[TI-H] Re: Link port drain

To: ti-hardware@lists.ticalc.org
Subject: [TI-H] Re: Link port drain
From: Adam Thayer <krevinek@matrix-z.net>
Date: Sat, 9 Jun 2001 13:05:03 -0700 (PDT)
Reply-To: ti-hardware@lists.ticalc.org


>>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.


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