Signal size after Splitter???

[Bill Mountain]

Here's a question I've always pondered but never thought to ask until now.

If I put a 1 volt signal into an active buffer/splitter would I get 2 ea. 1 volt signals out?  (Minus standard losses in the buffer).

I feel like the answer is no but I've never seen it addressed so maybe it's yes?

[merlinb]

If I put a 1 volt signal into an active buffer/splitter would I get 2 ea. 1 volt signals out?  (Minus standard losses in the buffer).

Yes, usually that is what you get.

This is a known issue with many hifi products that provide both balanced and single-ended connections- the balanced signals end up being 6dB louder than the single ended ones. A few manufacturers go out of their way to fix this and make them both have the same total amplitude (equal loudness), to match the way it is done in pro-audio gear.

[Bill Mountain]

When you say a balanced is 6dB louder is that because your're using a transformer or an opamp balanced out?

Is single ended just a simple passive split?

Edit:  Never mind.  I get it.

But to expand on my original post...why does it work this way?  Even though it is buffered, the signal is still being cut in half right?

[merlinb]

But to expand on my original post...why does it work this way?  Even though it is buffered, the signal is still being cut in half right?

Well, not really 'cut in half'. You have your original signal, and then you create a copy except it is phase inverted, so now you have two signals of the same amplitude.

[Bill Mountain]

But to expand on my original post...why does it work this way?  Even though it is buffered, the signal is still being cut in half right?

Well, not really 'cut in half'. You have your original signal, and then you create a copy except it is phase inverted, so now you have two signals of the same amplitude.

Aha...I'm not specifically talking about balanced signals.  Hence my confusion in my last post.

I mean something as simple as a buffer with two outputs at the end.  Sorry if I wasn't clear.

[merlinb]

Aha...I'm not specifically talking about balanced signals.  Hence my confusion in my last post.

I mean something as simple as a buffer with two outputs at the end.  Sorry if I wasn't clear.

In that case you would still normally get two output signals, both identical to the input signal. 1V input means two 1V outputs going down different wires.

[Bill Mountain]

Aha...I'm not specifically talking about balanced signals.  Hence my confusion in my last post.

I mean something as simple as a buffer with two outputs at the end.  Sorry if I wasn't clear.

In that case you would still normally get two output signals, both identical to the input signal. 1V input means two 1V outputs going down different wires.

Cool.  Thanks!  I'm working on a 1 knob bass distortion that is a simple blend between a low pass filter and a high passed fuzz circuit.  I want the low pass to be unity gain and I was afraid that splitting the signal after a buffer would cut it in half requiring a slight boost at the end.

Thanks!

[ashcat_lt]

It may not be exactly intuitive in the case you've described, but I think it's just in the way you've envisioned/drawn the circuit.  

It's a fundamental principle that parallel loads will always have the same voltage across them.  Consider the case of daisy-chaining power off of a regulated supply.  This could not work if things divided down the way you were thinking.

Of course, we're talking about a real world circuit here, not an ideal theoretical textbook thing.  The source has some non-zero impedance, and each of the loads will have some finite impedance, and there will be some voltage division in there.  If all of the load Zs are bigger enough than the source, then they will all be close enough to the source V to be called unity.  If even one is small enough to see meaningful division, then all of them will get dragged down with it.  Also, if the paralel total of the loads is small enough...

[Bill Mountain]

It may not be exactly intuitive in the case you've described, but I think it's just in the way you've envisioned/drawn the circuit.  

It's a fundamental principle that parallel loads will always have the same voltage across them.  Consider the case of daisy-chaining power off of a regulated supply.  This could not work if things divided down the way you were thinking.

Of course, we're talking about a real world circuit here, not an ideal theoretical textbook thing.  The source has some non-zero impedance, and each of the loads will have some finite impedance, and there will be some voltage division in there.  If all of the load Zs are bigger enough than the source, then they will all be close enough to the source V to be called unity.  If even one is small enough to see meaningful division, then all of them will get dragged down with it.  Also, if the paralel total of the loads is small enough...

This brings up another point I forgot to ask.  I think you answered it but...If both load impedances are different, would the signal get split evenly?  What if they are big enough to not matter (at least 10x the source impedance)?  Will there be any difference between the signals?

[ashcat_lt]

If there aren't any reactive components involved, then all of the loads hanging off of the thing should see exactly the same voltage input, not matter how far mismatched they are.  They all come off of the node where the "top resistor" of the source meets the "bottom resistor" of the parallel loads.  That voltage is what it is across every one of the individual resistors which make up that "bottom resistor".  One may draw a different amount of current than another, but they all have the same voltage.

Of course, there are reactive components involved and it's quite possible to see frequency response difference at the individual inputs.  They will, in fact, interact with one another with results that are difficult to predict unless you can control at least some of the variables...

It is best to make the loads as similar as possible, and significantly larger than the source.

[induction]

Of course, there are reactive components involved and it's quite possible to see frequency response difference at the individual inputs.  They will, in fact, interact with one another with results that are difficult to predict unless you can control at least some of the variables...

In a worst-case scenario of differing loads, could you double-buffer?  That is, one buffer to do the splitting, then another buffer on each of the inputs?

[Bill Mountain]

Sometimes when I want to over think something like this I look at the Big Muff tone stack.  This is essentially a passively split signal (after an active gain stage of course).  The signal is split into a lpf and a hpf then blended back together.  There is a volume drop of course.

How does this relate to what we're discussing?  Do I really need better performance for my circuit?