Hi all,
Just got our second O gauge loco today a RMT Beep. Its a great
little loco and a very good runner.
Just one problem is that the reverse system does not work. I checked
the switch under the loco but it is on.
My thought is this is a CW-80 problem, any one else had this happen?
Regards
Nick
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).
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%.
Might this work to make running smoother?
Does it matter which motor lead it is installed in inline (+ or - side)?
Since the cap is polarized, how do I determine which way it should be installed?
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!
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.
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.
See 2.
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.