Alpha 9500 HF Linear Amplifier Operating Manual Alpha Radio Products
Preparing Your Station Product Release 1
DOCNUMBER 9500
Document Issue 1, Revision 7
Page 4–4 June 2009
44
NOTE: The FCC requires users to check their installations for
compliance with published values for allowable exposure to RF fields.
This information is available in ARRL publications, FCC printed rules,
and on the web. We strongly recommend that you do this for any
installation, both fixed and at an expedition or contest site.
If you have any questions regarding engineering your amplifier into
your amateur radio station, visit our technical-support website at
www.alpharadioproducts.com.
Step 5 Provide surge protection.
Induced energy from nearby electrical storms or other power transients
may damage components. Such damage is not covered under warranty. It
is therefore important to use a good lightning arrestor. However the only
lightning-proof solution available is to disconnect antenna feedlines and
AC power when the equipment is not in use.
NOTE: Whenever the amplifier is online — either off, in standby
(STBY), or in warm-up with the WAIT LED lighted — the amplifier is
bypassed and the exciter is connected directly to the antenna. The
throughput limit in all cases is 1500 W.
4.2 Limitations of Operation at 90–130 VAC
Electrical-power equipment draws twice as much primary current from
120 V mains as from 240-V mains. Therefore, if you operate the
ALPHA 9500 on typical 120 V/20 A household circuit without exceeding
the 20-A circuit rating, you 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:
Po max = (Vline x Iline) / 2.3 (4-1)
For example, if the amplifier operates from a circuit that delivers 115
VAC at a maximum current of 20 A with no other loads connected to the
circuit, maximum peak RF output possible without tripping the 20-A
breaker (or fuse) is approximately:
Po max = (115 V x 20 A) / 2.3 = 2300/2.3 = 1000 W (4-2)
If the same circuit also supplies a transceiver drawing peak line current of
5 A and a lamp drawing 1 A, only 20-5-1 = 14 A is available for the
amplifier and maximum possible output is about:
Po max = (115 V x 14 A) /2.3 = 1610/2.3 = 700 W (4-3)
Following are some considerations at the high and low ends of this
voltage range that are rarely encountered.
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