QRZ! Ham Radio 4/QRZ Ham Radio Callsign Database - Volume 4.iso/digests/equip/940032.txt

Internet Message Format | 1994-11-14 | 18KB

Date: Mon, 14 Feb 94 04:30:24 PST
From: Ham-Equip Mailing List and Newsgroup <[e-mail]>
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Subject: Ham-Equip Digest V94 #32
To: Ham-Equip


Ham-Equip Digest            Mon, 14 Feb 94       Volume 94 : Issue   32

Today's Topics:
                      Alinco DJ580 and car power
                    Help/info about Standard C158A
                      Looking for LOGIKEY keyer
                Needed: 440MHZ plug for Bird Wattmeter
                         Need SW + AM?FM unit
                     Packet with an HT and Newton
                           Ten Tec Delta I
                           TH78A any good?
                      Vertical Antennas (3 msgs)
                    Wanted: YK 88C 500Hz CW filter

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Problems you can't solve otherwise to [e-mail].

Archives of past issues of the Ham-Equip Digest are available
(by FTP only) from UCSD.Edu in directory "mailarchives/ham-equip".

We trust that readers are intelligent enough to realize that all text
herein consists of personal comments and does not represent the official
policies or positions of any party.  Your mileage may vary.  So there.
----------------------------------------------------------------------

Date: 10 Feb 1994 19:58:51 GMT
From: concert!news-feed-2.peachnet.edu!umn.edu!gaia.ucs.orst.edu!ucs.orst.edu![e-mail]
Subject: Alinco DJ580 and car power
To: [e-mail]

In article <2jb09s$[e-mail]> [e-mail] (Larry Rubin) writes:
>My Alinco DJ580 is my only piece of ham equipment and thus I do not want
>to do anything that might destroy it.  But I would like to be able to
>hook it into the car's cigarette lighter to both save the battery and get
>more wattage out of it.  I have the necessary plug that fits into the DC
>in socket on the side of the radio, as well as a plane-jane cigarette
>lighter plug.  What type of fuse is recommended for this, and is there
>any other circuitry that is advisable to connect between the radio and
>the car?
>...

Larry, I also used a DJ580 for a long time as my only ham rig.  It was
in the mobile most of the time.  Most likely, from my experience, you
will get lots of engine noise through the cig lighter.

I recommend that you buy some 14 gauge wire and wire your plug straight
to your car battery, + and -.  Fuse it with no less than a 3A fuse.

For more power, I used a Ramsey kit amplifier for 2M -- 5 in 30 out.
This may cost you between $60 and $100.

Good Luck!          --Ray Stein
                      Oregon State University, Dept of E. Engineering

------------------------------

Date: 11 Feb 94 22:16:41 CST
From: agate!howland.reston.ans.net!vixen.cso.uiuc.edu!moe.ksu.ksu.edu!engr.uark.edu!news.ualr.edu!athena.ualr.edu![e-mail]
Subject: Help/info about Standard C158A
To: [e-mail]

Hi, I am in need of some info for the new Standard C158A 2 meter handheld
radio that is advertised in the Feb 94 QST on page 19... What I need is
the specific part number for some of the outer parts, knobs etc.. if anyone
has the radio and the book with the part numbers, please let me know..
I REALLY need this info.. Thanks..
********************************************************************************
*  -  -                         Peter M Stuckey                          -  -  *
*  0  0             University of Arkansas at Little Rock             0  0  *
*    )                  Internet "[e-mail]"                     (    *
*  \___/               Amatuer Radio Call Sign                      \___/  *
***KEEP**               KB5WCE - 146.55 MHz             **SMILIN'**
********************************************************************************

------------------------------

Date: Thu, 10 Feb 1994 19:19:15 GMT
From: netcon!bongo!netcomsv!netcom.com![e-mail]
Subject: Looking for LOGIKEY keyer
To: [e-mail]

Hannes Hogni Vilhjalmsson ([e-mail]) wrote:
: Can anyone tell me the present address of the Logikey Company,
: or any other outlet for their LOGIKEY microprocessor based morse
: keyer?

If I'm not mistaken, the LogiKey is the commercial version of the CMOS
Super Keyer II which was first described in the November 1990 issue of QST.
That keyer is available in Kit Form (i.e. parts, pcb, but no switches,
boxes, or batteries) from:

   Idiom Press
   Box 583
   Deerfield, IL 60015

When I bought mine (it is a WONDERFUL  keyer), I paid $45 + $3 for domestic
USA shipping.  Foreign orders were $45 + $5.

Sorry, that's all the info I have.
73 de Sandy WA6BXH/7J1ABV   [e-mail]   WA6BXH@N0ARY


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Date: 11 Feb 94 23:38:44 -0500
From: elroy.jpl.nasa.gov!usc!sol.ctr.columbia.edu!usenet.ucs.indiana.edu!master.cs.rose-hulman.edu!rosevc.rose-hulman.edu![e-mail]
Subject: Needed: 440MHZ plug for Bird Wattmeter
To: [e-mail]

Does anyone have or know where I can get a 440 MHz plug for the Bird wattmeter?

Dave

------------------------------

Date: Sat, 12 Feb 1994 06:30:30 GMT
From: mentor.cc.purdue.edu!mace.cc.purdue.edu![e-mail]
Subject: Need SW + AM?FM unit
To: [e-mail]

[Please excuse me if I posted this to an inappropriate news group.]

I am looking for a good receiver to integrate into my system. I want
a receiver that will have a good bandwidth SW  PLUS  AM & FM in one unit.
Unlike in Asia (and probably Europe), such units are hard to find here
in the USA. I have inquired at a number of stores and they don't carry them.

Any information regarding where I can find such units, approx. pricing,
alternatives (I am an absolute amateur putting together my sound system
for purely personal pleasure), obstacles to importing (if that's an option)
will help me greatly.

PL. E-MAIL ME. Many thanks in advance,

Gowri Narla
[e-mail]

------------------------------

Date: 9 Feb 94 20:53:54 GMT
From: concert!news.duke.edu![e-mail]
Subject: Packet with an HT and Newton
To: [e-mail]

I have a neat dual band radio, and a neat Apple Newton, and my thoughts
go as follows: If I can run packet off my calculator (HP48SX) with a
normal TNC, I should be able to find something for the Newton! Has
anyone tried this? Are there any products around that can run out of
the Appletalk or PCMCIA ports that are either TNCs or will allow me to
use an existing TNC with Newton? I'm not sure if I could use a TNC out
of the Newton appletalk (serial) port with just a cable adaptor, and
even then terminal software might be a pain. If any Macintosh users
have packet experience, that may be of help, too. Any
advice/references/opinions/prices would be greatly appreciated.

73's! N5SKQ  <-- hardest call to pronounce in the world

________________________________________________________________________
_____
                 J. D. Carter - [e-mail]


------------------------------

Date: 9 Feb 1994 13:29:35 -0600
From: ihnp4.ucsd.edu!sdd.hp.com!math.ohio-state.edu!cs.utexas.edu![e-mail]
Subject: Ten Tec Delta I
To: [e-mail]

Anyone know anything about this rig? Anyone have one for sale or trade?


73

Jeff,AC4HF

------------------------------

Date: Fri, 11 Feb 1994 22:34:11 GMT
From: sgiblab!cs.uoregon.edu!news.uoregon.edu!netnews.nwnet.net!raven.alaska.edu!acad2.alaska.edu![e-mail]
Subject: TH78A any good?
To: [e-mail]

I was just curious.. is the Kenwood TH78A HT any good?  I've been looking
through a number of magazines and catalogues and so far, it's the best I can
find.  Do any of you have any experience with this transceiver (140/440, 5w)?

Is there anything any of you are partial to?  I'm looking for a dual-band unit
with DTMF and small size (like a handheld, or even a portable).  I'd also like
something that can be easily recharged and is able to plug into a wall socket
so I don't need to run the batteries down.  Have any of you used something that
fits this description (and like it)?

Any help you could give would be great... E-Mail to this account, if you can.
My name is David Nesting (this account is a friend's).

Thanks again...

------------------------------

Date: Thu, 10 Feb 1994 23:03:16 GMT
From: mvb.saic.com!unogate!news.service.uci.edu!usc!howland.reston.ans.net!europa.eng.gtefsd.com!emory!kd4nc!ke4zv![e-mail]
Subject: Vertical Antennas
To: [e-mail]

In article <[e-mail]> [e-mail] (Alan Bloom) writes:
>Gary Coffman ([e-mail]) wrote:
>: In article <[e-mail]> [e-mail] (Alan Bloom) writes:
>: >Gary Coffman ([e-mail]) wrote:
>: >: In article <[e-mail]> [e-mail] (Alan Bloom) writes:
>: >: >Consider a vertical dipole in free space.  You could insert a horizontal
>: >: >infinite ground plane at the feedpoint without changing the radiation
>: >: >pattern.  Now you have two verticals, one pointing up, one pointing down.
>: >: >Each vertical radiates half the power of the original dipole.
>: >
>: >: True because each has half the current that flows in the entire dipole.
>: >
>: >No, the current is the same, but the power is halved.  There are (at least)
>: >two ways to see this:  1)  Only 1/2 the voltage is applied to each 1/4-wave
>: >element.  Since power = voltage times current, the power is 1/2.
>: >2)  The element is only 1/2 as long.  So the same current results in
>: >only 1/2 as much power radiated.
>
>: Dipole split by infinite ground plane.
>
>:                     |
>:                     |
>: -------/\/\/\/\---o | o---/\/\/\/\-------
>:                  E1 | E2
>:                     |
>:                     |
>         ^^^ 36.5 ohms       ^^^ 36.5 ohms
>         ------------ 73 ohms ------------
>
>: If we apply drive to E1-E2, equal currents are driven into each element's
>: impedance. So the halves of the dipole have equal currents flowing in them,
>: but 180 degrees out of phase. With the infinite ground plane isolating the
>: halves, one half has half the total current flow.
>
>Let's call the voltage applied between E1 and E2 "V".  Since there are
>equal and opposite voltages on the two terminals, the voltage applied
>to each is V/2.
>
>If, for example, V = 73 volts, the current in the dipole is 1 A (since
>the radiation resistance is 73 ohms.)  With the ground plane, the
>impedance of each 1/4-wave element is 73/2 = 36.5 ohms.  Since you have
>half the voltage (37.5 V) applied to each half, the current is still 1A
>in each 1/4-wave element.

Yes, yes, I understand that, but look at what you're saying, "the
current is *still* 1A in *each* 1/4-wave element." Since the dipole
has *two* elements, 1+1=2, it's instant flow is twice the current
of a single element. (I understand what phase does to *net* current
at the *feedpoint*, but that's really a different issue. When the
barrier of the infinite ground plane comes into play, it *isolates*
the two branches so we can treat them separately. Hence we can see
the individual 1 ampere flows at the feedpoints of the two halves
without phase combinations.)

Let's examine *why* an antenna radiates for a moment to see what
I'm getting at here. Radiation occurs when an electric charge is
accelerated. The relevant factors are the amount of electric charge,
RF current, the accelerating potential, RF voltage differential over
the charge path, and frequency, the rate of change of voltage along
a current path. These three are all intimately related, but in most
antennas, the instanteous current is a key to predicting radiation
field shape, and hence gain. The 1/4-wave monopole in the example
has half the instant current of the dipole, and half the total
end to end electrical potential. So crudely it would seem to have
1/4th the field strength, but it's length is 1/2 as great (frequency
effect) so the accelerating gradient is the same. That leaves the
*signs* of the current flows that make the field of a free space dipole.
These vector sum to the same field strength as the monopole over an
infinite ground plane. The dipole's currents generate fields which
vector sum in a way that makes 1+1 appear to equal 1. The ground plane
should be seen as a *shield* to prevent this summing, not as a mirror.

>The resulting field is the same for the ground-plane case as for the
>dipole in free space.  It is as if the other half of the dipole were
>still present.  That's where the concept of the "image" antenna
>extending below the ground plane comes from.

Uh huh, but an "image" antenna extending below the ground plane
is not reality. It's a visualization trick that's sometimes useful,
but the currents that actually flow are induced currents flowing
along the surface of the conducting plane. They are *not* the same
as the currents that would flow in an "image" antenna. They are the
currents a *field* generated by the image antenna would induce in
a perfect conductive sheet. This is important to understanding the
effects of *real* ground planes which are neither perfectly conducting,
nor infinite in extent. And is the reason *real* 1/4-wave monopoles
over *real* ground planes have less gain than vertical dipoles, or
1/2-wave vertical monopoles.

Gary


--
Gary Coffman KE4ZV          |    You make it,     | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems |    we break it.     | uunet!rsiatl!ke4zv!gary
534 Shannon Way             |    Guaranteed!      | emory!kd4nc!ke4zv!gary
Lawrenceville, GA 30244     |                     |

------------------------------

Date: Fri, 11 Feb 1994 02:05:52 GMT
From: mvb.saic.com!unogate!news.service.uci.edu!usc!sdd.hp.com!col.hp.com!srgenprp![e-mail]
Subject: Vertical Antennas
To: [e-mail]

Gary Coffman ([e-mail]) wrote:

: Yes, yes, I understand that, but look at what you're saying, "the
: current is *still* 1A in *each* 1/4-wave element." Since the dipole
: has *two* elements, 1+1=2, it's instant flow is twice the current
: of a single element.

If you installed RF ammeters in each element, they would read the
same no matter whether the ground plane is present or no.  (Since
the RF generator and both elements are in series, the current must
be the same in each.)  Each 1/4-wave element radiates 1/2 the total
power no matter whether the ground plane is present or no.

(Is anybody else still following this convoluted discussion?)

AL N1AL

------------------------------

Date: Wed, 9 Feb 1994 20:03:31 GMT
From: foxhound.dsto.gov.au!fang.dsto.gov.au!yoyo.aarnet.edu.au!news.adelaide.edu.au!basser.cs.su.oz.au!news.cs.su.oz.au!metro!dmssyd.syd.dms.CSIRO.AU!dmsperth.per.dms.@@munnari.oz.au
Subject: Vertical Antennas
To: [e-mail]

Gary Coffman ([e-mail]) wrote:
: In article <[e-mail]> [e-mail] (Alan Bloom) writes:
: >Gary Coffman ([e-mail]) wrote:
: >: In article <[e-mail]> [e-mail] (Alan Bloom) writes:
: >: >Consider a vertical dipole in free space.  You could insert a horizontal
: >: >infinite ground plane at the feedpoint without changing the radiation
: >: >pattern.  Now you have two verticals, one pointing up, one pointing down.
: >: >Each vertical radiates half the power of the original dipole.
: >
: >: True because each has half the current that flows in the entire dipole.
: >
: >No, the current is the same, but the power is halved.  There are (at least)
: >two ways to see this:  1)  Only 1/2 the voltage is applied to each 1/4-wave
: >element.  Since power = voltage times current, the power is 1/2.
: >2)  The element is only 1/2 as long.  So the same current results in
: >only 1/2 as much power radiated.

: Dipole split by infinite ground plane.

:                     |
:                     |
: -------/\/\/\/\---o | o---/\/\/\/\-------
:                  E1 | E2
:                     |
:                     |
         ^^^ 36.5 ohms       ^^^ 36.5 ohms
         ------------ 73 ohms ------------

: If we apply drive to E1-E2, equal currents are driven into each element's
: impedance. So the halves of the dipole have equal currents flowing in them,
: but 180 degrees out of phase. With the infinite ground plane isolating the
: halves, one half has half the total current flow.

Let's call the voltage applied between E1 and E2 "V".  Since there are
equal and opposite voltages on the two terminals, the voltage applied
to each is V/2.

If, for example, V = 73 volts, the current in the dipole is 1 A (since
the radiation resistance is 73 ohms.)  With the ground plane, the
impedance of each 1/4-wave element is 73/2 = 36.5 ohms.  Since you have
half the voltage (37.5 V) applied to each half, the current is still 1A
in each 1/4-wave element.

The resulting field is the same for the ground-plane case as for the
dipole in free space.  It is as if the other half of the dipole were
still present.  That's where the concept of the "image" antenna
extending below the ground plane comes from.

AL N1AL

------------------------------

Date: Fri, 11 Feb 1994 14:58:00 GMT
From: mvb.saic.com!unogate!news.service.uci.edu!usc!cs.utexas.edu!swrinde!sgiblab!pacbell.com!att-out!att-in!cbnewsm![e-mail]
Subject: Wanted: YK 88C 500Hz CW filter
To: [e-mail]

Hi ya all.
I need the 500 Hz cw filter that goes in a TS-430, 440, etc.
Kenwood xceiver (8.83 MHz).  I am not putting it into a xceiver so any
physical condition is OK as long as it works and I can solder
coax to the leads.

Anybody got one that I can buy?

72/73 de ND3P
Scott McLellan
days:610-391-2161
eves (EST before 9 pm) 610-756-6992
or email to me.

Thanks!!!

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End of Ham-Equip Digest V94 #32
******************************