Any Guideline for how many amps for watts?

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Caesar

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I am wanting to get a 10amp or 25amp psu for running a mobile rig in the house, i think max output is 45 watts on 1 radio and 80 watts on another, would the 10 amp work the 45 watts no problem, or better to check into the 25amp? i know that i've gotten about 20 watts out on a 3amp, but i think that was straining it, just courious if their are any sorta rule of thumb for this as i am sure the radio's amp drain varies from maker to maker...

thanks...
 

Caesar

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hehe, well, their is the cost difference, hmm, that and i just found out that the 25amp is a homemade one, i will have to see how it looks....
 

jay427

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Rule of thumb,

remeber this is rule of thumb and will get you kinda of close


Take your outputs watts (20watts) then take times 2 (2 x 20watts = 40watts) then divide it by your voltage (40watts / 13volts = 3 amps).


Just a quick rule of thumb, if anything it will be a lillte high.

You also need to look at the start up current/peak current.

I agree with the above post, go for the 25amp supply. Beside the price difference between a 10 amp and a 25amp is not a whole lot.


Hope this helps

Jay, n0mdf
 

MacombMonitor

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Caesar said:
hehe, well, their is the cost difference, hmm, that and i just found out that the 25amp is a homemade one, i will have to see how it looks....

Don't exclude it on the basis it's homemade. If it was made right, it's could be a lot better than the ones you can buy.
 

Concrete1

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If you check the Mfgr's specs of all modern 50W mobile radios, they all are rated at 10 or 11 Amps draw at full output, & all come with a 15A fuse in the line.

*** There are TWO Different wattage ratings when looking at a tranceiver....

*The Power wattage draw from the power supply, AND the RF output wattage from the tranceiver to the antenna.
You can't plug the "RF output" figure into a caluculation using the input voltage, that would be looking for the power consumption wattage figure....
For example, A New Yaesu or Icom radio Drawing 10 amps at 50W rf output, using ohms law (Volts x Amps = watts), 12v x 10a = 120watts - That radio is drawing & requires 120w from the power supply, to develop 50w of RF output, which makes sense since heat, dial lights, circuits, & component ineffencies ect are present. Those are your 2 different "watts" numbers for a tranceiver.
Some people erroneously plug the "Rf output" wattage into the voltage/power consumption calculation, & come up with far fewer amps than the Tranceiver actually requires at full output.
 
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N_Jay

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Concrete1 said:
If you check the Mfgr's specs of all modern 50W mobile radios, they all are rated at 10 or 11 Amps draw at full output, & all come with a 15A fuse in the line.

*** There are TWO Different wattage ratings when looking at a tranceiver....

*The Power wattage draw from the power supply, AND the RF output wattage from the tranceiver to the antenna.
You can't plug the "RF output" figure into a caluculation using the input voltage, that would be looking for the power consumption wattage figure....
For example, A New Yaesu or Icom radio Drawing 10 amps at 50W rf output, using ohms law (Volts x Amps = watts), 12v x 10a = 120watts - That radio is drawing & requires 120w from the power supply, to develop 50w of RF output, which makes sense since heat, dial lights, circuits, & component ineffencies ect are present. Those are your 2 different "watts" numbers for a tranceiver.
Some people erroneously plug the "Rf output" wattage into the voltage/power consumption calculation, & come up with far fewer amps than the Tranceiver actually requires at full output.

And I was just going to post a quip about "Ohms Law AND the efficiency factor", but you covered much nicer. :D
 

kb2vxa

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Hi again,

My overly simplistic "Ohm's Law" answer was designed to stimulate curiosity in it prompting a bit of study. Then you guys HAD to muddy the waters so here comes MY dose of the river bottom, headroom. Now let's say you're the average 6' tall man and you walk into a room with a 6' ceiling, you bump your head on the chandolier, right? That's why the standard is 8' which gives you 3' of headroom.

This must also be considered for the power supply. OK, you total up the load but it doesn't end there, the supply needs headroom so it doesn't strain to provide the proper voltage at full load and not develop excessive heat. Yeah, they rarely provide adequate cooling as it is. Most hams use 13.8VDC regulated supplies rated at 20A continuous and 30A intermittent which gives 100% headroom, the supply being able to deliver 200% of what the load requires. The supply then loafs along, provides more than adequate regulation and barely gets warm, especially since roughly 90% of the load is intermittent.

Please note that we don't consider "watts" because that is confusing, the voltage is understood so amps is the only thought we give to the matter. I have yet to see "watts" on the ratings plate, but some "wall warts" and other related DC supplies may have the AC INPUT rated in watts. I guess they want us to think in terms of the light bulb or something. (;->)

Now if I REALLY wanted to screw you up I would explain why "watts" have nothing to do with AC circuits being it's a DC measurement. The AC measurement is volt-ampere because VAR (volt-ampere reactive) is the AC equivelent of Ohm's Law and all that stuff about efficiency becomes power factor. I have only tantalized you, the next lesson is in polyphase power systems and vector calculations.

"You enter these halls with a mind full of MUSH, and leave thinking like electrical engineers."
Now go chase an electron. (;->)
 
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N_Jay

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kb2vxa said:
Hi again,
. . . Now if I REALLY wanted to screw you up I would explain why "watts" have nothing to do with AC circuits being it's a DC measurement. The AC measurement is volt-ampere because VAR (volt-ampere reactive) is the AC equivelent of Ohm's Law and all that stuff about efficiency becomes power factor. . . .

Watts are in fact both and AC and a DC measurement, VA are different from Watts in an AC system due to the phase shift between currentand voltage and the difference is measured as the Power Factor.

Watts and VA are the same in a DC circuit (during steady state operation).


Learn first then teach! :D :D
 

Al42

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kb2vxa said:
Now if I REALLY wanted to screw you up I would explain why "watts" have nothing to do with AC circuits being it's a DC measurement. The AC measurement is volt-ampere because VAR (volt-ampere reactive) is the AC equivelent of Ohm's Law and all that stuff about efficiency becomes power factor. I have only tantalized you, the next lesson is in polyphase power systems and vector calculations.
Actually watts DO figure into AC circuits, because VARs don't generate power, but that.s SO off-topic ...
 

rja1

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A real simple way to determine xmit current draw would be to check the manual.

bob
n2oam
 
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