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Voltage reduction for lighting


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I have some Pullman coaches with table lamps that are powered by 2 rechargeable batteries and I want to change them to be powered from the track, I have put in pickups on each bogie and have made bridge rectifiers to get DC to power the lights, my question is what size resistance do I need to reduce the voltage from 13.6 to 2.4 volts. There is plenty of room where the rechargeable batteries are for the wiring so can use a fairly large power resistor to do the job.

Have looked at the LED posts on this site but it just leaves me confused!

 

regards mike 

 

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Are you planning on permanently converting your models to track power?

 

Also, how many lights (presumably LEDs) are actually fitted inside each coach?

 

The reason I'm asking is that, if these were my models, I'd probably be tempted to rewire them from scratch - with the LEDs wired in series, perhaps in groups of 3 or 4, each group having a series resistor - in other words, a similar arrangement to what you'd find on LED light bars and tape lighting (albeit with significantly higher value series resistors).

 

This sort of thing would obviously involve more work than some people might like - but I'm not convinced that "in parallel", via one resistor, is necessarily the most effective (or efficient) method of connecting LEDs.

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How about some cheap function decoders? e.g. Lais.

Then you can switch the lights on & off.

I've just done a Lima Railcar, (but using a 4 output loco decoder already in place ) you can switch them on & off (and also vary the brightness by changing a CV).

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1000 ohm resistor is frequently used for the purposes you wish.

 

alternatively, you could have a look at some of the lighting strips on amazon, e.g. Govee

 

It's a 12v light strip that you can cut to length.  It's designed for home lighting, but it's bang on for lighting coaches.   Copper tabs are where you can solder to.

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1 hour ago, Ouroborus said:

1000 ohm resistor is frequently used for the purposes you wish.

 

alternatively, you could have a look at some of the lighting strips on amazon, e.g. Govee

 

It's a 12v light strip that you can cut to length.  It's designed for home lighting, but it's bang on for lighting coaches.   Copper tabs are where you can solder to.

Not much good for lighting Pullman table lamps.

If they are the standard Hornby illuminated Pullman Cars IIRC there is just one actual light source and light pipes to each table lamp.

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1 hour ago, Ouroborus said:

1000 ohm resistor is frequently used for the purposes you wish.

 

alternatively, you could have a look at some of the lighting strips on amazon, e.g. Govee

 

It's a 12v light strip that you can cut to length.  It's designed for home lighting, but it's bang on for lighting coaches.   Copper tabs are where you can solder to.

 

In the interests of encouraging people to support model railway traders during the current circumstances, the LED strips are also available from Kytes Lights.

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Thanks for your replies although helpful I think I did not explain what I was doing, the coaches are scratch built in O gauge and all the wiring is not accessible and would not be easy to change. The existing 2xAA batteries are accessible and the lights are switched with a reed & magnet all I am proposing to do is replace the batteries with a track supply using a bridge rectifier and resisters connected to the wiring that is coming to the batteries, as they have a DC 2.4 volt output all I need is the same.

 I hope that I have made it clear what I am trying to do.

 

regards mike 

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I doubt that "Scratch Built", "0 gauge" and "Pullman car" came in anyone's train (sic) of thought.:)

 

To do what you want to do you need to know how much current the lights consume.

Do you have a way to measure that?

 

The alternative is a simple voltage regulated supply to drop from track voltage to lamp voltage.

You could do that with a three terminal regulator and a few components.

 

Something like this:

LM317-typical-adjustable-regulator-ckt.p

 

Work out the value of R1 from the formula and use a fixed resistance.

Edited by melmerby
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Keith, thanks for your reply but I am just trying to replace the batteries, as they give a soft yellow light I would have thought that the bridge rectified DC volts only need to be dropped to the voltage of the batteries ie 2.4 volts with a single resistance.

At the moment I am not in a position to get any more items but have a good supply of resisters, I have a multimeter but not sure how I can measure the current draw from the LED’s.

 

thanks for your help so far, mike 

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You need to know how much current is consumed else you'll get nowhere.

With batteries in situ can you connect the meter on a DC current range (start high and work down) in series with one of the battery leads.

If that's not easy connect the meter across the reed contacts when it's not operated to complete the circuit.

 

With that reading you can then calculate the resistance needed to drop the volts.

 

N.B. Resistors only drop voltage if there is current flowing through them, the voltage drop is proportional to the current, the more current the more voltage drop.

 

 

Edited by melmerby
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Mike you have made the classic presumption that LEDs are like bulbs in that altering the voltage affects their brightness. LEDs are current driven devices and the series resistor that you get in LED circuits is a current limiting resistor. Hence as Keith has pointed out you need to establish the current draw of the LED circuit and as he has pointed out your meter needs to be in series with the battery and LED(s).

 

As we are looking at 2 AA batteries which should be giving you 3V or 2.8V if they are rechargeable (2.4V seems odd unless the batteries are well used). The LEDs are probably wired in parallel with individual resistors already. A single white LED will draw about 30mA max. and probably less if as you say they are emitting a less than white light so with parallel LEDs you will be looking at a max of 30mA or less x the number of LEDs in the coach just to give some idea of the range of current you can expect.

 

Now since voltage, current & resistance is related by Ohm’s Law ie V=IR. The calculation you need to make is:  R = V supply voltage - V led voltage/ I
 

So in your case 13.6 - 2.4(?) divided by the measured current.

 

Since your current value will in mA the answer you get will be in KOhms

 

Richard

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Surely the first thing to ask is what the actual track voltage and waveform is ?

 

So mikeg is that 13.6 volts measured as a maximum voltage from a DC  controller? 

 

Is it a measured voltage from a dcc type supply ? 

 

Is it 13.6 volts as an AC supply as used by some O gauge stock? 

 

We are assuming that the light source in the coaches is LED. Is it ? Could it be an incandescent lamp ? 

 

Just asking . 

 

 

 

 

 

 

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If it was I doing this, I think I would start with a circuit like the ^, or at the very least, a 5V version of it.  (bridge rectifier, voltage regulator and cap).  That way, the circuit should provide you with a steady 5V supply, and if you use a decent sized cap, a stay alive too.  You have the volume of the battery box to work within...which should be lots of space.  After the above, I'd start out with a fairly high (2k) resistor, and then work my way down till the lights light up about as bright as they are.

 

But that's only one way to fry this fish...

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This is probably the easiest solution, easy to adjust.

I'm feeding  in 12 vdc, this is to power some small motors, bought 5 for £4.5. Although it states 3v it goes down to almost

.5 volt.

https://www.ebay.co.uk/itm/Super-Mini-3A-DC-DC-Step-Down-Buck-Converter-3V-5V-16V-UK-Stock-From-0-65p/254490668546?hash=item3b40d36e02:m:mXspK3qZ0Zmk8BnjWvIQ7Rg

 

Simon

 

IMG_2571.JPG.d7ec9cf71b530191d30d4f9b610c6b43.JPG

 

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Hi all thanks for all the help, I have opened the coach and have charged up the batteries the out put voltage is 2.8 each battery measures 1.37v. I cannot get to the reed switch to measure the current but have found that there are 10 red LED's all in parallel. I have taken a picture which I have uploaded.

 

 

008.jpg

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I should have added in the previous post that I am using DCC with a track voltage of 13.6 volts measured with a rampmeter between aZ21and the bus wiring. I have found that on some of the tracks that the voltage drops to 12.6 volts which I think is due to LED lights on buffer stops!

 

regards mike 

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55 minutes ago, mikeg said:

I should have added in the previous post that I am using DCC with a track voltage of 13.6 volts measured with a rampmeter between aZ21and the bus wiring. I have found that on some of the tracks that the voltage drops to 12.6 volts which I think is due to LED lights on buffer stops!

 

regards mike 

If the voltage on the track drops from 13.6v to 12.6v because there are LEDs on the buffer stops there is something wrong with the circuitry.

A few LEDs should not make any difference.

 

And you still haven't said how much current the table lamps consume, without that it is pure guesswork.

Also, are you positive they are LEDs, all in parallel? They could be tiny filament lamps.

Edited by melmerby
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Keith they’re definitely LED’s as they are 1.5mm with 1mm square sticking out both sides with wiring soldered to a 1mm thick double sided copper clad that is 1mm wide and they form the table lamps. Each LED is attached to a copper tape either each side of the floor and one coach that has 16 LEDs the copper strip is down the centre. 

I have not been able to measure the current as the wiring is difficult to get at where there is any bare metal, whoever’s the builder was he’s done a good job covering any bare wires in the actual construction. I have charged the batteries in each coach and the voltage of each pair of battery is 2.38 to 2.47 volts.

 I think that for now I will leave well alone until I can get to the club that I belong to and discuss it with a friend that is probably able to advise me as he is an electronic expert.

 

Thanks for all your help I just wish that I was able to sort it all out. Mike 

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Right. They are definitely LEDs then.

 

if the LEDS consume 5mA each the total for 16 would be 80mA, therefore the resistor would need to drop 11.2volts (13.6-2.4) at 80mA which would be 140Ω (150 is a common preferred value) and it would need to be rated at 1W

However if the LEDs consume 10 mA that is 160mA, the resistor would be 70Ω and rated at 2W, that will be hot.

If they consume 20mA total is 320 mA, resistor is 35Ω and rated at over 3.5W, not recommended inside your coach without someway of dispersing the heat.

 

My suggestion

Try 270Ω 0.5W resistor, see how bright (or not) they are and then work down, say around 200Ω next and so on. Hopefully the current will not be too high.

N.B. this is for a track voltage converted to DC with just a bridge rectifier*. If you put a smoothing capacitor on the output of the bridge the voltage will be somewhat higher.

Note this is all approximations as there are various other things to take into account but not worth troubling with for a simple voltage dropping resistor.

 

BTW I'm not sure what your "electronic expert" will say that's different as I worked in electronics (repairing, servicing, modifying etc.) for most of my career!

I still have a full set of test gear and a stock of components.

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Keith has outlined several possibilities, but at the end of the day it is all guess work and you really need to get an accurate reading of the current draw. If as you say you are having trouble finding a point where you can connect your meter in series, why not insert a thin piece of insulating material - plasticard say between one battery terminal and the contact in the battery holder then simply put one probe on the battery terminal and the other on the contact to complete the circuit.

 

Richard

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Keith I think that I am going to leave the coaches as they are for now but thanks for all the help you have given, my main trouble is that each coach has a different number of LED’s and is wired differently which I only discovered yesterday when I dismantled them all. 

Richard the batteries are not separate ones in a box, they’re welded together and have strips as terminals that are installed so that they go under the inner floor and the wiring is nearly all hidden hence being hard to measure the current without damaging the build of the structure and fittings, you can just see the wiring disappearing in the picture above.

 

Thanks for both for  your help I just wish that I was able to do more to carry the mods out. Mike 

 

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  • 3 months later...

An update, I have managed to find the end of the connections between the batteries and the copper strips and have unsoldered it and have been able to measure the amps for the lights.

A coach with 10 LEDs measured 0.2 amps, another with 14 LEDs was 0.33 amps. I have 3 others that have 8, 12 & 13 but have not been able to measure the amps as the batteries are flat and I thought that as they are all wired up the same the amps will be between the 0.2 & 0.33, so can a solution be given.

 The track voltage is 12.9v DCC which I have checked and the batteries are 2 AA in series and measure 2.7 to 2.9 volts they are approximately 20 years old but have not been used for long periods, I have used them 5 to 7 times in the 6 years I have had them.

 

 I hope that you now have all the information you require, the lock down period has been very useful for sorting out my models and layout.

 

regards mike 

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Due to the DCC voltage being a hybrid AC most multimeters will not read the voltage correctly even on the AC setting, so your voltage numbers are highly likely suspect. Since you need a DC voltage for the LEDs anyway, connect a bridge rectifier to the coach pick ups, then measure the voltage on the output of the rectifier on the meters DC setting. Now a little maths, divide the voltage you have measured by the respective current readings you have for each set of coach lights that will be the value of the resistor you need to put in series with the LEDs. You will probably not get the exact value calculated since depending on which series of resistors you can get hold of, so simply use the nearest preferred value usually the nearest higher value. One final point is to multiply the voltage reading from the rectifier by the current readings which will give you the power requirement in Watts and make sure the resistors you use can sustain that power.

 

Richard

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Thanks Richard, I have put in a bridge rectifier across track supply and it’s output voltage is 12.5 so I can workout what I need from you answer. Having now checked all the coaches I have been wondering if it would be possible to connect them all together from the one set of pickups, I think I can get small 2 pin plug & socket connectors for between the coaches and as I can now get to the copper strips in each coach I could easily join them all together. 

Can you see any problem with this approach and can I just add all the amps together and use one resister to the supply in the pickup coach?

 

Hope I am not being a pest but electrics is not something that I fully understand so thanks for your help and time.

 

Regards mike 

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