Ion Thruster Lab

An ion thruster ionizes a propellant like xenon and accelerates it through an electric field set by the beam voltage, reaching an exhaust velocity of tens of kilometers per second — far beyond any chemical engine. Thrust is just beam current times the ionized mass flow times that exhaust velocity, so even a fully charged beam only produces a few tens of millinewtons. How well electrical power turns into that tiny push is total efficiency, three multiplied factors: electrical efficiency in the power supply, mass-utilization efficiency (how much of the injected propellant actually gets ionized and thrown out the back), and beam-divergence efficiency (how much of the beam stays pointed straight along the axis instead of spraying sideways). Raising beam voltage raises specific impulse but lowers thrust-to-power — the tradeoff behind every electric-propulsion design. Because so little propellant has to move so fast, a small xenon load spread over a long burn adds up to an enormous Δv. Set the beam voltage, beam current, ion mass, mass utilization, electrical efficiency, and beam divergence, then watch it all play out live. It all runs on your device.

You are in the Physics lab.

Phase / elapsedIdle · 0 d
Thrust0 mN
Total efficiency0%

Ion thruster

Electrical power in, three efficiency losses out — and how far a little xenon can push.

Exhaust velocity
Specific impulse
Total power
Thrust-to-power
Δv accumulated
Propellant left
Phase
Efficiency delivered
Loss split — electrical power vs thrust power
Thruster schematic
Δv & propellant over the burn
Specific impulse & thrust-to-power vs beam voltage
Total efficiency is a product of three factors, and each is a real physical loss. Electrical efficiency is what the power supply loses turning input power into the ion beam; mass-utilization efficiency is the propellant that gets injected but never ionized, so it never gets thrown out the back; beam-divergence efficiency is thrust lost to ions that leave at an angle instead of straight down the axis. Multiply all three and you get total efficiency — the fraction of electrical power that survives as delivered beam thrust power. Raising beam voltage raises exhaust velocity and with it specific impulse, but since exhaust velocity only grows with the square root of voltage while power grows linearly, thrust-to-power falls as voltage climbs — the central tradeoff of electric propulsion. Because exhaust velocity is already so high, only a little propellant needs to flow to build real Δv, so a small xenon load can deliver an enormous Δv over a long enough burn.

Reading the simulation

What the thruster schematic, the loss-split bar and the two plots are telling you — then what the model says about voltage.

1

The schematic and the efficiency-delivered gauge

The schematic shows neutral propellant ionized in the discharge chamber and accelerated by charged grids into a high-speed exhaust beam, with a neutralizer injecting electrons so the beam and the spacecraft don't build up opposite charge. The gauge beside it tracks total efficiency — how much of the electrical power actually survives as beam thrust power once electrical, mass-utilization and divergence losses are subtracted (the same split the loss bar shows).

2

The loss-split bar and the three efficiency factors

The loss-split bar breaks electrical power into three losses that never show up as a single number — electrical loss, mass-utilization loss, and beam-divergence loss — plus what's left as delivered beam thrust power. Widen any slice and total efficiency falls, since it's the product of electrical, mass-utilization, and divergence efficiency together.

3

The burn plot and the voltage tradeoff curve

The first plot tracks Δv accumulated and propellant remaining as a long burn runs from throttle-up to depletion. The second plot draws specific impulse and thrust-to-power against beam voltage and marks where the current design sits, showing why high-Isp designs trade away thrust density.

How It Works

Three multiplied efficiency factors, a velocity that trades against thrust, and why so little propellant goes so far.

Electrical power (in) Discharge & grids (losses subtracted here) Electrical loss (power supply) Mass-util. loss (un-ionized) Divergence loss (off-axis ions) Thrust power delivered Raising beam voltage lifts exhaust velocity & specific impulse, but lowers thrust-to-power
Electrical power enters the discharge chamber & grids, where three losses are subtracted before what's left reaches the beam as thrust power: electrical loss in the power supply, mass-utilization loss from propellant that's never ionized, and beam-divergence loss from ions that leave off-axis. Raising beam voltage raises exhaust velocity and specific impulse, but lowers thrust-to-power for the same electrical input.
1

Thrust: mass flow times a very high velocity

Thrust is the ionized mass flow rate — beam current times the ion's mass-to-charge ratio — multiplied by exhaust velocity, the speed a beam voltage gives each accelerated ion. Because exhaust velocity is enormous but mass flow is tiny, the product comes out in millinewtons: an ion thruster trades raw thrust for extreme efficiency per kilogram of propellant.

2

Total efficiency: three multiplied factors

Total efficiency is electrical efficiency times mass-utilization efficiency times beam-divergence efficiency. Electrical efficiency is the power supply's conversion loss; mass-utilization efficiency is how much injected propellant actually gets ionized instead of drifting out neutral; beam-divergence efficiency is how much of the beam's momentum stays pointed straight back instead of spraying off-axis. Each factor is a real, physical loss, and losing ground on any one lowers the whole product.

3

The voltage tradeoff and why a little xenon goes far

Exhaust velocity, and with it specific impulse, rises with the square root of beam voltage, but the power needed to drive the beam rises linearly with voltage — so thrust-to-power falls as voltage climbs, the central design tradeoff of electric propulsion. Because exhaust velocity is already so high, the rocket equation's Δv — exhaust velocity times the log of how much the spacecraft's mass falls as propellant burns away — adds up fast from very little propellant, which is why a small xenon load can push a spacecraft for months and still deliver a Δv a chemical rocket could never match with the same mass.

How does an ion thruster produce thrust?
Thrust equals the beam current times the ion's mass divided by its charge, times the exhaust velocity the beam voltage gives each ion — in practice, beam current times ionized mass flow times exhaust velocity, corrected for how much of the beam sprays off-axis. Because the exhaust velocity is enormous but the mass flow is tiny, that product comes out in millinewtons rather than the newtons or kilonewtons a chemical engine produces.
What is total efficiency, and why does it split into three factors?
Total efficiency is electrical efficiency times mass-utilization efficiency times beam-divergence efficiency. Electrical efficiency is what the power supply loses converting input power into the ion beam. Mass-utilization efficiency is the share of injected propellant that actually gets ionized and accelerated rather than drifting out unused. Beam-divergence efficiency is how much of the beam's momentum stays pointed straight back instead of spraying off-axis. Losing ground on any one factor lowers the product.
Why is thrust so tiny if exhaust velocity is so high?
Thrust is mass flow rate times exhaust velocity, and an ion thruster pushes that velocity to tens of kilometers per second by accelerating ions through a strong electric field — far beyond what any chemical reaction can reach. But building that field only pushes a very small mass flow rate at a time, since the beam current is limited by the electronics and the propellant ion's mass-to-charge ratio. The product of a huge velocity and a minuscule mass flow comes out as millinewtons of thrust, even though the propellant is being thrown out extremely fast.
What is the thrust-to-power tradeoff, and why does a little xenon deliver such a huge Δv?
Exhaust velocity rises with the square root of beam voltage, so specific impulse climbs the same way, but the power needed to run the beam rises linearly with voltage — so thrust-to-power falls as voltage climbs, the central design tradeoff of electric propulsion. Because Δv from the rocket equation scales with exhaust velocity times the log of how much the spacecraft's mass falls as propellant burns away, and exhaust velocity is already so high, even a small xenon propellant load spread across a long burn accumulates an enormous Δv that a chemical rocket would need far more propellant to match.

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