A Question for the Electronics Boffins.

[Potterton]

I wish to reduce the light output of a 12 volt grain of wheat bulb by around 25 - 33% so the brightness is reduce to 2/3rds - 3/4. What value of resistor do I need to insert in the circuit?

I have had a look at the Guide to Electronics someone on here recommended, but although I understood some of it, most is beyond me.

Many thanks.

[What About The Bee]

Hi Potterton

This is fairly straight forward.

Step 1: Find the resistance of the bulb. From the internet, I see quoted amperage draws of 50 milliamps and 70 milliamps for grain of wheat bulbs. Divide the voltage by the amperage to get resistance.

12volts/0.050amps = 240 ohms of resistance

Step 2: place a resistor in SERIES with the bulb, using mathematical ratios. To cut the light output by a factor of 2, place a 240 ohm resistor in series. By a factor of 4? 240 ohms for the bulb, 720 ohms on the resistor. Total resistance =960 ohms. 240/960 = 1/4.

A couple of quick cautions, light bulb output isn't strictly linear with voltage. Further, the resistance calculation assumes you know the current draw. You could measure the resistance, understanding you are at the mercy of your meter's accuracy and resolution.

But for just dimming the light, the above steps should be plenty good enough.

Bee

[gilbo2]

I found that putting 2 grain of wheat bulbs in series creates a lovely warm glow, then I spaced them as needed on the layout (eg street lighting).

[96RAF]

I powered my filament lights from the controlled output of a redundant analogue controller. This allowed any level of dimness to be achieved.

As stated by Bee the resistance of a filament lamp is odd as it changes as the bulb warms up.

[P-Henny]

Putting GoW bulbs in series to reduce brightness is more effective and power efficient than using a resistor. Any power reduction provided by a resistor has to be dissipated by the resistor in the form of heat.

These days GoW bulbs IMO should be relegated to the bin in favour of LEDs.

• LEDs are longer lasting.
• LEDs run virtually cold to the touch compared to bulbs.
• LEDs can be controlled more effectively.
• LEDs require a fraction of the current required by bulbs, so a relatively small PSU can operate far more LEDs than bulbs.
• LEDs offer a wide range of colours.
• LEDs are physically smaller.

[AndyMac1707817969]

You could place a potentiometer in series with the resistor and adjust until you get the required brightness. You could then measure the resistance of the two in series and replace them both with an equivalent single resistor.

[Potterton]

Thank you all for your responses.

What I'm doing is replacing the bulbs in an Eckon signal. I think the easiest solution will be to replace them with LED's. If I try to use the resistors supplied for use with the LED fitted signals, the bulbs do not illuminate at all. Without a resistor they are far too bright, and it is not recommended by Eckon anyway as the signal heads may distort with the heat.

[What About The Bee]

Hello Potterton

I found this thread which may be of use to you.

https://www.newrailwaymodellers.co.uk/Forums/viewtopic.php?t=47895

It discusses the resistor value supplied with Eckon signals, which have the grain of wheat filament bulbs. One individual there states 270 ohms [270R resistor] was supplied with the older Eckon kits.

Its worth a try!

Bee

[Topcat2018]

An additional comment. You cannot calculate the series resistor needed to give a particular current through a filament lamp based on measuring the cold resistance of the filament because the resistance of the filament increases with temperature, so the current when the lamp is lit will be less than what you calculated. Also, as What About The Bee mentioned, light output does not increase linearly with current and your eyes' response to increasing light intensity is logarithmic anyway.

[Brew Man]

As others have mentioned, much better to switch to LEDs.

[Ged_K]

I wouldn't be using grain-of-wheat bulbs in this day and age. LED use much less power (OK can't run them straight off 12V, need at least a dropper resistor). For either illuminator, two solutions, both mean the brightness can be kept fairly constant over a range of voltages.

A voltage regulator (usually look like a power transistor, usually need a heatsink like one of those), or a Pulse Width Modulation (PWM) module, this can vary the brightness to some extent

[Topcat2018]

Simplest solution if you want to set the brightness of multiple LEDs is probably an adjustable linear voltage regulator regulator such as a LM317 which (from an an appropriate input voltage) can provide output voltages from 1.25 V to 40 V. If you choose a suitable input voltage and resistor values (note that each LED or string of LEDs in series needs a resistor, you can't connect LEDs directly in parallel) then most of the power can be dissipated in the LEDs and resistors and not in the voltage regulator.

PWM is more complicated to use and probably overkill unless you have a really large number of LEDs.

[Potterton]

@ What About The Bee You said,

"I found this thread which may be of use to you.

https://www.newrailwaymodellers.co.uk/Forums/viewtopic.php?t=47895

It discusses the resistor value supplied with Eckon signals, which have the grain of wheat filament bulbs. One individual there states 270 ohms [270R resistor] was supplied with the older Eckon kits."

Thank you very much. that is exactly the answer I was looking for.

Thanks also though to everyone else who took the trouble to reply. You're a great bunch.

[Ged_K]

Got the LEDs (ultra-bright 1.8mm) have the voltage controller and bridge-rectifier (nearly forgot those) on order, think if run at 7.5V (most loco not move at less than 5V) the ultra bright would still give a good output and it would not tax the voltage regulator operating up to the 12V to volts.