Basic electrics question..(photo's!!)

[daan]

That's right, you have 110 volts, we have 220. Lowest to get here is 6 amps, but in my ods and ends drawer I found one of 1 amp, one of 4 and one of 6. I'll use the 4 amp one in the first place to be on the safe side. If it switches too fast (150% overload) I'll use the 6 amp.
I also found a few airfans in it which belonged to my old computer once. Something to cool the resistorplates?

[lionelsoni]

That's a good idea, Daan. I wouldn't have thought of it, since ours don't come with such low ratings. The smallest that is allowed here is 15 amperes, to protect an AWG14 circuit, which is about 2 square millimeters.

[daan]

Measurements where ok. It was solid at 24 volts from 0 to 4 amps (Couldn't find a bigger motor to drive) I also found a way to get a power cutt off if there is a shortage, simply with cirquit brakers used in every home. They are rated on amps only and can be used at 20 volts. A 6 amp cirquit braker should prevent the resistors from overheating and cut off the power when a shortage occurs. I've tested it and as long as I choose a fast acting cirquit braker it should be sufficient.
I thought they needed the watts (voltage x amperage) to work, but they only use the amps to brake the cirquit. Anyway, one step further again.

[daan]

That's right Bob, but since I was advised to get a bit of overrating in the resistors.. though that's not the power I have to get rid off.
About the transformer unwrapping; yesterday I drove the F3's and they drove fantastic (for half a minute) on the 24 volt overvoltage on my maerklin transformer... The machine needs that power somehow.
I'm going to measure everything first before I start unwinding. May be I'm only unwinding to 22 volts..
I first have to see what voltage the transformer has with no connections made; voltage drop will occur when power is consumed, so I'll measure that too.

[lionelsoni]

Daan, if you're planning to unwrap 6 volts from the transformer, the maximum power you can draw at 6 amperes is (34 - 6) x 6 = 108 watts; and, at the highest position, that will all be dissipated by the train.

[daan]

In the electronic shop they had not the resistors I cam for, but I have something for a nice setup (in my eyes). I have 6 1ohm 5 watt resistors, 8 1 ohm 9 watt resistors, 4 1ohm 7 watt resistors and 7 0.5 ohm 9 watt resistors.
I'm planning the following setup. 3x 1ohm 5 watt resistors parallel giving 0.33 ohm 15 watts in the first 2 contacts. The next four contacts are 2x 1 ohm 9 watts, giving 0.5 ohms 18 watts. Then 2 contacts 2x 1ohms 7 watts, giving 0,5 ohms 14 watts. the last contacts are made with the 0,5 ohms 9 watts resistors.

The idea behind it is that if a train drives, most of the time the last resistors are not in use; they are only there to gain speed. Giving that the train will not draw any currence when it is in standstill (e-unit will fall, but that's a drawback I have to take in this case) and that if the train is at speed only a few of the upper resistors are in use.
In this way only the first 8 contacts can handle 6 amps at minimum. If I drive the second train, again only the first few resistors are in use for the second train.
If I use only one train it can be up to 4 amps and the voltage will be sufficient to drive from the 6 amp resistors. If I use a train which uses only 2 amps or less, I need all the resistors I have to get running from standstill.
It's difficult to explain what I have in mind.. I have to find out if it's working. Otherwise I have to get other resistors.. the expense was $30 for the resistors. The contactplates where made from things thrown away and some stuff from the "odds and ends drawer".

next thing to make is the cooling for the resistors. I have 183 watts of maximum power to get rid of.. making it an ideal extra heating for my hobbyroom. [:D][:D]

[daan]

The transformer is rated for 24 volts and 12 amps. So will be (after rewind) a 20x12= 240 watts. Should be ok for running one or two trains..
I know about nichrome wire, I also used it to cut styrofoam wings for RC planes.. There is a big disadvantage on it. It changes in length when heated. that means that I can't make small resistors by winding it across an isolator. Heating it up will snap the wire.
To make a resistor I could make a 1 foot section with a spring on the end to catch the changes in length. Another disadvantage is that nichrome can't be soldered. I remember that the problem of connecting the wire when cutting styrofoam was only solved with crocodile clamps on the end.
To keep it small and most important, to keep the heat in control, I have to use resistors. They are not that expensive. In powerratings under 20 watts each they have porcelain resistors in about every value between 0,1 and 2,0 ohms at $1,50 each.
For the experiment I use only 1 setup of resistors. (15x1,50= 22,50, which is not a big expense to get a good powerregulator)
Those ceramic resistors can be mounted on a metal U profile of 4mm thickness and only 40x200mm. If I use conducting paste and add a few ribs to the U-profile (weld a few strips to it) the cooling of the resistors will be sufficient and the design is compact.
I can't make it that small with nichrome-wire and the nichromewire won't be cooling in such an efficient way which is essential if I want to use it continuously.
there is a good thing for nichromewire though. The change in length can be used to throw switches, signals, open doors in a locomotive shed etc.

[jkerklo]

Yes, 6 square mm, not 6 mm square. Still a nice transformer.

John Kerklo
TCA 94-38455
www.Three-Rail.com

[lionelsoni]

John, he said 6 square millimeters, not 6 millimeters square. That is about the same as AWG 10. In Europe, wire is measured by the cross-sectional area. AWG 20, which you mentioned, would be about .5 square millimeter.

Daan, 2 ohms at 5 amperes will drop 10 volts, putting you at 14 volts. You need 1.6 ohms to get 16 volts and 1.2 ohms for 18 volts, at 5 amperes.

If you would be satisfied with 5 amperes, as it seems now, and an 8-volt drop to 16 volts, you wouldn't need as much lamp power rating, 220 watts, or 3 75-watt lamps instead of 4.

[jkerklo]

I, too, am curious about how it works. Knowing too much about the electrical characteristics, I have dismissed resistive control in my mind, but now wonder if there might actually be some advantages.

Finding the resistors you need could be a problem. I would guess that they cost too much new (e.g. more than a KW). Finding used or surplus in the values you need will be by purest chance.

Let me re-suggest something from my first post on this topic: Nichrome Wire.

Nichrome wire is a high-resistance wire. It is what power resistors are made from.
There are many alloys, any will do.

For example, 20 gauge nichrome wire has a resistance of 0.65 ohms per foot (you do the metric). About 3 inches gives you your .25 ohms.

I use nichrome wire for cutting foam. Recently bought 30 feet for $5.95 off e-bay. It is used in a number of hobbies: rockets (igniter), RC airplanes (cutting foam wings), modelling (shaping foam scenery).

If you are interested, I will get you started with the computations. It is a balancing act between resistance, power, and a temperature limit.

John Kerklo
TCA 94-38455
www.Three-Rail.com

[daan]

John, that's exactly what I want to realize. I made 16 contacts. the 0 setting will be under the last contact, so basicly on the plastic. I have room for 15 resistors, so if I use 0.5 ohms per contact on 5 amps I have too many contacts on my contactplate. But if I use 2 amps the voltage on the lowest contact is 5 volts. The difference between a 5 amp train and a 2 amp train simply is solved by lowering the lever more down.
I realized from the start of the planning that it would be impossible to keep the same levertravel for all engines, so I've overdone the amount of contacts (and resistors) by 2.
The resistor values will be chosen on trying out what's best. Probably 0.25 on the upper 2 contacts and 0.75 on the lower 4. The upper contacts will be about 10 watts, the 0.5 will be 15 or more watts and the 0.75 have to be around 18 watts.
Depends on what there is offered at the electronic store.. I'm really curiuos how it will work when it's finished..

[jkerklo]

daan,

Good idea, removing some turns. One of us should have thought of it sooner.
If the secondary wire is 6mm square, you have one heck of a tranformer. Could run many trains.

I think your resistor design should account for a range of trains: say 2 amps to 5 amps.
If it is designed for just 5 amps, a single motor engine may run too fast at even the minimum settings. Perhaps, the range needed is worthy of a little experimentation.

I would start by determining the minimum track voltage you want when a train is running, say 8 volts. Dropping 18 - 8 = 10 volts at 2 amps means a total resistance of 5 ohms. At 5 amps, 2 ohms. So, the sum of all the resistors on your contacts should be 5 ohms for 2 amp trains and 2 ohms for 5 amp trains.

This illustrates one of the problems with using resistors for speed control.

Not to worry. this can be solved with a series resistor from the 18 volt post that is bypassed with a switch. Say 2 ohms. The resistor is in-circuit for 2 amp trains, bypassed (shorted) for 5 amp trains. If transformer construction allows, you could also provide a "tap" at, say 14 volts or so, and switch between 18 and 14 volts for different trains.

I think I counted 14 taps on your "resistorplate." With one for "full" and one for Off, that means 12 resistors. Say you want a spread of 3 ohms (just for example). That means 0.25 ohms per tap.

Note that resistor power is the same for all the resistors. At 5 amps and 3 ohms, total power dissipated is 75 watts , so each resistor needs to be (75 / 12) = 6.25 watts. With the control arm set for only one resistor in the circuit, power dissipation is 6.25 watts. How convenient.

Not all resistors need to be the same ohm value; you may want finer control in part of the range.

John Kerklo
TCA 94-38455
www.Three-Rail.com

[daan]

@ Bob, thanx for the calculation on this. I guess with 6 amps a 300 watt bulb is a bit overdone to keep it simple. my layout is not that big, and having such an amount (4x75 watts) on lamps just to cover up the 6 volts from the transformer.. I'm not sure that this is the way I want to go.
To me a resistor of 2 ohms giving between a 6 and 8 volt drop @ 5 amps will be a better solution together with cirquit brakers.
May be I can rewind the transformers last windings to get it down to 18 volts without lamps or resistors. That would be the best solution possible and that will be the way to go in the first place. The coilwire of the transformer in the secondary winding is a thick 6 square mm solid copper wire which can be wound off without taking anything more away then the outside insulation. It has 48 loops for 24 volts, so every 2 loops is 1 volt less. I have to get rid of 6x2=12 loops in order to get my 18 volts.
I didn't plan to do this, but the lightbulb solution and the resistor solution to get down to 18 volts are no good options. If getting rid of a few windings in the transformer (difficult job, because possible damage on the other wires will ruin it) I can leave the lamps and the big resistor which is a big plus to the powerconsumtion of my setup..

[lionelsoni]

Daan, you have just rediscovered an old engineers' "rule of thumb" (approximation) that the resistance of an incandescent lamp changes by a factor of 10 between off and on. But experiment is not the only way to solve the problem.

It is possible, actually without too much trouble, to predict what different lamps will do in your circuit. First the full-brightness current and resistance, which I'll call Imax and Rmax. Imax is just the full-brightness power (Pmax, which is the power rating on the lamp, divided by the rated voltage, Emax, which is 24 volts in your case. That is, Imax = Pmax / Emax. (The full-brightness resistance is Rmax = Emax / Imax.)

Now, the actual current at less than full brightness:

I = Imax * (E / Emax) ^ .55

where the ^ symbol means exponentiation.

To find the power rating of a 24-volt lamp that will provide a certain minimum voltage to the train as a certain load current, for example 6 amperes, subtract the required voltage, for example, 18 volts, from the 24 volts available, that is, 6 volts. So the lamp voltage is 6 and the current is 6. We can solve the equation above for Imax:

Imax = I * ((Emax / E) ^ .55) = 6 * ((24 / 6) ^ .55) = 6 * (4 ^ .55)

Pmax = Emax * Imax = 24 * 6 * (4 ^ .55) = 309

So it looks like a 300-watt lamp (or 4 75-watt lamps, or some other combination) would be about right.

[daan]

@Lionelsoni, I figured that too. Now the next problem are the bulbs. I've measured all kinds of bulbs for their resistance. BUT the calculation doesn't work with it. On a 24 volts 5 watt bulb I measure 12 Ohms. But on the calculator I come up with (P=VxI => P/V=I so 5/24= 0.2 amps. With that R=V/I R=24/0.2= 120 Ohms)
That means while heating up the bulb gets a 10 times higher resistance.
The problem is only solved by experimentation.. Since I still have to go to the electroshop to get my resistors, or order them, I still am stuck with the project for now. After finishing my resistorplate I can start fiddling with lights to solve this and hopefully drive trains.

[lionelsoni]

It might not be a bad idea to use both circuit breakers (I assume that is what you mean by "mechanical switchoff") and the lamps. The lamps will limit the current immediately, even with a dead short, and the circuit breakers can open the circuit a fraction of a second later.

The lamps also provide a resistance that stays in the circuit even at the highest setting, dropping the too-high transformer voltage a little so that more contact positions are available for normal control. Of course, you could do that with conventional resistors too; but I think that the lamps would be much more entertaining!

As for the size of the resistance steps, as I said before, you may want to use larger steps at the bottom, smaller steps at the top. However, I would expect that the effect of changing steps would be rather gentler with your 24-volt transformer compared to the 18-volt transformer that Roy used. Keep in mind that you will have enough resistance in series to drop 24 volts to no more than 18 even at top speed. When you move up a step, the current through that resistance increases somewhat, and so does the voltage drop, partly offsetting the voltage increase that you might otherwise expect. This effect should be much more pronounced with the large voltage drop from your transformer to the train than with Roy's transformer.

[daan]

@wrmcclallen; it could be a good idea to use the top resistors at 0.25 ohms. It will decrease voltage by 0.9 volts. My train will probably draw about 3 amps. The cirquitry in my maerklin transformer is laid out for 2 amps, and keeps the power on for a few minutes. I guess the train needs about 3 amps in this. But however, it is in my meaning to get a full train with my 2353 and 5 lit aluminium cars, so the current draw will be about the 5.6 amps you mentioned.
The lamps in parallel would be a good idea too, I emagine that they don't light fully if the train runs, but will light up when there is a shortcut somewhere. Those lamps can handle that stress, a resistor wouldn't. The problem is that the power isn't shut off to the track, which I start to think would be nice. With 5 amp @ 20 volts (100 watts) there is a serious threat of loosening soldered joints when something blockes or giant sparks burning into the track and roller pickups.

The "down side" of the contactplate is a dead spot. There are no contacts on that part of the plate and to shut off power I simply move the lever down far enough to reach beyond the contacts.
I'll visit the local electronic shop tomorrow or next week (they're closed at the weekend) to find out what resistors I can get there and if they have those mechanical switch-offs for about 5 or 6 amps.. If they don't I'll have to use the lightbulbs or make a mechanical switchoff myself.
I must say I'm getting inspired a lot by all the posts on this. It really encourages me to go on and try further. Thanx so far!!

[lionelsoni]

Four of those 75-watt lamps will provide 18.4 volts with a 5.6-ampere current, or 18 volts with a 5.8-ampere current.

[wrmcclellan]

Oops,

My mistake. I got several numbers mixed up. Hard to read everything in the Message window.

The resistors in series were 3.2 ohms, not 4.2.

Thus the voltage drop per 0.5 ohms was 1.7 volts. Amperage at this point was 3.4 amps (lights and e-unit).

At full speed the train was pulling 5.6 amperes.

I edited the previous post to correct these mistakes.

[wrmcclellan]

Daan,

I just completed a quick test.

With some 10 watt wirewound resistors I had (two 0.5 ohm and two 1.1 ohm) and my Lionel Great Northern Empire Builder set (1 conventional two motored Pullmor F3 and 5 aluminum lighted passenger cars) I achieved the following.

The resistors were in series for a total of 3.2 ohms. I used a ZW that produces a maxmum of about 17-18 volts into the 3.2 ohm and GN train load.

At 3.2 ohms, the voltage to the track was about 6 volts, just barely enough to activate the e-unit. Current draw was approx 2.6 amps (e-unit and lights) for an 11 volt drop (approx 1.7 volt drop per 0.5 ohms of resistance).

I left the 3.2 ohms in series with the load (train) and used a jumper lead to short from the transformer hot terminal (A) to different resistor junctions - thus I could go from the 6 volt minimum to a full 18 volts to the track. Leaving the 6 volts as a constant kept the e-unit energized (althought 6 volts is a bit low).

This quick experiment worked quite well. As I jumped between resistances the forward momentum of the train only slowed slightly (but noticably) if I was increasing voltage. Obviously if decreasing voltage to the train there was no noticable change.

I did notice that just switching 0.5 ohms at this heavy load (1.7 volt drop) did make a large difference in speed.You may want to obtain a few 0.25 ohm resistors to use in place of some of the 0.5 ohm resistors to have finer control at some speed ranges on your controller. This will take some experimenting on your part with the typical trains you intend to use.

And as predicted, the resistors get extremely hot!

I think your scheme is going to work very well and will be a fine piece of industrial art!

Regards,
Roy

[lionelsoni]

Daan, how about two of those lamps in parallel? They are 7.68 ohms at full power, but the resistance will go down quite a bit at lower power. You probably won't be drawing near 6 amperes, so two of them will have considerably less than 4 ohms at, say, 3 amperes.

In fact, since it is known that lamp current varies as the .55 power of voltage, we can predict that at 3 amperes, two such lamps will drop just 6.3 volts, reducing the 24 volts of the transformer to a reasonable 17.7 volts.

If you put three lamps in parallel and assume a 4-ampere maximum draw, as you did above, the drop is 5.1 volts, to 18.9 volts, also reasonable.

[jkerklo]

daan,

Looks really good. The pickup shoes are a perfect solution. You must have a great toolbox.

I couldn't tell from the pictures, but there should be one contact without a wire for "OFF."

Did the furry thing in the photos help with construction?

Have you determined the range of resistors needed? How will the resistors be arranged in the circuit?

John Kerklo
TCA 94-38455
www.Three-Rail.com

[daan]

This is the solution for the power pickup.. I used 2 isolated levers on an axle, with a graphite power pickup. These unit's are sprung and calculated for high voltages and amp-ratings. Besides that they have a huge pickup surface, which means they touch at least 2 contacts at once. The e-unit problem is solved for the rheostat. If I whi***o keep the direction I can do 2 things. 1: shut the e-unit off, 2: keep some resistors working by not pulling the levers down completely.

This is the complete rheostat, only the resistorplate has to be made. Levers, and pickup contacts are ok.

The front side, the pickups are visible

The power pickup shoe. A complete unit captured from a scrapped travelling platform about a year or two ago, which I found in my toolbox. Of course the pickups need some adjustment.. Both levers are identical, but in mirror.
(I've linked the photo's from my own website, since shutterfly is down for maintenance)

[daan]

Those lamps is not a new idea. In my maerklin transformer there is a light indicating short cirquits. If the lamp is dim, it's okay, but if the lamps is bright, the trafo is feeding a shortcut. I know it also has a mechanical switchoff however. I was thinking about lightbulbs of a truck. Those are 24 volts and up to 75 watts. Meaning 3 amps. The resistance in these lamps is about 8 Ohms. A voltage drop of 15 volts at 2 amps.
But since it's such a huge Ohm resistance, it is more suited for the e-unit resistor.
Probably I'll try to get the transformer a bit down, or put some normal resistors in front of it, in combination with a mechanical switchoff.
I'll post the photo's of the pickup arms, with a very good solution and no rollerball balls..

[lionelsoni]

John, it's that very resistance increase that makes me think that they might work for current limiting. The lamp should be chosen for minimal effect at normal currents (but, in this case, a little resistance is to the good, to take the edge off the 24 volts), but substantial resistance at high currents, so that a short circuit draws less current than if the lamp resistance stayed the same.

I admit that it would take some work to find a suitable lamp or lamps. They should be rated at 24 volts, for the short-circuit case, which makes me think of automotive lamps in series. But getting the resistance low enough may be a problem, since even automobile headlights use only a few tens of watts. I would recomment a higher-voltage lamp (230 volts in Daan's case) in series with the primary instead, since it would be much easier to get a high-powered, ergo lower-resistance, lamp at that voltage. Unfortunately, the two controls that he is building would interact then.

I wouldn't think the effect on the trains would be adverse, since Daan would be continuously compensating for any varying voltage across the lamps as he sets the control for the train current that he wants.

[jkerklo]

Bob,

I think you are right about the e-unit resistor and removing the plastic instead; easier. I would like to see how well the e-unit resistor works, but I guess I will have to try it myself.

Light bulbs are tricky. Their resistance increases as the filiment heats up. Thus, the current driving the train decreases. Getting just the right bulb might also be a problem.
Probably would need more than one bulb in parallel to get the right current for the train.

Varying light bulbs would be neat, but would likely adversely affect the train.

Seems to me I have seen the light-bulb-for-current-limit mentioned before. Anyone tried this?


John Kerklo
TCA 94-38455
www.Three-Rail.com

[lionelsoni]

John, I think that removing the plastic is a surer solution.

I hope that Daan does try putting a lamp in series which would both provide fault protection and drop a few volts in normal operation. A couple of 12-volt automotive lamps in series might do the trick. I imagine the lamp filaments glowing inside the clear glass bulbs as Daan moves the control up quickly and then fading back to a dull red as the train starts and he backs off on the throttle. I think it could be very dramatic.

[jkerklo]

daan,

The e-unit resistor limits the maximum current that can be drawn. With an 8 ohm resistor, even with the track shorted, at most 3 amps (24 / 8) can be drawn. With an e-unit and motor, the draw must be less than 3 amps. The 2 amps I suggested in the previous example is a starting point for experimentation, rather than a final value. There are too many unknowns, such as what trains you are running, might run in the future, etc.

It is true that some consideration must be given to varying current draw for different engines. This is probably best established by trial and error, running your trains.

Note that all the resistors need to have higher power ratings because at least 6 volts of the 24 volt transformer always needs to be "wasted." A lower voltage transformer would need to waste less power. I think, however, that we have been working to allow use of what you have at hand, rather than what might be a perfect solution.

We are only concerned about e-unit dropout in the time it takes to move the arm from one contact point to another. E-units are a bit slow to respond to power interruptions, otherwise trains would frequently stop on rough track or over turnouts. The e-unit resistor doesn't have to hold the e-unit indefinately.

John Kerklo
TCA 94-38455
www.Three-Rail.com

[lionelsoni]

In a motor or generator of any size, the brushes have a braided wire embedded in them to carry the current. Small motors, like traditional Lionel motors, can get away with letting the current occasionally pass through the springs. Your spring will have to carry the current for several motors and train lights, which might not be enough to damage it; but the real worry is what happens when there is a short circuit and very heavy currents flow for a short time. I have had several freak accidents in which the knuckle springs of couplers (briefly) carried the fault current (from one truck still on the track to another contacting the center rail) and one in which a fairly substantial link in a passenger-car truck fused. The motor's brush springs are not exposed to such dangers; but your controller's will be.

I hope you get it to work. I'm anxious to read your reaction to the way it feels to operate with a rheostat, which seems to me very different from a transformer.

You might consider the possibility of using larger resistors toward the bottom of the control range, since fixed resistance steps become a decreasing fraction of the total resistance at the low end. Are you still thinking about putting a lamp in series with the resistor string?