| Author |
Message |
    
jyd
| | Posted on Saturday, October 12, 2002 - 10:52 pm: | 
|
is this a good amp?how much does it draw? |
Tech671
| | Posted on Sunday, October 13, 2002 - 8:40 am: |
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Old or new? There's a big difference. The older Pals were pretty hardy. keep input carrier and drive low, they did about 100w peak and draw around 10a. The new ones called "80" aren't so hot. I'd much prefer the KL60 or 203. |
Taz
| | Posted on Sunday, October 13, 2002 - 11:53 am: |
|
Like I have said before.
10 amps per 100 watts.  |
jyd
| | Posted on Monday, October 14, 2002 - 5:28 pm: |
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i was thinking of getting a new one,i guess a tuned vr-9000 would kill it?just looking for something good and cheap maybe a texas star 1 pill modulator.you dont have to have alot of watts to talk skip if your stuff is set up right,copper dont carry the texas star modulator, why? i have had 100 watt amps draw more than 10 amps.i boubht another vr-9000 and i am sending it to 307.he does an awsome job plus does the frequency counter mod and quick turn around time. |
Taz
| | Posted on Monday, October 14, 2002 - 6:59 pm: |
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jyd, like I have said before its also an estimate |
jyd
| | Posted on Monday, October 14, 2002 - 11:49 pm: |
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its all good taz |
Yosimity
| | Posted on Tuesday, November 19, 2002 - 7:42 pm: |
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Do the math, if the power output is 100 watts. P= IxV P=100 V=13.8 I(amps)= P/v =100/13.8=7.25amps A 10amp supply should do well if it onle swings 100 watts. |
ChillyDog
| | Posted on Friday, November 22, 2002 - 10:30 am: |
|
Yosimity, Taz,
Another factor to include in your amperage draw calculations is amp efficency.
Depending on the amp bias, final stage efficency can run from less than 40% to 70% or so. Additional current is drawn for other amp circuitry and even more is pulled if there is a driver.
As a rule of thumb, if using peak output for calculating amp current requirements, a 50% efficency factor is usually adequate. This means that after using the Power equation based on a 100% modulated output, you double the results to represent the correct current draw.
Best Regards,
Bob |
|
