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 amps. 5+3=8 amps. What mode are you using and what will the duty cycle be?
73, Tim W2UI
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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. : )
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On 1/5/25 at 9:05?PM, Howard Lester via groups.io wrote:
From: "Howard Lester via groups.io" <howardlester12@...>
Date: January 5, 2025
To:
[email protected]
Cc:
Subject: [K2DLL - Saratoga County ARA] How to measure current drawn by a 100w transceiver at 25w
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 I transmit at, say, 25 watts? (No, 23A x 1/4 = 5.75A is not the correct answer; I wish it was that simple. It should be more than that.)
In practical terms, let¡¯s say I want to take the transceiver to the middle of nowhere and power it from a Bioenno battery for one hour. Again, let¡¯s say I¡¯ll be transmitting at 25 watts. (Yes, I know I won¡¯t be operating ¡°key down¡± for an hour straight.) What amperage battery would I need? (Never mind ¡°the biggest one you can afford!¡±) This is sort of an exercise in¡something, and I really do want to know.
Thanks,
Howard N7SO
Tim W2UI