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Old 02-10-2014, 02:09 PM
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craig3972 craig3972 is offline
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Join Date: Oct 2005
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Will splitting RCA's drop the HU voltage output?

I saw this on another forum.... and wondered if its true. I have a Alpine CDA-118 that is 10,000 ohm output impeadance at 2V, but i cannot find the input impedance anywhere for my JL amp MD750/1.



Lets take a typical Alpine HU. Lets pick this one.

CDA-9856
Preout: 2 volts / 10,000 ohm output impedance..

Connect this HU to an Alpine amp with 10,000 ohm input impedance,
the voltage is now 1 volt, cut in half.


Instead of buying an Alpine amp, you found a good amp with a 47,000 ohm
input impedance, the loading effect is minimal, you get 1.95v, much better
than 1.0v when you interfaced an Alpine HU to an Alpine amp.

I was shocked years ago when I found this.



Connect two Alpine amps with splitter, you now have 0.66v.

Hint: A good HU will have a 50 ohm - 200 ohm output impedance. Even 1000 ohm is ok,
but 10,000 ohm is stupid circuit design.

Moral of story: Always check equipment specs.




One HU & one amp;

HU output voltage * (amplifier input impedance / (amplifier input impedance + HU output impedance)).

Example.

HU preout voltage: 5 volts
HU output impedance: 200 ohms
Amplifier input impedance: 20,000 ohms.

5 * (20,000 / (20,000 + 200)) = 4.95 volts.

One HU & two or more amps

Step 1. Find the total parallel impedance of all your amps since the same input signal will be driving all the amp inputs.

1 / (1 / (amp#1 input impedance) + 1 / (amp#2 input impedance) + 1/ (amp#3 input impedance ... etc))

Example:

Amp #1 input impedance: 10,000 ohms
Amp #2 input impedance: 20,000 ohms
Amp #4 input impednace: 47,000 ohms

1 / (1 / (10,000) + 1 / (20,000) + 1/ (47,000)) = 5838 ohms

Step 2. Put that 5838 ohms back into the voltage divider formula.

HU output voltage * (amplifier input impedance / (amplifier input impedance + HU output impedance)).

Example.

HU preout voltage: 5 volts
HU output impedance: 200 ohms
Amplifier #1, #2, #3 parallel input impedance: 5838 ohms.

5 * (5838 / (5838 + 200)) = 4.83 volts.

Last edited by craig3972; 02-10-2014 at 03:33 PM.
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