DC or DCC? What kind of capacitor is it?

Trent-

Basically, you will need to add a resistor, and a diode as well as a capacitor. Forgive my crude drawing technique.

If you are running DC power, the diode will protect your capacitor when the track voltage is reversed (and the LED will not operate in that case).  If you are running DC and you want the light to operate even when the loco of the train is running in reverse, a more elaborate circuit will be needed.

If you are running DCC, the diode will turn the AC DCC voltage waveform into positive-only pulses, which will properly charge the capacitor (regardless of train direction).  Also, I do not know the details of the capacitor you retrieved from the TCS decoder.  If it is a polarized electrolytic capacitor (probably it is), it will have a small "+" sign on the can, identifying one of the two leads.  If it is not polarized (for example, a ceramic disk capacitor), it probably does not have a large enough capacitance value to help you.  Also, the voltage rating of the capacitor is important, and it should be marked (in tiny letters) on the capacitor.

So, let me know more information about the capacitor (voltage rating, whether it is polarized or not, and hopefully the capacitance value, although that is less critical to know) and whether you are running DC or DCC.  Then,  I can confirm whether the circuit will work with the capacitor that you have.

Dennis

Hello Dennis thanks for your

Hello Dennis thanks for your response

Im running DCC

The Capacitor is marked with the following notes

"470HF 16V "

" VR(M) " 

"B1329" 

"85c"

it has a blue and black wire and per TCS decoder notes the black wire is the ( - )and the blue is ( + )

I have one other question but will await what you can decode from this.

Thank you very much

Trent Blasco

Comments on the capacitor

Trent-

The Capacitor is marked with the following notes

"470HF 16V "

" VR(M) " 

"B1329" 

"85c"

it has a blue and black wire and per TCS decoder notes the black wire is the ( - )and the blue is ( + )

This is good.  It means it is a 470 Microfarad (it is not "HF" it is "uF") polarized electrolytic capacitor.  It can operate up to 85 deg C, which is fine.  It is rated at 16 volts max.  The VR(M) and the B1329 markings are mystery to me, but they are not important.

The 16 v rating is a concern.  Here is an excerpt from the NMRA Standard S9.1, the DCC electrical standard:

" In no case should the peak amplitude of the command control signal exceed +/- 22 volts.  ... Digital Decoders intended for "N" and smaller scales shall be designed to withstand a DC voltage of at least 24 volts as measured at the track.  Digital Decoders intended for scales larger than "N" shall be designed to withstand a DC voltage of at least 27 volts as measured at the track. "

 If you expose a capacitor to a voltage higher than its rated value, bad things can happen- like, the capacitor will have a small explosion.  Generally, you want a capacitor whose voltage rating is 20% higher than the largest voltage it will encounter.  I am guessing that this capacitor with the TCS decoder was used in a place in the circuit where the DCC signal had been rectified and regulated to either +5 or +12 v to run the decoder electronics.

It is difficult for me to predict exactly what peak voltage the capacitor might see in the circuit I drew for you.  It will be somewhat less than the peak DCC signal that appears on the tracks, due to the loss of 0.7v from the diode and due to some voltage drop across the 100 ohm resistor.   The size of the voltage drop across the 100 ohm resistor is hard for me to predict- it depends on a lot of things (the detailed output circuitry of your DCC booster, the electrical characteristics of your track feeder buses, the other DCC decoder loads that might be active at the same time your passenger car drumhead is trying to be lit).

So, the 100% accurate detailed design is elusive.  But, the good news is, that if you get a 30 volt of higher rated capacitor, you can rest assured that things will be ok, no capacitor explosion.

Such a capacitor is not expensive.  For example, here is a link to a 470uf 35 volt capacitor from Radio Shack for $1.49.  http://www.radioshack.com/470uf-35v-20-radial-lead-electrolytic-capacitor/2721030.html

It won't be tiny- maybe 3/4" long and maybe 3/8" diameter, but I would guess that you don't have a serious space problem in your passenger car.

For $1.99, you could buy a 35 volt rated 1000 micro farad capacitor from Radio Shack.  It would be slightly larger in size that the 470 uF unit, but with twice the capacitance, it would significantly increase the ability to ride out temporary power disruptions to the passenger car without having the LED flicker.  You could start with a 470 ufd unit, and if you don't eliminate all flickering, you could add a second 470 ufd cap in parallel with the first to get a total of 940 uFd.

I hope I am not making this sound too complicated!  Bottom line: I don't think it is worth the risk to use the 16 volt capacitor that you have.  Some DCC systems on some layouts will not generate track signals which would harm the capacitor, but I'd play it safe, since not much money is at stake to get a new higher voltage capacitor.

Feel free to ask questions.

Dennis

Hi Dennis    Very well

Hi Dennis 

Very well written and a wealth of info. Thank you

Our dcc voltage is set at 14.5 volts but it sounds like you are absolutely correct that for a few dollars on a different 35v capacitor I will be much safer.

I will pick up a new capacitor and would like to re visit the wiring aspect to make sure I do this correctly. 

If all works well I may install these in some of my lighted cabosses as well. 

Thank you again for your help. And will report back when I have the parts in hand

Trent Blasco

Other numbers and letters on cap

The other letters and numbers are likely the type/series of the capacitor and lot/date code.  A google of "VR(M) capacitor" does return the Nichicon VR series of capacitor.  The B1329 is the manufacture date, I don't know what the B stands for, but the 13 is the last two digits of the year (i.e. 2013) and the 29 the 29th week of the year.

It would be good to know the current draw of the LED to help determine the proper size (Capacitance) of the capacitor.

Initially I was wondering what the need for the resistor in front of the capacitor was, then I remembered it is to prevent the DCC system from thinking there is a short. Without the resistor the first time the system is powered on to charge the capacitor the capacitor has a large current draw, the resistor lowers that current draw enough to prevent the DCC system from thinking there is a short.

Voltage Rating

Do not use a 16 volt rated capacitor ! I have my DCC system set to 13.8 volts and added a 16 volt capacitor to one of my passenger cars. It worked great for about five minutes, then POW ! It exploded covering the interior of the car with a yellow substance that took forever to get cleaned out. I rebuilt the circuit using a 25 volt rated capacitor and it's still working as it should a year later.

I used a version of this circuit ....

I used a bridge rectifier with the AC terminals connected to the truck pick ups. The circuit above was attached to the + and - terminals. The LED with it's resistor was attached to the same + and - terminals in parallel to the circuit. Most LEDs are rated at 20ma at full potential. I used a 25 volt 4700uf capacitor which netted me about 8 to 10 seconds of illumination to 4 LEDs before they went out.

Mark.

In rush current and also Mark's circuit

Hi everyone-

Yes, Jim B. is correct that the 100 ohm resistor is there to limit in-rush current to the capacitor.  As he points out, this will prevent some DCC boosters or circuit breakers from thinking that there is a momentary short when the system is first turned on.  Another situation is: DCC has been on, but the passenger car with the circuit is placed on the tracks.  As with the capacitor voltage rating, in some systems, the extra resistor wouldn't be needed, but it is just good practice to include it.

The 100 ohm value is just convenient, and will be in the right ballpark.  The series resistor with the LED is going to be about 1000 ohms, and the current through the LED will be somewhere around 15 milliamps, enough for it to shine brightly.  The addition of the extra diode and the 100 ohm resistor will reduce the LED current (and hence the brightness) somewhat from what it would be without adding the capacitor circuit.  The exact numbers are not super important here.  One could fine tune the values of the 100 ohm and LED series resistor to get exactly the LED brightness that you seek, but I would think it is not worth the trouble.  Let me know if you want more information on this tweaking of the resistor values to play with the brightness.

If I understand Mark R. correctly, he adds a full wave bridge rectifier across the track DCC feeds.  This will all work, but it would seem to me that since Trent is not trying to power a decoder in the passenger car, this extra set of components is not adding value to his situation.  Not trying to start an argument Mark- tell me if I am missing something.  BTW, love the D&H Shark!

I think we should re-title this thread "Operation Lifesaver: Protecting Passenger Car interiors from explosions"

Dennis

The Only Difference ....

By using a bridge rectifier, the LED will illuminate in both directions if being used on DC. I do a lot of work for customers and I prefer this method giving people the freedom of being able to use the car where-ever down the road. If the car will only be used on DCC, then the first circuit shown above will work fine. This animation basically shows how the bridge rectifier functions on DC .... 

http://www.reuk.co.uk/OtherImages/bridgerectifier.gif

Mark.

Slight disadvantage of bridge rectifier

This might be viewed as nit-picking, but it is all in the interest of "education and enlightenment."

The use of the bridge rectifier means that the peak charging voltage that the capacitor will see is the peak track voltage (the DC level in case of DC, the peak pulse height in the case of DCC) minus two diode drops, equaling 1.4 volt.

Without the bridge rectifier (and giving up on DC operation of LED in both directions), the first circuit I drew has the capacitor seeing a peak charging voltage of the peak DCC pulse height minus one diode drop, equaling  0.7 volts.

The difference in the voltage to which the capacitor is charged to translates into a difference in the current to the LED, and hence brightness.  How much?  For simplicity's sake, assume 14 v DCC peak voltage, 100 ohm and 1000 ohm resistors.  With bridge rectifier, the current through the LED will be about 10.8 milliamps.  With the single diode approach, the current through the LED will be about 11.5 milliamps.  This is about a 6% difference, and that is not significant.  In fact, the 1000 ohm resistor is probably a +/- 10% tolerance component, and the potential variability in the actual value of this resistor would cause a similar range in the LED current.  I doubt that you could see the difference in LED brightness.

Like so many things in model railroading, there is more than one way to get the job done, with equivalent results.

Dennis

Correcting a mistake in my previous analysis

Oh well, at least I realized that I made a mistake and need to clean it up.  My previous post showing the current through the LED for each of the two circuits got it wrong.  Here is the correction:

Assume DCC peak voltage is 14 v, and the resistor values are 100 ohms and 1000 ohms.

OK: for the circuit with the bridge rectifier:

The capacitor charges to 14v minus 2 diode drops (due to the bridge rectifier), or 12.6 volts. The current through the LED is 12.6 minus 2 diode drops (the diode in parallel with the 100 ohm and the LED itself) across 1000 ohms-- which is 11.2 volts across 1000 ohms, yielding 11.2 mA through the LED.  (Note: the diode in parallel with the 100 ohm clamps the voltage across that resistor at 0.7v at this current level, since the drop across the 100 ohm resistor at 11.2 ma would be 1.12 volts-- so the diode rules).

For the first circuit, with the single diode:

The capacitor charges to 14v minus 1 diode drop, or 13.3 volts.  The current through the LED is 13.3 minus 1 diode drop (that of the LED) across 1000 ohms-- which is 12.6 v across 1000 ohms, yielding 12.6 mA through the LED.

So, the first circuit provides 12.5% more current through the LED.  Still not enough to make a big difference.  Keep in mind that decreasing the 1000 ohm resistor to 820 ohms would increase the LED brightness in either circuit.  So, the tradeoff between the 2 circuits is: in exchange for slightly less bright LED and one extra component (the bridge rectifier is available as a single component), you get bi-directional DC operation in addition to DCC operation.  Name your poison.

By the way, to be complete, the analysis should verify that as the capacitor discharges as it powers the LED in between successive DCC pulse peaks, that the voltage drop does not reduce the current through the LED by a significant amount.  If we use 470 micro farads and 1000 ohms, the time constant of the discharge circuit is .47 seconds,  The longest interval between DCC pulse peaks would be around .012 seconds, which would occur with a so-called "stretched zero bit."  I'll spare you the exact calculation, but this means that a very small, negligible drop in the capacitor voltage occurs from the time it charges on a DCC peak until the next time it can recharge.  Even if you go with an 820 ohm resistor, lowering the time constant, the conclusion remains the same.

Circuit analysis can be fun!  Sorry about the previous mistake.

Dennis

Confused...

In the first circuit diagram, if Track A is at 12V, then isn't track B at -12V?  So wouldn't the capacitor be seeing in the vicinity of 24V?  (Or if the voltage is varyinbg a bit, say to almost +/- 15V, that the capacitor would see roughly 30V)?  Am I thinking incorrectly about the DCC signal?

Thanks,

-Ed

DCC waveform

Hi Ed- I believe you are thinking of the DCC signal incorrectly. The first oscilloscope photo in this link shows what the DCC waveform looks like:

http://members.shaw.ca/sask.rail/dcc/DCC-waveforms/DCC_waveforms.html

The scope trace is showing the voltage difference between the two rails at any one time.  First, there is a +14 v (or so) pulse, then there is a -14 v pulse.

With the first circuit, each positive pulse charges/recharges the capacitor.  Each negative pulse is "ignored" by the first circuit.  With the second circuit, using the bridge rectifier, both the positive and the negative pulses charge the capacitor.

The capacitor is only seeing the amplitude of the positive (or negative pulses) minus the relevant diode drops.  It never sees the voltage difference between the peak positive and peak negative value.

Dennis

Er, no...

If you do not have a defined ground reference, you must consider the total voltage swing. With only two tracks, the effective voltage of +/- 12 becomes 24 volts. The signs only indicate the direction of current flow at any moment. The capacitor rating and the current calculation must use the peak-to-peak voltage.

pqe

Er, yes

What matters is the voltage difference between the two rails.  The circuit sees only that.  It has no connection to some other arbitrary ground point.

If people do not believe what the oscilloscope trace is showing, here is another diagram of what the DCC waveform looks like:

http://modeltrainadvisors.com/digital-command-and-control.html

(scroll down a little bit to see the waveform diagram).

Note that the caption for the diagram says:

"If one were to monitor the electrical signal across the train rails with an oscilloscope, the following characteristics should be noted:

1)      The waveform is Bi-Polar with its voltage varying between + 14 volts and – 14 volts. It is symmetrical about zero volts."

The voltage difference at any instant in time between the rails is either somewhere between 0 and +14v or between 0 and -14 v.  It is not +28v or -28v.  The peak to peak voltage swing of the track waveform is 28v, but the circuit does not see the positive and negative peak at the same time-- look at the waveform.

35 volts is still the rating for the capacitor that should be used.  Although the voltage into the circuit is nominally +14v or -14v, design for the worst case- pulse distortions due to the DCC wiring system, out-of-spec DCC boosters, etc.  That is why the NMRA spec says that an HO decoder should be designed to withstand +27 to -27 v at its inputs without damage, even though the spec says that the DCC booster should not put out that large a signal.

Dennis

More discussion of differential voltage

The DCC signal is based on the RS422 specification, which is a balanced signal. Balanced signals are output on two wires, usually a twisted pair. When one wire goes positive, the other goes an equal amount negative. Thus we speak of a signal at +/- 2.5 volts. The receiver reacts to the difference between the two wires, thus it sees a 5 volt signal. The NMRA voltages are increased to provide appropriate motor power.

The NMRA spec is not particularly clear, but this page has a good explanation, with diagrams.

Here is a scope trace of my NCE PowerCab output-- the input range is set for 5 volts per division. (At least I think it was. I first posted this in 2012, and the site search feature doesn't go back that far, so I can't read what I said. I'm 2500 miles from my workbench, so I can't double check.)

The center line is 0 volts, the swing is +/- 14, and the peak-to-peak voltage is 28.

pqe

Still disagree

Hi Pelsea-

I am sorry, and I intend no hostility or such, but I still say you are not correct.

You said:

"The center line is 0 volts, the swing is +/- 14, and the peak-to-peak voltage is 28."

I agree.  But let's look at your scope trace.  The trace is showing the output of your NCE Power cab at various points in time.  This is the same as the differential voltage seen between the two track rails (not accounting for wiring loss and pulse distortion).  Look at the wave form at any single point in time.  The voltage is never +28 ort -28.  At any point in time, it is +14 when there is a positive pulse and -14 when there is a negative pulse.  

The peak to peak value of 28 volts does not occur at a single point in time.  The waveform shows this.

You discussed balanced signal, one side going up by the same amount that the other side goes down.  If you want to think this way, you can imagine that one rail goes to +7 v relative to a ground reference inside the DCC booster while the other rail goes to -7v at the same time.  This produces a total differential across the rails at that time of +14 volts.  Then, when the pulse reverses, one rail moves from +7 to -7, and the other vice versa, producing a total differential across the rails at that time of -14 volts. The key point is that there are only 2 wires going to the circuit- one from each rail.  The circuit can only "see" the relative voltage difference between the two rails at a each point in time, and this differential voltage is +14 for the positive pulses and -14v for the negative pulses.  The circuit has no other connection to the DCC booster, and it has no idea what the ground voltage is in the booster.  It only sees what is on the tracks, which is what is on your scope photo.

If, in fact, the DCC signal had a differential voltage of +28 or -28v at a decoder's inputs, then decoders built to NMRA specs, ie, able to withstand up to +/- 27 v at their inputs would be blowing up all the time.

Dennis

0 volts is a fiction

Voltage is the electromotive force potential between two points. The two leads of a meter or scope, for instance. 0 volts or "ground" is nothing more than a designated part of the circuit we use as a base for voltage measurements. Usually that designation is made by the designer, and represented by a symbol on the schematic. The practical definition of ground is "put the black lead here". The polarity of a voltage measurement indicates which way the current is flowing through the meter. (Volt meters actually measure the current through a very precise resistor.) One very common ground point is the negative side of a battery. Another common choice is the case or cable shield of an audio device.

The NMRA spec defines 0 volts as the midpoint of the differential signal. (See footnote 3 of this paper.) That's not a place I can connect a lead, so I put one lead on each rail and adjusted the trace so it was on the center line when the power was off. When the power is on, I get the display you see, alternating between +14 and -14 volts. The difference between positive and negative is just a matter of which lead is on which rail, so if they were swapped, the display would be flipped. If I had hooked up two oscilloscopes, you would be able to see that one rail is always positive when the other is negative. I'll post a dual trace shot when I get home in a couple of weeks.

The decoder doesn't care about ground definitions. All it knows is the voltage difference between the rails, and has to handle the current through its own impedance. If the A rail is +27 volts, the B rail will -27 volts, a difference of 54 volts. An impedance of 100 ohms will pull a current of 540 ma.

If you are still fuzzy on the concept, here is an analogy. I am currently in Algonquin Il, about 35 miles north of Chicago. If I drove to Joliet, about 35 miles south of Chicago, I'd have to drive 70 miles. When I come back, it's the same distance, but the sign has changed (+ = north, - = south).

pqe

Hello Again Dennis I

Hello Again Dennis

I ordered some 470 microfarad 35V capacitors and 100 Ohm 1/2 watt resistors.

is there a specific rating on the diode that I will need?

also Im not following where exactly I should install the diode in the drawing you provided me. could you please point to where the diode would be installed?

I apologize for the questions as Im not real savvy on electronics beyond the basics of resistors being added to LEDs. I am enjoying the conversation and this has intrested me in adding some lights into my cabooses.

thanks again for the help

Trent Blasco 

Clarifying question on which voltage value

So I am hearing two different takes on the voltage rating to consider for the capacitor.  Based on the two different indications, and some other research online, I guess I have questions to try to sort out understanding it.

First, I found this Digitrax reference that seems to suggest that in fact there is some degree of relationship to ground in the individual track voltages:

http://www.digitrax.com/tsd/KB909/track-voltage-measurement-on-dcc-layouts-with-dire/

This seems to suggest to me that in fact the total peak to peak will in fact be roughly 12 to 15 volts (plus the indicated tolerance).  But then I still ask myself if thaty is really representative of the electrical stress placed on the materials of the capacitor given that the capacitor has no actual connection to ground.  If the capacitor has charged to 14V, then the polarity of the rails goes to the other half cycle, what will the voltage difference be between the positively charged plate of the capacitor and the B rail during the reversed half cycle?  After all, the "charged" capacitor is in fact an excess of electrons on one side and a deficiency on the other.  When the rail polarities are reversed, it is unclear to me whether the difference between the missing electrons on the positive side of the capacitor and the actual voltage relative to ground (per the Digitrax article) would be 14V, 21V, or 28V.

-Ed

Ground I go again.

The procedure in the article is a nice trick, but it is limited to one thing-- checking that the signal on the rails is symmetrical. This is important, failing the test described indicates a serious fault (or a DC offset in systems that support that). However, there are some problems--

• A random piece of hardware is not necessarily ground (although it often is). The only thing you can trust to be ground is a connection labeled ground. You may have to disassemble the booster to find one.
• The voltage obtained this way will not be accurate, because multimeters show the average signal assuming the input is a sine wave, and the DCC signal is a pulse.
• The measurement does not show the instantaneous difference between the rails. If you put a meter lead on each rail, you would get a reading of about 0.7 of the actual value.

But the most important factor is the peak voltage across the capacitor leads, which will be around 28 volts. In a couple of weeks I'll post a tutorial about making these measurements. Meanwhile, a 35v cap will be fine up to O scale, and the only penalty is it's a little bigger.

pqe

Diode rating, cricuit clarification

Hi Trent- Glad to hear that you are proceeding.  Soon your drumhead will no longer flicker.  Sorry for the ambiguous labeling in my diagram.  Here is a clearer version.

The parts list:

R1: 100 ohm resistor, 1/2 watt power rating

R2: 1000 ohm resistor, 1/2 watt power rating (could lower this to 820 ohm for more brightness)

D1: small signal general purpose diode, such as 1N4004

D2: the LED, which is a diode and is shown in schematics as a diode

C1: 470 micro farad 35 volt capacitor

Also, the diode that you provide in addition to the LED (which is also a diode) can be any general purpose "small signal"  or "rectifier" diode.  Diodes are rated in terms of the peak reverse voltage they can withstand and the forward current they can carry without damage. Commonly available diodes that can be used with the circuit include 1N4001, 1N4004 and 1N4007.  Here are 2 from Radio Shack that you could use:

http://www.radioshack.com/micro-1-amp-diodes/2761104.html  (Small size, handles a forward current of up to 30 Amps.  When the capacitor is first charging, the current through the diode will be the greatest and quickly go to zero once the capacitor gets charged.  The current will never reach 30 amps, though, because your DCC booster will be unable to provide that large a current.  This diode can handle a reverse voltage of 600v, which is way larger than will ever been seen in this circuit).

http://www.radioshack.com/1n4004-micro-1-amp-rectifier-diode/2761103.html  Just another choice, with similar ratings.

If you look at the photos of the diodes, they have a black case with a white band around one end.  The end with the white band corresponds to the side of the diode in the schematics that has the vertical "bar."  In the schematic above, this is the right end of the diode.

You can also find appropriate diodes on Amazon or from any other source of electrical components.  In this example, the exact specs. of the diode are not critical.

As before, feel free to ask further questions and I will try to be helpful.

---------------------------------------------

OK, now to answer Ed's question:

The capacitor will see a maximum voltage across it of something slightly less than the 14.2 volts that Trent's DCC booster puts out.  It will see around 13.5volts maximum.  The 35v rating is substantial larger than the capacitor will see, and is there as a safety margin.

I reiterate all my previous comments that the capacitor will not see 28 volts across it, regardless of what Pelsea says.  He is confusing the peak to peak voltage of the waveform (which is 28 volts) with the value of the voltage difference between track rail A and track rail B at any one time.  The circuit is connected to track A and track B and it can only "see" the voltage difference between those two tracks.  The voltage of track A relative to track B varies from a maximum of +14v to a minimum of -14v.  The important thing is that although A relative to B undergoes a peak swing of 28 v, at any point in time it is never more than 14 volts above or below B.

To use the analogy that Pelsea provided previously, assume that our circuit responds to the distance from our current position to Chicago.  (Chicago in this case can be called track B and our car can be called track A).  When we are in Algonquin we are 35 miles (north) of Chicago.  When we travel south and reach Joliet, we are then 35 miles (south) of Chicago.  The greatest distance we have ever been from Chicago is 35 miles, not 70 miles.  We have traveled 70 miles from Algonquin to Chicago, but have never been more than 35 miles from Chicago.

I suggest that further questions or debates about the value of the voltage across the rails in a DCC system be moved to a different thread.

Dennis

Diode with shite stripe

Hello Dennis

is this how the diode will fit into the circuit? I want to be clear on what way the white stripe should face

Thank you again.

Trent Blasco

@ LMACKATTACK: Yes

HI Trent-

Yes, you have the diode facing in the correct direction.  The band on the diode case indicates the side which is represented by the vertical bar in the schematic symbol.  The LED has its polarity marked in a different way.  If you are having trouble identifying the LED's terminals, send me details on the LED and I will try to help you sort it out.

Note that in your diagram, you added a yellow rectangle and marked it as LED.  Just to be clear-- the diode that is drawn just to the left of that yellow block is representing the LED, and its other side is connected to the 1000 ohm resistor, labeled as R2 in my diagram.

Looking forward to seeing a picture of that Santa Fe drumhead!

regards,

Dennis

Thank you

Thank you Dennis

Yes I will post a picture of it when I get all the parts and install it. The LEDs I have are prewired and have a red and black wire with the resistor pre wired to the red lead. Im not worried about frying the Led out by mistake as I have about 50 of them. The other parts I just bought 10 each.

this  has been a great learning experince. pics to come soon!!!