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Please keep us informed of any new information/changes. Thank you Joe KD2ODA ________________________________ Sent: Tuesday, January 7, 2025 8:24 AM To: [email protected] <[email protected]> Subject: Re:

Thanks Mike! [email protected]> wrote:

Hi all: In case you haven't heard from other sources, there was a fire yesterday at Brian Szewczyk (NJ1F) home, causing a lot of damage. Both he and his wife Susan (K1RQ) are fine (and so are their

?Howard Your question really peaked my interest. I remembered reading in my K3s manual that it has the ability measure the current draw. Had to go back and read the manual to determine the location

Nice David Gealt KD2BVA COMT, AUXCOMM Warren County EC/RO ENY Northern District ADEC

They also have it unterminated if that works better for you David Gealt KD2BVA COMT, AUXCOMM Warren County EC/RO ENY Northern District ADEC

[email protected]> wrote:

David, that’s it!! Thank you! I’m not, um, “currently” using powerpoles for the FTDX10, but I should be able to find a new (i.e., replacement) cable from Powerwerx when I buy the meter. Tim,

Howard, I was calculating transmitting efficiency, so didn't include receive current, so 20 amps is correct. I added 3 amps back at the end to get 8. I like the in-line meter with powerpoles! Even

Or you could go with less expensive multimeter KAIWEETS HT206D Digital Clamp Meter T-RMS 6000 Counts, Multimeter Voltage Tester Auto-ranging, Measures Current Voltage Temperature Capacitance

I’m assuming you are using powerpoles. Powerwerx Watt Meter, DC Inline Power Analyzer, 45A Continuous, 12 Gauge, Powerpole Connectors powerwerx.com David Gealt KD2BVA COMT, AUXCOMM Warren County

Thank you all for your detailed responses! Yes, I wish it was all that simple. (Tim, it looks like you meant to put 23, not 20, in the formula you cited.) What would help is if I still had my Astron

Hi Howard, Must make several assumptions. The receive current is the same at 25W. The efficiency is 100/(20x13.8)=36.2%. The efficiency does not change at 25W. Therefore, 25W/.362=69W, 69W/13.8V=5

Howard, If I use the P over IxE with P = 13.8. E =25 then I would be 5.52? Or. Amps = Watts/ Volts. 25 / 13.8 =1.812 A Just guessing. Frank. : ) Sent using the mobile mail app

? ?Howard Howard, you’re 100% correct on the current draw versus output power. The current draw of you rig operating at an output power of 25 Watts is not as simple as taking the current draw

The simplest answer I can think of is test it with an ammeter. David Gealt KD2BVA COMT, AUXCOMM Warren County EC/RO ENY Northern District ADEC

My Yaesu HF transceiver is spec’d to draw 23A @ 13.8VDC when transmitting at 100 watts. For receive, it’s spec’d at drawing 3A @ 13.8V. How can I determine how many amps the transceiver draws if

Happy New Year This first weekend of the new year has a several Radio Sport and On Air activities to partake in. First off ARRL Kids Day. While not really a contest or radio sport is it

Just a follow up and a reminder that if you made some contacts in last weekends ARRL 10 meter contest you still have time to enter your log. The deadline is 6:59 pm local on Sunday 12/22. If you used

Do you know a graduation high school student that will be attending college, or a current college student that has a ham radio license? The ARRL Foundation has many scholarships to be awarded in