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I have done a lot of research about these kinds of thrusters, and they all seem to require a relatively large amount of voltage but low currents. I see that they produce a low thrust per watt of power, but in general it seems that there are usually kilovolts and microamps, or something along those lines. I know that because of this, the resistance is high, but a part of my question is why must the resistance be high in the first place?

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    \$\begingroup\$ I think it is based on the principle of ionization of a propellant with RF pulses which in a vacuum is open circuit before arc of the plasma magnetically accelerated via exhaust. With 1~10kW RF pulses the plasma becomes a negative differential resistance partially conductive enough to be magnetically directed to make thrust. ( I think.. but there are many kinds) In a vacuum due to Paschen's_law, it requires exceptionally high V field [xxx kV/mm] \$\endgroup\$ – Sunnyskyguy EE75 Dec 29 '16 at 22:23
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    \$\begingroup\$ I think ion thrusters are shooting a few electrons and positive ions out into space really really fast to create thrust. I think only a few (guessing millions a second) are shot out so the current is very low. But the speed is very great. In order to do that I am sure you need a large amount of voltage. The law is V=IR so if the voltage is high and the current is low then the resistance has to be very high. \$\endgroup\$ – st2000 Dec 29 '16 at 22:31
  • \$\begingroup\$ The resistance is high because the voltage is high and the current is low. Ohm's Law is the definition of resistance. \$\endgroup\$ – immibis Dec 30 '16 at 10:44
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Think for a moment : particle energies are measured in electron volts.

So to increase the energy output, you either need to increase the number of electron charges (i.e. the number of ions), or accelerate them across a higher voltage.

And Tsiolkovski's rocket equation makes it clear you want to impart the highest velocity (thus most energy) to the smallest possible mass. This is what gives you the highest specific impulse from a given reaction mass, assuming you can generate the electrical energy somehow.

Put this together and the ion current will be as low as possible for a given energy, and that, using electrostatic acceleration, means using the highest practical voltage.

(Even if the acceleration is magnetic, the final particle energy will be measured in eV, and the current emitted in the microamp range, though the acceleration coils may be low voltage, high current devices).

Unlike a chemical rocket where the energy source (fuel) and reaction mass (combustion products) are closely related, the energy per particle is divorced from the nature of the particle itself, and therefore potentially unlimited. This is why ion drives can deliver much higher specific impulse.

It's not rocket science...

EDIT : Nayuki's comment below adds another part of the picture : this process is actually less energy efficient; as you reduce mass you have to increase energy per unit of momentum imparted to the vessel.

However the tyranny of the rocket equation means (to a close approximation) that doesn't matter; energy is cheap, (solar or nuclear) reaction mass is expensive once you're out of the atmosphere, which is why you need huge tanks of it.

This explains some of the excitement about reaction-mass-less drives such as the E-M Drive or Cannae Drive despite their (so far) monumentally low thrust.

The rest of the excitement seems to be based on their inability to follow the known rules of physics, despite the best efforts of the peer reviewers to cancel out experimental error...

Notes:

  1. I have some suspicion that the reaction "mass" will turn out to be composed of photons; but that seems to be de-bunked as part of one paper (linked from here)
  2. Skeptics may note that in Scots, "Cannae Drive" means "Can't Drive". We shall see...

EDIT 2 : Re: Tony's question, apparently not all ion thrusters are pulsed, and it's easy to imagine how a continuous thrust electrostatic one would operate.

I've seen one small "pulsed plasma thruster" in (lab) operation, operating well below 1 pulse per second. One can infer from data here - thrust 40 micronewton-seconds, total impulse 44 Ns, a lifetime of around a million seconds at that power, and a pulse rate about 1 pps. (It can be throttled back so lower rates are possible).

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    \$\begingroup\$ ...oh but it is \$\endgroup\$ – DrFriedParts Dec 30 '16 at 0:26
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    \$\begingroup\$ Also keep in mind that kinetic energy is 0.5mv^2, whereas momentum is mv. To achieve the same momentum on half as much mass, you need twice as much energy. \$\endgroup\$ – Nayuki Dec 30 '16 at 1:51
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    \$\begingroup\$ @Nayuki Your comment is the key here to understanding; and a direct answer to the OP. Nicely added. \$\endgroup\$ – jonk Dec 30 '16 at 5:18
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    \$\begingroup\$ Does anyone know the pulse rate as ion thrusters span a wide range in powers not just"usually kilovolts and microamps" which implies <1W when up to 7kW has been used. It does not indicate the kJ so rep rates must be slow due to the relaxation oscillator charging up to ionization cell breakdown potential with Xenon gas in a small gap E-field. In any case no one explained the ionization impedance changes for Coulombic and Lorentz forces, which I think was the "thrust" of the question. \$\endgroup\$ – Sunnyskyguy EE75 Dec 30 '16 at 8:47

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