PWM Output Voltage
Calculate the average output voltage of a PWM signal, V_out = (duty ÷ 100) × V_sup, from the duty cycle (in %) and the supply voltage V_sup. The result, in volts, is the effective average voltage delivered to a load (motor, LED, heater) by rapidly switching the supply on and off. Varying the duty cycle from 0 to 100% varies the average voltage from 0 to V_sup, allowing power control without dissipating energy in resistors — the basis of motor speed and LED brightness control in microcontrollers. Enter the duty cycle and the supply voltage.
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PWM output voltage
How do you control a motor's speed or an LED's brightness without wasting energy in resistors that just heat up? The answer is PWM (Pulse Width Modulation): instead of lowering the voltage continuously (dissipating the excess as heat), the supply is switched on and off very rapidly (thousands of times per second), and what gets controlled is the fraction of time it stays on — the duty cycle. The average voltage delivered to the load is V_out = (duty ÷ 100) × V_sup: at 50% duty cycle with a 12 V supply, the load 'sees' on average 6 V; at 25%, it sees 3 V; at 100%, it gets the full 12 V. The genius of the method is its efficiency: since the switching transistor is always either fully on (near-zero resistance, no voltage drop) or fully off (near-zero current), it dissipates almost no power — all the energy goes to the load, unlike a linear resistive control that burns off the excess. That is why PWM is everywhere: speed control of DC motors and fans, LED dimming, switch-mode power supplies, servo control (where the pulse width encodes the position), heaters. Microcontrollers (Arduino and the like) generate PWM in hardware with ease. A few caveats: the load needs inertia (mechanical, in a motor; thermal, in a heater) or a filter (capacitor/inductor) to 'integrate' the pulses and respond to the average — otherwise it would 'see' the individual pulses; the PWM frequency must be high enough for that (and, with motors, above the audible range to avoid whine); and the average-voltage relation loses accuracy with inductive loads unless a proper freewheeling diode is fitted. But the basic concept — duty cycle times supply voltage gives the average voltage — is the foundation of all modern power control in embedded electronics. Enter the duty cycle and the supply voltage.
Related Tools
Buck-Boost Converter
Calculate the output voltage of a buck-boost DC-DC converter in continuous conduction, V_out = V_in × D ÷ (1 − D), from the input voltage V_in and the duty cycle D (0 to 1). The result, in volts, can be lower (D < 0.5) or higher (D > 0.5) than the input — the buck-boost converter steps voltage down or up depending on the duty cycle, with inverted output polarity in the classic topology. It is used when the input voltage can vary above and below the desired output (discharging batteries, universal supplies). Enter the input voltage and the duty cycle.
PWM Resolution (Levels)
Calculate the number of levels of a PWM signal, levels = 2^bits, from the bit resolution of the PWM generator (timer). The result is the number of discrete duty-cycle steps available: an 8-bit PWM offers 256 levels (0 to 255), allowing power adjustment in 256 steps; a 10-bit one, 1024 levels, finer control. Higher resolution gives smoother control of motor speed and LED brightness, but reduces the maximum possible PWM frequency for a given timer clock. It is a central parameter in microcontroller configuration. Enter the resolution in bits.
Square-Wave RMS Current
Calculate the RMS value of a pulsing square-wave current, I_rms = I_p × √D, from the peak current I_p and the duty cycle D (fraction of the period the current flows). The result, in amperes, is the RMS current that determines the actual heating (I²R losses) of a component that conducts in pulses — such as a transistor or winding in a switching converter. Unlike the average value, the RMS is what matters for sizing conductors, resistances and dissipation. The smaller the duty cycle, the lower the RMS for the same peak current. Enter the peak current and the duty cycle.
Output Voltage Ripple
Calculate the output voltage ripple of a switching converter, ΔV = I ÷ (f × C), from the output current I, the switching frequency f and the output capacitance C. The result, in volts, is the residual oscillation superimposed on the DC output voltage, caused by the filter capacitor charging and discharging each switching cycle. Higher frequency and capacitance reduce the ripple. Keeping the ripple within limits (typically <1% of the output) is essential to supply sensitive circuits. Enter the current, the switching frequency and the capacitance.
Boost Converter (Step-Up)
Calculate the output voltage of a boost (step-up) DC-DC converter in continuous conduction, V_out = V_in ÷ (1 − D), from the input voltage V_in and the duty cycle D (0 to 1). The result, in volts, is always greater than the input — the boost converter raises voltage by storing energy in an inductor and releasing it in series with the source. As D approaches 1, the output tends to infinity (limited by real losses). It is used in supplies that must step up voltage (LEDs, batteries, power factor correction) and in photovoltaic systems. Enter the input voltage and the duty cycle.
Rectifier Average Voltage
Calculate the average (DC) voltage of a full-wave rectifier, V_dc = 2 × V_p ÷ π, from the peak voltage V_p of the sinusoidal input. The result, in volts, is the mean value of the pulsating voltage at the rectifier output before filtering — about 63.7% of the peak. A full-wave rectifier (bridge or center-tap) flips the negative half-cycles, doubling the ripple frequency and raising the average voltage compared with a half-wave rectifier (V_p/π). It is the basis of designing DC power supplies from the AC mains. Capacitor filtering then raises the voltage further (close to the peak). Enter the peak voltage.
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