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Re: [TenTec] battery life

To: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Subject: Re: [TenTec] battery life
From: Robert & Linda McGraw K4TAX <RMcGraw@Blomand.Net>
Reply-to: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Thu, 30 Mar 2006 19:41:02 -0600
List-post: <mailto:tentec@contesting.com>
I've always figure the discharge of a lead-acid car battery at 50% of its 
rating.  Example:  150 AH battery at 50% is 75AH.  Now to draw 10 amps 
average, we must consider transmit at some 22 amps and receive at some 2 
amps average for about 50% of the time.  Now, 75/10 = 7.5 hours.  If you 
have a "hot shot" CW operator that runs 30 to 50 WPM then the duty cycle 
goes up and so does the average current. Therefore 10 amps may be a bit low 
for an average figure.  If the bands are dead and you spend 95% of the time 
listening then the average goes down.

In a 100% receive mode theoretically the battery would last, 75/2 or 37.5 
hours.  The 2 amps is with min audio so expect the real current drain to be 
a bit higher. So we will take a more realistic approach and say  75/3 = 25 
hours.

Other factors one must consider is the internal IR drop of the battery.  New 
batteries are lower, old batteries are higher and deep cycle batteries tend 
to produce a higher voltage longer but at the expense of less current. 
Keep in mind that this radio is not going to like a voltage below about 12.8 
to 13 volts.  It is rated at 13.8 VDC.  A lead-acid battery open terminal 
voltage not under charge will likely be 12.8 to 13.5 depending electrolyte 
temperature. That takes a HOT battery, one that is very new and preferably a 
deep cycle marine battery.  Therefore, a car battery is not the battery of 
choice as it is designed to be a "high current" device and that is not the 
application here.  Based on this scenario, one could not expect a battery to 
provide much more than 25% of its rated AH value and thus for a 150 AH 
battery we now have 37.5 AH available.  Again using our example, 37.5/10 is 
3.75 hours of use time before required re-charge or a radio and associated 
internal processors getting flakey.

OK, so one says we'll have two batteries, one to run on and one to charge. 
Not so fast on the draw cowboy.  The correct charge rate should equal 2X to 
3X  the discharge time so 5 to 7.5 hrs will be required to restore the 
battery to full charge.   Oh, one can do a fast charge but this raises the 
electrolyte temperature, plates and spaces expand and the capacity of the 
battery in terms of AH decreases.   This gets us to the point that we need 3 
to 4 batteries to support the operation.

In other words, the Orion II is a power hog and not really suitable for 
Field Day battery operation.

One solution, use a good fresh marine deep cycle battery to power the radio, 
keep a charger attached and power the charger off of the generator.  Using 
this method, the radio will have lots of current when needed, the battery 
acts as a big filter capacitor so humm from the charger is not a concern, 
and the battery acts as a voltage regulator to protect the radio from surges 
and generators as they "do their thing with changing loads", and when the 
generator dies or is down for refueling {you do shut it down don't you?} you 
can continue operating the station.

73
Bob, K4TAX
A connoisseur of fine radios


----- Original Message ----- 
From: "Larry DiGioia" <listacct@longwire.com>
To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>
Sent: Thursday, March 30, 2006 5:56 PM
Subject: [TenTec] battery life


>I am looking for guesses as to how long an O2 would run, connected to a
> reasonable large, fully-charged car battery, while operating CW for
> Field Day...?
>
> -- 
> Larry  N8KU
>
> w w w . l o n g w i r e . c o m
>    100% CW     100% HF
>
> _______________________________________________
> TenTec mailing list
> TenTec@contesting.com
> http://lists.contesting.com/mailman/listinfo/tentec
> 


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