Owner`s manual

4.7.1 Power Module Configuration
The power module output current is approximately 10A/volt when it is configured for 120VAC, 5A/volt when config-
ured for 240VAC. Figure 4-2 to 4-4 are block diagrams of the power module. DC power enters the module via bus
bars and is applied to a small EMI filter (called a “PI filter”). The filter is designed to reduce the high frequency
switching ripple and noise on the battery bus. It consists of powdered iron bus bar chokes, aluminum electrolytic
and polyester capacitors. A DC current sensor (“LEM” LA 125-P used to control the battery booster), inrush current
control, and low DC voltage turn OFF circuitry is also located on this printed circuit board (PCB). The battery
voltage, 48V, is transformed to two separate, isolated +/-212VDC outputs, which is used by the inverters to generate
the AC output sine wave voltage. This “booster” is a Current Feed, Center Tapped, Push Pull (CFCTPP) converter
operating at 20KHz. Circuit connection is shown in Figure 4-4 and is as described below.
4.7.1.1 Battery Input
The positive battery input, after filtering, is connected to an input inductor, L1. The output side o of L1 is connected
to the center tap of a ferrite core transformer. The outside primary legs of the transformer are connected to Insulated
Gate Bipolar Transistors (IGBTs). The two secondary center tapped windings on the transformer are connected to
bridge rectifiers, which are located on the “Primary Snubber” PCB. The rectifiers then connect to an energy storage
capacitor bank. The capacitor bank voltage is regulated to produce two separate plus 212VDC and minus 212VDC
outputs. The capacitor bank (Capacitor PCB) connects to the inverter output IGBTs, which in turn, connects to the
output filter inductor, then to the output connect relays before connecting to the output filter capacitors in the AC
raceway by a rack and panel connector.
4.7.1.1.1 Battery Booster Operation
We will attempt to describe a typical “Battery Boost” switching cycle when the unit is operating properly. Figure 4-4
is a block diagram of the “booster”. First, both IGBTs are turned ON, thus creating a short circuit on the primary
winding of the boost transformer. Under this condition, the center tap of the transformer looks like it is connected
to the negative DC input battery terminal through the IGBTs. Battery voltage is now applied to the input inductor.
Current through the inductor builds up linearly as a function of time. The formula, V=L x dI/dT applies, where “V” is
the voltage applied across the inductor, “L” is its inductance, “dI” is the rate of change in the current, and “dT” is the
time increment. The choke’s inductance is about 35uHy. Ripple current is set to be near 18A at 48VDC input.
Ripple frequency is 40KHz, twice that of the operating frequency of the boost transformer. Under normal operation
at 48VDC input, the average battery current is 72Amps for full output of 3000 watts. Thus the current in the input
inductor is sweeping up and down between 64.5A to 82.5A at a 40KHz rate. When the LEM sensor, LA125-P
located on the input “PI Filter” PCB measures a peak current of 82.5A, the control circuitry turns off one of the IGBTs,
the other IGBT remaining ON. Because one of the IGBTs is now OFF, the current in the inductor will cause the
center tap of the transformer to ramp up to the near +78 volts in about 0.5uSec (due to the snubber circuit). Due
to the boost transformer’s turns ratio of 4:11, the voltage reflected to the primary center tap from the 212.5V
secondary will be near 78 volts ((212.5+1.35)*(4/11)=77.76). The transformer now has voltage impressed across it,
so power can be transferred to the output to charge up the output bus capacitors. The capacitor voltage will be
regulated to near 212.5VDC. Since the voltage on the boost choke is now reversed (output side at 78V, higher than
the input side at 48V), the current in this choke starts to decrease. Note that when one IGBTs is turned OFF, the
voltage on the collector of the OFF IGBT will be two times the center tap voltage, that is, 156V. When the choke
current becomes less than 64.5A, the IGBT that was OFF is turned back ON, thus both IGBTs are again ON and
the choke currents starts to build up again. This completes one half of the switching cycle. Next, the other IGBT
that was ON in the previous power transfer cycle is now turned OFF because the choke current reached the 82.5A
point, so power is again transferred to the inverters DC bus capacitors. The total time for the above switching cycle
is 50uSec, or 20KHz. The waveform applied to the boost transformer is a duty cycle modulated voltage, which
produces +/-212V on its output.
Owner’s Manual
Theory of Operation page 4 — 7