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In Reply to: Re: Same Set Up...Same Results! posted by amandarae on October 3, 2006 at 22:06:59:
The cartridge "produces" half the voltage of the unloaded condition. It is not a case of the "full voltage" hitting a "resistor brick wall" and being reduced.The secondary "sees" the load impedance and reflects it to the cartridge through the transformer primary. The load on/at the coils of the cartridge is X or 1/2X or whatever the transformer presents it.
There may be some good reasons (that I don't know) to put the additional parallel resistor at the preamp but that is not one of them.
Sorry. :-)
Follow Ups:
At least I got straighten out. I was assuming maximum power transfer since a 14 ohm impedance source driving a 14 ohm load. So now I know that it is not really the case. How do you compute the resulting voltage depending on the load values?thanks
Hi Abe! (it's earlier than I thought in SoCal)here's the formula:
Va = (Vi / Is+Il) * Il
Va = voltage actual
Vi = voltage initial
Is = Impedance source
Il = Impedance loadSo you can see for instance 2 / 28 * 14 = 1
As you point out that is max power but not max voltage. But who cares about voltage? Voltage is only loudness.
I really appreciate it.
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