Signal splitting question

[Top Top]

If you have one signal that is split into two separate transistor gain stages (NPN transistor and a three resistors), and you have voltage divider volume controls after each one (at the output of each transistor), will changing the volume on one affect the other at all? Or will the transistor keep the flow going in the right direction, so to speak?

[Top Top]

A visual aid - the input is coming from a signal which is working fine into one transistor gain stage as it is now. I'm out of breadboard space or I would just try it...

[petemoore]

  Inputs are common, outputs are separate, one volume controls that output only.
 A transistor output is really a 'copy' or analog of it's input.
 I ran out of Jfets mostly, they have very high input impedances...not that I know it has to matter, they just seem well suited to such application.

[R.G.]

If you have one signal that is split into two separate transistor gain stages (NPN transistor and a three resistors), and you have voltage divider volume controls after each one (at the output of each transistor), will changing the volume on one affect the other at all? Or will the transistor keep the flow going in the right direction, so to speak?

The answer is:

Ohm's Law.

You left some things off your diagram. Those are the source impedance of the signal driving the transistors, and the load impedances of the volume pots and the inputs they drive. If the source impedance is high, adding more transistor circuits in parallel to drive will load it down and you'll lose signal level. If the values of the output pots are more than 20 times the output impedance of the transistor stage, and the loads they drive from their wipers are more than 50 times the output impedance of the transistor stage, then you can just drive it from one transistor.

This is because the output impedance of that transistor stage is pretty close to the collector resistor. Any loading over 10X that resistor on it will not noticeably change its output voltage. So if the parallel loading of the output pots is greater than 10x the resistor, the collector resistor will dominate the output signal, and the volume controls will not noticeably interact. The output impedance of a volume control is at most 1/4 of the pot value, so if the outputs the volume controls drive are more than 10X the 1/4 of the volume control, they will not load the volume control noticeably.

You can get away with other situations, but this is a rough and ready guide to it.

[Top Top]

If you have one signal that is split into two separate transistor gain stages (NPN transistor and a three resistors), and you have voltage divider volume controls after each one (at the output of each transistor), will changing the volume on one affect the other at all? Or will the transistor keep the flow going in the right direction, so to speak?

The answer is:

Ohm's Law.

You left some things off your diagram. Those are the source impedance of the signal driving the transistors, and the load impedances of the volume pots and the inputs they drive. If the source impedance is high, adding more transistor circuits in parallel to drive will load it down and you'll lose signal level. If the values of the output pots are more than 20 times the output impedance of the transistor stage, and the loads they drive from their wipers are more than 50 times the output impedance of the transistor stage, then you can just drive it from one transistor.

This is because the output impedance of that transistor stage is pretty close to the collector resistor. Any loading over 10X that resistor on it will not noticeably change its output voltage. So if the parallel loading of the output pots is greater than 10x the resistor, the collector resistor will dominate the output signal, and the volume controls will not noticeably interact. The output impedance of a volume control is at most 1/4 of the pot value, so if the outputs the volume controls drive are more than 10X the 1/4 of the volume control, they will not load the volume control noticeably.

You can get away with other situations, but this is a rough and ready guide to it.

Ok... well... trying to digest that.

I am guessing I am going to just have to find some little slice of breadboard to use because other than the value of the pots (which could be any value - whatever works), and the input impedance of the 386 (which is 50K), I don't know the answer to the questions - the signal going into the transistors is a few signals generated previously in the circuit mixed through resistors, but I have no idea what the impedance is or how I would find out.

Edit: Reading through again though, I wonder if I miscommunicated something - the question is not whether having two transistors would lower the overall level - it is whether volume controls after the transistors would affect the volume level of the output after the OTHER transistor. If that is what you answered than I am sorry... I am still learning on a lot of aspects and may have misunderstood...

It is actually fine if the overall level got dropped by some degree as the output before the transistor was decently hot to begin with, BEFORE even being amplified. I wanted to put the transistors in there to keep the split of the signal into the headphone amp from dragging on the line out.

 Inputs are common, outputs are separate, one volume controls that output only.
 A transistor output is really a 'copy' or analog of it's input.
 I ran out of Jfets mostly, they have very high input impedances...not that I know it has to matter, they just seem well suited to such application.

Maybe I will give a JFET a chance...

I am thinking though that I want to replace with just one volume control (no reason I have to have separate ones), which would come before the transistors then - seems it would throw a monkey wrench into the equation.

[R.G.]

the question is not whether having two transistors would lower the overall level - it is whether volume controls after the transistors would affect the volume level of the output after the OTHER transistor.

That's one part of what I said, but it was well buried.

If the volume pots and loading on their wipers are each >10X the transistor's collector resistor, the cross-effects from one pot to the other will be neglegible.

[PRR]

1) DC-tying the Bases together like that defeats the self-bias scheme. (Though with the large emitter resistors, it may work out.... never thought about it.)

2) Why have two transistors? Their outputs are nominally identical. You "may" be able to split after the output and get tolerable interaction. Or use your 2-transistor budget for a more robust single amplifier which may tolerate load changes.

3) will changing the volume on one affect the other

Yes.

Enough to notice? Often not.

How to know?

There are ways to use a transistor so that load-change has less than 1/500th effect on the input. And ways to make the output load directly affect the input loading.

A similar example: my house has several fuse-circuits. I put a dimmer-lamp in each room. Will brightening the living room change the brightness in the bedroom?

For small lamps in a properly wired house, not enough to see except on a 3.5-digit DVM.

For large loads in -my- skinny-wired house: yes, you can tell when a load is powered-up suddenly. When the RadarRange comes on, voltage drops 4V. 2% or 2.4V is the usual guide for "good" wiring. But being 400 feet off the street, "good" is costly.

What R.G. said. Figure it out.

Taking your initial question: A true "voltage divider volume control" is UN-loaded so has the same input impedance at any volume setting. That would be nice, but never happens. With tube or FET loading the voltage-divider, the change may be negligible, from 1Meg to 0.990Meg. But the '386 is a 50K input. If you use a 50K pot, then the total is 50K full-down and 25K full-up. If your source is 250 ohms, this may cause a 1% droop, quite negligible. But your source is something like (not exact) 10K. This gives ruffly 15% or 1.3dB drop. You can hear 1dB drop, if you expect it. It is rarely a "problem".

And are you sure you want NO interaction? That's sometimes a goal. But if one out is monitor and the other is mains, there may be good reason to sag the monitor just a bit when the mains start to blare. Not that you can know "how much drop" to allow, but to say that maybe a little drop is fine.

There are many ways to skin cats. Here's how I might approach it.



I stole your Q1 for gain. Then I wired the 2nd transistor as an Emitter Follower. The output impedance is under 100 ohms. For small signals, loads as bad as 1K will not cause more than 10% drop which is generally negligible. Because I used to do Live Recording, I am paranoid, and put a 1K stopper in each output. To preserve the low impedance and good isolation, I use a large coupling cap.

Even if the "Line Out" user puts a dead-short in his jack and turns the pot full-up, the "HP Out" (and Aux etc outs) probably won't notice.

[Top Top]

Thanks guys for the additional info, and I will try something along that line PRR... You say the second transistor is a follower. Does that mean it helps to sort of "fortify" the original signal going into it, but with no gain?

I bought a second breadboard by the way... so it is back to experimenting...

[R.G.]

This is quibbling, but I'd do either a two-transistor feedback pair and get almost to an opamp, or for very, very nearly the same "cost" in number and cost of parts, number of pins, etc, I'd use a real opamp. It's really very hard to justify the extra complication and parts count for a two-transistor circuit unless  you already know an opamp will not work. Or you are religiously opposed to opamps. 

[R.G.]

This is probably as good a place as any to stick in a comment that's been bouncing around in my head for a while.

"Splitting a signal" is a misnomer as it's usually applied. Driving more than one source without interaction often just means making the signal so low an impedance that any of the loads intended to be connected to it will not interact. Sometimes signals need split to avoid hum, but most "splitters" are just buffers. Low impedance cures a lot of voltage-mode ills.

[Top Top]

This is quibbling, but I'd do either a two-transistor feedback pair and get almost to an opamp, or for very, very nearly the same "cost" in number and cost of parts, number of pins, etc, I'd use a real opamp. It's really very hard to justify the extra complication and parts count for a two-transistor circuit unless  you already know an opamp will not work. Or you are religiously opposed to opamps. 

That is a good point. I had originally had just one transistor gain stage in there to give it a nice strong signal at output, so it took off from that. Transistors are also a little more flexible in terms of putting them in a layout (which I have been trying to keep compact), BUT, I will take a look... might be the best option, I am not religiously opposed. Most of the rest of the circuit is CMOS, so it's not some mojo discreet thing...

[R.G.]

When I'm trying to find out what is the more complicated layout, I count up the following:
- how many passive parts are needed (two terminal things, like Rs, Cs and diodes to stretch the point)
- how many active device pins are there?
- for the active device pins, I count power and ground as one pin, for a number of semi-abstract reasons.

The R/C count for an opamp gain stage comes out remarkably the same as for two transistors in many cases. And if you have need of two amplifiers, a dual opamp in an 8 pin package has six active pins and one power/ground pair as opposed to six active pins for two transistors.

What you really lose with opamps is the two rows of holes under the body of the opamp.

You want compact layout AND good performance? Get a dual opamp in the single inline package. Like f'rinstance the 513-NJM072BL. This is our old friend, the TL072 made by NJR, but in a single-in-line eight pin package. US$0.50 at Mouser Electronics. This lets you use all the holes.

[PRR]

> quibbling, ...feedback pair and get almost to an opamp, or ... use a real opamp.

No quibble with that. Feedback pairs were my meat. Today I would turn to a chip.

For tutorial and parts-on-hand purposes, re-tangling the existing parts gives a circuit which is:

1) easy to understand (or teaches a new basic concept: sort of "fortify" the original signal),
2) keeps the same gain (in case that has been set in stone),
3) gives a vast increase in performance (Zout dropped from ~~10K to ~~100).

-------------------

I did, from habit, start to consider a more "holistic" pair than just a CC tacked to a CE. But I then realized that the "tack-together" met all the specs I could infer, and was right down at the minimum-cost level (excluding the chip; and that's not radically lower-cost).

The proposed plan has "flaws". In its defense, it is widely used and does work well.

For a "factory" job, I'd like to cite the collector voltage as a test-point. With this bias, the voltages vary with hFE which varies between devices. It will work with 3V to 7V at collector, but some service techs are uncomfortable with broad ranges, and some odd faults will give an in-range voltage yet not be happy.

The output overload varies with hFE. If we needed a full 3V output from a 9V supply, this plan won't assure it (can't even do it; few can). However I see a LM386 over there and it never needs over 0.2V input. Allowing 14dB reserve in Volume pot, the buffer still only needs to pass 1V.

A complementary CE CE pair might give somewhat higher input Z, somewhat lower output Z, more stable bias-point, higher overload. But at 9V supply, it might not be a ton better. And teaching how-it-works is non-trivial. And we don't need the performance, and if we did, the '072 whups the 2-Q (but needs a 3rd cap, though this Vref may be shared with other corners of the total plan).

[Top Top]

Well golly I feel silly... I tried just splitting (that word again) the signal from after the one transistor I already have on there (at the point of what was previously the output) - and just sent both the line out and the input of the 386 from that same point, and it works fine. Adding in the 386 had no (perceivable) effect on the output of the line out when I popped the wire in and out on the breadboard.

Popping the line out's wire in and out did affect the 386 a little (drops when the line out is hooked up), but not enough to make me worry about it as the headphone/speaker gets plenty loud as is.

[PRR]

> "Splitting a signal" is a...

Use a "froe". This is a tough piece of steel, sharp on the edge, with a right-angle handle. Examine the grain in the area where you want to split. Set the froe along grain-line with light taps from a wooden maul. Re-check the split direction......

Oh, wait. That's splitting firewood into chair-legs.

Electrons are electrons. They don't even stick together. How do we "split" electricity?

I have one power wire into the house. I have 13 lamps and 47 clocks (well, it seems like that many). How do I "split electric power"? As TipTop says: "just sent both ... from that same point". I twist all my lamps and clocks to the wire from the street.

OK, household power is a little more complicated. Overload and electrocution are safety problems, we use fuses, outlets, plugs. But if you open-up a $5.99 "power strip", you will see it is just strips of metal connecting the cord to a bunch of holes/outlets.

Sometimes it is that simple.

Sometimes it is not that simple. If I plug-in too many lamps for the size of my street-wire, they all get dim. In Power systems, if we aren't getting "full power to load", there's real problems like hot wires. So before my house gets real dim, the fuses blow.

If I connect "too many loads" to an audio circuit, "they all get dim", lose volume or distort. However unlike many-kiloWatt Power systems, audio lines are teeny fractions of a Watt, so there's no danger of fire, and generally nothing burns out.

Your plan has a 5K collector load, and we may approximate the output of the stage as a 5K resistance. When you connect just the 50K of the '386 input, there's about 10% sag, who cares? When you add 50K pot here, 10K pot and 10K line-out load, you will have more sag. But as long as it does not change mid-performance ("Popping the.. wire in and out"), and you still have "enough", who cares?

Splitting is more a problem when you do not have "lots" of signal. Right off a guitar pickup or a dynamic mike, you only have whatever power was captured from string or air, typically very low. Typically scaled to be just-enough to drive one load, the input of one amplifier.

[Top Top]

Very good, practical advice, thanks.