Why (not how) does phasing work?

Two power supplies can share the same ground. Ground is only one side of the circuit.

It's when you hook up both the hot and ground of two power sources to each other that things get fussy.

Thanks RoyBoy, but that's not really what I'm asking.  I'm trying to understand why phasing works.

As to your point, I'm using a time-delay relay circuit that requires 12VAC.  If I park a train over a sensor and the relay is activated for 20 minutes with 18VAC, my benchwork will catch on fire ;-)  And yes I could use a resistor -- but that's not necessary when I can simply use a small 12VAC transformer (and there are going to be about 60 sensors.)

Okay you just edited your post after my last reply ;-)  Again, I understand that they can share the same ground -- again, not what I'm asking.

I'm hoping somebody who understands what the AC signal looks like can help me understand what happens when two very different voltages coming from two different transformers share a common "ground."  Why does this work?

The answer, in the simplest terms, is that the electrons seek the easiest route back to "home" (ground) and ignore any other more difficult path.

Here are a few things to reflect upon. When you put them together, your brain MAY suddenly understand.  I will use non-electrical terms, along with official terms, to try to display what happens.

When a device is powered by any type of electricity, the power in the circuit is dissipated at the device. There is a voltage drop across the device, so, in a sense, all the oomph of the feed electricity is "used up" at the device.

Electricity will always try to get back to its source, rather than try to go "upstream" or "across the street" to some other destination.  If a supply has a ground reference, the return flow of electrons will seek the lowest impedance path. That would be the ground reference.

Since an 18-volt circuit is looking for its ground reference, it will "ignore" a different path that presents itself, if that path has a higher impedance to ground.  Similarly, any other circuit with a common ground will also seek the easiest route back to ground.

Multiple circuits with common grounds are present right in your home.  You have a 240-volt circuit breaker panel, and along each of the two 120-volt supply rails, you have circuit breakers that feed 120-volt devices in your home.  When you turn on a lamp in your living room, fed by the left side of your panel, and a lamp in your bedroom, fed by the right side, they do not interact. They each seek the low impedance of the neutral buss bar. Electrons do not "go rogue" and try to buck the system.  They just do what physics says they will do, which is get back to the neutral bar and ignore any other route.

This theory works for AC and for DC, or any combination of the two.

An example is a telephone company power supply (with which I am intimately familiar, having worked with them for 50 years).  Older telephone systems used the following flavors of electricity to make them work:

10 Volts AC for line lamps

18 Volts AC for intercom buzzers

24 Volts DC (unfiltered) for relay operations

24 Volts DC (filtered) for intercom voice connections

90 Volts AC @ 30 Hz for ringing the bells in the phones

All of these various supplies used the same ground reference.  They did not interfere with each other, as long as their respective ground connections were intact.

NOW...let's use the "what if" method of trying to grasp this.

WHAT IF a common ground connection DOES NOT present a very low impedance?  Here's an example that any guy over the age of 25 will understand. Have you ever seen a car with its tail lights OFF, (day time operation) and the driver puts on the left turn signal, and instead of the turn signal lamp getting bright, both the tail lamp and the turn lamp come on dimly?  THAT is an example of a bad ground, resulting in the electrons that fed the turn signal filament trying to get back to ground, but failing to do so, and sneaking back through the tail light filament.  The bad ground symptom was because some older cars were susceptible to getting water and crud in the tail light assemblies, causing rust and corrosion, resulting in a bad ground connection for the lamps inside the fixture. Instead of a very low impedance ground, the socket became a big resistor, and the returning electrons had a choice of tow paths, taking both, in fact, which caused the weird symptom.

I hope that you will think about these things, and maybe get a better grip on how it works. I sympathize with people who understand some electrical theory, but hit a mental wall at a threshold.  I have had the same thing happen to me in my career. There are some electronic theories that I simply cannot get my mind around, no matter how many experts try to explain them.  I just take them as gospel, and keep going.

EDITED TO CHANGE TAIL LAMP EXAMPLE.

Randy, if you draw a picture of the circuit, you will understand it.  You almost understand it now.

I have a masters degree in Electrical Engineering and 51 years of experience.  I have been playing with Lionel trains since 1950, so here goes an answer:

Using two transformers, one 18 volt to run the train and one for your 12 volt relays, here is what happens.  You connect one side of each transformer to the outside rail of the track.  You phase the transformers so that if you measure the voltage between the "hot" sides of the two transformers, you measure 6 volts AC (18 - 12 volts).  If you phase the transformers backwards, you will measure 30 volts between the hot sides of the two transformers (12 + 18 volts).  Now if you measure the voltage between the isolated rail and another outside rail, you will measure 12 volts and the relays are not operated.  If you then put a loco or car on the isolated track section, the voltage on the isolated rail will drop to zero compared to any other outside rail, and the relays will operate.  It doesn't matter if there is 18 volts on the center rail or zero volts or any other voltage.  Does all of this make sense?

Arthur, you forgot 48 volt office battery which is the voltage on a subscriber line.  The office battery was also used for the relays in automatic switches like Number 1 Crossbar, Number 5 Crossbar, Number 4 Crossbar, Crossbar Tandem, Step by Step, and Panel.  

With an AC voltage, the voltage rises and falls in a certain time frame, based on the frequency of the signal. When 2 transformers are in phase, both signals rise and fall at the same time, irregardless of their voltage. When the transformers are out of phase, the waveforms rise and fall out of sync with each other. Now you have a voltage that is at its peak for a longer period of time, effectively raising the voltage.

This is the principal that Pulse Width Modulation power regulation uses to control motor speeds. The width of the AC pulse is varied to control speed, although the voltage amplitude does not change. A wide pulse is a faster speed, and a narrow pulse is a slower speed. The amplitude of the voltage remains constant at say 18 volts, but the width of the pulse changes, effectively changing the speed of the motor.

When you phase the transformers, you are in effect giving the track a "narrow" pulse, and out of phase transformers in effect give the track a "wide" pulse.

(This is my recollections from ages ago, and may be a bit rusty. It may be updated by those with more current knowledge).

Larry

Servo:  I was talking specifically about a local (key system) supply, like the 20-B-2. There is no interconnection between the T & R of a CO line and the local supplies used for the system. The CO line is always isolated, right through the line circuit.

You are correct that in a CO, the -48 and the 90/20 share a common ground.

Until there is a set of wheels across the outside rails, the isolated rail is just a random piece of metal, with voltage relative to nothing...until and unless it is connected to the rest of a circuit...right?  I think we are just misunderstanding each other here.

Anyway that's a minor point I think.  Is my analogy of the amplitude of i.e. 3VAC vs the amplitude of i.e. 18VAC, and those two signals sharing a common...is that completely off base?  Am I confusing voltage with current?  Is the amplitude of the sine wave even relevant?

Randy, if you substitute the word "resistance" for "impedance" you may see it a little easier.

http://telephonecollectors.inf.../9974-167-440-201-i6

Hit "view" then when it opens, scroll down to Figure 5.  The fuses ("hot") run vertically. The grounds run horizontally, and are all connected inside the power supply.

This statement is correct:

"Now if you measure the voltage between the isolated rail and another outside rail, you will measure 12 volts and the relays are not operated."

Try it and see.

Impedance gets a little complicated.  To express the impedance of a capacitor or an inductor, you have to use complex variables.  Charles P. Steinmetz was the first to use complex variables for impedance.  For an inductor, the impedence is 

      j * omega * L

where j is the square root of -1, omega is the frequency in radians per second, and L is the inductance.  You can also write the impedance as 

      j * 2 * pi * freq * L

where the frequency is in Hz (cycles per second).

For a capacitor, the impedance is 

      1 / (j * omega * C)

and once again you can substitute 2 * pi * freq for omega.

You can see from the forumulas that the impedance of an inductor or capacitor is not constant but varies with frequency.  

RoyBoy said it: you can connect the ground side of all power supplies and it is find, the other side, however, say connecting 3 V and 18 V in phase, well, that usually produces interesting results including ruined equipment.  

As an electrical engineer, I recommend connecting the grounds of every power supply and track you have together.  Makes for fewer really strange problems down the road.  I do, all 12 supplies I use, including DC and AC.

That is how the bell and whistle button work.  The DC signal is generated by a rectifier to change a portion of the AC voltage into a DC voltage and sends it down the same line.  If you draw your 12V AC sine wave, you than draw a -5 or +5 V straight horizontal line across the graph.  That is you DC signal, which circuitry inside the engine can detect and respond to.  For PW trains a simple coil/relay.  Modern trains microchips and components.  G

I drew a picture. Hopefully a picture will save a thousand or so words....

Yes Randy, amplitude (the height of the sine wave) represents voltage.

In the top diagram, they are in phase because they both go positive in "sync".  These waves (from point A to C) repeat 60 times in one second, that's frequency.

Note that in the top diagram, voltage measured between point F and G is 6 volts.  In the bottom diagram, voltage measured between point F and G is 30 volts.

(For tech types...I left "peak-to-peak" out of the discussion)(also pardon the not-quite-sine-waves I drew)

Hope this helps...

Ed

GGG,

My understanding is that (in conventional operation if the voltage to the train was 18V), the 5Vdc is connected to the sine wave in series when the horn button is pressed.  So it would add to the sine wave voltage at all points along the wave.

Like this:

Then again, I may be completely wrong! 

Ed

eddiem,

The DC whistle voltage does not add to the AC signal. It is superimposed on top of it. On the diagram above, the DC voltage would be represented by a straight horizontal line at the +5Vdc level. The 12 volt waveform would remain unchanged and would remain at its present level.

Larry

Au contraire, the baseline of the AC waveform is shifted by the DC voltage.  It's not possible to "independently conduct" AC and DC like you suggest, that can be easily demonstrated if you dig out your 'scope.  The resultant waveform will be as Ed suggests.

The red line in the diagram seems to indicate that the AC voltage would increase. The bottom of the red waveform would need to be shifted up the same amount as the top of the waveform has. Then the entire AC waveform would be offset by the same 5 volts.

Larry

The measurements are not precise, but yes, the entire waveform (voltage) increases by 5V.  When a bell button is pressed, the 5V is connected in reverse in series, so the entire waveform (voltage) drops by 5V from the normal centerline.

Your PWM explanation is clear, but when you say, "When you phase the transformers, you are in effect giving the track a "narrow" pulse, and out of phase transformers in effect give the track a "wide" pulse.", I'm not sure that's right.

When you phase transformers, you are synchronizing the waveforms so they rise and fall together (top diagram), as opposed to one moving positive while the other is moving negative (bottom, out of phase diagram).  Waveform width is not affected when phasing "real" transformers.  With all the new "electronic switching transformers", the sine wave isn't quite as "sine-like" in shape.

Ed

ps. I hope this is helpful to Randy's original question...

That is correct. We used to make whistle controllers that didn't slow the trains using "diode rings" of six diodes (you did get an overall voltage drop of about 4 volts across the six diodes). AC flowed through because three were forward and three were backward, but if you shifted the point of connection for the track voltage, it gave the AC wave a positive or negative bias similar to Ed's diagram. Actually saw it on a scope years ago.

Ed. Don't get us started on Electronic Switching Transformers. I made a couple of those -- easy to make; a pain to clean up the ersatz sine wave.

Randy,

a picture...

It works because the transformers which are phased, share a common ground, and the voltages from each transformer do not.  It takes two wires to make a circuit, and you can see that although the blacks from the two transformers are connected, the reds are not.  That's why they don't interact, or affect each others' voltage.

We usually switch the "hot" wire to an accessory, light, etc., but in this case, the hot is always connected to the accessory and it only works when the isolated rail provides it with a ground (black) connection.

Ed

No, the voltage from the compensating winding will still raise the voltage 5 volts, the DC bias changes from positive to negative(above line - +, below, -)

Rob,

compensating windings aside...

If you take a basic transformer (with no compensating windings) and run a train at 18V, then add a DC source in series with the AC from the transformer, the horn will blow! (or the bell will ring, depending on which way you connect the DC).  The diagram will look like the one I provided.

You will see a slight surge in engine speed as the additional voltage is added. I'm guessing that the compensating windings in some transformers compensate for the surge.

Ed

The compensating windings(and chokes in 167 style/family controllers) were to compensate for loss through the rectifier the extra load of the whistle motor on steam locomotives.

Any additional outside voltage added in would raise the output voltage.

A half wave rectifier alone would reduce the output voltage. A full wave rectifier would raise the voltage an amount depending on the efficiency of the diode array and the total capacitance of the circuit.

K-Line actually made a whistle controller that did exactly that - it was powered by a DC wall-wart type AC adapter.

Rob,

Although I'm sure you know what you're talking about relative to transformers with compensating windings, I fail to see what it has to do with the question asked, the posts that followed, or the diagrams and statements I offered.

Sorry, but it's like someone asking about how to fix the crank up window in a car, and you saying, "no, that's not how my power windows work!"

Ed

Rob,

I recall someone posting on the forum that you could actually use a switch to connect a D cell in series with the power to the train to blow the horn!.... pretty neat!

Ed

It might have been me. It's an old trick from the 40's-50's to automate crossing whistles/horns using sections of isolated center rail, powering the isolated sections through a D cell. The 147 whistle controller does this exactly with the use of a fast-acting snap switch to prevent power drop-out from triggering the E-Unit.

This was the 147.

Larry

The two sine waves are referenced to common, so there is no potential in the common wire. The potential between 18 & 12, as referenced to common, is 6 volts, and is actually usable, but in the toy train world it is not advisable to do so as there is often no overload protection(due to larger transformers having the common protected by a breaker, instead of the "hot").

Done that, too. It works, but I was never comfortable with the idea of even momentarily feeding 12+ VAC serially through a dry cell battery. The thought of the battery going "poof" was always in the back of my mind.

Randy; The Neutral/Ground wire stays at ZERO volts, tho current does flow in it. It is the reference for all other voltages.

In your marbles in a pipe the Voltage is the pressure as you expect.

But the marbles run out the other end of the common / ground pipe with no resistance, thus no volts there. tho marbles / current is flowing.

As mentioned before, the Power entering your house is 220VAC peak to peak on 2 wires and a neutral/ground. In the Breaker panel, The power is split in half with one half and the neutral going to each outlet or light. The other half and neutral go to other outlets and lights.

This is one reason folks talk about Phasing. 2 Transformers plugged into separate outlets may end up on opposite sides of the 220VAC and thus be out of phase to each other.

I avoid this and give myself cutoff ability by powering the layout with a remote control power strip. Switch in hand or pocket, entire layout in phase as it's all powered from the same outlet.

Randy,

You're almost there....!

When we phase our transformers, the common (black) wire doesn't have a sum of the voltages (ie. 18+12=30), but it does have to carry the CURRENT (amps) of BOTH transformers.

We wire in phase so that voltages do not ADD, as on my out of phase diagram.  If two 18V transformers are NOT in phase, the voltage between the two red terminals is 36V.  In our train environment, we don't want ANY connection to two wires to have a voltage of 36V, so we keep everything in phase.

On the other hand, if we had two 18V transformers, capable of providing 5 amps each, in phase, feeding different items on our layout, all going to a common black wire that goes back to the two transformers, the common wire must be sized to carry the CURRENT (amps) of both transformers or 10 AMPS.

In house wiring, we can use two 120V feeders from a 3 wire cable with black WHITE (corrected, thanks guys!) as the common because the two 120V wires MUST BE connected in the fuse panel so they are out of phase!  So if one house circuit is using 15 amps (hair dryer) and the other circuit is using 2 amps (tv), the common black wire they share will have a total current (amps) of 15-2=13 amps.  If they were both feeding 15 amps, the current on the common would be 0 amps.

Bottom line:

Trains:  We wire all transformers in phase to insure against high voltages, but be sure the common can carry the current of all feeds.

Houses:  We wire the two circuits in a three wire cable out of phase, so that the total current the common carries will never be higher than the current required on the higher of the two feeds.

And, as Russell says, it's best to plug all your train stuff into one house circuit!  If your layout needs two circuits, make sure that the outlets you use are in phase with each other!

Ed

Thanks for all the help guys, and especially thanks to Ed for those great diagrams!

Randy