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Hollow Cathodes 261 Since the electron temperature is given by the solution to Eq. (6.4-1) in the insert region (as shown above), Eq. (6.4-15) can be solved for the cathode sheath voltage as a function of the discharge current if the radiation and conduction heat losses are known. The insert heat losses are found from thermal models of the cathode, which will be discussed in Section 6.6. Equation (6.4-15) can be greatly simplified by realizing that in most cases TeV / 2 << (U + + s ) , and the right-hand side is essentially equal to one. Equation (6.4-15) then reduces to a simple power balance equation, and the cathode sheath voltage is s=H(T)+5TeV + wf IeR. (6.4-16) Ie 2 Figure 6-10 shows the calculated sheath voltage from Eq. (6.4-16) for the NSTAR cathode at a fixed 3.7-sccm xenon flow rate as a function of the discharge current for four values of the combined radiated and conducted power loss. From Fig. 6-9, the electron temperature is taken to be 1.36 eV for the 7.8 torr measured at 13 A of discharge current and this flow. A thermal model of this cathode [30] indicates that the insert heat loss is about 13 W at 13 A of discharge current, resulting in a sheath voltage from the figure of only about 3.6 V. In this case, a significant fraction of the 1.36 eV electron-temperature plasma electrons can overcome the sheath voltage and be collected on the insert to provide heating. The balance of the power required to heat the insert in the NSTAR cathode comes from orifice plate heating [30], which will be discussed in the next section. Fig. 6-10. Insert sheath voltage versus discharge current for the NSTAR cathode for four values of the radiated and conducted heat loss.PDF Image | Fundamentals of Electric Propulsion: Ion and Hall Thrusters
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