Electrical loss budget

Follow every watt.

Propagate input power through each voltage boundary. See what reaches the GPU, what becomes heat, and how current rises near 0.7V.

Pout = Pin × efficiency; heat = Pin − Pout; current = Pout ÷ Vout. Efficiencies are editable assumptions, not supplier benchmarks.

Load the sample or enter ordered stages.
—delivered
—heat
—cumulative efficiency
—final current

Delivered power and cumulative loss

Energy balance
—Input must equal delivered power plus heat.
Largest thermal socket
—Lower voltage raises current for the same delivered power.

Track delivered power, conversion heat and current

Read the explanation

The synthetic chain starts at one million watts. The first stage has efficiency zero point nine nine five, so output is nine hundred ninety five kilowatts and heat is five kilowatts. The next stage multiplies that remaining power by zero point nine seven five, leaving nine hundred seventy point one two five kilowatts. The final zero point eight eight stage leaves eight hundred fifty three point seven one kilowatts. Efficiencies multiply rather than simply subtracting percentages. For each stage, heat equals input power minus output power. The three stage losses are five, twenty four point eight seven five, and one hundred sixteen point four one five kilowatts. They sum to one hundred forty six point two nine kilowatts. Delivered power plus heat is exactly the original one thousand kilowatts. The final stage is the largest loss in this declared fixture, not a measured supplier benchmark. The model calculates current as stage output power divided by output voltage. At the final zero point seven volt boundary, eight hundred fifty three thousand seven hundred ten watts corresponds to roughly one point two two million amperes. At twice the voltage, the same power would require half the current. This is the tool’s aggregate idealized chain calculation, not proof that a single conductor carries that current or that a physical rack can realize the declared efficiencies.

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