daft question on LED indicators and high VDC

This might be daft but I'm trying to wire two LED indicators (http://www.farnell.com/datasheets/629825.pdf) and I'm struggling a bit. I've got one which is 230VAC, this is pretty easy as it's simply just wires directly into a 230VAC source, which in my case powers an SPS705 68VDC PSU. The other one was supposed to be used as a DC 'power on' indicator for the 68VDC PSU, so I wanted to wire the 68VDC into a 24VDC version of the indicator. This is what's proving tricky as the voltage is plenty to just dissipate over a resistor. Any ideas on what can be done to drop the 68VDC to 24VDC so it's safe to use on the 24VDC LED indicator?

Regards,
T.

Well a series resistor could be used.. 68-24 = 44v @ 20mA = 44/.02 = 2200ohm and 44 * .02 = 0.88W so need a 2200ohm, 2W resistor.
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Alternatively could use a 43v zener diode like this which might actually be cheaper since can be bought 1 off whereas resistors generally need to be bought 10 off. When using the zener the end with the band goes to the + side of the supply. Get it the wrong way round and both zener and LED will fry faster than you can say magic smoke!
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For heat dissipation purposes the zener should ideally be mounted on a bit of single sided circuit board approx. 60mm x 30mm. Make a single 2mm cut across the copper 40mm from one end to create two electrically separate areas. Form the leaas of the zener so it sits flat and solder the lead from the banded end to the larger area, keeping the lead full length. the body of the zener should just straddle the cut. Solder the other lead to the smaller area. If possible mount the board vertically to aid convection.

These usually take between 12mA and 20mA, as you say a series resistor would do.
Say it's 12mA
R=V/I
(68-24)/0.012 = 3.7 k ohms
watts = IxIxR = 0.54 watts

Say it's 20mA
R=V/I
(68-24)/0.02 = 2.2 k ohms
watts = IxIxR = 0.8 watts

So I'd start with a 3.7k 5watt resistor and measure the voltage across the lamp, if it's too low then try a lower value resistor.

Ah ha, Irving just beat me to it :encouragement: I chose 5 watts having come through the 'brick shit-house' school of engineering.

Thanks for the replies, I was thinking resistor, but that's a lot of watts, so thought that maybe there's a better idea. Another way of approaching this is ditching the 68VDC and getting a 42VDC as this is for a pretty small Nema 23 (SY57STH56_2004A) with a DM856 driver, this would then mean less dissipation needed, although also means more monies to spend. With the Zener costing 36p I might just give that a go and leave the 68VDC PSU:)

Cheers,
T.

Quote:

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Originally Posted by irving2008

Alternatively could use a 43v zener diode like this which might actually be cheaper since can be bought 1 off whereas resistors generally need to be bought 10 off. When using the zener the end with the band goes to the + side of the supply. Get it the wrong way round and both zener and LED will fry faster than you can say magic smoke!

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Irving, good idea using zener but should the connection go like this ?

supply +ve to lamp +ve
banded end of zener to lamp -ve
plain end of zener to supply -ve

I thought there's two ways:

http://www.mycncuk.com/attachment.ph...id=13490&stc=1

I might be wrong though:)

T.

Quote:

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Originally Posted by dsc

Thanks for the replies, I was thinking resistor, but that's a lot of watts, so thought that maybe there's a better idea. Another way of approaching this is ditching the 68VDC and getting a 42VDC as this is for a pretty small Nema 23 (SY57STH56_2004A) with a DM856 driver, this would then mean less dissipation needed, although also means more monies to spend. With the Zener costing 36p I might just give that a go and leave the 68VDC PSU:)

Cheers,
T.

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Or you could use one of these:- http://www.amazon.co.uk/99-9V-Digita.../dp/B008KXMNTU ..Clive

Quote:

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Originally Posted by dsc

I thought there's two ways:

I might be wrong though:)

T.

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Umm.. the first way is correct - the resistor and zener are interchangeable in this diagram, just different ways to drop the volts.

The second is unnecessary as uses two components when one will do. LED doesn't need regulated supply so the zener is unnecessary.

Quote:

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Originally Posted by EddyCurrent

Irving, good idea using zener but should the connection go like this ?

supply +ve to lamp +ve
banded end of zener to lamp -ve
plain end of zener to supply -ve

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You could, but convention is that indicators always have one side grounded as often, in times past, the panel was the ground.

Quote:

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Originally Posted by Clive S

Or you could use one of these:- http://www.amazon.co.uk/99-9V-Digita.../dp/B008KXMNTU ..Clive

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You could, but a pain to mount, unless you're CNC'ing the panel and overkill :lol:

Quote:

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You could, but a pain to mount, unless you're CNC'ing the panel and overkill :lol:

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Mount like thishttp://www.mycncuk.com/attachment.ph...id=13491&stc=1 ..Clive

Quote:

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Originally Posted by irving2008

Umm.. the first way is correct - the resistor and zener are interchangeable in this diagram, just different ways to drop the volts.

The second is unnecessary as uses two components when one will do. LED doesn't need regulated supply so the zener is unnecessary.

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The second provides a solid voltage on the zener regardless of the load, which is different to a standard resistor divider. I'll go with the first one, it's 36p for the Zener vs. 9p for a resistor but as you said before minimum order quantity is 10 (doesn't really matter as I need to go over £5 for the order anyway).

Those 7seg displays are pretty neat considering the voltage range on them. I'll have to stick with a small LED indicator as I'm tight on space as it is.

T.

Just a quick update, I've tried both the resistor and the diode, both get rather hot, although the diode seems to handle it better. Done some checks with a TC and the resistor sits at around 80degC whereas the diode goes up to around 60deg. The ideal solution would be to change the PSU to a 48VDC, but with the current solution costing less than a pound I don't think I'll bother:)

T.

Did you mount diode on a bit of board as I suggested? It's running a little hotter than I calculated. Junction temperature will be around 120 degC, still less than the maximum allowed of 150 degC, but 20 degC more than I'd like.

Hi you say your wanting to connect a LED to a 68VDC PSU. you can use a LED with a resistor in series If your machine is on permanently, I would suggest to drop the current to 15mA, LEDs work at 20mA but run constant shortens their life expectancy. If your LED is a 24 volt version check the spec on it the resistor already in situ may already drop the current to the LED to 12~15mA, just remember the power supply you are connecting to will either be PWM or linear my guess PWM either way the voltage (68VDC) will be constant

This was only a quick test with a choc block, I'll need to cut up a piece of laminate and give it another go.

T.

Hi, you should not need a diode your power supply will be polarized +ve & -ve your problem is you are trying to drop 44 volts which creates a lot of heat, if your voltage was a bit lower you could have used a LM317 or similar the best you could input to these would be 32vdc its a weird voltage drop to are trying to use my suggestion is to try to get hold of a NUD4011 http://www.onsemi.com/pub_link/Collateral/NUD4011-D.PDF
this should handle your voltage drop

Quote:

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Originally Posted by Binary

Hi, you should not need a diode your power supply will be polarized +ve & -ve your problem is you are trying to drop 44 volts which creates a lot of heat, if your voltage was a bit lower you could have used a LM317 or similar the best you could input to these would be 32vdc its a weird voltage drop to are trying to use my suggestion is to try to get hold of a NUD4011 http://www.onsemi.com/pub_link/Collateral/NUD4011-D.PDF
this should handle your voltage drop

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Hi and welcome to the forum. Have you read this complete thread because I think you have missed the point and purpose about the diode also all the math has been done. I am not trying to put you down in any way but this guy has stated he wants a simple way and he is short of space:very_drunk: ..Clive

Thank you for the welcome. I understand he has a problem with space but if you are using a 5Watt resistor you need a lot of space the chip i suggested can be mounted on a 15mm x 8mm pcb depends how you proceed, that would include a smd resistor also i use its brother the nud4001 so i know how small it is, the chip is a current limiting chip that is why it can be used with such high voltages up to 200Volts input as for the price i pay around 30p for the nud4001 the price for the nud4011 is a little higher..its only a suggestion he does not have to use it but many brains make light work lol i'm not sure it may even fit on the pads of a piece of vero board,
One other point " a LED is a Diode" a zener can be used to "clamp" the voltage" you just need to current limit it, I also have another chip in mind and the size of that is only 3mm looks like a smd diode 2 pins thank you clive

I've made a small PCB 62mm x 40mm, it's got a larger section of copper on the banded area, roughly 38mm x 40mm. Just ran some tests and it levels at around 56degC, so better than before, but not massively.

I do like the NUD chip, the issue is it's SMD (can't solder those) and for 68VDC and a single LED it's over it's power limit anyway. I guess I might have to get that 48W PSU anyway...

Cheers,
T.

BTW. I'm not limited for space, I've got loads:)

Quote:

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Originally Posted by dsc

I've made a small PCB 62mm x 40mm, it's got a larger section of copper on the banded area, roughly 38mm x 40mm. Just ran some tests and it levels at around 56degC, so better than before, but not massively.

I do like the NUD chip, the issue is it's SMD (can't solder those) and for 68VDC and a single LED it's over it's power limit anyway. I guess I might have to get that 48W PSU anyway...

Cheers,
T.

BTW. I'm not limited for space, I've got loads:)

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This is what you quoted [Those 7seg displays are pretty neat considering the voltage range on them. I'll have to stick with a small LED indicator as I'm tight on space as it is.] This is why I mentioned that ..Clive

Should've been more clear, I've got no space on the front panel, but loads in the guts, which basically means I need to stick with a single panel mounted LED and whatever internals it needs.

T.

I didn't think of using a voltage regulator at first, given the OP's requirements, but it's a good idea.

If its because the chip i mention is a smd can i say don't be frightened to try just make sure your not using a blow lamp to solder with, I have soldered these i.c.'s using a soldering iron, there are a few ways of removing the solder if you mess up, solder sucker for one, then check by running a stanley blade between the contact legs to make sure no "whiskers of solder is left".
Can i say if you can buy leds at a good price like Meeee. if your using the LED panel light which will have a resistor in using red leds as a zener could be good. Red LEDs have a 2.2 volt drop times this by 10 in series, yes you've got it 22 volts so 20 leds will give you 44 volt drop with your 24 volt panel light also in series your resistor in the panel light current limiting you leds you will have a 68 volt led ...and a lot of additional light ..but no heat
this is food for thought.

I can see that your going to be another valuable member of this forum. :beer: ..Clive

Most of the thinking seems based on 20ma current through the LED. It might be worth experimenting with resistor values that give about 5ma and see if the brightness of the LED is acceptable. So then a single series resistor would dissipate less power and could be a cheaper component.
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Just had a play on the breadboard and LED in series with 4 x 4.7k ohm (18.8k) still gives a reasonable light. I tried it on 30v 15v and 5v (5v is probably too dim). I used a standard 3mm red in a clear capsule LED.

Cropwell has a good point and running the LED at a lower current reduces the heat generated - assuming sufficient brightness.

You can't change the laws of physics - any linear analog solution needs to dissipate 0.88w @ 20mA LED current and using more esoteric devices doesn't change that. The NUD chip is a great solution for driving strings of LEDs and/or where input voltage varies. It's simply overkill for this requirement and adds unnecessary complexity. Incidentally the NUD chip would need 2sq in of copper (the same 40 x 30mm board) to stay within ratings and from the data sheet would run a junction temperature of 113degC and a lead temperature of 46degC due to marginally better thermal characteristics than the diode, but for 10degC difference its not worth the hassle.

@binary. Your assertion that a string of LEDs would generate no heat is wrong. LEDs are around 50% efficient and do generate heat in the junction. They don't emit IR like incandescent bulbs so don't feel hot through radiated heat. But a standard 5mm 2.2v red LED run at 20mA would consume 0.02 x 2.2 = .044W of which 50% is light and 50% is heat. A string of 20 would indeed drop 44v but would still generate 4.4W of heat, the other 4.4W being light energy. But bulky overkill !!!

Re. putting more diodes in series, maybe this is the next trend for control cabinets :welcoming:
http://www.custompcreview.com/review...eview/12513/3/

Btw this is the LED I'm using:

http://cpc.farnell.com/apem/q8f3cxxr...56?Ntt=SC09356

Get a grip !!!

Quote:

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Originally Posted by EddyCurrent

Re. putting more diodes in series, maybe this is the next trent for control cabinets :welcoming:
http://www.custompcreview.com/review...eview/12513/3/

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Set me thinking though, I have a reel of white smd Led's bought for a job that never happened.:devilish:

Maybe this thread would be useful ? :joker: :hysterical:
http://www.instructables.com/answers...er-dc-voltage/

I have just made and installed a control panel for a glass blower friend. The labels for the modules are fibreglass circuit board. If you put LEDs behind them the lettering shines though beautifully. Next phase of project ??? I might even go full steampunk. maroon and green with gold striping and ornate Ibis carvings in the blank areas of the panels and , of course I would have to change the stainless metric button head screws to whitworth bronze slothead.

http://www.mycncuk.com/attachment.ph...id=13619&stc=1
The picture btw is a screenshot of an IP camera set up to view the panel from anywhere.

Nice work on the furnace control, just for those still not disillusioned with LEDs have a look at http://www.onsemi.com/pub_link/Colla...I45020JZ-D.PDF http://www.onsemi.com/pub_link/Colla...SI45030A-D.PDF and there are versions of this with different current and voltages
I'm still trying to work out how to use a illuminated led as a switch if any one has done this let me know

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Originally Posted by Binary

I'm still trying to work out how to use a illuminated led as a switch if any one has done this let me know

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What ? do you mean with an LDR or something ?

I think you might mean as a touch switch - I presume you could use a metal cased led on an insulating panel as the sensor on a touch switch circuit
https://www.google.co.uk/search?q=to...55%3B761%3B504

Pick one !

no, not a metal can version a basic 5mm LED .....my thoughts the LED stays on, capacitance on one of the LED pins is the switch so just touch the resin or get close and the switch action operates the led to turn off or any output can be turned off

You won't get enough pick up from the LED lead to give you a reliable operation, if you ramp up the sensitivity you will get switching from any electrical noise, so again it won't be reliable.

I don't want to mislead you this has been done, how its been done and what with I'm not sure
I was going to use an Arduino which will easily give a output for a LED it can also be used as a sensor
the method in combining both switch with LED would make or break it.
this is for only a prototype, then code for 8052