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I would appreciate help on working out values of resistors for LEDs.
Voltage 10v, required to drop down to 5v - 4.5v. Using the table of 0.02 as divider, to calculate value of 1/4 watt resistor.


5mm LED 5v should require a 260R resistor - meter reading 260R with 10v = 6.5v
470R with 10v = 6.0v
590R with 10v = 5.6v
To save further buying of resistors, what value of resistor is required to drop voltage to under 5v.
 

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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.
 

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Discussion Starter · #3 ·
10v fed into resistor, meter read across resistor. LED is a Maplin 24BY 5v Super Bright Blue - spec has a 5v reverse.
Incorrect resistor value has blown several white clear LEDs - reason for requiring correct value. One source stated that a 1/4 watt 470R resistor be used - this shows a reading in excess of 5v.
 

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Double00, most LED's like around 2v - some a bit higher but all should work well on lower voltage.
Some data available has these figures for LED with 2v, 20mA
6v - resistor od 220 ohms
9v = 330 ohms
12v = 560 ohms
24v = 1200

I use 9.6 volts & with bright blue LED's use 3000 ohms & they still are OK.
My normal red/green LED's - I use 680 ohms. I increase the resistance to protect the LED & they still have sufficient illumination

The formula is R= (E-Vf) x 1000/i where R= resistance; E = DC supply voltage; Vf = forward voiltage drop of LED - typically 2v & i = LED current in mA
 

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doubleOO, I think the reason you are getting conflicting voltage readings across the LED is because you are using a 5v LED. By using a 5v LED this means that you are using an LED to which the manufacturer has already fitted an integral current limiting resistor making it suitable for operation on 5 volts. What you don't know however is to what value the manufacturer has limited the current through the LED. You are basing your calculations on the LED passing current are 20mA (the divider value of 0.02 you quote) whereas the 5v LED may (and probably is) taking less current than that which means your additional current limiting resistor is going to drop less voltage than you expect. To be able to accurately calculate the value of your addition series resistor for operation on 12volts you will need to measure the current the 5v LED is taking when connected to a 5volt supply and then use that figure as you divider (as you call it). To be honest though this is all a bit academic as the important thing is that you don't allow too much current to pass through the LED ie. keep it below 20mA. As SOL has said, a normal LED usually requires about 2volts and for use on a 12volt supply the calculation would be 12-2=10/.02=500ohms but as 500 is not a preferred value a 560ohm resistor would normally be used. However, as its not a good idea to drive an LED at its maximum I would suggest using a divider (again your word) of .01 (1mA) which will give a resistor value of 1000ohms (1kohm). The difference in brilliance will be barely noticeable and the LED will last just about for ever. In view of all this, I would simply put a 1kohm resistor in series with you 5volt LED and see what it looks like in the safe knowledge you're not going to do it any damage.

As regards the reverse voltage figure, unless you're likely to reverse the DC supply to your LED or attempt to operate it on an AC voltage I would disregard it.

Steve
 

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Discussion Starter · #6 ·
What I am contemplating in doing is to reduce KPC 10v AC rating transformer output to 5v AC.
I have tried several rated 1/4 watt resistors from 200R to 580R, none of these values reduce the voltage below the required value of 4v-4.5v (voltage reading taken across the resistor).
One resistor that I presume to be correct rating of 1.2K though the coloured rings are rather hard to decipher - brown/red/black and perhaps another black coloured ring, reduces the voltage to 4.5v.
 

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doubleOO, a resistor with the bands brown/red/black will have a resistance of 12 ohms (the forth ring indicates tolerance rating). See this site if it helps : http://www.uoguelph.ca/~antoon/gadgets/res...rs/rescalc.html

What's concerning me now is that you are supplying your LED from an AC source which is of higher voltage than the maximum voltage your LED is designed to withstand. This means you should be incorporating a diode to protect the LED from the reverse voltage as the series resistor won't. The reason for this is that the resistor only has voltage dropped across when it is passing current, hence the calculation you do to work out what value resistance to use. Current will only be flowing in your circuit when the LED is forward biased. When an LED is reversed biased (as it will be for 50 percent of the time when connected to an AC voltage) the current through the LED is effectively zero so the voltage dropped across the series resistor is also effectively zero. This means that your LED will see the source voltage when it is reversed bias and it is more than likely this which is damaging your LEDs and not excessive forward current.

Steve.
 

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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?
 

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MUS052, the diode could be either parallel to the LED (as you say, connected in reverse) so as the voltage is still dropped across the resistor or simply wired in series with the rest of the circuit (in which case it would need to be in the same bias as the LED). If in series it might be worth remembering that some voltage will be dropped across the diode (.7v) when working out what value resistor to use of course. Personally I'm not in favour of the capacitor idea if only because it starts to complicate the issue somewhat.

Steve
 

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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
 

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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!!!!!
 

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Hi Mus

I must admit I didn't bother looking at the Maplins site and just took doubleOO word on the spec. As you say, it does seem that the N24BY is not a 5v LED at all. It also is apparent that it has a max forward current of 30mA and not 20mA so calculating the resistor for 20mA is safe enough. As regards the use of a series diode, it won't matter as I see it whether it's in the negative or positive lead of the LED as providing it's reversed biased on the same half cycle as the LED, it won't conduct and so protect the LED. Having said all this however, the parallel configuration is probably the better option. I agree with you that as voltage source is 10vac, the current limiting resistor should be calculated on 5v and not 10v. As a aside, I personally never feel happy running LED's from AC as it means they are constantly switching on and off and would much prefer to use a simple rectifier circuit complete with a fixed voltage regulator (78S series) to feed the circuit, but each to our own. I've no idea what doubleOO wants these LEDs for - he may feel moved to enlighten us after reading these posts. I just firmly believe the simpler it can be kept the easier it will be for everyone to understand (me included!).
 

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DoubleOO,

I'm sure you've read the above posts with interest if not a certain apprehension. Personally I think perhaps the simplest and best way forward for you is to halfwave rectify the output from your transformer with a single diode (1N5401) as this would then give you an approximate 5v DC source to work from. I say approximate because the actual AC voltage output of the transformer your using is really unknown although I know it's rated at 10vAC. In view of this the actual DC voltage would need to be measured after fitting the rectifier diode and before carrying out any resistor calculations. I realise this DC output would be unsmoothed and very choppy but at least the polarity would be constant and your LED's won't have to cope with reverse polarity issues. For the LED in question I see from the Maplins site that the typical forward voltage is 3.2 volts and the maximum forward current is 30mA. If we therefore adopt the 3.2 volt but use a forward current of 20mA rather than drive the LED flat out at 30mA the device should be perfectly OK. Using these parameters and assuming the halfwave rectified DC voltage is in fact 5 volts, the calculation for the resistor is now 5-3.2=1.8/0.02=90 ohms. The power rating of this resistor need be no more than .25 watt.

Steve
 

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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
 

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Discussion Starter · #15 ·
Gentlemen, no red herring was intended - Maplins staff member stated that the N24BY LED was rated at 5v.
I have now put into circuit a 1.2K 1/4 watt resistor, this has now reduced transformer 10v to required 4.5v.
Resistor coloured rings red/brown/black reads 120R and red/brown/black/black 1200R as according to my colour coding.
You suggest 1N5401 diode be connected between transformer and resistor as a precaution against burning out the N24BY LED.
The N24BY LEDs are used to illuminated various buildings on my layout.
All information given by you is most appreciated.
 

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QUOTE (MUSO52 @ 1 Jul 2007, 21:10) <{POST_SNAPBACK}>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.

Hi Mus

Tell me more about the constant current devices. I've not used them but I'm game to give them a go. Do you have any type numbers? I take when you say they're the same as a BC108 do you mean they're in a TO92 package?

Steve.

PS. I reckon your wrong about the series diode. No matter where it is in the circuit it will effectively act as a halfwave rectifier on the source and protect the LED from the reverse voltage.

Hi doubleOO

Your resistor colour coding seems a bit odd. Resistor can have three or four coloured bands to denote their resistive value (ignoring the tolerance band). Using the colours you quote, black = 0, brown = 1 and red = 2 so a three band resistor with the bands red, brown, black would have a value of 2(red), 1(brown) and 0(black) = 21 ohms. A four band resistor having bands red, brown, black, black would have a value of 2(red), 1(brown), 0(black), 0(black) = 210 ohms. For a three band resistor to have a value of 120 ohms the colours would be brown(1), red(2) and brown(1). For a four band resistor to have a value of 1200 ohms the bands would be brown(1), red(2), black(0) and brown(1). The last band is a bit confusing I know as it really denotes the number of noughts added ie. if it's black no noughts are added, if it's brown 1 nought is added, if it's red 2 noughts are added, if it's orange 3 noughts are added, etc, etc,.

Steve.
 

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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
 

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Discussion Starter · #18 ·
The output from KPC transformer is 10v - resistor is connected to the + side - meter + is connected to the other end of resistor and negative of meter to negative of transformer. A reading of meter shows 4.5v.
 

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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
 

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QUOTE (double00 @ 1 Jul 2007, 23:21) <{POST_SNAPBACK}>The output from KPC transformer is 10v - resistor is connected to the + side - meter + is connected to the other end of resistor and negative of meter to negative of transformer. A reading of meter shows 4.5v.

DoubleOO, this method will give you an incorrect reading. The voltage drop across resistor is dependant not only on the value of the resistor but also the current passing through it. The value of the resistor is fixed but in the configuration you've described above the only load drawing current is the meter, which is likely to have a relatively high impedance (resistance) - much higher than the LED you intend to use in fact. This means that less voltage will be dropped across the resister when using the meter as the load than would be if the LED was in circuit. This is why the value of the resistance must be calculated from the known (or desirable) current passing through the LED and the amount of voltage which needs to be 'lost' from the source. For example, if the source voltage is 5v DC and the LED requires 3.2 volts across it it will be necessary to 'loose' 1.8 volts. If the maximum current the LED can safely pass (or the current we want it to pass within it's specified limits) is 20mA then using Ohms Law formula R=V/I the value of the resistance comes out at 90 ohms (R=1.8/0.02).

Hope this all makes some sense.

Steve.
 
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