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From: Spehro Pefhany
Subject: Re: Don't do this! Just don't!
References: <7FYW9.1208$PR2.firstname.lastname@example.org> <email@example.com>
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Date: Mon, 20 Jan 2003 20:29:43 GMT
NNTP-Posting-Date: Mon, 20 Jan 2003 15:29:43 EST
On Mon, 20 Jan 2003 20:20:44 GMT, the renowned "Ken McDonald"
>"Spehro Pefhany" wrote in message
>> On Mon, 20 Jan 2003 20:06:27 GMT, the renowned "Ken McDonald"
>> >> +15
>> >> |
>> >> p
>> >> 10K o <--------------10K -----x--- small offset
>> >> t |
>> >> | 49R9
>> >> | |
>> >> | 0V
>> >> -15.
>> >> It's a bit nonlinear and the output impedance varies a teeny bit,
>> >> (from 12.5K||49R9 to 10.0K||49.9) but not bad.
>> >If the pot is adjusted to the rail, won't the 49.9 ohm resistor smoke?
>> Perhaps I should have mentioned that the 49R9 resistor should be rated
>> for a *minimum* of 1/8 mW.
>How much power will a 49.9 ohm resistor connected to 15 volts dissipate? I
>keep coming up with 4.5 watts.
You're somehow missing the 10K series resistor.
If you crank the pot (say) to 15V, the pot element sees 30V and
dissipates 90mW. The 10K resistor dissipates (15 * 10/10.0499)^2/10K =
22.3mW. The 49.9 ohm resistor has only 15 * 49.9/(10049.9) = 74.47mV
across it, and dissipates only 0.111mW.
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