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Is this feasible?

 

 I run three stop-and-go trolleys on a large loop.  Two are stopped at stations while the third runs over the insulated startup track section(s) and then it stops etc.  Everything works great but it would look much better if a trolley’s interior lights remained on when it stops.  

 

Idea:   1 – Remove interior bulb(s).   2 – Add a capacitor and LED lighting. 3 – Charge the capacitor with track power.  As the trolleys run, the capacitors come alive and lights remain on until the capacitor looses charge from power-off.  

 

 If this is doable, I would appreciate recommendations and/or a parts list. 

 

Thanks for reading,

 Paul

 

Long time with trains –First time on Forum  

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 That's welcome news. Thanks.

1.  minute or two.

2. Cost shouldn't be an issue.
3. Have done of fine soldering, but not with diodes.
4. Trolley's are current MTH models but no Pics at moment.**

* A trolley travels an 80' track back and forth (160') between 2 reversing loops where others wait.
** This is a club layout. I'm going out of town and won't have access for a while.
 

Regards

Originally Posted by 1942guy:

 

 That's welcome news. Thanks.

1.  minute or two.

2. Cost shouldn't be an issue.
3. Have done of fine soldering, but not with diodes.
4. Trolley's are current MTH models but no Pics at moment.**

* A trolley travels an 80' track back and forth (160') between 2 reversing loops where others wait.
** This is a club layout. I'm going out of town and won't have access for a while.
 

Regards

Welcome,

Looks like you've almost got this sorted on your own "42"

 

#1 being answered as close as possible will get a very exact part out of Stan to get you what you want.  I don't think two minutes is going to be hard. No longer than that?

 FYI -You have attracted some of the best help you can get with John and Stan.

 

 Soldering an electronic components leg off a board, is very much like soldering wire, but pre-tinning it can normally be skipped. Pliers as a heat sink at the leg base if you are really having to heat the job hard. Like anything, sometimes it will go better than others. But its normally easy. On boards you just fit the leg in the hole. Heat the tip and trace as you fill it. 

 Even with too much heat, you wont burn the components nearly as often as you'd first cook traces till they peel off a board.

 Plumbing solder, will only make you curse. Use electrical solder.

Flux is for "dirty" things, and old wire. It really helps too. Its a cleaner.(remove excess) 

 

 

Good luck!

 

Install constant voltage lighting using LEDs. Here are 2 different ways

 

LINK1

 

LINK2

 

This will give constant intensity lighting from about 5 volts up. Using multiple  1 farad super capacitors in link 2, added to the regulated 5 volt end,the lights will stay on when power is removed for 30 seconds or more. I use five,  1 farad button type,they mount easily to the fish paper linked. I do this on passenger cars and they stay lighted fairly well for up to 2 minutes,gradually fading. They still glow very faintly after an hour or more.

 

Or

 

 insert a diode string in series to the motor(s). Enough diodes so the trolley motor stalls at 5 volts input. Instead of cutting power completely to the trolley during a stop,leave on 5 volts instead. This can be accomplished with a relay spdt where the 5 volts is inputed into the NO contacts.

 

The diode string is shown here.

If the trolley only runs one direction you only need a single string.  for a trolley 3 amp or even 1 amp diodes are OK sometimes depending on the trolley

 

LINK3

 

Dale H

Last edited by Dale H
Originally Posted by 1942guy:
1.  minute or two.
2. Cost shouldn't be an issue.
3. Have done of fine soldering, but not with diodes.
4. Trolley's are current MTH models but no Pics at moment.**

* A trolley travels an 80' track back and forth (160') between 2 reversing loops where others wait.

As others point out, the 2 minute specification makes this very interesting. 

 

A few more follow-up questions.

 

5. So a trolley stops/waits on a reversing loop with no track power for up to 2 minutes. When a trolley runs on powered track, a) what is the minimum "guaranteed" track voltage and b) for how long will that trolley be on powered track before stopping?  This line of questioning goes to the ratio of charging vs. discharging time which applies if using some kind of energy storage device (capacitor, battery, momentum flywheel, whatever).

 

6. What are the MTH stock numbers of these trolleys? There's a rechargeable battery in the MTH Protosound 2 variants if that's what you mean by "current" models.  That said, is OK to tap into this battery as a source of backup power?  I'm not asking how-to, but as a logistical matter - for example, you may have taken out the PS2 electronics and put in a barebones motor controller. 

 

7. If using stock PS2 electronics, are the trolleys operating conventional-mode (track voltage controls speed) "locked" in forward direction?

 

8. Read Dale H's method of lowering the voltage to just the motor while keeping a higher voltage available to the PS electronics to keep the lights running.  So rather than turning off track voltage to a stop block/section, you would reduce it.  You have to think about it for a while and read his linked info.  Along the same lines, if you are using stock MTH electronics in conventional-mode with factory-default speed-control turned "on", note there is a small-zone of track voltage where the electronics is ON (lights, sounds, etc.) but the trolley is stopped.  It's around 6-8 V on the track.  Did you already try this?  For example, rather than turning OFF power to the station stop blocks/sections you would apply, say, 7V.  The trolley would stop moving but the electronics (lights/sounds) would be active.  Variants of this idea may include switching over to DC track power as it might be easier to generate the correct "stop" voltage with DC than AC as current MTH electronics can operate on either.

 

Last edited by stan2004

One thing about using a superCAP or several of them to add capacity.  You can charge them far faster than they discharge, unlike traditional rechargeable batteries. 

 

I just connected two 2.5F 5.5 volt superCAPs to a 39 ohm resistor and three parallel LED's.  Feeding that I have a diode (to block reverse flow to the power source).  I feed the network with 5.5 volts.  The caps easily charged up from a 1A current limited supply in about 25 seconds, and the LED's were burning brightly.  I killed the input power and the three LED's burned for over three minutes at a nice brightness.  At the two minute mark, I still had over 4 volts across the superCAP bank and the LED's were burning brightly.

 

The caps are $4/ea, a regulated power supply is $3-4, and you just have to wire up the LED's.

 

If you want more LED's, you could add another cap or two.

 

Originally Posted by BillP:

Suppose you wanted to drive 6 LEDs, headlight,  three overhead interior lights, two red tail lights.  

 

What at would be a good circuit?  What would be the function for computing light time from the capacitance?

A back-of-envelope engineered-design would convert your lighting power requirement into energy.  Energy (in Joules) = Power (in Watts) x Time (in seconds).

 

Let's say each LEDs is 3 Volts drawing 0.02 Amps (20 milliamps).  Power (Watts) = Voltage (in Volts) x Current (in Amps).  So each LED requires 0.06 Watts.  Now say you want to power your 6 LEDs for 10 seconds.  The energy needed is 6 x 0.06 Watts x 10 seconds = 3.6 Joules. 

 

The theoretical energy stored in a capacitor is 1/2 x Capacitance (in Farads) x Voltage x Voltage (in Volts).  So a 1 Farad capacitor (a typical supercap value that runs about $5) charged up to 5 Volts stored 1/2 x 1 Farad x 5 Volts x 5 Volts = 12.5 Joules.

 

So that supercap, fully-charged, has enough energy to power the LEDs for 10 seconds.

 

Then you start dealing with practical issues such as you can't really squeeze all the energy out of the supercap.  For example, as the supercap discharges, its voltage drops.  At some point the voltage drops below 3V (or whatever the LEDs need to operate) and the stored energy is for naught.  Then you design a circuit to boost that low voltage up above 3V to squeeze out the last ounce of energy.  Or you play with the variables (add another supercap to have more energy to begin with).  Or, and so on.

 

There are other design considerations - for example, on the charging side there are practical issues of how fast you can charge a supercap.

 

 

Originally Posted by gunrunnerjohn:

With six LEDs, I'll bet three of the 2.5F caps will get it done for two minutes.   Even if they get a bit dimmer, they're still light, and the slow decay would probably not be that noticeable.  It's simple and fairly cheap to build.

 

I stand ready to throw darts at any circuit you propose!

 

In your previous example of the supercap dropping to 4V, I calculated an approximate drop of 30-40% in LED brightness.  This is occurring over many tens of seconds you so probably don't notice it.  And it does make for a simple circuit.  But if the OP wants constant-brightness, then additional regulation is required.

 

However, I'm curious how you make a simple and cheap rapid charging circuit.  A 1 Amp constant-current charger that cuts off when the supercap(s) reaches 5V is "easy" if you have a bench power-supply.  But a hand-built circuit? 

 

Also, when the supercap is charging for tens of seconds, I'd think you want full-brightness on the LEDs.  Or maybe not?  If so you need a bypass circuit - could be as simple as diode OR'ing - to power the LEDs to full-brightness whenever track power is available.

 

The 1A supply is actually easier than it sounds.  You just need the 5.5 V adjustable supply, the diode, and a low value resistor.  I'd consider a 5-6 ohm to start.  Dropping 5V across the 6 ohm during the initial charging state is possible and it ends up less than an amp surge current.  It actually charges the caps pretty quickly, and remember that we're not ever totally discharging them, we just want to run 2 minutes and then we'll have the "charger" working again.  When we're running under power, the few ohms of the series resistor is insignificant in lighting the LED's. 

 

The only time you'd ever notice the caps charging is on initial power on, after that they should already be charged to a reasonable level.

 

It's the KISS method.  No switching, just a 1A adjustable supply, a diode, and a couple of selected resistors.

 

Fire those darts, be make sure you hit the target.

 

 

Now, if he's anal about truly constant intensity, you'd probably have to add some complexity.  Truthfully, I think this would do the trick and nobody that didn't know the secret would realize that the intensity changes some during the power off cycle.

 

Originally Posted by stan2004:
Let's say each LEDs is 3 Volts drawing 0.02 Amps (20 milliamps).  Power (Watts) = Voltage (in Volts) x Current (in Amps).  So each LED requires 0.06 Watts.  Now say you want to power your 6 LEDs for 10 seconds.  The energy needed is 6 x 0.06 Watts x 10 seconds = 3.6 Joules. 

 

The theoretical energy stored in a capacitor is 1/2 x Capacitance (in Farads) x Voltage x Voltage (in Volts).  So a 1 Farad capacitor (a typical supercap value that runs about $5) charged up to 5 Volts stored 1/2 x 1 Farad x 5 Volts x 5 Volts = 12.5 Joules.

 

So that supercap, fully-charged, has enough energy to power the LEDs for 10 seconds.

 

Then you start dealing with practical issues such as you can't really squeeze all the energy out of the supercap.  For example, as the supercap discharges, its voltage drops.  At some point the voltage drops below 3V (or whatever the LEDs need to operate) and the stored energy is for naught.  Then you design a circuit to boost that low voltage up above 3V to squeeze out the last ounce of energy.  Or you play with the variables (add another supercap to have more energy to begin with).  Or, and so on.

 

There are other design considerations - for example, on the charging side there are practical issues of how fast you can charge a supercap.

 

 

I question the calculations here.

 

The energy needed to light the LED's for 10 seconds is 3.6 Joules.

The fully charged capacitor has 12.5 Joules of stored energy.

 

Why can it only light the LED's for 10 seconds?  I know you can't squeeze all the energy out, but surely you can do better than less than 30%.

 

What am I missing?

Originally Posted by gunrunnerjohn:

The 1A supply is actually easier than it sounds.  You just need the 5.5 V adjustable supply, the diode, and a low value resistor.  I'd consider a 5-6 ohm to start.  Dropping 5V across the 6 ohm during the initial charging state is possible and it ends up less than an amp surge current.  It actually charges the caps pretty quickly, and remember that we're not ever totally discharging them, we just want to run 2 minutes and then we'll have the "charger" working again.  When we're running under power, the few ohms of the series resistor is insignificant in lighting the LED's. 

 

The only time you'd ever notice the caps charging is on initial power on, after that they should already be charged to a reasonable level.

 

It's the KISS method.  No switching, just a 1A adjustable supply, a diode, and a couple of selected resistors.

 

Fire those darts, be make sure you hit the target.

 

 

Now, if he's anal about truly constant intensity, you'd probably have to add some complexity.  Truthfully, I think this would do the trick and nobody that didn't know the secret would realize that the intensity changes some during the power off cycle.

 

KISS is fine, but a fixed-value resistor is a very slooow way to charge a supercap.  So when you say 5 ohm resistor, that is a peak current of 1 Amp if from a 5V source and only occurs if the supercap starts totally discharged.  When the supercap reaches 3V, the charging current drops to 0.4 Amps.  When the supercap reaches 4V, the charging current drops to 0.2 amps.  And the current keeps dropping.  Basically you have the so-called RC time-constant which is the product of R and C, or in this case 5 ohms times 2.5 Farads which is 12.5 seconds.  Starting from a discharged cap, it takes about 4 time constants or 50 seconds to charge up the cap to 99%.  OTOH, if you had a constant-current charger which is what you'd find in engineered rapid-charger circuits (more complex), a 1 Amp charger would fully charge the cap 4 times faster.

 

So with your basic KISS circuit, the LEDs would slooowly dim over a minute or whatever when no power is available, and then slooowly get brighter over a minute or whatever when power is restored.  It's not a question of right or wrong but just need to be clear on the alternatives.

Originally Posted by gunrunnerjohn:
Originally Posted by stan2004:
So that supercap, fully-charged, has enough energy to power the LEDs for 10 seconds.

I question the calculations here.

 

The energy needed to light the LED's for 10 seconds is 3.6 Joules.

The fully charged capacitor has 12.5 Joules of stored energy.

 

Why can it only light the LED's for 10 seconds?  I know you can't squeeze all the energy out, but surely you can do better than less than 30%.

 

What am I missing?

You're missing nothing.  Just semantics.  If you're going to buy me a beer that costs $5 and you have a $20 bill, then your $20 bill will pay for the beer.

 

For hobby purposes, when all's said and done, when you do the back-of-envelope calculations on how many Farads you need, or how many seconds you'll run for, etc. then double/halve the calculation to be conservative.  In other words, you might be able to recover 50% of the theoretical stored energy. If you're engineering the circuit using power management devices such as step-up or step-down switchmode devices,  an aggressive design (which comes at a cost) might shoot for recovering 80-90% of the stored energy.

You fellas are blowing me away. Even experimenting?  I'm really amazed at everyone's abilities and appreciate all the interest and time.

 

Feel obligated to get into the club tomorrow, run the trolleys and get some actual times. Also take a few pics. All Trolleys are similar to MTH #30-2545 (Not PS 2.0).

 

I confess, some of the technical information is overwhelming but I understand the concepts.

 

Thanks again.

Stan, the 50 seconds doesn't appear to be an issue with the lighting, so I don't see the problem.  In actual bench tests with three LEDs and two 2.5F supercaps it looked great.  I had two minutes off and one minute on, and the dimming of the LED's was noticeable but not that significant.  The fact that it happens slowly is a bonus IMO as it lessens the impact of any intensity change.

 

It's up to 1942guy to say if this is sufficient.   It's simple and cheap...

 

GRJ, OK by me. Let's call your design the baseline.  What is the parts list?  And did you happen to note the voltage of the LED and supercap at the start and end?  Presumably you just have a resistor between the supercap and LEDs so the voltage measurements will also give us start and end current.

 

 

Separately, I was thinking about how geeky using Joules is.  I figure kW-hr as a measure of energy is more familiar since that's how you pay your electric bill.  OK, so here's the math. 1 kW-hr is 1000 Watts for 3600 seconds.  That's 3.6 million Joules! 

 

So here we're talking about, say, 50 Joules of energy going in and out of the supercap for each lighting cycle.  That's about 0.00001 kW-Hrs.  With electricity costing about 10 cents per kW-Hr from a utility, we're talking 0.0001 cents!

 

 

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