I have a question about the constant lighting circuit (for rolling stock) on page 48-49 of the May 2008 issue of MR. I was wondering if this circuit would power 3 - 4 miniatronics, 3mm yeloglo LEDs? I would rather build this circuit myself than buy any overpriced, ready-made, lighting kit, as I have several passenger cars to upgrade. Also I don't want a circuit that uses batteries, as I don't want to have to disassemble the cars to change out the batteries.
Thanks in advance for any replies.
I'm not familiar with the circuit you're asking about, but if LEDs will work okay I think you'd be much better off using inverted cone LEDs from moreleds instead of the yellow glow ones. Inverted cone LEDs radiate light 360 degrees instead of being a pinpoint source like the ones from Miniatronics.
http://www.moreleds.com/railroad.htm
Sorry it took me so long to get back to you on this. They don't mention the output from the circuit in the article. The "critical" items that were listed in the article for this circuit are as follows:
Bridge Rectifier (50V - 1A)
Capacitor (4.7UF - 25V)
Resistor (4.7K - 1/4w)
LED super-bright white (5mm / 3.5V / 20mA)
There was only one LED connected to the circuit in the article, with no mention of adding any additional LEDs. I was hoping that I could connect 3 or 4 LEDs & would also like to use yeloglo LEDs on this circuit (I personally think they look better than super-bright white LEDs.)
Yeloglo LED (3mm / 3 - 4V / ??mA)
Thanks again for any replies.
I, of course, can find every MR issue from Feb 2009 to Nov 2009 EXCEPT the May issue, so I can't see the circuit. From the part's list I would guess they are rectifying the DCC into a straight DC current and then filtering it with the capacitor. The 4.7K is a load, but that seems high to me. Is the diode in series with the resistor? If so I would guess that no it won't handle another LED. To put another diode in parallel with the existing one, one would need to increase the current coming through the circuit by lowering resistance.
But I am guessing.
Why don't you just build the circuit. Add the extra LED and see what happens? If it doesn't work adjust accordingly, then add another, etc.
Yelo Glo LEDs have pretty much the same rating as most other white LEDs - 3.5v, 20ma. So it's a direct substitute in this circuirt. I don't have the issue handy to see what it actually looks like as far as adding additional LEDs. With the limited number of parts it sounds like the LED is a key part of the circuit, not just the load applied to an otherwise independent circuit, so just plugging in more LEDs with the circuit as-is may not be possible.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
This is the wiring diagram of the circuit that I'm questioning about, from the May 2008 issue. I drew this diagram using Microsoft Paint (following the July 2008 "corrected" magazine illustration) and added the values of the components that were listed in the article. I would build the circuit and try it out, but the local Radio Shack here where I live, does not carry the particular bridge rectifier or capacitor that is required by this article. The only place I can find them is online and I do not want to have to order multiple components and pay $10 or more in shipping fees unless this circuit is going to do what I want it to do.
http://www.flickr.com/photos/69815348@N00/3987511229/
The circuit does not appear correct. I don't have that issue of MR handy, but I will look for it.
The numbers imply that if you put 3.5 volts across the LED it would draw 20 milliamps. The same current must pass through the 4.7K resistor. Ohm's Law says the voltage across the resistor would equal the resistance times the current. 4700 X .02 = 94. I don't think the circuit develops 97.5 volts. If the circuit developed 12 volts, 470 ohms would be a realistic value for the resistor. 470 X .02 = 9.4 volts.
9.4 volts across the resistor plus 3.5 volts across the LED would be 12.9 volts. This could make some sense.
If you put 2 LEDs in parallel, they would draw twice the current. You would have to cut the resistance in half. The resistance equals the desired voltage divided by the current. 12.9 / .04 = 322.5 ohms.
Oops
You still want to drop 9.4 volts. 9.4 / .04 = 235 ohms. You need to watch the power. Power equals the voltage times the current. 9.4 volts X 40 milliamps = .376 watts. You need a half watt resistor.
If you put two LEDs in series, you need to drop 12.9 - 7 = 5.9 volts. 5.9 / .02 = 295 ohms. 5.9 volts X 20 milliamps = .118 watts. you need a 1/4 watt resistor.
I think there were some comments made about this circuit and drawing power from the power for locos. The current for lamps can add up quickly.
Why not get a TCS FL2 decoder and program the address of the car # into it? You could use the one lighting circuit for the main cabin and put another LED in the lavatory and control it with your throttle. I put an FL4 decoder in a friends car with 4 drawing rooms and each room was lit with a different LED. An FL2 can be had for around $12.
Pete
I pray every day I break even, Cause I can really use the money!
I started with nothing and still have most of it left!
The circuit shown is very simple. The bridge rectifier changes the DCC to DC for the LED power. Any current rating in excess of 100ma will do (for 4 LEDs). You can also use 4 diodes wired as a bridge to do the same thing (25PIV or better, 100ma or higher). The capacitor filters out any ripple in the power - 25 volt rating is minimum, 4.7ufd can be varied substantially without effect. The higher the value in ufds, the better the filtering of low freq components.
The resistor will have to be changed to handle the number of LEDs if the LEDs are wired in series. A much better arrangement is to use a resistor for each LED, and wire the LED-resistor combos in parallel. The resistor value given seems a little high, you might want to go lower if the LED is not bright enough. Somewhere between 1K and 1.5K is usually about right, depending on actual input voltage. Shoot for 10-15 ma through the LED.
Measure the voltage out from the rectifier/capacitor combo with track power in. Then subtract the voltage drop of the chosen LED from the output voltage. Divide the result by 0.012 amps to get the ideal resistance. Drop down to the next commercially available resistor size and you should be around 15ma. A 1/2 watt resistor should be fine unless the track voltage is higher than 14 and/or you are pushing the full 20ma through the LED. If you want a dimmer LED go to the next higher available resistor size.
hope this helps
Fred W
The capacitor in a full-wace rectifier circuit 'pumps' up the voltage quite a bit. Say you feed it with 14.5V DCC, there will be a drop across the diode bridge but that will more than be compensated for by the capacitor. I'd expect 15-15V at the terminals of the capacitor, depending ont he exact stream of DCC commands goign out - at least. Build it with no LED or resistor, measure the volts across the cap, and then plan for LEDs and resistors based on that voltage.
Radio Shack 276-1152 is a reasonable sub for the bridge, it's 1.4 amps at 100 volts.
272-1024 is a 4.7uF 35 volt electrolytic capacitor - use that one.
Both those parts are typically available in the stores. In the case of the rectifier, you can use one with a higher amp rating and/or higher voltage - just the more amps and more volts they get physically bigger. As for the capacitor, the same value at a higher voltage is always safe.
fwright...The resistor will have to be changed to handle the number of LEDs if the LEDs are wired in series. A much better arrangement is to use a resistor for each LED, and wire the LED-resistor combos in parallel...
rrinkerThe capacitor in a full-wace rectifier circuit 'pumps' up the voltage quite a bit. Say you feed it with 14.5V DCC, there will be a drop across the diode bridge but that will more than be compensated for by the capacitor...
I was thinking of wiring the circuit like this. The only problem is, I don't know what the rating of the bridge rectifier or capacitor should be. I was hoping I could use 1K resistors, either 1/4 or 1/2 watt, as I have several left over from a previous project. The voltage input that is shown in this diagram, I measured from my MRC Prodigy Advance (the original one, not the newer Squared version.) With power to the track (no engines or rolling stock on it,) I measured 13.5V. With power on and running my newest engine (MTH GS4 Daylight,) I measured a peak of 13.9V. I put 14.5V max. on the diagram just for a little protection against any spikes in power from any other engines.
http://www.flickr.com/photos/69815348@N00/3989277076/
macmackI was thinking of wiring the circuit like this. The only problem is, I don't know what the rating of the bridge rectifier or capacitor should be. I was hoping I could use 1K resistors, either 1/4 or 1/2 watt, as I have several left over from a previous project. The voltage input that is shown in this diagram, I measured from my MRC Prodigy Advance (the original one, not the newer Squared version.) With power to the track (no engines or rolling stock on it,) I measured 13.5V. With power on and running my newest engine (MTH GS4 Daylight,) I measured a peak of 13.9V. I put 14.5V max. on the diagram just for a little protection against any spikes in power from any other engines. http://www.flickr.com/photos/69815348@N00/3989277076/ Thanks again for any replies.
As was discussed in the previous couple of posts, your meter reads an RMS AC voltage. In the case of a sine wave, the RMS is 0.707 times the peak voltage (or the peak voltage is 1.414 times RMS). DCC is not a sine wave, and calculating the peak voltage is rather difficult - the easiest way is to measure with an oscilloscope.
The full wave rectifier will have a voltage drop of 1.4 volts from whatever the peak input is. The capacitor, if of reasonable size, will charge to pretty close to the peak voltage it receives, not the RMS. Again, if DCC were a sine wave (it's not even close), the output of the power circuit would be 1.4 times the input RMS minus 1.4 volts dropped by the rectifier. In your case of a DCC input, we don't know what the DC output of the power circuit will be - it will be somewhere between (1.4*14)-1.4=18.4 and 14-1.4=12.6 volts. Correct resistor size is going to vary, depending on what the true output voltage is. Please measure it.
The experience of friends who have installed LED headlights and marker lights on their DCC locomotive (function outputs from a decoder) is that resistors of 1K to 1.5K seem to work best. Less than 860 ohms tends to overdrive the LED and may cause it to self-destruct, although Circuitron supplies smaller resistors (usually 560 ohms) with their "12V" LEDs. My friends usually try several different size resistors in the suggested range to get the LED brightness that suits them.
To answer your question, the rectifier has to have a current rating high enough to drive the current of all of your LEDs, so 100ma or higher is fine (I think 1 amp is the smallest common rating). The PIV (peak inverse voltage) rating has to be higher than the peak voltage, so a 25 PIV or higher is fine.
The suggested capacitor size of 4.7ufd is good, and voltage rating again needs to be higher than any peak. 25V would be the minimum. In both the capacitor and rectifier, ratings higher than 25 volts will protect against any surges and spikes, at the cost of larger physical size.
Here are some links to this issue.
http://www.members.optusnet.com.au/nswmn1/Lights_in_DCC.htm
http://www.dccwiki.com/LED_Lighting_and_Resistors
http://www.awrr.com/lighting.html
http://www.opamplabs.com/eirp.htm
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
To fwright, thanks for all your help, this info is greatly appreciated, I'm sorry that I missed this info in the earlier posts, I have been checking this forum in the late hours of the night (12-3AM Eastern.) I am not sure how to measure the "true" output voltage of my DCC system, but I've just found (I've been looking for it for 3 years now) the book to my DCC system. In it, it says the output to the track is "DCC signal with 14.5V amplitude" (don't know if that helps.) I have installed LEDs in a few of my locos with 1K resistors, but I have never dealt with rectifiers or capacitors and wanted to be dead sure what to use, before I attempted anything.
Thanks again,
Chris M.
macmackTo fwright, thanks for all your help, this info is greatly appreciated, I'm sorry that I missed this info in the earlier posts, I have been checking this forum in the late hours of the night (12-3AM Eastern.) I am not sure how to measure the "true" output voltage of my DCC system, but I've just found (I've been looking for it for 3 years now) the book to my DCC system. In it, it says the output to the track is "DCC signal with 14.5V amplitude" (don't know if that helps.) I have installed LEDs in a few of my locos with 1K resistors, but I have never dealt with rectifiers or capacitors and wanted to be dead sure what to use, before I attempted anything. Thanks again, Chris M.
Again, you don't need to worry about your DCC output voltage for the purposes of installing LEDs. What does matter is the DC voltage across the LED/resistor combination. That DC voltage (output across the filter capacitor) is easily measured with a meter and is used to calculate the resistor size. There are several on-line resistor calculators for LEDs - Rich gave some links. 1K or 1.5K is usually a safe starting point in mr applications and LEDs, with final size being determined by desired brightness.