| Author |
Message |
    
Rumblefish
| | Posted on Saturday, December 07, 2002 - 1:14 pm: | 
|
I plan on using my shooting star 225 in the house on my base. Any suggestions for the proper amp rating on a power supply to run. I've heard the rule of ten's for this. 120 watts divided by ten would be 12 amps. Would this be enough to run the amp? nothing else will be run off the supply, just the amp. |
307
| | Posted on Sunday, December 08, 2002 - 8:51 pm: |
|
You will need about 35 amps to do the job...
225 Watts / 13.8V = 16.3A * 45% Eff = 23.65A Min
307 |
Taz
| | Posted on Monday, December 09, 2002 - 1:19 am: |
|
a rule of thumb, approx 10 amps per 100 watts |
Rumblefish
| | Posted on Monday, December 09, 2002 - 4:26 pm: |
|
Thanks 307 & Taz. I'm not sure if I understand the 45% eff rating though. |
307
| | Posted on Monday, December 09, 2002 - 7:58 pm: |
|
eff is the "Efficiency" factor..Most amps are only about 45% , 55% is heat transferred to the heat sink...a 100% eff amp would use super conductors or "close to 0 ohms" impedance would cause no heat..about 50K if they were available...
307 |
Taz
| | Posted on Monday, December 09, 2002 - 9:40 pm: |
|
hhmmmmm, cant they be made? |
Rumblefish
| | Posted on Tuesday, December 10, 2002 - 5:41 pm: |
|
Thanks for the clarification on efficiency. From what I understand with the super conductors, In order for them to operate properly they must be subjected to some seriously cold temps. -465 fahrenheit. bbbbbuuuurrrr..... |
Orion
| | Posted on Monday, May 26, 2003 - 6:19 am: |
|
Just to clarify this so I understand it:
Amperage*Voltage=Wattage is the formula unless I am mistaken. Is this correct so far?
Now, with the efficiency factor of 45%, I would multiply that wattage by 0.45 and that would be an accurate depiction of the usefull wattage I could get out of the power supply. And, the amperage factor in the equation would have to be the "continuous" amp rating. Ok, do I have a handle on this thing?
That's for amplifier issue. Would a radio be different. Lets say a radio can peak RF of 55watts. It takes a bit more to run the lights and other components. Is that extra power need negligible or does the formula need a general alteration when considering radio power requirements? For example, "the above formula for 55 watts plus 5%" or something like that. Thanks for the info in advance guys.
Thanks techs and knowledgable folks out there for response in advance. |
Orion
| | Posted on Monday, May 26, 2003 - 4:01 am: |
|
A late addition to ths string indeed, but, the LED lighting principle is the same. ncandescent(conventional) lighting see's the majorty of it's energy wasted in heat and just a smigeon go toward illumination. Ridiculously wasteful and inneficient. LED's throw NO heat and nothing but light. I've got a CMG Infinity Task light. One pure white LED will light illuminate ten feet wide and 15 feet ahead although not very brightly for 43 hours on one AA battery. The future is LED, but right now they're not relatively cheap. WWWith all the talk about heat and efficiency that's what came to mind. Oughta put them in radio lighting; no heat and negligable power drain. |
|
