How much power is required to drive 4-1.5v incandescent bulbs. I need to power a display model where there is not outlet and want to use a battery (1.5v dry cell). Will that work? Is it sufficient power?
RicZ
Ric,
The AA or AAA battery should work just fine for a 1.5V supply. The total amperage of your incandescent bulbs (usually measured in milliamps or mA) will determine how much of a draw it will be on and how often you replace your battery. Most 1.5V incandescent bulbs are either 20 or 30mA.
Tom
https://tstage9.wixsite.com/nyc-modeling
Time...It marches on...without ever turning around to see if anyone is even keeping in step.
As long as the bulbs are wired in parallel.
You can also wire your batteries in parallel (as many as you want) to increase the time between battery changes.
You can figure how long a typical battery will run a load by dividing the battery's capacity in mAh by the load. A typical AA battery has a rating of about 1200 mAh. To run four 30 ma bulbs (for a total of 120 ma, you would be looking at about ten hours. Two AA batteries in parallel would almost double that.
Keep in mind though, a standard battery when new puts out slightly more than 1.5 volts, and only about 1.25 volts at 50% capacity. So the bulbs will begin to dim in about half the calculated time.
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
I have always had a hard time understanding these calculations. So, I find this thread helpful.
What is the key consideration here? It seems to be voltage. If the incandescent bulbs are 1.5 volt each, then the power source cannot be greater than 1.5 volts, correct?
Rich
Alton Junction
It depends.
With an incandescent bulb, the voltage rating is the critical factor. Exceed that, and you shorten the life. Exceed it enough and you make a flashbulb out of it. An ideal power supply for a 1.5V light bulb is a rechargeable battery - most rechargable types put out 1.2 volts. Slightly under voltage will make the bulb blow without being a super bright and will extend the life. 14-16V bulbs are good at 12V.
In parallel, voltage is the same to each device in parallel. In parallel, the total current drawn by each device adds up. So for 4 1.5V bulbs of 30ma each in parallel, it's 1.5V at 120ma.
In series, the voltage adds up but the current is the same across the entire string. So those same 4 1.5V 30ma bulbs in series would need 6V at 30ma.
A light bulb is an example of a resistive load. In simple terms, you cna supply that 1.5V 30ma light bulb with 1.5 volts at 200 amps and nothign bad will happen (unless you short out the connecting wires). The lamp will draw 30ma - amps don't 'pump in', the circuit draws what it needs, so you need to supply at least the required, supplying higher only means the pwoer supply can work less hard. ANother example - your house likely has a 200 amp incoming service. When you plug in your vacuum cleaner, it doesn;t get hit with 200 amps and explode, it draws whatever the rating is - probably around 4-5 amps, and works fine.
LEDs work a bit differently. They are current driven, not voltage driven. They operate within a specific current range, exceeding the upper limit will burn them out, just liek excessive voltage burns out a light bulb. There's also a minimum voltage required by LEDs, to overcome the the semiconductor junction that gives off the light., Once this point is reached, teh voltage across an LED is essentially constant. There IS an upper llimit which will break down the semiconductor junction (ie, destory the LED) so you can't just feed 200 volts in, but the point being that is the LED is say 3V, you can run it off a 12V power supply with no problem - the key factor being the current. The LED specifications will list the maximum current it an handle - depending on the LED, 20-30ma is common. SO how do you control the current? The easiest way is with a series resistor. Remember from aobve, in parallel voltages add, but currents are equal. We know a few things: the voltage of the power supply (12V example), the voltage across the LED from its rating (3V example). From this we can calculate the voltage across a resistor in series with the LED: 12(power supply) - 3 (LED) = 9 volts. Wel also know that the LED can stand up to 25ma. But we don;t want to run at the maximum, the LED will be plemty bright at less, plus if the power supply exceeds 12V and we are running at the maxmium, we will exceed the limit. So let's say we want 15ma in the LED, well within the limits. Since currents through devices in series are the same, that means the resistor will also see 15ma.
The next bit if Ohm's Law: Voltage = Resistance X Current. Voltage in Volts, Resistance in Ohms, Current in Amps. 1ma = .001 Amp, so our 15ma number is .015 amps. ANd we know the voltage, 9 volts. Solving for Resistance, we have Resistance = 9 Volts/.015 amps = 600 ohms. Now, reistors have a tolerance range, common cheap ones we use have a 10% tolerance rating, and so don;t come in every possible numeric value. Standard values are listed here: http://www.elexp.com/t_eia.htm So we need to pick somehting close to the calsulated value. It's generally safest to pick the next HIGHEST standard value, in this case 680 ohms. Plugging that 680 back in and calculating the current, you have Current = Voltage/Resistance = 9V/680 ohms = .013ma, still definitely a safe value.
Clear as mud now? The real key to understanding any of this stuff is the behavior of circuit elements in series and parallel with respect to voltage and current, these are derived from Kirchoff's Laws, and Ohm's Law, that's the Voltage = Resistance x Current equation. Please don;t look these up on something like Wikipedia unless you understand calculus, as they are expressed in calculus terms which is their true root meaning but it really all simplifies down to the simple terms I tried to state it in.
And a note to other EE's - I am an EE too, and I understand that in most cases I am oversimplyfying and assuming ideal circuits which don't really exist in nature., However, for the purposes of the use here, in model railroading and simple hobby electronics, it's more than accurate enough. Plus if these methods didn;t work, every headlight in every one of my locos would have blown up a long time ago
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Randy,
That is very helpful. Thanks for that additional and detailed information.
Uiing a 1.5 volt battery to power incandescent bulbs on a layout will result in having to buy a lot of batteries.
You're better off fronting the money for a power supply to do the job. You'll save money in the long run.
Springfield PA
Hamltnblue Uiing a 1.5 volt battery to power incandescent bulbs on a layout will result in having to buy a lot of batteries. You're better off fronting the money for a power supply to do the job. You'll save money in the long run.
I recently bought a couple of 6 volt wall warts for next to nothing. Can the lights be wired in such a way to utilize a 6 volt wall wart without blowing the bulbs or adversely affecting the bulb life? It sounds like it can be done.
4 in series is 6V, however cheap wall warts aren;t regulated and at 15-30ma total current (depending on thebulbs - also if you series wire them they must all be the same current, do not mix 15ma and 30ma bulbs). A couple of 1 amp diodes will drop the voltage by 1.2 volts (.6 volt per diode) and not get hot. Check the voltage coming out of the wall wart and see how much you have to drop.
You can also wire a resistor in series with each bulb nd wire each assembly in parallel. Measure the voltage of the wall wart, because if it truly is 6 volts regualted, then a 150 ohm resistor is erfect if the bulbs are 30ma. A 180 ohm resistor is probably better but if the voltage of the wall wart really is only 6V the bulbs will be too dim. If they are 15ma bulbs, a 330 ohm resistor should work.
One final question-what size resistor do I need for the bulbs if I use 3v (2-AA batteries) and then once the contest is over, what size resistor do I need for a 12v curcuit? Also, can I wire the 4 bulbs in parallel with one reistor between the bulbs and the power supply?
Hate to be such a bother.