LED'S ARE THE BEST!!!!!!!!!
- Thread starter 242bats
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242bats
Member
Quote:
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/80#post_3407739
Very good, Sir! I knew you could do it. Do you have the fuses in-line before or after the resistors?
after the resistors???????? do they need to be before or after????
btw
this will set on top of the hood for now, that is the reason for the trailer plugs so i can unplug the drivers, take that off and then remove the hood.
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/80#post_3407739
Very good, Sir! I knew you could do it. Do you have the fuses in-line before or after the resistors?
after the resistors???????? do they need to be before or after????
btw
this will set on top of the hood for now, that is the reason for the trailer plugs so i can unplug the drivers, take that off and then remove the hood.
242bats
Member
Quote:
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/80#post_3407744
Just looking at your pics again. It looks like you have the positive and negative outputs of the drivers reversed. I think the red wire is the positive. A quick swap should fix that.
after looking at the two threads on the other site one has the it the way i have and other has the other way ?????????
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/80#post_3407744
Just looking at your pics again. It looks like you have the positive and negative outputs of the drivers reversed. I think the red wire is the positive. A quick swap should fix that.
after looking at the two threads on the other site one has the it the way i have and other has the other way ?????????
2quills
Well-Known Member
Quote:
Originally Posted by 242bats http:///forum/thread/386777/led-s-are-the-best/80#post_3407745
after the resistors???????? do they need to be before or after????
btw
this will set on top of the hood for now, that is the reason for the trailer plugs so i can unplug the drivers, take that off and then remove the hood.
I would do the fuses first before the resistors. I like the plugs, I was planning on something similar with probably molex connectors since my drivers won't be on the actual fixture itself. I would definitely swap the positive and negative wires around. Otherwise your fuses aren't going to protect the strings if current is running backwards.
Originally Posted by 242bats http:///forum/thread/386777/led-s-are-the-best/80#post_3407745
after the resistors???????? do they need to be before or after????
btw
this will set on top of the hood for now, that is the reason for the trailer plugs so i can unplug the drivers, take that off and then remove the hood.
I would do the fuses first before the resistors. I like the plugs, I was planning on something similar with probably molex connectors since my drivers won't be on the actual fixture itself. I would definitely swap the positive and negative wires around. Otherwise your fuses aren't going to protect the strings if current is running backwards.
242bats
Member
Quote:
Originally Posted by meowzer http:///forum/thread/386777/led-s-are-the-best/80#post_3407706
I look at those wires and shiver with fear
have no fear, just go for it
Originally Posted by meowzer http:///forum/thread/386777/led-s-are-the-best/80#post_3407706
I look at those wires and shiver with fear
have no fear, just go for it
2quills
Well-Known Member
Quote:
Originally Posted by 242bats http:///forum/thread/386777/led-s-are-the-best/80#post_3407754
after looking at the two threads on the other site one has the it the way i have and other has the other way ?????????
Actually, JP's thread he has them the right way with the red hooked up to the fuses and resistors. That led zeppelin thread the guy had them backwards. On Kcress's thread he has them hooked up the proper way as well so that's how I would do it.
Originally Posted by 242bats http:///forum/thread/386777/led-s-are-the-best/80#post_3407754
after looking at the two threads on the other site one has the it the way i have and other has the other way ?????????
Actually, JP's thread he has them the right way with the red hooked up to the fuses and resistors. That led zeppelin thread the guy had them backwards. On Kcress's thread he has them hooked up the proper way as well so that's how I would do it.
242bats
Member
Quote:
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/100#post_3407758
ok not hard to move some things thanks
Originally Posted by 2Quills http:///forum/thread/386777/led-s-are-the-best/100#post_3407758
ok not hard to move some things thanks
scsinet
Active Member
Just to comment on the earlier posts about fuse locations...
Circuit design best practices dictate placing fuses as close to the power supply as possible. A fault occuring between the fuse and the power supply would not be protected, so the fuse should always be where it will protect the largest portion of the circuit, so your diagrams have them in the right place.
However, I'm not sure the fuses are necessary. Let me explain...
Your diagrams indicate that you are using constant-current LED drivers. Your resistors, coupled with these supplies, are functioning merely as impedence-based load distribution, to prevent one string of LEDS from "hogging" all of the current due to slight impedence differences between LED strings. Fair enough.
However, the supply will output a fixed amount of current, no matter what, even in a dead-short scenario, by virtue of your LED drivers, so the circuit shouldn't be at risk of meltdown. Where a fuse should definitely be located is on the power supply input on the line voltage side.
That said, extra fuses don't hurt... in order for them to be any good, the fuses must be rated at less than the dead-short current output of your supply, so I'd use a meter to determine exactly what your LED strings draw - without the fuses - then size your fuses at that value plus 20% or so.
Also of note on your diagram... you say you can have different LEDs of in the string... careful. All LEDS in a series connected string must have the same forward voltage and current ratings.
By the way... in reference to our earlier discussion about harmonics... I spotted the following enormous Aqua-Illuminations setup at an LFS when I stopped by for some ARM today, and thought of this thread. I had a few minutes to poke at it, all supplies go back to one circuit.
Circuit design best practices dictate placing fuses as close to the power supply as possible. A fault occuring between the fuse and the power supply would not be protected, so the fuse should always be where it will protect the largest portion of the circuit, so your diagrams have them in the right place.
However, I'm not sure the fuses are necessary. Let me explain...
Your diagrams indicate that you are using constant-current LED drivers. Your resistors, coupled with these supplies, are functioning merely as impedence-based load distribution, to prevent one string of LEDS from "hogging" all of the current due to slight impedence differences between LED strings. Fair enough.
However, the supply will output a fixed amount of current, no matter what, even in a dead-short scenario, by virtue of your LED drivers, so the circuit shouldn't be at risk of meltdown. Where a fuse should definitely be located is on the power supply input on the line voltage side.
That said, extra fuses don't hurt... in order for them to be any good, the fuses must be rated at less than the dead-short current output of your supply, so I'd use a meter to determine exactly what your LED strings draw - without the fuses - then size your fuses at that value plus 20% or so.
Also of note on your diagram... you say you can have different LEDs of in the string... careful. All LEDS in a series connected string must have the same forward voltage and current ratings.
By the way... in reference to our earlier discussion about harmonics... I spotted the following enormous Aqua-Illuminations setup at an LFS when I stopped by for some ARM today, and thought of this thread. I had a few minutes to poke at it, all supplies go back to one circuit.
2quills
Well-Known Member
Quote:
Originally Posted by SCSInet http:///t/386777/led-s-are-the-best/100#post_3410207
Just to comment on the earlier posts about fuse locations...
Circuit design best practices dictate placing fuses as close to the power supply as possible. A fault occuring between the fuse and the power supply would not be protected, so the fuse should always be where it will protect the largest portion of the circuit, so your diagrams have them in the right place.
However, I'm not sure the fuses are necessary. Let me explain...
Your diagrams indicate that you are using constant-current LED drivers. Your resistors, coupled with these supplies, are functioning merely as impedence-based load distribution, to prevent one string of LEDS from "hogging" all of the current due to slight impedence differences between LED strings. Fair enough.
However, the supply will output a fixed amount of current, no matter what, even in a dead-short scenario, by virtue of your LED drivers, so the circuit shouldn't be at risk of meltdown. Where a fuse should definitely be located is on the power supply input on the line voltage side.
That said, extra fuses don't hurt... in order for them to be any good, the fuses must be rated at less than the dead-short current output of your supply, so I'd use a meter to determine exactly what your LED strings draw - without the fuses - then size your fuses at that value plus 20% or so.
By the way... in reference to our earlier discussion about harmonics... I spotted the following enormous Aqua-Illuminations setup at an LFS when I stopped by for some ARM today, and thought of this thread. I had a few minutes to poke at it, all supplies go back to one circuit.
I'm not even gonna go there lol.
That isn't actually my diagram, it's just one of many floating around that people have been using a general guideline for some of these builds when running parallel strings.
As far as I know the main reason why folks have been using the 1ohm resistors is merely for a simple way to test current going to each string with out having to open up the strings. Since balancing the forward voltage on each string is pretty much recommended (not necessarily totally needed) but is a good idea to keep the fixture balanced. From what I understand is that small differences in voltage or resistance typically equate to larger differences in current passing through each string. And over time the voltage or resistance can stray a bit so balancing the fixture again sometime down the line may not be a bad idea. But I think I see what you're getting at and wonder if different sized resistors might even be a better idea.
In regards to the fuses. Agreed, in some cases they may not be necessary. Especially now that I'm wondering about different resistors. But with that said I think the idea of using them is primarily for folks running parallel strings. As you said, the power supply will supply a certain amount of current output no matter what. Say lets say you're running a power supply that puts out 5 amps, and you're running 5 strings in parallel that share those amps so you end up with 1amp of available current for each string. Now lets say you're running XR-E series cree leds rated for a maximum of 1amp. You use your variable voltage resistor (pot) or pwm signal to dim that output do approximately 90% or 4.5 amps total which breaks down to 900mA per string. Ok, everythings good, the fixture runs just how you want it. But all of a sudden one of your leds dead shorts (open) and you lose current passing through that string. Remaining current then gets distributed to the remaining 4 strings. Now you have 1.125 amp or 1125mA passing through your remaining led strings that are only rated for 1amp max. But you're in luck because you installed some 1amp fast blow fuses which blow and there for hopefully save yourself from losing or damaging a few dozen leds. I think that was the general idea for them.
But if I'm thinking what you were thinking about the resistors initially. Then if you used the proper resistors, would they not just simply allow a limited amount of current from being able to pass through those remaining strings? Thus negating the need for fuses? Or am I thinking about this all wrong?
BTW, XP-G leds are rated up to 1.5 amps. So I'm thinking that fuses may not totally be needed in such an event as the likely hood of losing a led in two separate strings at the same time would probably never occur. But you never know I guess. I think one really needs to look at the situation on a per build basis to be able to make the proper determination of what would be a safe back up or not and weigh the odds if you think something is really necessary or not. But if it doesn't hurt anything, and protects you from the "what if" factor than whey not eh?
I just saw recently an idea on another build that I'm seriously considering. And that would be utilizing PLED's as shunts. Much like they do in street signs or traffic signals. Adds a little more to the cost of the build but if you ever lose a led you won't actually lose the whole string. And it wouldn't be huge emergency to have to get the fixture fixed asap. Also, a nice thing about that is you'd know right away right where the problem is.
Originally Posted by SCSInet http:///t/386777/led-s-are-the-best/100#post_3410207
Just to comment on the earlier posts about fuse locations...
Circuit design best practices dictate placing fuses as close to the power supply as possible. A fault occuring between the fuse and the power supply would not be protected, so the fuse should always be where it will protect the largest portion of the circuit, so your diagrams have them in the right place.
However, I'm not sure the fuses are necessary. Let me explain...
Your diagrams indicate that you are using constant-current LED drivers. Your resistors, coupled with these supplies, are functioning merely as impedence-based load distribution, to prevent one string of LEDS from "hogging" all of the current due to slight impedence differences between LED strings. Fair enough.
However, the supply will output a fixed amount of current, no matter what, even in a dead-short scenario, by virtue of your LED drivers, so the circuit shouldn't be at risk of meltdown. Where a fuse should definitely be located is on the power supply input on the line voltage side.
That said, extra fuses don't hurt... in order for them to be any good, the fuses must be rated at less than the dead-short current output of your supply, so I'd use a meter to determine exactly what your LED strings draw - without the fuses - then size your fuses at that value plus 20% or so.
By the way... in reference to our earlier discussion about harmonics... I spotted the following enormous Aqua-Illuminations setup at an LFS when I stopped by for some ARM today, and thought of this thread. I had a few minutes to poke at it, all supplies go back to one circuit.
That isn't actually my diagram, it's just one of many floating around that people have been using a general guideline for some of these builds when running parallel strings.
As far as I know the main reason why folks have been using the 1ohm resistors is merely for a simple way to test current going to each string with out having to open up the strings. Since balancing the forward voltage on each string is pretty much recommended (not necessarily totally needed) but is a good idea to keep the fixture balanced. From what I understand is that small differences in voltage or resistance typically equate to larger differences in current passing through each string. And over time the voltage or resistance can stray a bit so balancing the fixture again sometime down the line may not be a bad idea. But I think I see what you're getting at and wonder if different sized resistors might even be a better idea.
In regards to the fuses. Agreed, in some cases they may not be necessary. Especially now that I'm wondering about different resistors. But with that said I think the idea of using them is primarily for folks running parallel strings. As you said, the power supply will supply a certain amount of current output no matter what. Say lets say you're running a power supply that puts out 5 amps, and you're running 5 strings in parallel that share those amps so you end up with 1amp of available current for each string. Now lets say you're running XR-E series cree leds rated for a maximum of 1amp. You use your variable voltage resistor (pot) or pwm signal to dim that output do approximately 90% or 4.5 amps total which breaks down to 900mA per string. Ok, everythings good, the fixture runs just how you want it. But all of a sudden one of your leds dead shorts (open) and you lose current passing through that string. Remaining current then gets distributed to the remaining 4 strings. Now you have 1.125 amp or 1125mA passing through your remaining led strings that are only rated for 1amp max. But you're in luck because you installed some 1amp fast blow fuses which blow and there for hopefully save yourself from losing or damaging a few dozen leds. I think that was the general idea for them.
But if I'm thinking what you were thinking about the resistors initially. Then if you used the proper resistors, would they not just simply allow a limited amount of current from being able to pass through those remaining strings? Thus negating the need for fuses? Or am I thinking about this all wrong?
BTW, XP-G leds are rated up to 1.5 amps. So I'm thinking that fuses may not totally be needed in such an event as the likely hood of losing a led in two separate strings at the same time would probably never occur. But you never know I guess. I think one really needs to look at the situation on a per build basis to be able to make the proper determination of what would be a safe back up or not and weigh the odds if you think something is really necessary or not. But if it doesn't hurt anything, and protects you from the "what if" factor than whey not eh?
I just saw recently an idea on another build that I'm seriously considering. And that would be utilizing PLED's as shunts. Much like they do in street signs or traffic signals. Adds a little more to the cost of the build but if you ever lose a led you won't actually lose the whole string. And it wouldn't be huge emergency to have to get the fixture fixed asap. Also, a nice thing about that is you'd know right away right where the problem is.
2quills
Well-Known Member
Question for you SCSI.
Wondering if this idea is doable.
Ohms law states that the voltage (V) across a resistor (R) is proportional to the current (I) where the constant of proportionality is the resistance. (I think I said that right)
Equivalently ohms law can be stated as: I = V/R.
If I have a 48v power supply. And I was running 5 strings in parallel series on that power supply. Could I just use 50ohm resistors and negate the need for fuses? Using the formula I come up with 48v / 50ohms = 960mA. So if I simply went with 50ohm resistors at the head of each of my strings then theoretically they would only allow 960mA to pass through them correct?
But this begs the question of what happens to the excess current that is resisted? If they were installed using terminal blocks like in the diagram above would it then be wise to shunt the resistors over to the negative wire going back to the driver?
Wondering if this idea is doable.
Ohms law states that the voltage (V) across a resistor (R) is proportional to the current (I) where the constant of proportionality is the resistance. (I think I said that right)
Equivalently ohms law can be stated as: I = V/R.
If I have a 48v power supply. And I was running 5 strings in parallel series on that power supply. Could I just use 50ohm resistors and negate the need for fuses? Using the formula I come up with 48v / 50ohms = 960mA. So if I simply went with 50ohm resistors at the head of each of my strings then theoretically they would only allow 960mA to pass through them correct?
But this begs the question of what happens to the excess current that is resisted? If they were installed using terminal blocks like in the diagram above would it then be wise to shunt the resistors over to the negative wire going back to the driver?
scsinet
Active Member
Quote:
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410221
I'm not even gonna go there lol.
That isn't actually my diagram, it's just one of many floating around that people have been using a general guideline for some of these builds when running parallel strings.
As far as I know the main reason why folks have been using the 1ohm resistors is merely for a simple way to test current going to each string with out having to open up the strings.
Yep, this is one other thing you can do. I occasionally build tube amplifiers and I use this method to measure bias current on the output tubes while the system is hot. 1 Ohm resistors work perfectly, because across a 1 ohm resistor, one amp = 1 volt on the meter.
Since balancing the forward voltage on each string is pretty much recommended (not necessarily totally needed) but is a good idea to keep the fixture balanced. From what I understand is that small differences in voltage or resistance typically equate to larger differences in current passing through each string.
Yes, it can make a big difference. Since LEDs are current driven devices not voltage driven, current makes all the difference. When you have a constant current supply, a finite amount of current is available, and when an imbalance exists, that can lead to significant differences in the lighting levels between strings, and even potential overloads on the string with the least resistance.
And over time the voltage or resistance can stray a bit so balancing the fixture again sometime down the line may not be a bad idea. But I think I see what you're getting at and wonder if different sized resistors might even be a better idea.
Sizing the resistor in this application depends on the maximum nominal differences you tend to encounter. In the case of this type of setup, if the wires are the only difference (slight variations in wire lengths, connections, etc), the differences are probably in the hundreths of ohms, in which case a 1ohm is more that sufficient. If the difference can be larger due to the LEDs, it may require something different.
Keep in mind that resistors have a voltage drop, and they dissipate that drop as heat, which means wasted power and efficiency, so you want to keep that resistor small.
In regards to the fuses. Agreed, in some cases they may not be necessary. Especially now that I'm wondering about different resistors. But with that said I think the idea of using them is primarily for folks running parallel strings. As you said, the power supply will supply a certain amount of current output no matter what. Say lets say you're running a power supply that puts out 5 amps, and you're running 5 strings in parallel that share those amps so you end up with 1amp of available current for each string. Now lets say you're running XR-E series cree leds rated for a maximum of 1amp. You use your variable voltage resistor (pot) or pwm signal to dim that output do approximately 90% or 4.5 amps total which breaks down to 900mA per string. Ok, everythings good, the fixture runs just how you want it. But all of a sudden one of your leds dead shorts (open) and you lose current passing through that string. Remaining current then gets distributed to the remaining 4 strings. Now you have 1.125 amp or 1125mA passing through your remaining led strings that are only rated for 1amp max. But you're in luck because you installed some 1amp fast blow fuses which blow and there for hopefully save yourself from losing or damaging a few dozen leds. I think that was the general idea for them.
I can see the argument here... I wasn't thinking about protecting the other strings, but if that's the goal, that's what you'd want to do. Sounds like you have this aspect (sizing the fuses and all) nailed down.
But if I'm thinking what you were thinking about the resistors initially. Then if you used the proper resistors, would they not just simply allow a limited amount of current from being able to pass through those remaining strings? Thus negating the need for fuses? Or am I thinking about this all wrong?
Yes and no. These constant current LED driver modules are a relatively new thing. Time was (and still is) that when driving an LED or a string of them, all you'd do is connect a series resistor that is sized to limit the LEDs current. As long as voltage stays constant - which any regulated power supply will do - then all you need to do is take the supply voltage, the forward voltage and current of the LED, and apply ohm's law to come up with the size resistor you need. In this manner, the resistor performed EXACTLY the same function as a metal halide ballast - to limit current.
However, what is going on here is the use of a constant CURRENT supply. The voltage output varies, it's the current output that remains steady. If you used series resistors for this, just as you pointed out above, if one goes out, the supply will increase the voltage to whatever is necessary to maintain the same current flow. Since resistor sizing is based on supply voltage, the resistors "become" too small and the strings can overload.
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410221
That isn't actually my diagram, it's just one of many floating around that people have been using a general guideline for some of these builds when running parallel strings.
As far as I know the main reason why folks have been using the 1ohm resistors is merely for a simple way to test current going to each string with out having to open up the strings.
Yep, this is one other thing you can do. I occasionally build tube amplifiers and I use this method to measure bias current on the output tubes while the system is hot. 1 Ohm resistors work perfectly, because across a 1 ohm resistor, one amp = 1 volt on the meter.
Since balancing the forward voltage on each string is pretty much recommended (not necessarily totally needed) but is a good idea to keep the fixture balanced. From what I understand is that small differences in voltage or resistance typically equate to larger differences in current passing through each string.
Yes, it can make a big difference. Since LEDs are current driven devices not voltage driven, current makes all the difference. When you have a constant current supply, a finite amount of current is available, and when an imbalance exists, that can lead to significant differences in the lighting levels between strings, and even potential overloads on the string with the least resistance.
And over time the voltage or resistance can stray a bit so balancing the fixture again sometime down the line may not be a bad idea. But I think I see what you're getting at and wonder if different sized resistors might even be a better idea.
Sizing the resistor in this application depends on the maximum nominal differences you tend to encounter. In the case of this type of setup, if the wires are the only difference (slight variations in wire lengths, connections, etc), the differences are probably in the hundreths of ohms, in which case a 1ohm is more that sufficient. If the difference can be larger due to the LEDs, it may require something different.
Keep in mind that resistors have a voltage drop, and they dissipate that drop as heat, which means wasted power and efficiency, so you want to keep that resistor small.
In regards to the fuses. Agreed, in some cases they may not be necessary. Especially now that I'm wondering about different resistors. But with that said I think the idea of using them is primarily for folks running parallel strings. As you said, the power supply will supply a certain amount of current output no matter what. Say lets say you're running a power supply that puts out 5 amps, and you're running 5 strings in parallel that share those amps so you end up with 1amp of available current for each string. Now lets say you're running XR-E series cree leds rated for a maximum of 1amp. You use your variable voltage resistor (pot) or pwm signal to dim that output do approximately 90% or 4.5 amps total which breaks down to 900mA per string. Ok, everythings good, the fixture runs just how you want it. But all of a sudden one of your leds dead shorts (open) and you lose current passing through that string. Remaining current then gets distributed to the remaining 4 strings. Now you have 1.125 amp or 1125mA passing through your remaining led strings that are only rated for 1amp max. But you're in luck because you installed some 1amp fast blow fuses which blow and there for hopefully save yourself from losing or damaging a few dozen leds. I think that was the general idea for them.
I can see the argument here... I wasn't thinking about protecting the other strings, but if that's the goal, that's what you'd want to do. Sounds like you have this aspect (sizing the fuses and all) nailed down.
But if I'm thinking what you were thinking about the resistors initially. Then if you used the proper resistors, would they not just simply allow a limited amount of current from being able to pass through those remaining strings? Thus negating the need for fuses? Or am I thinking about this all wrong?
Yes and no. These constant current LED driver modules are a relatively new thing. Time was (and still is) that when driving an LED or a string of them, all you'd do is connect a series resistor that is sized to limit the LEDs current. As long as voltage stays constant - which any regulated power supply will do - then all you need to do is take the supply voltage, the forward voltage and current of the LED, and apply ohm's law to come up with the size resistor you need. In this manner, the resistor performed EXACTLY the same function as a metal halide ballast - to limit current.
However, what is going on here is the use of a constant CURRENT supply. The voltage output varies, it's the current output that remains steady. If you used series resistors for this, just as you pointed out above, if one goes out, the supply will increase the voltage to whatever is necessary to maintain the same current flow. Since resistor sizing is based on supply voltage, the resistors "become" too small and the strings can overload.
scsinet
Active Member
Quote:
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410301
Question for you SCSI.
Wondering if this idea is doable.
Ohms law states that the voltage (V) across a resistor (R) is proportional to the current (I) where the constant of proportionality is the resistance. (I think I said that right)
Equivalently ohms law can be stated as: I = V/R.
If I have a 48v power supply. And I was running 5 strings in parallel series on that power supply. Could I just use 50ohm resistors and negate the need for fuses? Using the formula I come up with 48v / 50ohms = 960mA. So if I simply went with 50ohm resistors at the head of each of my strings then theoretically they would only allow 960mA to pass through them correct?
But this begs the question of what happens to the excess current that is resisted? If they were installed using terminal blocks like in the diagram above would it then be wise to shunt the resistors over to the negative wire going back to the driver?
You're about there, but you need to take the forward voltage of the LEDs into account, so you need a bit of a twist on Ohm's law. The LEDs drop some voltage so you need to effectively exclude that.
The formula to calculate an LED's dropping resistor is
R = (Vsource - Vled) / I
R = resistance in Ohms
Vsource = supply voltage
Vled = LED forward voltage
I = Current, in amps.
So let's say you wanted to run a string of (5) LEDs with a 3.6v forward voltage per LED, and you kept the original 960ma.
5*3.6 = 18 forward volts total.
R = (48 - 18) / .96 OR R = 30 / .96
R = 31.25, which would probably mean a 33 Ohm since that's going to be the closest common value.
As you've noticed however, the current can't just disappear... it has to do "work" to go away. That work comes in the form of lots of heat.
Again, apply ohm's law to look t the power being dumped by the resistor.
In your 50 ohm example:
W = V(2) / R
(watts = Vsquared / resistance) Keep in mind that the voltage is the voltage across the resistor, not the LEDs.
W = (48*48) / R OR W = 2304 / 50 Which give us W = 46.08W
That's a huge amount of wattage requiring a huge resistor... very huge. It's also incredibly wasteful.
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Now, with the 33 ohm example: W = (30*30) / 33 = 27w. Better, but still not good.
The somewhat obvious question is... if you have a 48 volt supply, why not use bigger strings of LEDS? If you use a string of 12 LEDs instead of 5, that puts your total forward voltage at 43.2 volts.
R = (48 - 43.2) / .96 Which gives us: R = 5 Ohms (which will likely be a 5.1 ohm resistor... I know... weird... )
W = (4.8*4.8) / 5.1 Which gives us: 4.5 watts. NOW we're talking!
Make sense?
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410301
Question for you SCSI.
Wondering if this idea is doable.
Ohms law states that the voltage (V) across a resistor (R) is proportional to the current (I) where the constant of proportionality is the resistance. (I think I said that right)
Equivalently ohms law can be stated as: I = V/R.
If I have a 48v power supply. And I was running 5 strings in parallel series on that power supply. Could I just use 50ohm resistors and negate the need for fuses? Using the formula I come up with 48v / 50ohms = 960mA. So if I simply went with 50ohm resistors at the head of each of my strings then theoretically they would only allow 960mA to pass through them correct?
But this begs the question of what happens to the excess current that is resisted? If they were installed using terminal blocks like in the diagram above would it then be wise to shunt the resistors over to the negative wire going back to the driver?
You're about there, but you need to take the forward voltage of the LEDs into account, so you need a bit of a twist on Ohm's law. The LEDs drop some voltage so you need to effectively exclude that.
The formula to calculate an LED's dropping resistor is
R = (Vsource - Vled) / I
R = resistance in Ohms
Vsource = supply voltage
Vled = LED forward voltage
I = Current, in amps.
So let's say you wanted to run a string of (5) LEDs with a 3.6v forward voltage per LED, and you kept the original 960ma.
5*3.6 = 18 forward volts total.
R = (48 - 18) / .96 OR R = 30 / .96
R = 31.25, which would probably mean a 33 Ohm since that's going to be the closest common value.
As you've noticed however, the current can't just disappear... it has to do "work" to go away. That work comes in the form of lots of heat.
Again, apply ohm's law to look t the power being dumped by the resistor.
In your 50 ohm example:
W = V(2) / R
(watts = Vsquared / resistance) Keep in mind that the voltage is the voltage across the resistor, not the LEDs.
W = (48*48) / R OR W = 2304 / 50 Which give us W = 46.08W
That's a huge amount of wattage requiring a huge resistor... very huge. It's also incredibly wasteful.
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Now, with the 33 ohm example: W = (30*30) / 33 = 27w. Better, but still not good.
The somewhat obvious question is... if you have a 48 volt supply, why not use bigger strings of LEDS? If you use a string of 12 LEDs instead of 5, that puts your total forward voltage at 43.2 volts.
R = (48 - 43.2) / .96 Which gives us: R = 5 Ohms (which will likely be a 5.1 ohm resistor... I know... weird... )
W = (4.8*4.8) / 5.1 Which gives us: 4.5 watts. NOW we're talking!
Make sense?
2quills
Well-Known Member
Yeah I think so.
So if I have a 48v power supply running strings of 12 xp-g leds at 3.2v a piece for a total of 38.4 v total I would get...
R =(48 - 38.4) / .96 = 10 So I would need 10 ohm resistors for those strings.
W = (4.8*4.8) / 10 = 2.304 watts?
And no need to shunt anything right? The resistors simply chew up the excess current and spit it out as heat?
I guess one could still theoretically employ fuses as a back up incase a resistor ever failed? Probably a little over kill...I know.
So if I have a 48v power supply running strings of 12 xp-g leds at 3.2v a piece for a total of 38.4 v total I would get...
R =(48 - 38.4) / .96 = 10 So I would need 10 ohm resistors for those strings.
W = (4.8*4.8) / 10 = 2.304 watts?
And no need to shunt anything right? The resistors simply chew up the excess current and spit it out as heat?
I guess one could still theoretically employ fuses as a back up incase a resistor ever failed? Probably a little over kill...I know.
scsinet
Active Member
I'm not sure where you got the 4.8 in your calculations. The "V" in the power calculation is the difference in voltage between supply and LED total. Your LED total is 38.4, so 48 - 38.4 = 9.6, so:
9.6(sq) / 10 = 9.21 watts. I think what happened is you used my voltage number from my hypothetical 43.2 volt string, instead of your 38.4 volt string.
Not really over kill on the fuses because you are not using a constant current power source, you are using a constant voltage.
First yes, the resistors simply dissipate the current as heat. The point of calculating the heat dissipated is to size the resistor's power rating. Since you arrived at 2.3 watts, you can use a 20 watt power resistor (a 10 may work, but you want to go a good margin over).
Say those LED strings shorted. In that scenario, the voltage across the resistor is 48 volts, not 9.6. So:
48sq / 10 = 230 watts. That would burn up your series resistor for sure.
Another scenario that would not affect the current supplies is that if an individual LED shorts, it's current consumption must be dissipated into the remaining LEDs. It's kind of like the little twinkle christmas lights, where if one goes out the string stays lit, but if you pull a bulb out, the string goes out. Those strings are series connected, and the bulbs are designed to auto-short when they blow, keeping the string lit. When one lamp goes out, it's current gets piled onto the other bulbs. When there are like 34 more bulbs (in a string of 35 lights), that's not a big deal, but eventually, another blows... then another... as each lamp goes out, the current rises a bit more (a bit more than it did when the previous lamp(s) went), and lamps start failing faster and faster until the whole string suddenly goes out rapidly as the additional current reduces the remaining lamps life from many hours to a few seconds. Cascading, catastrophic failure.
With a constant current supply, if an LED shorts, the supply compensates. With a regulated (constant) voltage supply, the current rises because the voltage is being held steady. So in a string of 12 LEDs, if one shorts, that's 8% more current piled onto each remaining LED. When one of the 11 remaining goes out, that's 9%. At this point, the 10 remaining LEDs are dealing with 17% more current than when they were all working, and a third LED is bound to go. When it does, the load on the remaining 9 is now 27% above nominal.
So fusing the LEDs with a very carefully chosen fuse is a good idea on a system that is using a constant voltage supply.
9.6(sq) / 10 = 9.21 watts. I think what happened is you used my voltage number from my hypothetical 43.2 volt string, instead of your 38.4 volt string.
Not really over kill on the fuses because you are not using a constant current power source, you are using a constant voltage.
First yes, the resistors simply dissipate the current as heat. The point of calculating the heat dissipated is to size the resistor's power rating. Since you arrived at 2.3 watts, you can use a 20 watt power resistor (a 10 may work, but you want to go a good margin over).
Say those LED strings shorted. In that scenario, the voltage across the resistor is 48 volts, not 9.6. So:
48sq / 10 = 230 watts. That would burn up your series resistor for sure.
Another scenario that would not affect the current supplies is that if an individual LED shorts, it's current consumption must be dissipated into the remaining LEDs. It's kind of like the little twinkle christmas lights, where if one goes out the string stays lit, but if you pull a bulb out, the string goes out. Those strings are series connected, and the bulbs are designed to auto-short when they blow, keeping the string lit. When one lamp goes out, it's current gets piled onto the other bulbs. When there are like 34 more bulbs (in a string of 35 lights), that's not a big deal, but eventually, another blows... then another... as each lamp goes out, the current rises a bit more (a bit more than it did when the previous lamp(s) went), and lamps start failing faster and faster until the whole string suddenly goes out rapidly as the additional current reduces the remaining lamps life from many hours to a few seconds. Cascading, catastrophic failure.
With a constant current supply, if an LED shorts, the supply compensates. With a regulated (constant) voltage supply, the current rises because the voltage is being held steady. So in a string of 12 LEDs, if one shorts, that's 8% more current piled onto each remaining LED. When one of the 11 remaining goes out, that's 9%. At this point, the 10 remaining LEDs are dealing with 17% more current than when they were all working, and a third LED is bound to go. When it does, the load on the remaining 9 is now 27% above nominal.
So fusing the LEDs with a very carefully chosen fuse is a good idea on a system that is using a constant voltage supply.
2quills
Well-Known Member
Quote:
Originally Posted by SCSInet http:///t/386777/led-s-are-the-best/100#post_3410426
I'm not sure where you got the 4.8 in your calculations. The "V" in the power calculation is the difference in voltage between supply and LED total. Your LED total is 38.4, so 48 - 38.4 = 9.6, so:
9.6(sq) / 10 = 9.21 watts. I think what happened is you used my voltage number from my hypothetical 43.2 volt string, instead of your 38.4 volt string.
Not really over kill on the fuses because you are not using a constant current power source, you are using a constant voltage.
First yes, the resistors simply dissipate the current as heat. The point of calculating the heat dissipated is to size the resistor's power rating. Since you arrived at 2.3 watts, you can use a 20 watt power resistor (a 10 may work, but you want to go a good margin over).
Say those LED strings shorted. In that scenario, the voltage across the resistor is 48 volts, not 9.6. So:
48sq / 10 = 230 watts. That would burn up your series resistor for sure.
Another scenario that would not affect the current supplies is that if an individual LED shorts, it's current consumption must be dissipated into the remaining LEDs. It's kind of like the little twinkle christmas lights, where if one goes out the string stays lit, but if you pull a bulb out, the string goes out. Those strings are series connected, and the bulbs are designed to auto-short when they blow, keeping the string lit. When one lamp goes out, it's current gets piled onto the other bulbs. When there are like 34 more bulbs (in a string of 35 lights), that's not a big deal, but eventually, another blows... then another... as each lamp goes out, the current rises a bit more (a bit more than it did when the previous lamp(s) went), and lamps start failing faster and faster until the whole string suddenly goes out rapidly as the additional current reduces the remaining lamps life from many hours to a few seconds. Cascading, catastrophic failure.
With a constant current supply, if an LED shorts, the supply compensates. With a regulated (constant) voltage supply, the current rises because the voltage is being held steady. So in a string of 12 LEDs, if one shorts, that's 8% more current piled onto each remaining LED. When one of the 11 remaining goes out, that's 9%. At this point, the 10 remaining LEDs are dealing with 17% more current than when they were all working, and a third LED is bound to go. When it does, the load on the remaining 9 is now 27% above nominal.
So fusing the LEDs with a very carefully chosen fuse is a good idea on a system that is using a constant voltage supply.
Ok, I think I follow you now. Yeah I saw the 4.8x4.8 in your calculation above and it kind of threw me off a bit. I wasn't really sure where that came from or why I coppied it. I should have just stuck to the formula.
So let me see if I got it now. Using the 12 xp-g voltage rating of 3.2.
48v power supply
12 leds 3.2v (12*3.2) = 38.4v
R = (48 - 38.4) / .96 = 10 ohm or 9.6 / .96 = 10 ohm
W = (9.6 * 9.6) / 10 = 9.21 watts or 92.16 / .96 = 9.21 watts
So I'd be looking for a 10ohm / 10-20watt resistor. But it's best to go with a higher safety margin so a 10 ohm / 20 watt resistor would be more appropriate. And this would limit the current able to pass through each string to 960mA. So in a case where I want to run parallel strings I could theoretically use resistors to to aid in keeping my strings/fixture balanced. Which I think you hinted to a few posts ago. And therefor the resistors would actually be serving more than one function as apposed to using them as just a simple way to test current going to each string instead of having to open up strings and hook up your meter in series to test current. And properly chosen fuses is still a good idea as a way to help avoid a cascading failure effect.
I think I got it dude.
I was actually under the impression from other posters that when a cree led fails it usually fails and shorts open thus losing all current being able to pass through the string. But I was reading from another source of information that was saying this actually rarely is the case.
Originally Posted by SCSInet http:///t/386777/led-s-are-the-best/100#post_3410426
I'm not sure where you got the 4.8 in your calculations. The "V" in the power calculation is the difference in voltage between supply and LED total. Your LED total is 38.4, so 48 - 38.4 = 9.6, so:
9.6(sq) / 10 = 9.21 watts. I think what happened is you used my voltage number from my hypothetical 43.2 volt string, instead of your 38.4 volt string.
Not really over kill on the fuses because you are not using a constant current power source, you are using a constant voltage.
First yes, the resistors simply dissipate the current as heat. The point of calculating the heat dissipated is to size the resistor's power rating. Since you arrived at 2.3 watts, you can use a 20 watt power resistor (a 10 may work, but you want to go a good margin over).
Say those LED strings shorted. In that scenario, the voltage across the resistor is 48 volts, not 9.6. So:
48sq / 10 = 230 watts. That would burn up your series resistor for sure.
Another scenario that would not affect the current supplies is that if an individual LED shorts, it's current consumption must be dissipated into the remaining LEDs. It's kind of like the little twinkle christmas lights, where if one goes out the string stays lit, but if you pull a bulb out, the string goes out. Those strings are series connected, and the bulbs are designed to auto-short when they blow, keeping the string lit. When one lamp goes out, it's current gets piled onto the other bulbs. When there are like 34 more bulbs (in a string of 35 lights), that's not a big deal, but eventually, another blows... then another... as each lamp goes out, the current rises a bit more (a bit more than it did when the previous lamp(s) went), and lamps start failing faster and faster until the whole string suddenly goes out rapidly as the additional current reduces the remaining lamps life from many hours to a few seconds. Cascading, catastrophic failure.
With a constant current supply, if an LED shorts, the supply compensates. With a regulated (constant) voltage supply, the current rises because the voltage is being held steady. So in a string of 12 LEDs, if one shorts, that's 8% more current piled onto each remaining LED. When one of the 11 remaining goes out, that's 9%. At this point, the 10 remaining LEDs are dealing with 17% more current than when they were all working, and a third LED is bound to go. When it does, the load on the remaining 9 is now 27% above nominal.
So fusing the LEDs with a very carefully chosen fuse is a good idea on a system that is using a constant voltage supply.
Ok, I think I follow you now. Yeah I saw the 4.8x4.8 in your calculation above and it kind of threw me off a bit. I wasn't really sure where that came from or why I coppied it. I should have just stuck to the formula.
So let me see if I got it now. Using the 12 xp-g voltage rating of 3.2.
48v power supply
12 leds 3.2v (12*3.2) = 38.4v
R = (48 - 38.4) / .96 = 10 ohm or 9.6 / .96 = 10 ohm
W = (9.6 * 9.6) / 10 = 9.21 watts or 92.16 / .96 = 9.21 watts
So I'd be looking for a 10ohm / 10-20watt resistor. But it's best to go with a higher safety margin so a 10 ohm / 20 watt resistor would be more appropriate. And this would limit the current able to pass through each string to 960mA. So in a case where I want to run parallel strings I could theoretically use resistors to to aid in keeping my strings/fixture balanced. Which I think you hinted to a few posts ago. And therefor the resistors would actually be serving more than one function as apposed to using them as just a simple way to test current going to each string instead of having to open up strings and hook up your meter in series to test current. And properly chosen fuses is still a good idea as a way to help avoid a cascading failure effect.
I think I got it dude.
I was actually under the impression from other posters that when a cree led fails it usually fails and shorts open thus losing all current being able to pass through the string. But I was reading from another source of information that was saying this actually rarely is the case.
scsinet
Active Member
Quote:
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410575
So I'd be looking for a 10ohm / 10-20watt resistor. But it's best to go with a higher safety margin so a 10 ohm / 20 watt resistor would be more appropriate. And this would limit the current able to pass through each string to 960mA. So in a case where I want to run parallel strings I could theoretically use resistors to to aid in keeping my strings/fixture balanced.
I was actually under the impression from other posters that when a cree led fails it usually fails and shorts open thus losing all current being able to pass through the string. But I was reading from another source of information that was saying this actually rarely is the case.
Yes, I'd go with a 15W resistor minimum. Operating a resistor continuously at more than 90% of it's rated power handling will likely result in reduced lifespan.
Keep in mind that this resistor will dissipate a considerable amount of heat... so you'll want to separate it from any combustible surfaces by at least a couple inches all around and at least 6-8 inches above. That, or you can use a Dale resistor which includes a heatsink and bolt it to a dedicated heatsink. Alterantively, you can reduce the power levels by adding another one or two LEDs to each string, which will downsize the resistors value and power dissipation.
You MUST use a resistor (or a more sophisicated means of current limiting) when using LEDS on a regulated voltage supply. Your remarks implied that they will help with multiple strings, but never forget that in this setup, they are mandatory. Your LED strings must have a total voltage of something less than the supply voltage, with a resistor included to make up the difference. Best practices say to get as close as possible with your LEDs to reduce waste in the resistor.
Semiconductors either fail open or fail shorted, and either scenario is possible. If you had a string of 100 LEDs on a 350V supply this is not as big of a concern because a "failed shorted" LED increases the current in the others by a smaller percentage, but when dealing with a smaller number, a single failure has a bigger impact and thus can cause additional failures in a much shorter timeframe, leading to catastrophic failure of the entire string, worst case.
Originally Posted by 2Quills http:///t/386777/led-s-are-the-best/100#post_3410575
So I'd be looking for a 10ohm / 10-20watt resistor. But it's best to go with a higher safety margin so a 10 ohm / 20 watt resistor would be more appropriate. And this would limit the current able to pass through each string to 960mA. So in a case where I want to run parallel strings I could theoretically use resistors to to aid in keeping my strings/fixture balanced.
I was actually under the impression from other posters that when a cree led fails it usually fails and shorts open thus losing all current being able to pass through the string. But I was reading from another source of information that was saying this actually rarely is the case.
Yes, I'd go with a 15W resistor minimum. Operating a resistor continuously at more than 90% of it's rated power handling will likely result in reduced lifespan.
Keep in mind that this resistor will dissipate a considerable amount of heat... so you'll want to separate it from any combustible surfaces by at least a couple inches all around and at least 6-8 inches above. That, or you can use a Dale resistor which includes a heatsink and bolt it to a dedicated heatsink. Alterantively, you can reduce the power levels by adding another one or two LEDs to each string, which will downsize the resistors value and power dissipation.
You MUST use a resistor (or a more sophisicated means of current limiting) when using LEDS on a regulated voltage supply. Your remarks implied that they will help with multiple strings, but never forget that in this setup, they are mandatory. Your LED strings must have a total voltage of something less than the supply voltage, with a resistor included to make up the difference. Best practices say to get as close as possible with your LEDs to reduce waste in the resistor.
Semiconductors either fail open or fail shorted, and either scenario is possible. If you had a string of 100 LEDs on a 350V supply this is not as big of a concern because a "failed shorted" LED increases the current in the others by a smaller percentage, but when dealing with a smaller number, a single failure has a bigger impact and thus can cause additional failures in a much shorter timeframe, leading to catastrophic failure of the entire string, worst case.
2quills
Well-Known Member
Understood. Thermal runaway is exactly what I'd be trying to avoid.
Actually if I was going to go with the 38.4 vF scenario I would probably drop down to a 42v power supply. And keep the 48v supply for the 43.2vF scenario for the two separate series of led series/colors. Since those are the two closest suited drivers that meanwell offers.
Crazy things running through my mind. Resistors would seem as though they should be a MUST have for running parallel strings after this discussion. Would you feel the same for a single series string? The vast majority of these diy builds that I see aren't using resistors for single series and that ones that are running parallel strings are simply using the 1ohm/5w resistors as recommended on one of the diy supply sites. And they are using those same resistors on a wide range of build sizes as a simple way to test voltage drop across strings. If you look at the pdf.'s on these drivers they have a minimum and a maximum voltage requirement. But in the case of the ELN drivers that the O.P. is using for instance...you could run 8-12 leds at 3.5v in series. So why isn't a resistor need to run either or? Or should it be and people just aren't doing that?
Should I be looking into current mirrors vs resistors or both? Are my questions annoying you yet? lol You don't have to answere them. I just enjoy the fact that I feel like I'm actually learning something here. And appreciate the input.
Actually if I was going to go with the 38.4 vF scenario I would probably drop down to a 42v power supply. And keep the 48v supply for the 43.2vF scenario for the two separate series of led series/colors. Since those are the two closest suited drivers that meanwell offers.
Crazy things running through my mind. Resistors would seem as though they should be a MUST have for running parallel strings after this discussion. Would you feel the same for a single series string? The vast majority of these diy builds that I see aren't using resistors for single series and that ones that are running parallel strings are simply using the 1ohm/5w resistors as recommended on one of the diy supply sites. And they are using those same resistors on a wide range of build sizes as a simple way to test voltage drop across strings. If you look at the pdf.'s on these drivers they have a minimum and a maximum voltage requirement. But in the case of the ELN drivers that the O.P. is using for instance...you could run 8-12 leds at 3.5v in series. So why isn't a resistor need to run either or? Or should it be and people just aren't doing that?
Should I be looking into current mirrors vs resistors or both? Are my questions annoying you yet? lol You don't have to answere them. I just enjoy the fact that I feel like I'm actually learning something here. And appreciate the input.