, load torque, and motion at several duty and frequency settings. Keep protection and current limits active.

7. Check yourself

What does 75% duty give from 12 V ideally?
Answer: Vavg=0.75×12=9.00 V.
Why does 2 kHz give ten times the rise of 20 kHz?
Answer: Its period and on-time are ten times longer at the same duty.
Does 4.20 V predict exact fan speed?
Answer: No. Load, back-EMF, resistance, driver loss, and mechanics still matter.
Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

5 kHz
Frequency, sample rate, or event rate
50%
Percentage, ratio, or gain
200 µs
Time, interval, or service-life value
100 µs
Time, interval, or service-life value
5 V
Voltage or voltage-step value
2.50 V
Voltage or voltage-step value
50 ms
Time, interval, or service-life value
250 cycles
Cycle, step, or position count
75%
Percentage, ratio, or gain
12 V
Voltage or voltage-step value
9.00 V
Voltage or voltage-step value
20 kHz
Frequency, sample rate, or event rate
35%
Percentage, ratio, or gain
50.0 µs
Time, interval, or service-life value
17.5 µs
Time, interval, or service-life value
4.20 V
Voltage or voltage-step value
0.175 A
Current or responsivity value
2.10 W
Power or power-loss value
1.00 mH
Inductance value
4.00 V
Voltage or voltage-step value
0.140 A
Current or responsivity value
2 kHz
Frequency, sample rate, or event rate
1.40 A
Current or responsivity value

They are first estimates, not a motor model.

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