3.6M north10.pdf - Dean-O's Toy Box

160 High-Power Microwave-Tube Transmitters
charge is increasingly intolerable. (Remember what we used to pick up on the
AM car radio when we drove near high-tension wires, especially on a foggy
day?)
We have discussed the fact that the removal of charge from a capacitor bank,
no matter how large, will result in its terminal voltage being smaller than it was
before the charge-removal interval, or pulse. In the interpulse interval, therefore,
it will be necessary to replace the charge removed during the pulse. In the
discussion of line-type modulators, much was made of the issue of recharging
without incurring energy loss if the current was limited by resistance. Is there a
similar concern when it comes to replacing partial discharge? The answer is not
nearly as much. The reason for this is shown graphically in Fig. 10-6, which plots
energy removed from the dc power supply, energy transferred to the capacitor
bank, and energy dissipated in series resistance versus the per-unit amount that
the capacitor is discharged each pulse. (This graph assumes that the resistance is
small enough so that the RC time-constant is less than approximately 1/3 of the
minimum interpulse interval, assuring that the capacitor will recharge to approximately
the power-supply terminal voltage between pulses.) If we look at
full discharge, or per-unit discharge of unity, the results are familiar. The perunit
capacitor energy transfer and resistor energy dissipation are both unity, and
the per-unit energy removed from the power supply is 2. Notice, however, that
only the energy removed from the power supply is a linear function of per-unit
discharge voltage. For less-than-unity per-unit discharge, energy transferred to
the capacitor falls off less rapidly than dissipation in the resistor. Therefore, the
efficiency of energy transfer increases as AV/Vdecreases.
The reason for this can be seen if we note the change in charge during a pulse
of current, AQ = CAV, and the change in capacitor energy, AW = AQ, times average
capacitor voltage during discharge, which can be expressed as
~Avx
V+ V–AV
2
or
~vAv _ C(AV)2
The energy removed from the power supply during recharge is AQ times the
power supply voltage, which can be expressed as CAV x V. Recharging efficiency
is the ratio of energy transferred to the capacitor to the energy removed from the
power supply, or
2“
+Av-a=,
——— l[AV)
CVAV 2V”