PWM

YOU WILL BE ABLE TO

Define duty cycle and compute a PWM signal's average voltage.
Explain why PWM controls power efficiently.
Design an RC filter to recover an analogue voltage from PWM.
Choose a PWM frequency for a given load.

A microcontroller has no analogue output — its pins are digital, and can produce only two voltages. Pulse-width modulation gets around this by switching fast between them and varying the fraction of time spent high. The duty cycle $D$ is that fraction, and over any interval long compared with the switching period the average voltage is simply the supply scaled by it.

V_{avg} = D \cdot V_{supply}

the average of a PWM waveform

Two duty cycles on a 3.3 V rail: the waveforms are always at a rail, but their averages are not.

Two quite different things are done with this, and confusing them causes real design errors. For loads that respond slowly compared with the switching — LED brightness perceived by the eye, motor speed governed by rotor inertia, a heater's temperature — no filter is needed at all: the load does the averaging itself, and PWM controls power directly. For producing an actual analogue VOLTAGE, a filter is required, and then it is CA-6's low-pass with a cutoff far below the switching frequency.

Recovering an analogue voltage: an RC low-pass averages the PWM. Ripple falls as R, C, and the switching frequency rise.

The residual ripple on the filtered output is what remains of the switching, and a useful approximation for a simple RC is:

V_{ripple} \approx \frac{D(1-D)\,V_{supply}}{f_{pwm}\,RC}

peak-to-peak ripple, worst at 50% duty

Note the shape of that result. Ripple is worst at $50\%$ duty and vanishes at either extreme; it falls in proportion to switching frequency, and in proportion to the time constant. So there are three dials, and only one is free: the load usually fixes what $RC$ can be (a large $RC$ makes the output slow to respond), and the supply is what it is, so raising $f_{pwm}$ is the usual lever. The RP2040 makes that easy — its PWM blocks divide a $125$ MHz clock through a 16-bit counter, so kilohertz to hundreds of kilohertz is routine.

WORKED EXAMPLE

Filtering PWM into a control voltage

Produce a $0$ – $3.3$ V analogue control voltage from a $3.3$ V PWM at $10$ kHz, with under $10$ mV of ripple.
Worst case is $50\%$ duty: $D(1-D) = 0.25$ .
Required: $0.25 \times 3.3/(10^4 \times RC) \le 0.010 \Rightarrow RC \ge 0.00825$ s.
Choose $R = 10$ kΩ, $C = 1$ µF: $RC = 10$ ms, giving about $8$ mV of ripple. ✓
Cost of that choice: the output takes about $5RC = 50$ ms to settle after a duty change (CA-5's five-time-constant rule). Fine for a setpoint, useless for audio.

R = 10

kΩ,

C = 1

µF — 8 mV ripple, 50 ms settling.

WORKED EXAMPLE

Choosing a frequency for a motor

A small DC motor is PWM-driven. Below about $100$ Hz the motor's torque pulses audibly and mechanically.
In the audio band the switching is heard as a whine — commonly the most complained-about property of cheap motorised products.
Above about $20$ kHz the whine is inaudible, but the driver's switching losses rise (each transition costs energy, ME-5's $\tfrac{1}{2}Cv^2$ again) and the MOSFET's gate drive matters more.
Typical choice: $20$ – $25$ kHz — just above hearing, below where losses become troublesome.

Just above the audio band, for reasons that are half acoustic and half thermal.

PRACTICE

A $3.3$ V PWM signal has a duty cycle of $30\%$ . What is its average voltage?

What duty cycle produces $2.0$ V from a $5$ V supply?

An LED is dimmed to half brightness two ways: by PWM at $50\%$ duty, and by a transistor in its linear region dropping $2$ V while the LED takes $10$ mA. Compare the heat dissipated in the controlling device.

A filtered PWM output has too much ripple. Name the three quantities you could change, and state the practical cost of each.