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Splitting One Line-Level Output to Two Amplifier Inputs

post #1 of 8
9/15/05 at 10:37am
Thread Starter 
If I want to drive two headphone amplifiers with the line-level output of my Adcom CD player, do I need some sort of splitter/combiner thingy? From my vague recollection of studying RF electronics (I've never studied up on audio and my RF experience is decades old) it's a big no-no to just splice two wires from the output of one stage to the inputs to parallel following stages. You need something to isolate the inputs from each other. That would seem to apply in this case since I belive there is usually a volume control somewhere around the input to the first stage of amplifaction in a headamp.

I think I'm picturing a little resistor circuit that acts like, for instance, a 6dB attenuator where half the power is wasted, a quarter goes to one input and a quarter goes to the other. Or something like that. To be specific, imagine the output of the CD player is rated 4.5V into 100 ohms and the inputs are those of two Portaphile amplifiers.

So question one is, would something like that work to isolate the CD player output from the effects of the two headamp? If so, what is the tradeoff between isolation and attenuation.

The other question is would, say, 6dB of loss before the headamp input(s) add too much to the noise figure? I think that may be a weak-signal RF problem that doesn't apply to line-level audio.
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post #2 of 8
9/15/05 at 10:53am
Just split them, straight to each amp
The high input impedence of each amp will pose no problems to your source and they wont affect each other
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post #3 of 8
9/15/05 at 1:01pm
The impedance of the two amps will be in parallel so the resulting impedance "seen" by the source will be :

"Amp A input impedance + Amp B input impedance divided by 2"

If the source has a low impedance output and good drive combined with a highish amp impedance for both amps you should be OK but even then you couldl conceivably have some strange interactions between the inputs of the two amps depending on their design.
Far better would be (in order of cost/complexity/quality)

1-a 1-to-2 resistor splitter (cheap,simple)
2-active 1:2 distribution amp (another active stage inline plus requires a power supply of some kind) which would isolate each output from a common input while allowing full drive levels
2-a 1-to-2 line level splitting transformer (best solution but expensive,added benefit of total signal ground loop isolation plus digital ground noise blocking for CDP sources )

All will add their own colorations,some subtle and some not so subtle depending on design but you really weant to avoid interaction between the two amp inputs if possible and the above solutions offer degree levels of this from "pretty good" to "total isolation".


Proper in/out signal selection and distribution in a mixed signal system is an area seriously overlooked by the average audiophile.Not as simple as "select A and send to Z" but needs to be thought of as "if I select A what will the effect be on X,Y or Z".
Passive signal isolation and/or active signal buffering is the only way to know the end result of the various load impedances and their effect on the driving source.

****BTW*****

Don't forget to add the cable R/C into the equation.There is a reason for cable drivers/receivers and that is to take them out of the equation totally
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post #4 of 8
12/17/10 at 12:43pm

I have a similar situation. My Placette passive has 2 outputs. If I run one output to the input of the power amp and the other output to the input of my headphone amp, thus resulting in the amps being in parallel, should I be OK with the following specs:

 

Source (CD player) output = 50 Ohms

Power Amp input = 31k Ohms

Headphone Amp input = 14k Ohms

 

In this case, the resulting impedance "seen" by the source would be 22.5k Ohms applying the formula in the preceeding post. Would this be considered to be in the "safe" zone to avoid/minimize untoward interactions? I would appreciate any advice. Thanks.

Ken 

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post #5 of 8
12/17/10 at 2:03pm

Wouldn't the resistance seen by the source be

A parallel B = A*B/(A+B) = 9.6k in Kenm's case or have I missed something?

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post #6 of 8
12/17/10 at 2:05pm

 

Quote:
Originally Posted by porthillsbomber View Post

Wouldn't the resistance seen by the source be

A parallel B = A*B/(A+B) = 9.6k in Kenm's case or have I missed something?


You got it.

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post #7 of 8
12/17/10 at 6:47pm

Sorry. I was just applying the math from the previous post to arrive at 22.5k,  not from my own knowledge. Given whichever is correct, though, would the specs of my gear compromise sound qualty in the configuration I describe? Thanks again in advance for any advice.

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post #8 of 8
12/18/10 at 5:15am

Would the resistance seen by the source in my example be likely to compromise SQ? Anyone? 

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