Bob Blick wrote:

> build) that take consumer signal(-10db unbalanced 10kohm) and make
> professional signals(0db balanced 600 ohm) out of them, and also convert

********

...now, after looking up this matter of db, I understand that when you say
"0dB balanced 600 Ohm" you have an unspoken reference of 1 mW into 600 ohm,
giving an RMS voltage of about .78V.

P =  I^2R, so...

1mW = I^2 Amps * 600 Ohms

I = 0.0129A

1mW / 0.0129A = 0.078V.


.....Now, about this "-10dB 10 KOhm"

This is 10 dB down from 1 mW also? If so then...

-10dB = 10 log (x/1mW)

-1 = log (x/.001)

x/.001 = 10^-1

x/.001 = .1

x = 100uW

100uW = I^2 Amps * 10KOhm

I = 100uA

100uW / 100 uA = 1V.

So 100 uA flowing through 10K and 1 V across it.

Is that right?

Mark

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