Running RMT Beep with CW-80

I had the same problem. Try using the direction button instead of the throttle. Depress the direction button for a full second or two.

Jim

Thanks Jim,
I will give that a try.

Nick

Otherwise, you might try always putting some sort of illuminated car behind the engine, or at least putting some sort of light bulb directly across the transformer terminals. A lighted lockon is ideal for this.

The CW-80 has an unusual design in that it has to see a certain amount of current before it will behave properly. It could be that your Beep isn't drawing enough current to make the CW-80 happy, and you need to add a little bit more. A light bulb of some sort is the easiest way to do this.

Nick,

The problem with the CW-80 is under no load or very light loads (like the BEEP or a Lionel Thomas loco), the CW-80 still outputs a high voltage. I just measured mine (got it a week ago to play with) and with no connections it reads 18 volts on an AC voltmeter with the orange control handle on zero..

If one looks at the postings by Dale Manquen over on the OGR forum, Mr. Manquen traced the circuits and one can see that the direction button goes to the microprocessor and does not physically break the circuit to the track terminals as older Lionel transformers. Thus pushing the DIRECTION button may have no effect. As a quick check, I just tried this with mine and with no load and the orange handle set to zero, the output voltage read by the voltmeter does not change, still 18 vac. The schematic shows a resistor and a capacitor that bypass the triac which regulates the output power to the track (and accessory output), thus it appears this unit will leak some ac to the output terminals all the time if there is no load. I do not know the value of the capacitor and Mr. Manquen did not show it on his version of the schematic.

I placed a 100 ohm 40 watt power resistor on the output to put a similar current load as a typical light bulb. At 18 volts the current would be 0.180 amperes (180 milliamps). Now the CW-80 output on the AC voltmeter goes to zero. When I turn the voltage up and press the direction button, the output goes to zero.

Many of the newer electronics in these locos draw very little current so when in neutral, the CW-80 output voltage may not go below the level necessary to trip the electronic e-unit.

So you may need to do as some others have suggested which is to place an 18 volt light bulb across the track terminals of the CW-80 to get it to work right with low current draw circuits.

Regards,
Roy

Hi Guys,
Thanks for the info. Yes I was just testing the loco so it was on its
own on the track. I did not have any lights loading the transformer. Doh
I should have remembered this![:)]

I have a Powerhouse on order as I already have a powermaster, Cab1 and
command base. I have the TMCC upgrade from Electric Railroad on order also so the beep will be our first TMCC loco.

Thanks again

Nick

Roy,

Placing that resistor on the outputs sounds like a great idea. I'm going to try that. I've heard the light bulb idea many times before, but didn't suggest it because it didn't work for me. I've placed 1, 2, 3 and 4 lighted cars behind Thomas and Beeps with no success. They still acted screwey with that transformer.

Any chance of going into more detail about where exactly the resistor goes, or perhaps even a photo of your CW-80 with one hooked up to the outputs. A picture can be worth a thousand words.

Jim

Roy,

I'm on chapter 6 of my Electronics Dummy book and I've read about capacitors. Is the light bulb trick sort of doing that? And, would a capacitor work as well?

I believe a capacitor would have the opposite impact and would keep the circuit voltage up.

Jim H

OK, then an inductor (as measured in Henrys). [:D]

(See, I TOLD YOU that learning a little bit about electronics is more dangerous than knowing practically nothing at all) [:D][:p]

Dave, here goes - a long winded answer for your reading enjoyment.

The light bulb essentially looks like a resistive and inductive load (The coiled filament is inductive and the filament itself is a resistor), but it is basically loading the output circuit. This gets into the voltage split across the CW-80 internal triac bypass resistor (just 10 ohms so I will ignore it for now) and capacitor (which is a 1 microfarad capacitor - 1 uF - and represents about 2,650 ohms impedance to 60 Hz – formula below) and the external load which in my case was a 100 ohm resistor.

In my initial post, I had no load and I kept the triacs off so there was no current flow (the voltmeter looks like a very high impedance so current flow into the voltmeter is negligible) and I measured 18 volts at the output. There was no voltage loss across the internal resistor and capacitor since there was no current flow.

Once I add a load of any type across the output, I have to account for it in the circuit network equations to calculate current flow and voltage into the load.

When I added the 100 ohm load, my total AC impedance load to the 18 volt transformer was the sum of the resistor (the 10 ohms I am ignoring), the 1 uF capacitor, and the 100 ohm resistor for 2650 + 100 = 2750 ohms. Triacs were off and I will keep them off to try to keep this simple.

Using V=IR and solving for current, this becomes I=V/R.

I = 18/2750 = .00654 amps (6.54 milliamps)

The voltage developed across the internal and external components becomes:

Voltage across the 100 ohm resistor: V=IR=.00654x100 = .65 volts (instead of 18 volts and is now low enough to trip electronic e-units – which is why the light bulb works and why the resistor worked in my test earlier).

Voltage across the internal components (resistor and capacitor): V=IR=.00654x2650=17.35 volts ac

So the voltage adds up to the original 18 vac (ignoring small rounding issues).

Theoretically I could pick a capacitor to put across the transformer output and determine the right impedance to provide the same 100 ohm load.

Impedance (the ac resistance of the capacitor) is: Xc=1/(2*pi*frequency*capacitance)

Since I need 100 ohms, solve the equation for the capacitance:

C=1/(2*pi*freq*Xc)=1/(2*3.14159*60Hz*100) = 2.65 x 10e-5 = .0000265 farads

So a 26.5 microfarad capacitor should provide the same load (remember the triac is still off).

Since this is AC, this requires a non-polarized capacitor. 26 uF is a large value for a non-polarized capacitor so one can achieve this by putting two polarized capacitors in series with the (typically) negative leads tied together (real non-polarized large value capacitors are built differently or use well matched pairs of polarized capacitors and are not recommended for extended use with high levels of ac). To get 25 uF requires two 50 uF caps in series (total capacitance of capacitors in series add as 1/C = 1/C1 + 1/C2 +1/C3 etc.). Just to have fun I took two 100 uF polarized caps I had and tied the negative leads together (50 uF total) and put it on the CW-80 and yes the voltage is appropriately reduced (0.3 vac) on the output.

So your answer Dave is yes, theoretically (in a perfect world), a capacitor (as would an inductor under similar analysis) would basically work (however, remember the triac is still off).

However, capacitors do not like large amounts of AC running through them. There are several issues such as internal heating (due to a characteristic called effective series resistance - ESR), dielectric punch-through, reverse voltage breakdown, temperature, and aging which reduces the capacitance and its voltage breakdown (both typically lower with age). The higher the ac voltages and currents the capacitor is subjected to the faster the capacitor degrades.

Now if we were to turn on the triac (which now by-passes the internal current and voltage limiting resistor and capacitor we discussed above) by moving the transformer handle the situation gets more complicated. We will have to take into careful account the ESR and quality factor of the capacitor, among several other factors). So at the end of the day, in the real world, having a capacitor provide the load is not the best idea.

As discussed, capacitors can heat up with unbalanced or large ac ripple currents and when they overheat they tend to explode. A lot of work has occurred in the last 50 years to make these things more reliable (pressure vents, etc.), but occasionally they do explode (particularly when a polarized capacitor is installed backwards in a circuit) and when they do, if the pressure vent does not work, the metal capacitor case becomes sort of a bullet. Just 3 weeks ago, a friend in our train club pulled out an old MTH PS1 loco and set it on the track to charge. After about 10 minutes it sounded like a firecracker went off and then smoke poured out of the shell (diesel loco). We opened it up and inside were the remains of a polarized (electrolytic) capacitor.

Your inductor idea can be analyzed in a similar manner and theoretically it would also work. But like the capacitor, the inductor will be sensitive to frequency and with a chopped sine wave, there are many other frequency components that exist that will not meet the impedance criteria for the inductor (and the capacitor) to look like a 100 ohm load. This situation will also result in overheating of the inductor.

So in the end, the light bulb is the simplest solution. The resistor is also OK, but keep in mind the resistor will be dissipating heat and can get quite warm to the touch if the voltages is set high, so it requires a certain type of resistor (wire wound ceramic for high temperature) and careful mounting to ensure the heat does not damage anything.

Just having light bulbs in the train consist is not always effective as the bulbs in a moving train are not always in good contact with the track. The bulb needs to be in a fixed position secured between the track and transformer. This is why some folks recommend the Lionel lighted lockons.

Regards,
Roy

You cannot simply add reactance and resistance to get impedance. Impedance is the root-sum-square of the two.

Very interesting. guys (and nicely technical), but I would just advise the fellow to try another transformer. If there ever was a finnicky and problematic transformer, the CW-80 is it.

Roy has sent me a schmetatic diagram of the CW-80, whose output circuit is as he describes, a triac shunted by 10 ohms in series with 1 microfarad.

The capactor would indeed leak current to the track, as he describes. In calculating that current, not only can the 10 ohms be neglected but also the 100 ohms of resistance that he put in series as a dummy load. The reason is that the voltage drop on the resistors is in quadrature to the total voltage, which is essentially the voltage across the capacitor, and, because of the root-sum-square rule that I mentioned, has virtually no effect on the capacitor voltage, not that it makes much difference in the numbers anyway.

He is also right in choosing a lamp as the best dummy load. Any low impedance would do the job; but the lamp has the advantage that it draws far less current when the voltage is turned up than the resistor would, since its resistance increases greatly as it heats up, typically by a factor of 10.

A properly-sized capacitor would draw the same amount of current as a resistor. However, the current would be in quadrature to the main load and would therefore add very little to its magnitude. But it might upset the transformer's operation, since the phase-control design used relies on the current's going to zero every half-cycle to shut the triac off. The capacitor's current would cause a phase shift in the current which could possible confuse the circuit that fires the triac by shutting the triac off early.

An inductor, like a capacitor, might cause a problem, but by delaying the triac shutoff. More importantly, when the whistle or bell is operated, the DC component of the waveform will produce a current flow limited only by the resistance of the inductor's winding, which is much less than its AC impedance. So it's not a very practical alternative.

Thanks much Bob, Roy, et al.

I've just got one tiny last question. [:D] (with 4 parts to the question)

As you know, I've added R/C to my toy trains, fed off a 7.2 3300mAh battery pack. I would like my loco(s) to run a bit more smoothly (operation going downhill under a lighter load is a bit jerky).

In reading my book, I noticed that a capacitor can be used to "smooth out voltage" so that it "stays at a nice, constant level." Since the battery pack is routed through an R/C motor boat receiver and then to my loco motors, my thoughts are that the motorboat receiver might be signalling current requirements that are a bit different from those used in the variable load conditions of the trains.

Well, I just returned from Radio Shack, where I spotted a large capacitor, a 4700uF polarized electroylic with axiel leads, 35 WVDC max + or - 20%.

1. Might this work to make running smoother?

2. Does it matter which motor lead it is installed in inline (+ or - side)?

3. Since the cap is polarized, how do I determine which way it should be installed?

4. If I decided later on to run a second battery in series (doubling the voltage and mAh stated above) to pull several locomotives in a consist, would the cap still be adequate to meet my needs (without melting)?

Much thanks!

1. I doubt it. The battery itself is much like a giant capacitor. It is possible that imperfections in the controller's drive circuitry degrade the load regulation however. So it might. But even if the locomotive is drawing as little as 1 ampere, the capacitor would discharge from 7.2 volts to zero in 34 milliseconds. So it might be more appropriate to say that a capacitor is much like a teeny-tiny battery.

2. Putting it in series won't work at all. It will block the DC current to the motor. Putting it in parallel is the only thing that makes any sense; but the motor's polarity reverses with direction; so you can't use a polarized capacitor there either. You can connect two such capacitors in series to get around the polarization; but the capacitance of the combination is halved.

3. See 2.

4. The 35-volt rating should be ample for that. Putting two batteries in series will not increase the charge available (3300 mA-hour, which is 11880 coulombs in metric units). The larger battery will however deliver that charge at twice the voltage, which means twice the energy.

I've had the same exact problem with my RMT Beep and CW-80. The Beep runs just fine with my other, cheaper transformers. I was wondering if there might be something wrong with my transformer, but looks like it's an understood quirk. I'll give the light-bulb load solution a try, and see if it helps.

My Beep also has problems with the uncoupler. It hangs so low, that it will bump some sections of my track, like in the switches.

Not much speed with the Beep, but I like it. I just wish it had a horn. I've actually purchased a diesel horn sound board from Lionel and was planning on installing it in the Beep, once I figure out a place to put it.

Jeff

Guys,

I run a couple Beeps on my layout using a MTH Z-750 or Z-1000 at between 10VAC to 15VAC. The reversing units cycle fine, however, I am having one problem. When cycling from forward to neutral both Beeps will creep forward in neural. The funny thing being that the directional lighting does not light in the direction of the Beeps movement. I have replaced one of the reversing units and the Beep still creeps forward. Any thoughts?
P.S.
Did RMT début any new products the York Meet?

Hi All,
Just fitted my Beep with the Electric Railroad TMCC upgrade with
sound commander. This is my first TMCC loco. It is great the sounds are
cool including the "beep-beep"[:)]. TMCC appears to work fine with my
inverter set up.
Here's a intresting point. The track power is provided by a transformer
powered from the inverter (60Hz US spec. power) but my command base
has the UK replacement transformer running from 50Hz UK spec. mains.
I was a bit concerned that this might cause some sort of problem but it
all appears to work fine.

Now I am driving everone mad with the coupler clank and beeping.[:D]

Regards

Nick

Hi, Newbie here, but lots of great info. Just unpacked a Thomas O-Gauge kit with my son, and we have the intermittent reversing problem. Solved it sort of by putting a bulb across the terminals, but it appears that the Fastrack terminal straight piece tries to put a load there with a mini built-in bulb?? I can't find reference to this piece (12016 is the regular fastrack terminal piece) so I can possibly just order a stronger bulb of that form factor?? Anyways, emailed lionel tech support in case they have a work around that doesn't involve the incandescent bulb I'm using.

CW-80
On the outside one of the nicest looking small tranformers there is.
On the inside............well, that's another story.

get yourself a 1033 and you'll never have a problem.

Mike Sacco

Hi Mike,

Well we'll see how much use we can get out of this CW80, but I have been looking at an MRC Tech II which seems a little more affordable than a clean used 1033 which I've also looked at and they seem pretty high (though possibly worth it).

Anyways, between the new terminal strip coming from lionel with embedded lamp, and my own voltage dropping DIY setup, I'm pretty sure the direction issue is solved. So the next steps are these intermittent freezes or CPU glitches remain, I'll work with the lionel folks, and actually send them video of the glitches occurring -- hiccup will usually occur within 10-15 minutes of use. If there just is no fix, I'll get another transformer, maybe a 1033 or MRC Tech II. If there is some hope for a fix, I'll work with the lionel folks.

I'm not sure how the Thomas engine ever got through the QA process with the CW80 without the voltage dropping terminal piece, unless changing directions was just not meant to be supported, but it is supported per the manual (but still won't work in circle mode, unless they've made a curved terminal strip with embedded lamp/diode/whatever circuit embedded) or unless I put my strip of light bulbs across the rails to drop the voltage down.

QUOTE: Originally posted by msacco CW-80 On the outside one of the nicest looking small tranformers there is. On the inside............well, that's another story. get yourself a 1033 and you'll never have a problem. Mike Sacco