[MUSO52]

Have a look at this site. http://www.kpsec.freeuk.com/components/led.htm. This may be better than me trying to explain!! If you have any specific problems in the use of led's then come back. 1 thing to note in their use however, not all leds have the same volt drop across them. For example, white ones require more voltage than red ones. This is due to the frequency of the light emited and the led's construction. Have you noticed how yellow led's light before white ones on DC diesel engines? Another thing of note is the beamwidth , which relates to angle of light emited. Read the specs for each led before purchase!

Where did you place your meter, across the led or resistor?! Also note that resistors normally have a (typical) tolerence of 10% although 5% ones can be purchased.
Hope this of use.

 

[MUSO52]

Gofer (Steve) is absolutly correct, but we must remember to connect the protecting diode in reverse to protect the LED. This 'protecting' diode could of course be a similar led and (again connected in inverse parrallel) with each illuminated on alternate half cycles. Steve, (if I may call you so) would it not be better to use a capacitor in this instance? We know the volt drop and current draw of his Maplin N24BY,
We can calculate the Xc (reactance) at an assumed 60hz and use this to drop voltage across the led (s)
With the simple inclusion of a resistor (might need to be .5w) to limit in rush to cap might be the way to go. What are your thoughts?

 

[MUSO52]

Steve, agree with what you say, but I would imagine that you would place, say, a IN4001 AFTER the led as this device can tolerate a higher reverse voltage (20v?) when the circuit is in the neg cycle. To place it before the led would still leave it (the led) suseptible to the full reverse voltage( - ) when there is no current flow, i.e. reversed biased. Otherwise the only thing the additional diode is doing is dropping the voltage by .7v No?

Personally, if I were doing this I would probably still use a capacitor but can understand the complexity issue from Double O's position. Maybe still better to use two leds in adverse parallel.
What does he want them for anyway??

MUSO52

 

[MUSO52]

And one more thing Steve, Double O threw in a red herring! The rated max Vf of the Maplin N24BY
is 3.6v and not 5v as was implied in his original post. Again, since he is sourcing his led from an ac supply, should we not halve this voltage when calculating series limiting resistors as it is only conducting in half the cycle? i.e. the voltage swings from -5 to + 5 which = 10v.

We'll sort it Double O! Bet you wished you never asked!!!!!

 

[MUSO52]

Hi steve and DoubleO,

On the whole I agree with Steves last post with however one reservation. (sorry!).
Using half wave rectification using a single diode is acceptable where the load (RL) is a resistive one, or possibly inductive, a dc motor maybe.
An led is, by its very nature is a rectifier in its own right! Its a diode which emits light --- yes? And like an ordinary diode will only conduct in one direction.

So, my only concern is still the reverse breakdown voltage of the N24BY, a known Maplin spec of 5v. As we stated earlier with no current flowing in the circuit, irrespective of where the series limiting resistor is placed, this reverse voltage of approx 5v is too close to the breakdown voltage of the N24BY.

Therefore Steve I still think it prudent to:

a) place the additional diode AFTER the led where its increased Vr threshold ( assuming the diode you stated has a higher Vr than the led), will act as a barrier to protect the LEDor,

place additional diode in parallel (but reversed) with led, and thus bypassing it when the dc is in neg half cycle. Using this we should remember to take out the volt drop across this additional diode (0.7v) when calculating series limiting resistor.

Sorry Steve, maybe I'm wrong , but I think that using the additional diode the way you stated do'es not negate the reverse polarity issue , Its still there. With no current flowing the assumed 5v of the neg cycle would still be present at the diode when its reversed biased. No?

Anyway, this tranny of double O's , 10v peak to peak, mean, rms, center tapped.... WE DONT KNOW!!
So because of this maybe we should err on the side of caution. That breakdown voltage of the led is too close to the neg value of approx 5v and is still a concern.

So Steve, what have you been doing today? Like me I reckon, stuck in doors cos its P*****g down!
Wheres Double O? Out in the sunshine I expect!! Well if he reads these posts when he gets back, he will probably go and buy some batteries for his LED's!! Come on Double O be more specific, what do you want a super bright blue led for anyhow?!? Let us know what application you want for your led then we can be more specific in sorting out the probs. ( agree Steve?)

Steve, you are quite right about the use of 78 series regulators and they are my prefered choice when sourcing more than one led. However have you used a constant current device? These are ideal led drivers , up to a max of 20ma. Irrespective of voltage change, and hence speed of an engine using DC, these will only source a constant current depending on the value of one additional resistor. I believe that Hornby use these devices in their current diesel locos. Although I have not opened up my Hornby loco's, these devices, which are in the same package as a BC 108 transistor, can be seen on the engine pcb. Basically, once the led reaches its full working voltage, the light remains constant irrespective of controller voltage, under dc control. However, my Bachmann diesels do not have this led circuit. The brightness varies with dc controller voltage, and looking at the pcb on them , only have a series limiting resistor. Hope this, as an aside, is of interest.

WHERES DOUBLEO?? Given up on his LED circuit and gone down T' Pub!!

MUSO 52

 

[MUSO52]

Many thanks for coming back so prompt DoubleO!
Forgive the reference to a red herring, no disrespect to you , sorry... But the stated working voltage did throw Steve and I a little. With the greatest of respect to Maplins staff, though very helpfull, in this case they got it wrong. It is up to us to fully understand the specs. of the components we wish to use, and there inteded application, so again I mean no disrespect to your goodself, as you looked to us (on the forum) for help in this matter.

So, is the circuit and led working? Great. But for how long?

You stated that your chosen resistor value dropped the voltage to 4.5v How was this measured?
Ok, I know with a multimeter, but where and how? I think this is important - we do not want to burn out anymore leds!! The reverse voltage implications are important.

Resistor codes.

Take a look at the following http://xtronics.com/kits/rcode.htm This will be of help in future projects.

Again, and with respect, you are wrong somewhere. Red = 2, Brown = 1, did you type in wrong order for value?

Now we know your intended application for these leds, I am sure we can come up with something a little more effeicent and cost effective. You still might blow them leds!

MUSO52

 

[MUSO52]

HHi Steve ( and dubleO) Notice the spelling? Sorry guys, had a few cans, all kids and grand children round for day ( lunch inclusive) Thats stressfull! (But I luv 'em) Start a new job in morning also, a company I've worked for 3 times before! I go to the company which wins the contract !! Never out of work though!! I know the companies product ( I developed it) Soory, after a few cans hope you see the humour!!

About the constant current device. At this time of night ( and a few cans) I cant remember actual device ID, Think it is a L344z ( in the same package as the BC108), TP whatever. Will come back tomorrow and let you know details.

Lets agree to differ for now Steve on the Diode issue. To late to discuss!! We could go over to the RMweb forum and turn it into a argument I suppose!!!???!!! Forgive me, my sense of humour makes me laugh at least!! Anyway, apparently, the RMweb guys have a meeting planned shortly. I'm going, promisses to be a right punch up after what i have read recently.

Sorry lads, its late, tired, had a few cans, going to bed. Pse do not think bad of me...........

MUSO52

 

[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.

 

[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.

 

[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