Dumb question:
I’m assuming the ohm value of a resistor applies to any light, LED or GOW or tiny incandescents. Or any appliance such as a smoke unit.
Correct or not?
Input voltage, desired number of lights and Ma of the light. (Or appliance.)
j
R = E/I
R is resistance in Ohms
E is Voltage you want to drop
I is Current drawn by device
Also:
W = E * I
W = Watts for the resistor
Pretty much, but there can be higher temporary inrush currents on incandescent bulbs, for a brief instant up to 10 times the steady state current draw. This is the kind of thing that decoder manufacturers have to take into consideration.
Also, inductive loads, like a relay coil, or a wirewound resistor in the inside of a smoke unit can also have other effects.
But, as Tom mentions, once things have settled down, V=IR.
Regards, Greg
Greg Elmassian said:
Pretty much, but there can be higher temporary inrush currents on incandescent bulbs, for a brief instant up to 10 times the steady state current draw. This is the kind of thing that decoder manufacturers have to take into consideration.Also, inductive loads, like a relay coil, or a wirewound resistor in the inside of a smoke unit can also have other effects.
But, as Tom mentions, once things have settled down, V=IR.
Regards, Greg
Your post reminds me of something to keep in mind whenever dealing with reactive loads (in this case, relay coils).
More and more solid state devices are apparently being used in this game. whenever you drive a relay coil with a transistor, TTL or CMOS gate, etc., it’s a good idea to put a reversed diode (1N914 is prefect) across the relay coil.
When the relay is deactivated, the magnetic field collapses into the core and coil, creating “Back EMF”; a nasty spike of voltage, sometimes going into the hundreds of volts for an instant, that can make toast out of a logic gate.
The so-called “snubber” diode across the coil eats the spike before it can eat the gate the coil’s connected to.
I assume everyone here already knows about this, but IMHO if it’s gonna save somebody’s control circuitry, it can’t be repeated too many times.
Mr. T.
A 1N914 is usually used as a signal diode. They have low current handling capability. The ideal snubber is a higher voltage higher current, low on resistance diode, like a Schotky, but they are expensive in several amps.
Just try a 3 amp 100 volt or higher diode from radio shack. I’m afraid you can destroy 1N914’s with a relay coil. They might be fine for a small reed relay.
Regards, Greg
Greg Elmassian said:
A 1N914 is usually used as a signal diode. They have low current handling capability. The ideal snubber is a higher voltage higher current, low on resistance diode, like a Schotky, but they are expensive in several amps.Just try a 3 amp 100 volt or higher diode from radio shack. I’m afraid you can destroy 1N914’s with a relay coil. They might be fine for a small reed relay.
Regards, Greg
Right!
I’m thinking in terms of relays that can be driven directly by logic gate chips… and the back EMF enegry from thise is small enough that a 1N914 can handle it.
For bigger relays that handle REAL power, you’re correct… but you’re not driving those directly with digital logic chips; for those, we’re talking fairly hefty transistor drivers. In that case… 1N4000 family diodes are the right medicine; they’re cheap and easily available.
Mr. T.
John Bouck said:
Dumb question: I'm assuming the ohm value of a resistor applies to any light, LED or GOW or tiny incandescents. Or any appliance such as a smoke unit. Correct or not? Input voltage, desired number of lights and Ma of the light. (Or appliance.)j
Yes … a 100 ohm resistor is still 100 ohms, regardless of what circuit you put it in. I don’t understand the question.
Just because somebody told you to use an X ohm resistor in series with an LED, doesn’t mean that applies to incandescent lighting, or smoke units. We need more details on what you are trying to accomplish John.
Del,
I want to power the entire front light board of a USA Geep.
It has two LED headlights, two number board lights(not LED’s) and two LED directional lights. (The latter two are not really needed.)
14.4 volts is the power source.
I already smoked one, by testing with various resistors on the input lead.
So I just yanked it, and put two LED’s in the headlight mounts.
But it would be nice to light the number boards as well.
jb
John Bouck said:Well for the LEDs, you need a resistor in series with each LED. 600 to 1000 ohms should work fine, with the lower values providing more current, thus more light. As for the number boards, it depends what the voltage rating of the bulbs are (perhaps someone else with USA locos knows). If they are 12 volt bulbs, you should be able to connect your power source directly.
Del, I want to power the entire front light board of a USA Geep. It has two LED headlights, two number board lights(not LED's) and two LED directional lights. (The latter two are not really needed.) 14.4 volts is the power source. I already smoked one, by testing with various resistors on the input lead. So I just yanked it, and put two LED's in the headlight mounts. But it would be nice to light the number boards as well. jb
Del,
It’s a nice, neat 1 piece circuit board, already equipped with resistors, LED’s, etc, and two leads running back to the main circuit board.
I guess I have to find out the voltage of the two leads. I’m sure they are lowered somewhere on the main board with a voltage regulator.
As soon as I find that out, I can then apply a resistor in the leads to the light circuit board.
All I know is it won’t take 12 volts–I blew the last one up testing it without a resistor.
jb
USA trains have complex electronics. I have installed a DCC decoder in an F unit, and I believe I found 3 different voltages supplied to various lights and LEDs in the loco. Often they use 3 or 3.5 volts regulated to boards that have leds.
The way I figured this out was to power the loco on DC, and varied the track voltage while measuring EACH power “lead” to a lighting board.
Read the last two paragraphs on this page:
http://www.elmassian.com/index.php?option=com_content&task=view&id=71&Itemid=82
on my web site.
Regards, Greg
Greg,
I figured that since there is only 1 lead (double wire) serving this light board, I’ll find the voltage that the main board sends to the light board and try to duplicate it.
jb
Yes, although part of my point was to check the voltage at varying track voltages, then you can see if it is regulated or really track power, or otherwise.
I hope you check it that way. (hoping for no more burned out stuff!)
Regards, Greg
John Bouck said:Gents, The main circuit board in a USA Geep provides 3.5 Volts to the front headlight/number board assembly. Just tested it with track power. Motors, sound, and smoke off.
Greg, I figured that since there is only 1 lead (double wire) serving this light board, I'll find the voltage that the main board sends to the light board and try to duplicate it. jb
With four LED’s and 2 incandescents on that light board,
Whats the Ohm value of the resistor I need?
The on-line calculation I find says 22 Ohms.
http://ledcalc.com/
Hmm… still seems to have something missing. V=IR means you need to know 2 items to solve for the third.
If you know that the voltage to the board is 3.5 volts, that’s one of the 2 items. You need to know the current draw at 3.5 volts. That will be the second item. then you can solve for the resistance by transforming V=IR to R=V/I …
so 3.5 volts divided by the current you measure will give your resistance needed. You also should calculate the wattage, P=I(squared)R… square the current and multiply by the resistance. That will give you wattage. You should pick a resistor at least DOUBLE the wattage you calculate, or you will have one HOT resistor.
In my USAT F3, the headlights were GOW incandescents, and were supplied 3.57 volts regulated from direct measurement. The 2 bulbs drew 100 ma. In my case, (I use a higher voltage than you) this resulted in 1.6 watts, so a very hefty resistor was needed. My resistor calculated to 164 ohms.
Be careful.
Regards, Greg
Sounds like magic smoke time, to me.
Steve Featherkile said:Hey steve, is it line of sight from your place? :) ;)
Sounds like magic smoke time, to me.
Steve Featherkile said:
Sounds like magic smoke time, to me.
That’s ok, I have a supply from my old British car days:
Bob,
Where can I get some of that? I need it to get my MGA back on the road.
I’ve seen that MG.
You need a miracle…