[TI-H] Re: Link port drain


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[TI-H] Re: Link port drain




>>>Bingo... although your positive voltage should be whatever your load
>>>will drain in mA * X kOhms.  If you plan on using 6V, then you will want
>>>6V / X kOhms = desired mA drain... this will ensure you get proper
>>
>>So that would be 0.1 kOhms x 6 V = .6 A ? No, you said 90 mA. Which kOhms
do
>>you mean?
>>
>>>current through your device(s).
>
>Okay, lemme be a bit more clear: Voltage = Current * Resistance. You have
Voltage,
>and you have desired Current (X and Y). Your numbers now should be: X = Y *
Resistance.
>Solve for resistance and that will be the resistor, in Ohms, to put between
the battery, and
>your load. It doesn't hurt to put 2 resistors of half this value on either
side of the transistor,
>to prevent a base current problem, but with the link port only draining 90
microamps, it
>won't burn out your transistor. Although better safe than sorry.
That's a lot clearer, thanks. So I don't need a resistor between the port
and the transistor?


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