Signal split and inpedance

[AM]

Hello,
I have a (probably easy to answer for the more knowledgeable) question. I want to split my signal from a buffered pedal to two amps without using an extra A/B or splitter box. Output impedance of the pedal is 1KOhm. What would be the theoretical output impedance if I just use one of those Y splitter mono adapters and two cables from there to the two amps? Would it be the equivalent of 2K? Amps vary according to different places I play. Thanks in advance.

[antonis]

Every input should "sees" buffer's output impedance (1k) in parallel with other input impedance..(load for your buffer..)

From amp input side of view, total impedance "seen" consists of every path to AC ground..

e.g. if you drive two amp inputs of 10k input impedance each one, any individual input "sees" a total impedance of 910Ohms..

You'll just have to consider about output current dubling (in case of equal split impedance values)

[GibsonGM]

Every input should "see" buffer's output impedance (1k) in parallel with other input impedance..
(load for your buffer..)

e.g. if you drive two amp inputs of 10k input impedance each one, any individual input "sees" a total impedance of 900Ohms..

You'll just have to consider about output current dubling (in case of equal split impedance values)

Would separating the 2 split outputs with some sort of resistance help isolate them better, and prevent the doubling?

[antonis]

Would separating the 2 split outputs with some sort of resistance help isolate them better, and prevent the doubling?

Naughty boy... 

Any resistance in series with buffer's output is an obstacle on current drive..
(it's considered in series with drive source resistance hence deviate more it from "ideal" voltage source....)

Otherwise, we'll could place a huge resistor on any driven circuit input and raise its impedance without buffering troubles..


P.S.
Are you trolling me, Sir..?? 

[GibsonGM]

Ok, what I was picturing was feeding the wiper of a pot, as if we were going to use this to pan...but keeping it centered (2 resistors...).   For some form of "isolation".   But I see your point, that we would be 'closing down' the output's ability to drive the next stage, and increasing impedance!

I was picturing SOMETHING like this (although this is for headphones, not amplifiers!):  http://www.celestial.com.au/~rosswood/diy/headphonedist/images/headphonedistschem_10.JPG     

I think I see....In our case, the amplifier input impedance is enough to maintain stability, though?   No 'troubles' if the impedances are different, or unknown?

No, I'm not trolling, ha ha!  I saw a place to ask a question! 

[R.G.]

I'm still not up to my target coffee rate this morning, but isn't it simpler than that?

Most guitar amps have input impedances of ~ 1M. If you have a buffer with an output impedance of 1K, just Y-cable them up. They will have no apparent signal interaction. Even for 10K PA amps, there's no practical problem.

However, there is highly likely to be a hum problem with numbers of unspecified amps. Different amps leak different amounts of AC power line current into their chassis, and connecting the grounds of two such amps very often makes one or both of them have hum. One of the big aims of advanced splitters is isolation of the grounds so that you don't get this hum.

[antonis]

No, I'm not trolling, ha ha!  I saw a place to ask a question! 

But now see what you've done.. !! 

(you've drive R.G. out of his target coffee rate..) 

[GibsonGM]

Ha ha!  I suspect that R.G. likes the challenge!

Ok, I see - the buffer will have no problem driving the two inputs, as long as we understand that they are in parallel, so the load may double, or whatever, according to those inputs...I was concerned about the buffer itself, but for no reason, since it can source more than what a typical amp will draw...it was the wrong place to think about this   

[R.G.]

I'm still a cup or two below operational status. 

[GibsonGM]

It's almost lunch time where I am, and I know I need 1 or 2 more cups to be normal, ha ha!

[AM]

...Most guitar amps have input impedances of ~ 1M. If you have a buffer with an output impedance of 1K, just Y-cable them up. They will have no apparent signal interaction. Even for 10K PA amps, there's no practical problem...

Thanks everybody for answering.
So, R.G., basically there would be no audible problem, signal degradation, high-end loss, etc. It would be pretty much the same as connecting a single cable to one amp. Am I understanding you right?

The hum problem is something I will address if it occurs by getting an isolated outputs splitter.

[GibsonGM]

The hum problem is something I will address if it occurs by getting an isolated outputs splitter.

This is what I was thinking, but not for the same reason (noise).   Looks like you don't (or may not) need it...

[PRR]

> I want to split my signal from a buffered pedal to two amps without using an extra A/B or splitter box

I've got one wall-outlet and two lamps I want lit.

Do I have to buy a power company? Wind-mill? Kerosene lamp?

Usually I just use a power strip, multiple outlets on one plug.

This works OK because one lamp is say 240 Ohms and the wall-outlet is typically under 2 Ohms.

Assuming 120.0V supply:

One 240 Ohm load on a 2 Ohm source gives 0.9917 of voltage, 119V.

Two 240 Ohm loads on a 2 Ohm source gives 0.9836 of voltage, 118V.

Nobody knows the difference 119V or 118V.

Yes, the current doubles. One 240 Ohm lamps is 0.5 Ampere. Two 240 Ohm lamps is 1 Ampere. Typical wall-outlets can provide 15 Amperes. We are in no danger.

If you repeat the math with your 1K source and typical 50K-1,000K guitar-cord loads, the drop is probably less.

The guitar cord loads typically draw micro-Amps of current. (1V in 100K is 10uA.) A pedal-source with 1K output is typically a chip op-amp or similar. Most will deliver 30 milli-Amps, 30,000uA, if you want.

With the setup you propose, you get into "trouble" around *hundreds* of loads. Not two.

As said, ground differences between multiple loads IS a real problem with no 10-cent solution.