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Current Mirror for LEDs


M@rine_lover

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M@rine_lover

Hi all,

 

I have got this Current Mirror with Over Current protection circuit from website, ledsmagazine.com.

I found that it’s very useful and would like to in-corporate into my LED lighting to prevent current overdrive in case of some LED failure -_- .

 

Currently, I have one Meanwell LED driver of 27V, 2300mA on hand and will be running 2 strings of 6 LEDs in parallel. Should I follow the same circuit with same component value or should I need to change the value of the resistors to my configuration :o ? Value 680ohm and 1.5ohm to?

 

Please kindly advice :bowdown:.

 

Quoted from ledsmagazine,

"The addition of a small signal transistor as a current monitor protects the LEDs from being overdriven in the case of any LED failures. If LED1‐LED5 fail open circuit, then the current in the second string falls to zero as before. However, if LED6–LED10 fail, then the current increases in the first string until the voltage developed across the 1.5 Ohm emitter resistor reaches around 0.7V, thus turning on the BC337 transistor and pulling the base voltage of the power transistor to ground and limiting the current. With the component values given in the circuit, the measured current limit was 445mA with String 2 open circuit."

 

Recom_Fig4.jpg

 

more info on this circuit: ledsmagazine

 

Thanks.

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You shouldn't need to change the resistance values, but you do need to make sure the wattage of the 1.5 ohm resistor is adequate. The 680 ohm resistors are there just to bias the transistor.

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That circuit will NOT work at 1.whatever amps. Replace the 1.5s with ~.47s, the BD139's with something bigger (anything with a DC current gain over like 80 at 1A and a maximum continuous current of 2A or more) and the 680s maybe drop to 470... Now that I think about it this circuit kinda sucks, if you get a transistor with a lower end gain its going to eat up a buttload of efficiency.

 

I say just put a 1 ohm resistor in series with each LED string, measure to make sure there isnt a massive mismatch, and call it a day.

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Ocean Flyer
Hi all,

 

I have got this Current Mirror with Over Current protection circuit from website, ledsmagazine.com.

I found that it’s very useful and would like to in-corporate into my LED lighting to prevent current overdrive in case of some LED failure -_- .

 

Currently, I have one Meanwell LED driver of 27V, 2300mA on hand and will be running 2 strings of 6 LEDs in parallel. Should I follow the same circuit with same component value or should I need to change the value of the resistors to my configuration :o ? Value 680ohm and 1.5ohm to?

 

Please kindly advice :bowdown:.

 

Quoted from ledsmagazine,

"The addition of a small signal transistor as a current monitor protects the LEDs from being overdriven in the case of any LED failures. If LED1‐LED5 fail open circuit, then the current in the second string falls to zero as before. However, if LED6–LED10 fail, then the current increases in the first string until the voltage developed across the 1.5 Ohm emitter resistor reaches around 0.7V, thus turning on the BC337 transistor and pulling the base voltage of the power transistor to ground and limiting the current. With the component values given in the circuit, the measured current limit was 445mA with String 2 open circuit."

 

Recom_Fig4.jpg

 

more info on this circuit: ledsmagazine

 

Thanks.

 

It seem ok to me :) . May use your circuit if It work :D !

 

That circuit will NOT work at 1.whatever amps. Replace the 1.5s with ~.47s, the BD139's with something bigger (anything with a DC current gain over like 80 at 1A and a maximum continuous current of 2A or more) and the 680s maybe drop to 470... Now that I think about it this circuit kinda sucks, if you get a transistor with a lower end gain its going to eat up a buttload of efficiency.

 

I say just put a 1 ohm resistor in series with each LED string, measure to make sure there isnt a massive mismatch, and call it a day.

 

Is it a good practice for LED application B) ?

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Wow another vote that the circuit looks legit at 2300ma?

 

If built as shown, string 1 is going to be clamped at 460ma or so, read the blue paragraph as to exactly how. String 2 is going to take the rest of the current (1840ma or so) assuming there is enough headroom to cover the increased Vf at that current. Ok I take it back the circuit will work in that it lights up all the LEDs... Just make sure you heatsink the crap out of string 2 :)

 

You *need* the changes I noted above to scale this thing up and still have it balance the load properly.

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Upon looking a little further into this, you are right. You do need to change some parts. I don't think it's as drastic as you make it out to be though.

 

The point of this circuit is to save the remaining LEDs in the case of a failure. That could get quite expensive. The second itteration of the circuit on that page will keep either side of the parallel array within the set current range set by the smaller resistor (about 0.57 ohms to keep the current at no more than 1.2A), regardless of which side fails.

 

cptbjorn, everyone here is here to learn. If you know that something doesn't work, educate us as to why and how to fix it, or show us a better way to do it opposed to taking a somewhat condesending tone towards those of use that don't know. I won't say that I know everything. I learn new things every day. Help us learn something new ;)

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Sorry bro didn't mean to sound condescending, I just wanted to make sure someone didn't fry their $$ LEDs. I definitely appreciate all you do around here, I guess I'm just used to my real life where people specifically do NOT want to know the boring nerdy details of how I fix stuff.

 

Anyways here we go, full version. First of all the BD139 is going to have issues running at 1150ma. While a good device for the original circuit, Its gain drops drastically after 4-500ma and could be as low as 8-10 at 1150ma (down from 80 or so originally). For 1150ma at gain of 10 it will need 1150/10=115ma of base current through the 680 ohm resistor, which equates to a .115x680=78v! drop. We need a higher gain transistor or a smaller resistor (or both).

 

The second problem is what actually happens when the overcurrent protection kicks in. When the voltage across the 1.5R resistor goes above ~.7v, the BC337 starts conducting. This limits the current through the leds by increasing the voltage drop across the left BD139. The Meanwell driver will try to compensate to make the circuit "take" the full 700ma by pegging its voltage output to the maximum it can. If you have 10v of headroom, that BD139 is going to be dissipating 460ma * 10v, or 4.6 watts, which if not well heatsinked it will fail in seconds. When you scale it up by a factor of 3 like we are planning to there is a potential for a whole lot of heat when the protection kicks in. The two transistors should be screwed down right next to each other on a large heatsink (such as the one your LEDs are on) with an isolating thermal pad.

 

A quick mouser search found me the KSD1408, which looks like it should work much better at this current. Also it is in an insulated package so it can be mounted straight to the heatsink without an isolating pad, just some good thermal grease. It should have a gain of 65 or so, which means for 1150ma we want around 17ma of base current. I would shoot for 3v drop across the resistor, so R=3v/.017ma = 176 ohms. 180 should work just fine. The other resistors need to drop just under .7v at operating current so that the protection circuit works properly. .7v/1.15A= .608, I would use probably .5 ohms for these to allow for tolerances.

 

tl;dr I don't like this circuit and I'm sending the editor an email about all the issues with it, but if you must use it with a meanwell driver at 2300ma I would use .5R and 180R resistors and replace the power transistors with KSD1408s or similar and make sure to heatsink them well.

 

Also just using 2 separate drivers will be far superior IMO.

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Sorry bro didn't mean to sound condescending, I just wanted to make sure someone didn't fry their $$ LEDs. I definitely appreciate all you do around here, I guess I'm just used to my real life where people specifically do NOT want to know the boring nerdy details of how I fix stuff.

 

Anyways here we go, full version. First of all the BD139 is going to have issues running at 1150ma. While a good device for the original circuit, Its gain drops drastically after 4-500ma and could be as low as 8-10 at 1150ma (down from 80 or so originally). For 1150ma at gain of 10 it will need 1150/10=115ma of base current through the 680 ohm resistor, which equates to a .115x680=78v! drop. We need a higher gain transistor or a smaller resistor (or both).

 

The second problem is what actually happens when the overcurrent protection kicks in. When the voltage across the 1.5R resistor goes above ~.7v, the BC337 starts conducting. This limits the current through the leds by increasing the voltage drop across the left BD139. The Meanwell driver will try to compensate to make the circuit "take" the full 700ma by pegging its voltage output to the maximum it can. If you have 10v of headroom, that BD139 is going to be dissipating 460ma * 10v, or 4.6 watts, which if not well heatsinked it will fail in seconds. When you scale it up by a factor of 3 like we are planning to there is a potential for a whole lot of heat when the protection kicks in. The two transistors should be screwed down right next to each other on a large heatsink (such as the one your LEDs are on) with an isolating thermal pad.

 

A quick mouser search found me the KSD1408, which looks like it should work much better at this current. Also it is in an insulated package so it can be mounted straight to the heatsink without an isolating pad, just some good thermal grease. It should have a gain of 65 or so, which means for 1150ma we want around 17ma of base current. I would shoot for 3v drop across the resistor, so R=3v/.017ma = 176 ohms. 180 should work just fine. The other resistors need to drop just under .7v at operating current so that the protection circuit works properly. .7v/1.15A= .608, I would use probably .5 ohms for these to allow for tolerances.

 

tl;dr I don't like this circuit and I'm sending the editor an email about all the issues with it, but if you must use it with a meanwell driver at 2300ma I would use .5R and 180R resistors and replace the power transistors with KSD1408s or similar and make sure to heatsink them well.

 

Also just using 2 separate drivers will be far superior IMO.

 

Thanks for explaining...however, I'm not an EE and 70% of what you said is lost. I understand the purpose of using a current mirror in parallel strings of LEDs, and the general purpose of each component. I lack the "tools" to determine the accurate components. That said, and now that I have time, I'm going to set up the following an ELN-60-27 with two // strings of 8 Cree LEDs. This is the same PS as above, but only more LEDs. Does this change things?

 

What do you suggest. Evil, your advise/input is welcome too,

 

Thanks, Nick

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M@rine_lover
Sorry bro didn't mean to sound condescending, I just wanted to make sure someone didn't fry their $$ LEDs. I definitely appreciate all you do around here, I guess I'm just used to my real life where people specifically do NOT want to know the boring nerdy details of how I fix stuff.

 

Anyways here we go, full version. First of all the BD139 is going to have issues running at 1150ma. While a good device for the original circuit, Its gain drops drastically after 4-500ma and could be as low as 8-10 at 1150ma (down from 80 or so originally). For 1150ma at gain of 10 it will need 1150/10=115ma of base current through the 680 ohm resistor, which equates to a .115x680=78v! drop. We need a higher gain transistor or a smaller resistor (or both).

 

The second problem is what actually happens when the overcurrent protection kicks in. When the voltage across the 1.5R resistor goes above ~.7v, the BC337 starts conducting. This limits the current through the leds by increasing the voltage drop across the left BD139. The Meanwell driver will try to compensate to make the circuit "take" the full 700ma by pegging its voltage output to the maximum it can. If you have 10v of headroom, that BD139 is going to be dissipating 460ma * 10v, or 4.6 watts, which if not well heatsinked it will fail in seconds. When you scale it up by a factor of 3 like we are planning to there is a potential for a whole lot of heat when the protection kicks in. The two transistors should be screwed down right next to each other on a large heatsink (such as the one your LEDs are on) with an isolating thermal pad.

 

A quick mouser search found me the KSD1408, which looks like it should work much better at this current. Also it is in an insulated package so it can be mounted straight to the heatsink without an isolating pad, just some good thermal grease. It should have a gain of 65 or so, which means for 1150ma we want around 17ma of base current. I would shoot for 3v drop across the resistor, so R=3v/.017ma = 176 ohms. 180 should work just fine. The other resistors need to drop just under .7v at operating current so that the protection circuit works properly. .7v/1.15A= .608, I would use probably .5 ohms for these to allow for tolerances.

 

tl;dr I don't like this circuit and I'm sending the editor an email about all the issues with it, but if you must use it with a meanwell driver at 2300ma I would use .5R and 180R resistors and replace the power transistors with KSD1408s or similar and make sure to heatsink them well.

 

Also just using 2 separate drivers will be far superior IMO.

 

Hi cptbjorn,

 

Thanks for your explantion. I'm also abit lost as I'm new in this industry, however, I would like to try out.....!

 

Thanks again :P !

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Sorry bro didn't mean to sound condescending, I just wanted to make sure someone didn't fry their $$ LEDs. I definitely appreciate all you do around here, I guess I'm just used to my real life where people specifically do NOT want to know the boring nerdy details of how I fix stuff.

 

Anyways here we go, full version. First of all the BD139 is going to have issues running at 1150ma. While a good device for the original circuit, Its gain drops drastically after 4-500ma and could be as low as 8-10 at 1150ma (down from 80 or so originally). For 1150ma at gain of 10 it will need 1150/10=115ma of base current through the 680 ohm resistor, which equates to a .115x680=78v! drop. We need a higher gain transistor or a smaller resistor (or both).

 

The second problem is what actually happens when the overcurrent protection kicks in. When the voltage across the 1.5R resistor goes above ~.7v, the BC337 starts conducting. This limits the current through the leds by increasing the voltage drop across the left BD139. The Meanwell driver will try to compensate to make the circuit "take" the full 700ma by pegging its voltage output to the maximum it can. If you have 10v of headroom, that BD139 is going to be dissipating 460ma * 10v, or 4.6 watts, which if not well heatsinked it will fail in seconds. When you scale it up by a factor of 3 like we are planning to there is a potential for a whole lot of heat when the protection kicks in. The two transistors should be screwed down right next to each other on a large heatsink (such as the one your LEDs are on) with an isolating thermal pad.

 

A quick mouser search found me the KSD1408, which looks like it should work much better at this current. Also it is in an insulated package so it can be mounted straight to the heatsink without an isolating pad, just some good thermal grease. It should have a gain of 65 or so, which means for 1150ma we want around 17ma of base current. I would shoot for 3v drop across the resistor, so R=3v/.017ma = 176 ohms. 180 should work just fine. The other resistors need to drop just under .7v at operating current so that the protection circuit works properly. .7v/1.15A= .608, I would use probably .5 ohms for these to allow for tolerances.

 

tl;dr I don't like this circuit and I'm sending the editor an email about all the issues with it, but if you must use it with a meanwell driver at 2300ma I would use .5R and 180R resistors and replace the power transistors with KSD1408s or similar and make sure to heatsink them well.

 

Also just using 2 separate drivers will be far superior IMO.

 

Thank you for the explanaition. This makes a lot more sense now that it is explained out completely, and I see the flaws with it. It's still useable, but definitely not as simple as the author of the article made it out to be. It can still be implemented effectively, but adding it to the heatsink adds a little more headache to the matter.

 

Npain316, if you are going to run that driver, use the values of components that cptbjorn listed at the end of his post. This will set your max current to a little over 1A. Make sure the resistors are sized correctly. The 0.5 ohm should be 1W or greater, and the 180 ohm can be a 1/4W.

 

This is one reason why I only sold the ELN-60-48 in the group buy so you can run as many LEDs in series as possible at 1A.

 

 

cptbjorn, I am curious to hear what your opinion is on how Meanwell sets up their parallel arrays in the datasheet with a series-parallel gridwork to help equalize failures.

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Thank you for the explanaition. This makes a lot more sense now that it is explained out completely, and I see the flaws with it. It's still useable, but definitely not as simple as the author of the article made it out to be. It can still be implemented effectively, but adding it to the heatsink adds a little more headache to the matter.

 

Npain316, if you are going to run that driver, use the values of components that cptbjorn listed at the end of his post. This will set your max current to a little over 1A. Make sure the resistors are sized correctly. The 0.5 ohm should be 1W or greater, and the 180 ohm can be a 1/4W.

 

This is one reason why I only sold the ELN-60-48 in the group buy so you can run as many LEDs in series as possible at 1A.

 

 

cptbjorn, I am curious to hear what your opinion is on how Meanwell sets up their parallel arrays in the datasheet with a series-parallel gridwork to help equalize failures.

 

 

Thanks Evil.

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I just checked out that meanwell datasheet and that series/parallel grid is very simple and elegant; it should work perfectly. It will work best with 4 or more parallel strings though as a single failed LEDs current is spread over the remaining 3 good ones. Wouldn't work with just 2 strings and 3 is cutting it close especially for the royal blue XR-Es as I have yet to find any data on how they do at 1500ma.

 

The version with the series resistors doesn't make much sense as only the first parallel set of LEDs will "see" the resistors. In order to balance current mismatches using resistors you would need a smaller series resistor on each individual LED. This shouldn't be necessary though, as I assume the Vf's of a group of LEDs from the same run will be close enough to not need it.

 

And if they DO have large Vf variances it would be much more effective to match the LEDs by their Vf values into groups of 4 or however many parallel strings you are doing. All it would take is a LM317 on a small heatsink, a 1.25-3.5 ohm resistor, a 5v/1+amp power supply and a multimeter and you could match your set of LEDs into groups. The schematic is in the LM317 datasheet, basically all you do is run each individual LED at 350ma-1A and measure the voltage across it, then either make piles or stick post-it notes.

 

Actually the series/4*parallel web with Vf-matched LEDs would be just about as good as a single long series. It looks like they have some 24v/4A power supplies which should run 24 leds each in this configuration, would be a good solution for larger tanks.

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  • 4 weeks later...

I had a pm asking how to adapt this current mirror to run 2 LED strings in parallel from one of the 1.3A max output Meanwell drivers. Since the XR-E LEDs do just fine up to 1500ma the overcurrent protection part of the circuit isn't really needed, so the BC transistor can be removed and the 680 ohm resistors can be shorted/replaced with wires/circuit board connections. The 1.5 ohm resistors can be removed as well but I would recommend just replacing them with 1 ohm resistors as this will help with current matching. Also as an added bonus since V=I*R and R=1, you can measure the voltage across these resistors and it will be equal to the current running through that LED string.

 

Just like the original circuit the transistors need to be fairly closely matched and thermally connected, so make sure to get both at the same time (99% of the time this will make them from the same batch and matched close enough for our purposes) and that they are bolted together with a thermal pad. Alternatively you can get thermally isolated transistors like the KSDs I recommended above and just bolt them together with a small heatsink between them.

post-44503-1244247647_thumb.png

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  • 5 months later...

Been a while since anyone replied to this thread, but I have a question on current mirrors. I want to run 2 strings of LEDs on the ELN-60-48D, at 700 mA each. I think I can easily power 24 LEDs on one driver this way. I'm not sure of the components to get, though. I want to retain the overcurrent protection, so I'd like to build the original circuit, or something similar:

Recom_Fig4.jpg

But the original schematic was for 350 mA LEDs, so I think I need different components, but am not positive what to change out. With my limited knowledge, it seems that all I need to do is change the 1.5 ohm emitter resistors to something that will turn on the BC337 transistor in the event of a blown LED. I would like the transistor to turn on at about 1 amp, so I think I need 0.75 ohms resistors instead of the 1.5 ohms. Is that right? Will everything else work if I do that, or am I missing something?

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But the circuit in post 13 removes the overcurrent protection, so if an LED in string 2 dies, then 1.4A will flow through string 1. I would rather not drive the LEDs at 1.4A if possible. Or am I misunderstanding the circuit?

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Combine the two then. The 1 ohm resistor is what sets the current limit per string. The 680 ohm resistors are just to bias the transistors, so they don't need to change.

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