P
Pooh Bear
- Jan 1, 1970
- 0
Don said:Where do you obtain the 0.5 value and what is the "potential" divider?
The potential divider is formed by the source Z ( 600 ohms in this case ) and the
load Z ( also 600 ohms ).
Nother more complicated than high school math / science should illustrate that forms
a potential divider with a value of 0.5.
When an amplifier's input and output impedances are the same (for example
600 Ohms), and it's gain is adjusted with a 600 Ohm 0 dBm input signal, to
provide a 0 dBm output into a 600 Ohm meter, There is neither gain nor loss.
Only an impedance mismatch or an uncalibrated oscillator or meter will
provide a different "view."
No-one works that way any more actually but I've been there, done that, got the
diploma. Old ATRS ( analogue tape recorders ) that we used at Sound Developments for
example ( toob based Ampex 350Cs and 351s ) had a selector switch that chose between
10k ( bridging ) load and 600 ohm ( terminated ) load. When on a 600 ohm circuit,
select terminated and the level drops by 6dB relative to bridging assuming nothing
else is on the circuit. You can even see it on their VU meters !
Btw - try working out the power supplied by the signal *source*.
You may get a surprise ! And I mean *work it out* not believe what the faeries at
the end of the garden say !
Graham