User's Manual
ALPHA 9500 Interim OPERATING MANUAL
www.alpharadioproducts.com
24
4.2.4 Power Cord Connections
WARNING!
To avoid the hazard of a potentially fatal electric shock and/or severe damage
to the ALPHA 9500 and other equipment, always use an AC plug that is
appropriate for the primary mains voltage, current rating and configuration.
NEVER use 120V-type plugs to connect to power receptacles for 190-250V
circuits.
ALWAYS use grounding type AC connectors which conform to local codes and ensure that the
green wire in the Alpha 9500 power cable is wired only to the AC mains safety ground (or to
neutral, as may be necessary with a 240V circuit configured 120V-N-120V without a separate
ground, commonly found in the US).
The green conductor in the power cord is wired to the ALPHA 9500 chassis. It MUST be
connected only to the power source safety ground or neutral. The black and white power cord
wires connect to the two “hot” wires of the AC source; either wire may be connected to either side
of the line. For best results use a dedicated 200-240 V branch circuit of #10 AWG copper wire or
equivalent, rated at 20 A, to feed the amplifier.
4.2.5 Important Information About Operation from 90-130V AC
Electrical power equipment will draw twice as much primary current from 120 V mains as from
240 V mains. Therefore, operating the ALPHA 9500 on a typical 120 V/20 A household circuit
without exceeding the 20 A circuit rating will limit maximum peak power output to about 600-
1000 W. Maximum possible RF output power for any particular primary AC voltage and current
capacity may be estimated as:
P
o
max = (V
line
x I
line
) / 2.3
For example, if the Alpha 9500 operates from a circuit that is capable of delivering 115 V AC at a
maximum current of 20A, with no other loads connected to the circuit, maximum peak RF output
possible without tripping the 20A breaker (or fuse) is approximately:
P
o
max = (115V x 20A) / 2.3 = 2300/2.3 = 1000 W
If the same circuit also supplies a transceiver drawing peak line current of 5A and a lamp drawing
1A, only 20-5-1 = 14A is available for the amplifier and maximum possible output is about:
P
o
max = (115V x 14A) /2.3 = 1610/2.3 = 700W