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How to drop voltage from 4 v dc to 3v dc

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Help required please.

I’ve belatedly realised I cannot see the setting of a small number of points from my preferred operating position. As I have already constructed a mimic panel purely for uncoupler buttons and infra-red sensor train position indicator LEDs, I now do not want to disturb that, by adding point indicators.  I am using DCC to operate point motors, and I realise I can always check/change points from that controller, but what I would like to do (as the scenery is still under construction) is add an LED indicator in the proximity of a said point, but of a scenic nature, eg bonfire, brazier, arc welder, etc. So that they light up when point is set to straight. The locations of said points allow me to do that.

My dilemma is this. The Gaugemaster PM10D digital point motors I am using kick out 4.1v dc from the designated LED terminals on them. Most commercially available devices (ie the bonfires, braziers etc) from Kytes Lites, Train Tech, Layouts4U etc, have LEDs that require either 3 volts or 12 volts. So I think I need to do something to either drop the 4.1v to 3v (resistors?) or step it up to 12v.

Any advice please on what value resistor I may need, and where it fits. Or how I might lift to 12v?

 

or, indeed, will 3v LEDs be able to cope with 4.1 v, with no resistors?

thanks

ian

Edited by ITG
Added question

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27 minutes ago, Sol said:

Use  560 ohm resistors on the 4.1v output but if in doubt , use 1K ohms

https://www.dccconcepts.com/manual/leds-and-how-to-get-the-best-from-them/

There are two assumption made in the above post.

One is that the device is an LED, from the list given it could be a 3V bulb.

The other is that the current drawn is about 2mA, it may be higher so more voltage will be absorbed.

 

If the current drawn by the device is measured then the correct resistor can be selected to drop the 1.1V required.

One other option is to place an ordinary diode in series with the device. This will drop about 0.6 to 0.7 volts.

A second diode, in series with the first, will drop double this. This would give 2.7 to 2.9 volts on the figures given, and under running a bulb is preferable to a slight over run.

Some experimentation will be needed to see if this gives sufficient brightness.

Edited by Tony Cane
spelling

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Thanks all.

the devices I’m contemplating using state they are LEDs, either requiring 3v or 12v for differing models from various sources.  GM themselves say the output is 4.1v and measurements I’ve taken myself agree with this.

I’m going to an exhibition tomorrow where hopefully at least one or more of the possible suppliers plan to have trade stands, so I will also pick their brains.

as a solution unfolds, I’ll let you know how it goes.

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There aren't 4.1v LEDs, so Gaugemaster designed it to work with normal LEDs which drop about 3v, which is what they show in the manual.   The actual voltage, within reason, doesn't really matter, what matters to the LED is the current which is allowed to flow through the LED.   

As its designed to work with panel LEDs, I expect it works fine with any reasonably normal "3volt" LED of about 20mA max current, and the circuit on the motor/decoder includes current limiting to the LED outputs.    

If bothered about things, or if the LED is too bright, then a series resistor will further reduce the current, and thus the LED brightness.  Start at 100-200 ohms for resistor value.      

 

 

A "12v LED" is a normal LED plus resistor in a single package.  The LED is still an approximately 3v device, the manufacturer has fitted a resistor within the package to allow direct connection to 12v. 

 

 

 

- Nigel

 

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6 hours ago, ITG said:

The Gaugemaster PM10D digital point motors I am using kick out 4.1v dc from the designated LED terminals on them.

 

If Gaugemaster say you can connect LEDs with no current limiting resisitor, I would hope they have installed resistors on the PCB itself - it could be that if you're measuring 4.1V from the LED terminals with a volt-meter and no load it is reading high because with virtually no current flowing there won't be the expected voltage drop across a resistor that you'd get if there was an LED in the circuit.

 

You might want to ask Gaugemaster whether there is a resistor built into the indicator output and what its value is - you may want to add more if your track-side effect is too bright. 

 

Remember, white LEDs and blue have a typical Vf between 3 and 3.5V (you might choose these for an arc welder) but red, yellow and orange very often have a lower typical Vf - often somewhere around 1.8-2.2V (these colours might be more suitable for a bonfire or brasier). In either case, to make your indicator effect more realistic, you might also want to look into flicker effect circuits. 

Edited by sharris
suggestion of flicker effect.

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type in 'led resistor calculator'  input:- source voltage, led current & forward voltage and click 'calculate'. This will give you the correct resistor value required. Good luck. 

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4.1 volts for LEDs sounds a bit odd.    Too much for Red/ Green LEDs not enough for Blue /Clear LEDs  I would put a couple of 1N400 diodes in series between the Gaugemaster and the LEDs, That should give circa 2.5 volts.   if the LED is too dim try just one diode, if it don't light the diodes are probably the wrong way round. If the LEDs light but are still too dim try it without a diode.   Diodes usually drop voltage 0.7 volts or thereabouts more or less irrespective of the current.

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I suspect what you are dealing with here a 5V logic level output from the decoder processor, followed by a diode (and probably a resistor) in series with the output  to prevent people accidentally connecting an external power source to it and causing damage. That would explain the off-load voltage that is being measured.

I would suggest you ignore the posting immediately above, and try it with a conventional LED and resistor first.

If you can, put a DMM in series with it, set to measure milliamps,  and see what comes out.

Anything in the 2 to 20mA range should be OK, then adjust the resistor value to suit the brightness you need.

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 Could be a constant current circuit. This would give you a high voltage reading on a multi meter as the constant current circuit would raise the voltage attempting to get the design current flowing. 
 

constant current solutions are much better then a resistor As they don’t vary because of forward voltage 

Edited by Junctionmad

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Thanks all again. Some varying responses here. I’m hoping that those who suggest that the 4.1 v is showing because of no load are correct, as that would simplify the solution. That’s what I wanted to hear! That would mean - if I understand correctly - I could use standard 3v LEDs connected directly. Which would make sense, as why would the point motor instructions mention connecting LEDs (thinking of a mimic panel) directly would be ok. Now I’m beginning to grasp that all LEDs run on 3v, and that it’s current which is the key factor, the penny is beginning to drop. 

I think I have some LEDs laying around somewhere so it’s worth trying. I didn’t do so before, because I was mistaken in thinking the bonfire/brazier etc effects were different LEDs, and thus any trial run wouldn’t prove anything. 

Its a learning experience for sure.

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Led forward voltages vary from around 2.6B to about 3.3v. Somewhere between 2.7 and 3V is typically taken as the “nominal “ value most modern leds will light quite brightly on 1-2 mA 

Edited by Junctionmad

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5 hours ago, ITG said:

Now I’m beginning to grasp that all LEDs run on 3v, and that it’s current which is the key factor, the penny is beginning to drop. 

 

It's dropped the wrong way :)

 

Just over 2V is typical for a red or green LED more like 3 - 3.5V for a blue or white.

 

The real thing to remember is that LEDs do not behave like resistors and you cannot use ohms law to calculate current form voltage or vice-versa.

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2 hours ago, Junctionmad said:

Led forward voltages vary from around 2.6B to about 3.3v. 

 

It really depends on the colour and chemistry -e.g.  red LEDs often have a Vf between 1.8V and 2.2V at their specified If.

 

(Crosland sneaked in a similar comment seconds before me)

Edited by sharris

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Posted (edited)
On 06/12/2019 at 08:52, Tony Cane said:

..... One other option is to place an ordinary diode in series with the device.

This will drop about 0.6 to 0.7 volts........

 

Many thanks for this tip :rolleyes:
At the time I had been trying to use a string of 20 Christmas light LED bulbs supplied/bought to be worked from a 2 x AA battery source, but I wanted them on a external decoration* 
The decoration (a Nativity Scene) is powered on a time clock (5pm - 11pm) and there are various transformers etc., contained within the IP67 Supply Box, including the transformer for this item.
I had plenty of spare 5.5v (500ma) transformers from some 'Falling Snow' decorations that had been scrapped, and at the time you posted, I was wondering how to reduce to circa 3v DC. 
I tried both 4 and 5 x 4N1000 diodes in series, test set up through a choc-bloc and although the 4 gave a bright light, the 5 in series gave me the less bright result I wanted.  
Yes, I know the LEDS are for indoor use, but with plenty of marine varnish soaked into the back of each LED connection etc., they have survived some harsh weather in our coastal Village in west Cornwall.  I have some others (with a mains transformer etc.,) that have survived 4 Christmas' on our harbour wall, so the varnish bit works - I also have a 100 bunched close together around the Cradle in the Stable, again with Varnish weathering.
* The 20 LEDS were divided into 3 x 5 of a specific colour (Amber, Blue or Green)  plus a red one, the remaining 2 red ones being not used, all connected in parallel for 'Gold, Frankinscense and Myrrh' on a seasonal display :) 
Again, many thanks Tony Cane.
The follow on from this is I now have some LED's on a new control panel and I was wondering how to power them, fortunatly 4N1000 diodes come in packs of a hundred, I can see the ex. Falling Snow transformers having another life.

Edited by Penlan
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