[Brian]

Hi all
Either I'm missing something (which is quite likely

) or what's been said about installing this super bright LED is rather mind blowing and perhaps a little confusing!!
Maplins quote the forward voltage for the LED as Min 3.0 and Typical as 3.6 v dc at 20Ma Maplins Tech sheet
I have wired in LEDs into industrial control panels for work and many model railway layouts using many thousands of LEDs (Wish I had a pound for every one!) and never have I fitted a series Diode into the power supply leg. While I can see why it being recommened in threory, in practice I have not found it necessasary, even on some very poor quality bridge rectifer power supplies. All the LEDs I have installed have lasted for many years without failure.

So, all calculations for the resistor needed should be based on the typical values of 3.6 @ 20Ma If you require to run the LED on ac then place a diode such as a 1N4148 across the LEDs terminals in inverse parallel and calculate the resistor needed and then reduce that value by 50%
Your calculation is simply .... supply volts minus the normal forward voltage of the LED divided by 0.020 Assuming a 10v PSU then its 10-3.6/0.020 = 320 OHMS
Use any resistor above 320 ohms to a max of around 600 OHMs to maintain a good light output. If PSU is ac then after fitting a reverse diode use any resistor above 160 ohms to a max of 400 OHMS

 

[Gofer]

QUOTE (Brian @ 2 Jul 2007, 10:34) <{POST_SNAPBACK}>and never have I fitted a series Diode into the power supply leg. While I can see why it being recommened in threory, in practice I have not found it necessasary, even on some very poor quality bridge rectifer power supplies. All the LEDs I have installed have lasted for many years without failure.

That's the problem Brian. The PSU in question doesn't have a rectifier at all leave alone a very poor bridge rectifier. By fitting a series diode any number of LEDs could be run from the PSU (within the power output constaints of the unit of course). Without the series halfwave rectifier diode I'm suggesting each LED would need it's own reversed biased diode.

Steve

 

[Brian]

Hi Steve
Agree with you. Simple to do, but I personally would look at wiring in a bridge rectifier and let the LEDs run on full wave rectifed dc. Maplin sell these for 30p for a 2.0 amp bridge or use 4 x 1N5400 diodes in bridge formation. This fully rectified supply could be used soley for all LED lighting etc?

 

[double00]

Gentlemen, am I correct in thinking that LEDs N24BY as at present wired with a 1K 1/4 watt resistor is to high rated - is LEDs brilliance being considerably reduced or as said wrong assumption?
My meter is an Avo 7, old technology but reliable. Also I have, a Maplin hand held digital meter - more like a gaming machine, as it take ages to finally stop and give a reading, that again jumps a further few times - is the last registered number the correct reading - only God knows!
I downloaded the software that calculates required resistor value, and shows coloured rings appropriate to value of resistor - certainly a most useful piece of software to have at hand.

 

[Brian]

Hi Double00
If your PSU is giving 10 volts dc then a 300 to 600 OHM resistor per LED will be fine. If you're output is ac at 10v then you will need to reduce the resistor even further to around half the calculated value as a diode will be needed across the LEDs terminals in reverse direction to the LED. (Though I would opt to install a bridge rectifier if my LED supply was ac!) The 1K resistors you're using is ok but will result in lower current flow through the LED and consequently a reduced light output Though this is often hard to see.
With your AVO or digital meter set on volts, measure off load the output of the transformer. Use this as the basis for all calculations.
The fluctuations on your digital meter are quite normal. You have to learn to read the average. Some digital meters have an "Average" button which will freeze the display and give the average reading. The AVO 7's needle and coil are far to slow to react quickly to minor voltage fluctuations. Lovely meters though!

 

[MUSO52]

Good Evening Double O, Steve and a welcome aswell to Brian!

Well once Brian had indeed realised he had missed the point ( only joking!) , the problem as he later realised being the use of a AC source, he is quite right in what he says, and indeed I am sure Steve would agree also. The use of a full wave rectifier (with a smoothing cap Brian/Steve?) is recomended.

If DoubleO intends using more than one led, the use of an inverse parrallel diode on the latter and any additional leds is not really cost effective and is time consuming. Might aswell buy a Full Wave rectifier as Brian stated and be done with it.

STEVE.

May I, ( and again with all due respect ) return to our discussion of yesterday with reference to protecting the diode? Forgive me, but it certainly got me thinking if I was right or wrong! (Senior Moment?) Of course the voltage would have been halved from the tranny, BUT ONLY if the transformer was centre tapped! We both missed that one!! Basically each side of the transformer would see 5v referenced to this centre tap, which is normally tied to earth. DoubleO's is'nt center tapped I think!
Without the transformer being centre tapped , this would further compound the reverse voltage issue we discussed. Why? Your idea of placing a diode prior to the led to rectify the ac seems logical, but as I said, the led itself is a diode, and like the additional diode you included, would only conduct on one half cycle depending which way round it was placed. So what happens when we have the corresponding neg half cycle and the diodes are reversed biased? With no current flow, this neg voltage would be present at the p-n juction of the led in all its glory! Not just 10v either. That would have been the rms value so the PEAK voltage at the junction of the device would be more like 14v, and this exceeds the max reverese breakdown voltage of the N24BY by 9 v's!

This was the reason I suggested placing the ' protecting diode' after the led as with a higher tolerence to reverse voltages (and hence breakdown), though not ideal would have protected the led.

I believe Brian would possibly concur, and I for one would welcome his input on this discussion ( he will have to read all the posts though!). As he stated, when using it with an ac source, a diode placed in inverse parallel to the led would be required to protect it from reverse biasing and possible breakdown.

Anyway, not an argument, I told you so, or even a 'I know more than you' thing, please except my words in the friendly way I am not very good at conveying! If I am wrong ( and I am sure there are others who know more than I) please let me know, then I will have learnt something from all this also!!

Constant Current Device

Of interest to not only Steve but others also I hope, here is the Constant Current Device I mentioned previously .Visit this link for a full description and application.
http://www.pollensoftware.com/railroad/index.html

Finally, should you read the above notes on the device Steve, even others, like ourselves make mistakes! What he did not mention was the voltdrop across the device. No gain without pain, dont get anything for nothing! Yes it will drive a typical 2v led but this dosnt include the voltdrop across the device which I measured at 2.2v hence 2.2 v + 2v = 4.4v

Hope this of interest to you all.

 

[double00]

Gentlemen, perhaps it maybe of interest to know my set up of feeding six LEDs from my 10v KPC cased transformer.
I use three strips of six wire connectors 3 amps from Maplins (12 connectors = 1 strip). I use two strips, halving each strip to give strips of six. One half of each two strips inner side connected in series, one strip to the positive, one negative - and connected to transformer -/+ outputs.
Third strip connector is connected individually to the positive connector using 1k 1.5 Watt resistor to bridge across. Positive of the LED is connected individually by use of length of wire to the connector connected to the individual resistors. LED negative, connected to the connector strip, that is connected to the transformer negative output.
If instruction can be followed, if not apologies - where does the suggested diode or bridge rectifier be wired into which of the three wire connector circuit.
I note Express Models advertise their LEDs complete with resistor - I asked EM what value resistors were - negative reply!

 

[Gofer]

doubleOO


a couple of variations on a suitable circuit for you :

The values of the resistances would need to be calculated once the actual voltages are known. Also, as said earlier, although your LEDs will work on AC it would be much better if they were supplied from a smoothed (and preferably stabilised) DC source.

MUSO52

as regards the series resistor conversation, consider this sketch:

Obviously I've ignored all losses in the circuit and across components.

Sorry the sketches are a bit rough but hopefully you'll be able to decipher them.

Steve

 

[double00]

Gofer, many thanks for the trouble you have taken in showing the various circuits.
Am I correct? that the "Bridge Rectifier" is soldered across the +/- terminals of the KPC 10v transformer, and what is recommended to obtain ie Maplins N400C? It is now years since using electronic components - "Rectifier" is it similar to LED anode/cathode as if soldered into the circuit incorrectly can "Blow" or burn it out? Again my thanks.

 

[Gofer]

DoubleOO, I'm a bit confused by your statement that your KPC unit has terminals marked + and -. These identification marks are normally reserved for DC supplies only. AC supply units would normally have the terminals marked with a sinusoidal (waveform) sign. Are you absolutely sure unit you have has an AC output, as being marked + and - does suggest the output has already been rectified and as such is DC? With respect, I'm reluctant to advise you further until we can establish exactly what output your unit has. If you are sure your unit has a AC output then a bridge rectifier should be connected as I've shown in the sketch with the sinusoidal waveform signs connected to the output terminals of the power supply (doesn't matter which way around) and the + and - connected to your LED's as shown (it does matter which way around).

Equally I'm just as confused with the Maplin part number you quote (N400C). A search of their site only reveals that number as being a laptop battery.

There are dozens of different bridge rectifiers available today for either for fitting to a PCB or having wire terminals. Alternatively it's a simply enough exercise to make you own bridge rectifier using four standard 1N4001 diodes. If you have to purchase any bits and pieces the only criteria you should worry about is the current the rectifier will be required to pass and the operating voltage (you should choose one that is designed to operate on at least double the output voltage of your power supply). With LED's (even six) the total current is going to be very small (0.02 x 6 = 0.12A) so a bridge rectifier capable of passing 1 amp is going to be more than adequate and unlikely to get warm let alone hot. If you are unsure as what to order, email me with your address and I'll send you one for the cost of a stamp plus 50p

Steve.

 

[double00]

Gofer, firstly I examined the KPC cased transformer - details shown, 1 10v output plus name of KPC and address. A two core cable + three pin 3 amp plug connect to the mains, another two core cable black PVC extends from the rear of the case (red + black coloured inner cable), reading from this two core cable reads 10v, no indication or markings show if voltage is AC or DC other than a red and black cable.
Maplins reference code has been wrongfully shown - 1N4001S QL73Q again perhaps I have read the wrong part number.
Your kind offer to obtain the correct bridge rectifier is certainly most appreciated, please advise me as to total cost incurred. To suit my already built voltage drop to LEDs, perhaps the B R with wire terminals will be more suitable.

 

[double00]

Gofer, I ventured into the KPC web site and apparently the KPC transformer is a 10v AC 2 amp.

 

[Gofer]

doubleOO, as they say on the TV, 'you now have two choices'

Either I can send you a basic 'off the shelf' bridge rectifier which will require you to solder wires to it in order to connect it to your circuit or I can make you a bridge rectifier unit with screw terminals and simple instructions to enable you to connect it to your circuit. Either way the cost is small - all I ask is that you cover my costs and to this end I'll be happy with £0.50 plus the cost of a stamp for the basic bridge rectifier or £1.50 plus the cost of a stamp for the unit with the screw terminals. Email me with your address and let me know which option you would prefer and I'll sort it out for you.

Steve.

 

[MUSO52]

Hi Steve and DoubleO,

Just read todays installments!!

I was about to suggest the same thing Steve, have components vero etc, across room!
Muso

 

[MUSO52]

Steve.

Thanks for the sketches, look like bras on a washing line! (only joking).

Fully understand what you are trying to illustrate, (must learn to upload sketches).
Maybe I wasn't fully correct after all, but on the right 'track'!
My understanding now is thus. With ref to the half wave + led, When the supply to the reg diode is +v
with respect to the -ve on the other side of the led, both forward biased, current and volt drop and led lights as we know. With Ov on the front end of reg diode, 10v will be at Neg leg of led. No current flow, reversed biased. However, this voltage would not break down led as this same potencial will also be on the pos leg of led. There being no current flow, the Half wave rectifing diode ( now with a 0v on its pos leg) will not conduct providing this voltage do'es not exceed the Vr or breakdown voltage of the diode. However if we exceeded this Vr, then the p-n juction will faill, current will flow and destroy both diode and led. Avalanche effect, I think its called. I think this is what you tried to explain in your sketches, Yes

Back to DoubleO.s Circuit.

The full wave reg is the way to go as u say but one thing I note from sketch, would not the output be closer to 10v than 5? Minus a drop across the F/W reg of course.

I think it safe to assume that the tranny is 10 rms. If the output was smoothed, this would lift voltage to near a peak Voltage of approx 12.5 to 13v ( by smoothing out top of half waves Yes or no?}

Again full wave reg still way to go, with half wave might see 50c/s ripple on led, and even unsmoothed
full wave would give a 100c/s ripplebut the ye might not see it. What do you think?

Did you look at current limiter, I think its ideal in Bachmann diesels, not so good for carriages as can only source 20ma.

 

[Gofer]

QUOTE (MUSO52 @ 3 Jul 2007, 20:03) <{POST_SNAPBACK}>Steve.

Thanks for the sketches, look like bras on a washing line! (only joking).

Fully understand what you are trying to illustrate, (must learn to upload sketches).
Maybe I wasn't fully correct after all, but on the right 'track'!
My understanding now is thus. With ref to the half wave + led, When the supply to the reg diode is +v
with respect to the -ve on the other side of the led, both forward biased, current and volt drop and led lights as we know. With Ov on the front end of reg diode, 10v will be at Neg leg of led. No current flow, reversed biased. However, this voltage would not break down led as this same potencial will also be on the pos leg of led. There being no current flow, the Half wave rectifing diode ( now with a 0v on its pos leg) will not conduct providing this voltage do'es not exceed the Vr or breakdown voltage of the diode. However if we exceeded this Vr, then the p-n juction will faill, current will flow and destroy both diode and led. Avalanche effect, I think its called. I think this is what you tried to explain in your sketches, Yes

The junctions cannot fail because there's effectively a PN junction between a PN junction which can never get a potential so can it cannot fail (within the limits of sensibility of course). The reverse potential is on the P of one device and the N of the other device.

Steve

 

[double00]

Will my made up circuits require the resistors valued at 1k require changing to lower rated resistor? if this be the case will require twelve resistor changes

 

[Gofer]

QUOTE (MUSO52 @ 3 Jul 2007, 20:03) <{POST_SNAPBACK}>Back to DoubleO.s Circuit.

The full wave reg is the way to go as u say but one thing I note from sketch, would not the output be closer to 10v than 5? Minus a drop across the F/W reg of course.

No. The diode will only conduct on one half of the waveform so although I'll agree the voltage won't be 5v (I did say the voltage will need to be measured), it won't be anywhere near 10 volts either. If a smoothing capacitor is included in the circuit then yes, the voltage will rise to somewhere near the peak voltage of the supply but whether that peak voltage could be maintained would depend on the values of the capacitance and the current the load is taking. If a full wave or bridge rectifier was used then the voltage even without smoothing would probably be somewhere around the 10 volt mark but of course there would now be a volt drop across two diodes as opposed to one in a half wave circuit. Again, if the circuit was smoothed the peak voltage would be realised.

Steve.

PS. I think you've been looking at too many washing lines.

 

[Gofer]

QUOTE (double00 @ 3 Jul 2007, 20:42) <{POST_SNAPBACK}>Will my made up circuits require the resistors valued at 1k require changing to lower rated resistor? if this be the case will require twelve resistor changes

DoubleOO, with the greatest of respect I think you're making too much of this resistance lark. LED's by there very nature will not display a marked difference in brilliance even if the value of the resistance is changed appreciably. The important thing is to err on the side of safety and ensure that not too much current passes through your LED's. The 1k ohm resistors will be fine but if you're concerned I'll include a dozen (why a dozen? I thought you only had six LED's) 470 ohm free of charge when I send you the rectifier unit you choose.

Steve.

 

[MUSO52]

No. The diode will only conduct on one half of the waveform so although I'll agree the voltage won't be 5v (I did say the voltage will need to be measured), it won't be anywhere near 10 volts either.

Surely, if the transformer was 10 volt peak to peak then I would agree, on one half of the waveform 5v.
But I would believe that the tranny is 10RMS and therefore on one half cycle 10v rms equates to about 13.5v peak value yes?

MUSO