LED Current Mirror

R3 is for base current compensation. it makes the current mirror more accurate, but it's not connected in the right spot to do that in your figure. The resistor R3 needs to be in series with the base of Q2. The issue is that the base current of Q1 & Q2 pull current from the collector of Q2 so the reference current is in error. If beta is about 100, then you get a 1% error. Probably fine for this. If you really want to get picky, then the resistor will help if moved to the connection I described. In this case, the base current of Q2 will increase the voltage on the base of Q1 so the output current gets the correct voltage despite the base current loss at the collector of Q2.
 
Has anyone on here ever had the commonly-used style of fuse NOT work? Do you have any basis for your nervousness?

Not trying to be rude, just trying to understand. I've never heard someone say something like this before.

I've never known a fuse to fail closed.
There are roughly 2000 small fuses in each SCR building that I am working on right now, and this is my 3rd building. That's 6000 sub 2A fuses. And then I have seen countless 1000's of higher current fuses in my Electrician Career, and never had a fail closed.


I just know LEDs tend to fail faster then even quick blow fuses. And this seems like a minimal effort to add alot of protection to the system.

I gotta ask, how do you know? I've never seen any equipment fail past the fuse.
 
If I were running a factory that was churning LED arrays out on an automated production line, something like that might be desirable. For a beginning electronics hobbyist who's soldering the array on his workbench at home, balancing the strings "manually" by measuring current and rearranging as required is probably less effort, cheaper, less confusing, and just as (if not more) effective.

Can someone explain the 'balancing' part of this thread? Are we assuming parallel strings (with or without the extra circuitry) that need to pull the same amount of amps, and that not every LED will pull the same? You wouldn't care if you were doing a single non-parallel string, would you?... Just measure your total amp draw?
 
R3 is for base current compensation. it makes the current mirror more accurate, but it's not connected in the right spot to do that in your figure. The resistor R3 needs to be in series with the base of Q2. The issue is that the base current of Q1 & Q2 pull current from the collector of Q2 so the reference current is in error. If beta is about 100, then you get a 1% error. Probably fine for this. If you really want to get picky, then the resistor will help if moved to the connection I described. In this case, the base current of Q2 will increase the voltage on the base of Q1 so the output current gets the correct voltage despite the base current loss at the collector of Q2.

Now this makes much more sence. I'm far from an electrical engineer, so I wonder why every version of current mirror I have found shows R3 connected as in figure 2 above?

for the intended use of this circuit on tank builds the slight error doesn't seem like a problem and one less component makes it that much easier to build.

Would hooking up R3 in the manner you suggested have eliminated the need for having a second diode?
 
Can someone explain the 'balancing' part of this thread? Are we assuming parallel strings (with or without the extra circuitry) that need to pull the same amount of amps, and that not every LED will pull the same? You wouldn't care if you were doing a single non-parallel string, would you?... Just measure your total amp draw?

yes, the point of this thread is for parrelel strings. most people balance strings manually by taking measurements of each string and each LED, and swaping them around to get equall voltage on each string and therefore equal current draw on each parrellel string.

the current mirror just does the work for you by adding volgate drop to the lower string, if your up to the challenge of building and implementing it. for me it's much faster to solder a few extra components together than waste hours ballancing multiple LED strings. I also don't mind the little bit of power used by the circuit to do it's thing.
 
Now this makes much more sence. I'm far from an electrical engineer, so I wonder why every version of current mirror I have found shows R3 connected as in figure 2 above?

for the intended use of this circuit on tank builds the slight error doesn't seem like a problem and one less component makes it that much easier to build.

Would hooking up R3 in the manner you suggested have eliminated the need for having a second diode?

I would have to look at it closer, but I don't think the diode (D2) could be removed if that is what you are referring to. The issue is that if the string on the output of the mirror is an "open", then the output NPN Q1 will saturate. In this mode, the base becomes very low impedance and the base-emitter looks like a diode. So, the reference string of LED's will get all the current. It may not be that bad depending on how much voltage is applied across the emitter resistors.

The diode D2 acts to allow current flow into the output string in normal mode, then if the output string fails, it will allow the PNP base to go high and cut off the current. I don't particularly like that structure as it requires about two diodes drops on the output leg and one diode drop on the reference leg. That is just wasted power in the normal case which hopefully is 99.99% of the time.

I would try to use MOSFET (N-type for this application). Make sure the Vds voltage rating can handle your power supply and the Id max rating can handle your intended load current (with some margin on both). In this case, if the output leg(s) are open, then the reference leg is unaffected (gate will not pull any current), so the output shuts down nicely.
 
No, acutally I meant to say I had to add another diode (D3) in series with stirng two before the LEDs in the diagram of figure 2 I posted. the circuit didn't work correctly otherwise. so that's why I asked if using the R3 as you suggested would have done the same thing with greater efficiency?

I'm not very familiar with working with MOSFETs but if there's a better more efficient way to accomplish the mirror's function I'm all ears.

Thanks,
 
Ok, yes D3 in series with string 2 is required to keep the drop on the reference string correct. Diode D2 under normal operation is forward biased with virtually all of the current in string 1. This results in the top of string 2 being higher voltage than the top of string 1 due to the voltage drop of D2. You are matching that drop in string 1 by adding a diode D3 in series.

I'm not sure if the matched MOSFETs are readily available. I design IC's and I'm used to using FET mirrors on the chips but off chip the matching might be more of an issue. If I think of a good idea, I'll post it for you...
 
Good Stuff

Good Stuff

Zachts:

Thank you so much for working through this. The reliability factor here is huge and repair might be very much safer as well.

I've been monitoring this led stuff for a while and really want to do it right.

It may be that with the LDD ---- h drivers being so cheap and with diy pcb designs from O2, that this problem has been solved in another way.

I will learn enough to get started very soon. This type of careful dissemination of info. is the helpful thing I've seen.

Thanks > Barry
 
So I did this mirror with BD139's on an existing circuit. It's an aluminum PCB for the Ecoray lights. Everything works great as long as I dont have the jumper from the base to the reference string. When I do have the jumper there I have a near 100mA difference between the two strings. What is causing this variation? And is it fine to keep it without the jumper? As of right now every led has the same mA running for nearly 2 hours(+ or - like .004mA)...is it safe to say it will stay like this...or will the absence of the jumper cause issues in the long term?
 
So I did this mirror with BD139's on an existing circuit. It's an aluminum PCB for the Ecoray lights. Everything works great as long as I dont have the jumper from the base to the reference string. When I do have the jumper there I have a near 100mA difference between the two strings. What is causing this variation? And is it fine to keep it without the jumper? As of right now every led has the same mA running for nearly 2 hours(+ or - like .004mA)...is it safe to say it will stay like this...or will the absence of the jumper cause issues in the long term?

Would you mind posting pictures of your circuit?

Assuming your using the circuit from the figure in post #15 and by jumper you mean the connection from the string with LED1,LED2,and LED3 to the base legs of the two tansistors then it is possible that your error is caused by a number of things.
1. Something is not wired correctly, because if the bases are just connected to one another and there is no reference voltage being supplied then the transistors should both be off and none of you strings should light up.
1. Transistors could be mismatched. What value did they measure at? They should have identical or nearly identical values when measured on your multimeter.
2. Possibly your strings are perfectly balanced to begin with and the current drawn from the reference string to the base legs is resulting in the imbalance. Though 100mA is really high if this were the case. (is it at all possible there is already a current mirror on the pcbs somewhere? Or any other circuitry on the boards that could be interfering with the mirror?
 
Zachts, thank you very much for replying. I can guarantee that there are no other circuits within the pcb itself. There is nothing on the backside of it, and on the face side you can see where everything goes. From there they are just connected to the wires from the drivers. I think maybe your final suggestion might be accurate...but then again you know more about this stuff than I do.

http://imgur.com/uk46m88

Please don't judge the quality! This was just the first attempt to make it work...I'll pretty it up later!
 
Well, from the looks of it, I think you may have bad transistors. The circuit should not light up with it the way it is connected in that picture you linked. With no reference voltage to the bases the transistors should be off and no current should flow thru either string. There is either a shorted solder that I can't see or the transistors are bad.

On a side note did you epoxy down the two LEDs you soldered the circuit into? If not they are going to separate from the heatsink with only one leg soldered down and will over heat and quickly burn out.

I probably would have scratched the mask off a portion and broken the traces on the pcb to make new solder points for the connections rather than how you've wired it, but that works too as long as the chips are glued down.
 
yeah i used thermal epoxy to fix the leds down for the reason you pointed out. Your idea for scratching the mask is something I will do later...just that for right now I wanted to see that this would work with the leds I wanted on there.

So do you think the entire batch of transistors are bad? I bought them from a U.S. source...they were inexpensive, but not the most inexpensive I could've gotten. And these are number 5 and 6 ive tried since the first 4 did the exact same thing.

The other thing to note is that something is happening. Without these transistors installed, the circuits are imbalanced and they increase over time...with them installed I can run it for 24 hours with no fan and both sides of the circuit are 370mA no matter what led I try.
 
Very interesting indeed, I will have to try a circuit as you have it connected. In the mean time perhaps the activity on the thread has caught the attention of one of the Electrical Engineers that frequent RC, and perhaps they can chime in on what might be causing the circuit to work as you've observed.

I guess the worst that could happen is that something could burn out and stop working if you kept running it this way.

Do the transistors get warm? and have you measured the voltage drop across each of the transistors to see what they are actually doing? I've very curious now so will have to play around with this now, but if it stops raining I need to get out side and change the brakes on my car so might not get to this today......
 
When the transistors aren't affixed to the pcb they get very warm indeed. However now that they are fastened in place they dont even get warm.

I just did the measurements and this is what i get

Voltage Drops:

E->C (1&2): 6.5
B->C (1&2): .69
E->B #1: 5.82
E->B #2: 5.77

when i measure the mA between E and C on both 1 and 2 I get 540mA, and when I do this measurement all the lights get brighter...obviously because the transistors arent doing work anymore.
 
Hi!

I would like to build one as I have a 1.5amp cc source, and want to drive two strings of 700mA. How would one calculate the values of the resistors?
 
Hi!

I would like to build one as I have a 1.5amp cc source, and want to drive two strings of 700mA. How would one calculate the values of the resistors?

Not possible, 1.5 amps divided into two strings will yield 750mA per string if perfectly balanced. Figure plus or minus a few percent, you won't get an exactly even split. If you LEDs are only rated for 700mA this is not a very good solution as one string will likely end up over driven if not both. Otherwise you only need 1.5 ohm resistors on the output leg per the diagrams, the various other resistors are kind of irrelevant unless your aiming for a dead on matched current which doesn't really matter for most applications.
 
@Aqualund, I still haven't had time to fiddle around with your observation and circuit to see what's up with the function you noticed. Hopefully over the holidays.......
 
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