Re: AR 2 bench testing

I think everyone has the formula's right, but need to locate their calculator's
;^) (I had to ;^)

Ohms Laws: I * R = E

Where I is current, R is resistance, and E is electromotive force (whatever,
it's just Volts), so,

   Volts / Amps = Resistance, and,
   Volts / Resistance = Amps, and,
   Amps * Resistance = Volts

And power is just as stated below:

   Volts * Amps = Watts

So....

On Tue, 29 May 2001 15:03:43 -0500, tom mcclintock <tomm@mgcap.com> wrote:

>Matt McCullar was kind enough to remind me of a little formula for
>calculating a required load.
>
>volts/amps = resistance (ohms)
>volts*amps = power (watts)
>
>So, assuming we have an Atari board that needs 5 volts and draws 3 amps,
>that gives us an equivalent load of 2 ohms at 15 watts.

5v / 3a = 1.666 [2 ohms is close, but not 'equivalent']

5v / 2ohms = 2.5a [Which is probably a pretty good value to test a 3amp supply]

5v * 2.5a = 12.5w [15 watts is fine, just don't touch it during tests]

>I guess if you wanted to check the 22V output of the ARII (staying
>strictly vector), you could use a 7 ohm, 65 watt resistor as well.

22v / 7ohm = 3.14 amps
22v * 3.14 = 69 watts [65w resistor might start turning brown on you]
>
>I hope I got my math right... :)
>
>
>
>tom
>
>John Robertson wrote:
>>
>> That is a very light load - only about 200ma @ 5VDC, I use a 1 ohm 25
>> to 50 watt resistor as a load for testing powers supplies. That is 1
>> amp @ 5VDC, and is a reasonable load for any power supply.

5v / 1ohm = 5a! [Might be a bit high for a 3amp supply]

5v * 5a = 25w [25w to 50w is correct, for a 5a shunt]

-Zonn

>I think it
>> would be good to just short the sense resistors out, as they only seem
>> good to burn up, and you can adjust the supply manually for the
>> correct voltage.
>>
>> John :-#)#
>>

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Received on Tue May 29 17:02:21 2001