Minatronics 1.5V Incandescent Bulbs

[TwinStar]

I'm going to use the Minatronics 1.2 mm 1.5V 30mA bulbs for my E unit headlights. I'm needing assistance in ensuring that I don't blow these things up. I'm assuming a resistor of some sort and while I'd like to just ask for the answer I would prefer to learn how to calculate the value of the resistor that I need for these.

Thanks.

[Alan]

Those are incandescent lamps. No resistor needed. An incandescent lamp is a resistor. One that glows brightly!

[TwinStar]

Alan:

How do I step the voltage down then? I'm assuming the RP LM-2S light output is track value of approx 14V. Won't that damage the 1.5V bulb?

[Alan]

My apologies. I thought you had the supply voltage available already since you bought 1.5v lamps and were asking if a current limiting resistor is needed like with an LED.

You need what's called a resistor network i.e. two resistors connected in series. Since a bulb is a resistor it acts as of of the two resistors in the series. There is a voltage drop across each resistor proportional to its percentage of the overall circuit resistance.

We know the bulb passes 30mA of current at 1.5V. Using Ohm's Law we can determine its on resistance R = V / I which is 1.5 / 0.030 = 50 ohms. [EDIT] Oops, had the decimal point in the wrong place. Fixed now.

Now that we know the bulb has 50 ohms of resistance we can plug that value into a resistor divider formula to determine the value of the second resistor when the supply voltage is 14.8 volts.

Bulb voltage = ( Supply voltage x bulb resistance ) / ( Unknown Resistor + Bulb resistance )
1.5 = ( 14.8 x 50 ) / ( R + 50 )
1.5 = 740 / ( R + 50 )
1.5 * ( R + 50 ) = 740
75 + ( R * 1.5) = 740
( R * 1.5 ) = 665
R = ( 665 / 1.5 )
R = 443
The series resistor needs to be 443 ohms to create a 1.5 volt drop across a 30mA bulb at 14.8 volts. The closest standard value is 470 ohms.

Now that we know the resistor value we calculate the total power so we can choose a resistor of sufficient wattage. Once again using Ohm's Law we calculate the current flowing through the circuit I = V / R.
(We already know this because the bulb spec told us it is 30mA but for sake of making the math complete...)

Current = Voltage / ( Bulb resistance + Series resistor )
I = 14.8 / ( 50 + 470 )
I = 14.8 / 520
I = 0.028

Power (watts) = Voltage * Current
W = V * I
W = 14.8 * 0.028
W = 0.41
The series resistor needs to have a power handling capacity of 0.41 watts. The closest standard value is 0.5 watts. Typically power ratings are halved for cool operation so the resistor really should be 1 watt.

Final answer: On a 14.8 volt system a 470 ohm 1 watt resistor will cause the bulb to glow at full 1.5 volt brightness.

There will be a quiz. 

[TwinStar]

Alan:

Thank you! And of course I only have 1/4 and 1/2 watt resistors in stock. Time for Amazon.

The bulbs were selected solely for their size of 1.2 mm. I had tried to use LED's and fiber rod but the size was too big. Another E unit era modeler had successfully used these bulbs so it was a case of size over anything else. These bulbs fit perfectly in the dual light housings that I've installed.

These units have only been worked on for over 5 years. Maybe now I can finish them.

Thanks again.

[Alan]

You could use two of your 1/2 watt resistors in parallel. Use 1K paralleled to equal 500 ohms.

[TwinStar]

I might if space weren't an issue and I have enough. There are 4 bulbs in each nose and I think I'm correct in assuming that I'll need a resistor for each bulb.

[Alan]

4 bulbs changes the equation drastically. Don't order those resistors just yet.

If you wire the bulbs in series then the total resistance is 200 ohms. Let's do the math...

Hold on. Wife is calling me to dinner. Back in about an hour.

[TwinStar]

No worries.

There are:

1 clear Mars light

1 red emergency application light

2 headlights in the lower housing

No rush, enjoy the dinner.

I will add that 3 may be on at the same time, or just two, or just two on dim.

[Alan]

And we're back. With a full tummy!

I'm glad you listed the lights. So much for wiring them in series.

Yes, since you will be operating the lamps individually you will need a resistor (or resistor parallel pair) for each lamp. The headlights are the exception since both are operated together. If you wire the headlamp bulbs in series then the math for them looks like this:

Combined bulb voltage = ( Supply voltage x Combined bulb resistance ) / ( Unknown Resistor + Combined bulb resistance )
( 1.5 + 1.5 ) = ( 14.8 x ( 50 + 50 ) ) / ( R + ( 50 + 50 ) )
3 = ( 14.8 x 100 ) / ( R + 100 )
3 = 1,480 / ( R + 100 )
3 * ( R + 100 ) = 1,480
300 + ( R * 3 ) = 1,480
( R * 3 ) = 1,180
R = ( 1,180 / 3 )
R = 393
The series resistor needs to be 393 ohms to create a 1.5 volt drop across each 30mA bulb when two bulbs are in series at 14.8 volts.

In the end it doesn't change things since the next standard value above 393 is our original 470 ohms. Wiring the bulbs in series just means they will be ever so slightly less bright but you need only a single resistor instead of one for each. Should you go the parallel resistor route then you can use two 810 ohm resistors in parallel to get 405 ohms. That will make your series wired headlights full brightness and you can use 1/2 watt resistors.

You should wire up a bulb and a resistor outside of the loco. Turn it on for 15 minutes. Check to make sure the resistor isn't so warm it will melt plastic.

BTW Unless you are a devout Amazon fan, it is the last place you want to buy electronic components. Very expensive. eBay from China will be a fraction of the cost but last I checked Beijing doesn't offer free 2 day shipping! 

[Alan]

Speaking of testing, you may want to wire up a bulb with one of your 1/2 watt resistors to see how hot the resistor gets. Technically, you are within its operating range but being so close to its upper limit I suspect it will get very warm. Try it and see. If it is not hot enough to melt plastic then you can use them. You could also use a higher value resistor so it passes less current. All depends on how bright you want the bulbs to be.

It is important to remember a resistor is a heater. Your toaster toasts bread using resistor wire. Little carbon resistors are the same. They are so small and create so little heat we tend not to think of them as heaters but they are.

Case in point. Switch mode power supplies are notorious for having poor voltage regulation when there is no or very little electrical load on them. I ran into this problem head on with my railroad. There are some electronic gizmos that run on track power yet they draw only a very trivial amount of current when in stand by. I noticed the power supply voltage drifting when there are no locomotives on the track to create an electrical load. This gave me concern about my gizmos possibly being damaged by over voltage. To remedy the situation I placed a dummy load on the rails so the power supply is always loaded.

I did the math, installed the resistors, and checked the system voltage. Nice and steady. Job well done. Until I touched the resistors. Ouch, they were very hot. Not hot enough to damage them, they were operating within their power rating, but hot enough to concern me. There is 2" insulation foam scenery base mere inches above them.

Long story short, today I have forced air cooled heat sink mounted resistance loading on my train track and a lesson well learned - never underestimate the heat a resistor can generate.

[TwinStar]

Amazon and me get along very well! Here is what I typically purchase when I need these guys since Radio Shack in gone.

https://www.amazon.com/MonkeyJack-Pieces-Resistors-Assortment-Active/dp/B077JW144B/ref=sr_1_8?ie=UTF8&qid=1535719888&sr=8-8&keywords=470+ohm+1+watt+resistor

A pack of 50 for less than $4 may not be the best deal but I don't know that I'd ever need more. I'll look for the 870 ohm as well.

I will most certainly test them as I've learned that lesson before the hard way. Burns hurt.

[Alan]

I'm not anti-Amazon. I too buy a lot from Amazon. Just making you are aware in case you didn't know there are some things that Amazon is not a good deal, with electrical components often being one of those things.

A 1 watt resistor typically sells for 4 to 5 cents so your Amazon find at 7 cents is not grossly out of line. Although, a $4 buy isn't going to break the bank regardless of what you purchase. I tend to be more sensitive to the pricing because for commonly used components like resistors I usually buy in quantities of 1000. That's the industry sweet spot where volume pricing starts kicking in and it means I never have to buy that particular component again. But then I tend to use a lot more than the typical model railroader.

Good luck with the lights.

[Alan]

Jacob, here is a thought for you.

The problem with using a series resistor to control the brightness of a bulb, other than needing hefty resistors, is the bulb brightness will vary with voltage. If the voltage on the rails varies for whatever reason so too will your headlight brightness. There is a simple solution that will provide uniform bulb brightness regardless of voltage and uses small 1/4 watt components.



LM317 is a basic, no frills, adjustable voltage regulator in a tiny package. Using the resistor values shown your bulb will always see 1.5v so long as the track voltage is 3.7v or higher. LM modules shut off below 10v so it is safe to say the bulb will always be the same brightness under all conditions. If you were to construct this using surface mount style components it could be made into a very small package.

Just a thought.

[TwinStar]

Alan:

I assume I'll need 3 of these? One for the mars, one for the emergency stop, and one for the headlight pair?