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Dropping 1V from a Regulated 6V Wall Wart
On Mon, 29 Dec 2003 14:47:10 -0600,
said...
On Mon, 29 Dec 2003 19:54:47 +0000, John Woodgate
wrote:
I read in sci.electronics.design that John Fields jfields@austininstrum
ents.com wrote (in ) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 19:07:59 +0000, John Woodgate
wrote:
I read in sci.electronics.design that John Fields jfields@austininstrum
ents.com wrote (in ) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
wrote:
R1/R2 = 5
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R2/R1 = 5
Not always. (;-)
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I'm sure there's a joke in there somewhere... :-)
I meant that for 6V in and 5V into Rl, R2 = 5R1.
OK, let's go on in the same vein. R2 is not always 5R1; it might be 5R6
or even 47K. (;-)
GROANNNN...
OK, it arteried: "In order to assure a voltage of 5VDC across Rl the
resistance of R2 should be five times the resistance of R1." ?^)
Ah! So, you mean R2 = 5*R1 or 5.R1 in K.A. notation.
--
Best Regards,
Mike
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