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Re: How to measure current drawn by a 100w transceiver at 25w

 

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 when sending CW, the duty cycle is only 50%. Add receive time during a QSO and you're down around 25%, add casual listening and you're maybe 5%. I like the LiFePO4 batteries. High capacity, lightweight, good voltage during discharge. Good luck, 73, Tim W2UI

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On Jan 6, 2025, at 8:37?AM, "Howard Lester via " <gmail.com@groups.io target=_blank>[email protected]> wrote:

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 RS-35M with its meters. Then I could get at least an approximation of amperage draw at different power outputs. Or, is there is such a thing as an ammeter that can be installed between the radio and the power supply? I know there is such a thing as a clamp-on ammeter, but would that be the right tool for the job? (I don¡¯t like to ¡°play¡± with electricity!)

Brian, the Yaesu FTDX10 radio draws a constant 3A with a received signal, far more than a QRP radio like an IC-705. I don¡¯t know what the lowest voltage is that the Yaesu can tolerate. That¡¯d be another thing I'd have to find out.?

I run both SSB and CW, no digital. At home I mostly listen. If I were out, I¡¯d want to be making contacts. Duty cycle? I don¡¯t know, but surely it¡¯d be more listening than sending. I know that CW draws more than SSB. It¡¯d be a lot easier if I just had a QRP-type radio. ?:-) But, I¡¯d like to be running 20 or 25 watts, not 6 or 10. I suppose the easiest thing for me to do, if I wanted to operate ¡°out there somewhere,¡± is to just get a 12 or 15A battery and have at it. ?

Howard ?N7SO



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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