[Gofer]

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

 

[Gofer]

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.

 

[Gofer]

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

 

[Gofer]

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!).

 

[Gofer]

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

 

[Gofer]

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.

 

[Gofer]

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.

 

[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

 

[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

 

[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.

 

[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.

 

[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

 

[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.