Resistor in constant current circuit? (EE Q) - The Planted Tank Forum
Old 04-04-2018, 07:26 PM Thread Starter
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Resistor in constant current circuit? (EE Q)

I am evaluating this current mirror. The only potential problem is that the BC337 part has a max limit of .8A at collector. This works for a 700mA driver, but my drivers are 1500mA. There is a 680 Ohm resistor in series with the collector, but I'm not sure how a resistor works in a constant current circuit. (Everything in school was constant V) Does it limit the voltage or the current? If the full current of the driver goes through the collector, then this design is only applicable for up to 700mA drivers.

(Yes, there is another current mirror that doesnt use a BC337 and instead uses a diode in the current splitter. That circuit is far second choice because the diode siphons off close to a volt. Besides potentially pushing my voltage beyond what the driver can provide, using that circuit entails a ~3% power loss.) https://www.edn.com/design/led/43681...om-overcurrent

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Old 04-04-2018, 10:34 PM
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Originally Posted by ChrisX View Post
There is a 680 Ohm resistor in series with the collector, but I'm not sure how a resistor works in a constant current circuit. (Everything in school was constant V) Does it limit the voltage or the current? If the full current of the driver goes through the collector, then this design is only applicable for up to 700mA drivers.
I don't see your exact design on that link but the current wouldn't go through the BC337 since there is a 680 ohm resistor in series with the collector. The driver will have a maximum output voltage, once it is exceeded it will no longer maintain the rated constant current. The drivers maximum output will be the input voltage minus a small drop (usually ~3V), so if we assume we're using a 36V supply the maximum output voltage will be 33V, so the current would be:
I = V / R
I = 33V / 680ohm
I = 0.049A

So no worries on exceeding the BC337 current rating.

Edit: For what it's worth I couldn't find any references to your circuit and tried simulating it but I still got full current through string 1 if string 2 opened so not sure what the BC337 is supposed to add over the standard current mirror circuit.

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Last edited by MrMan; 04-04-2018 at 11:10 PM. Reason: edit
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Old 04-05-2018, 12:23 PM Thread Starter
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Originally Posted by MrMan View Post

Edit: For what it's worth I couldn't find any references to your circuit and tried simulating it but I still got full current through string 1 if string 2 opened so not sure what the BC337 is supposed to add over the standard current mirror circuit.
This was originally from one of the Recom constant current driver datasheets. This article describes it, somewhat: http://www.ledsmagazine.com/articles...-magazine.html

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Old 04-05-2018, 03:37 PM
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This was originally from one of the Recom constant current driver datasheets. This article describes it, somewhat: http://www.ledsmagazine.com/articles...-magazine.html
Gotcha. So in their example the BC337 is used to limit the current in string 1 to 445mA which is more than half the driver current but protects the string from the full 700mA. Since you're using 1.5A drivers you'll need to change the resistors from 1.5 ohms to 0.7 ohms (should limit to 1A if I understand their design). I tried simulating the circuit with their numbers and it didn't limit the current as they said it would, and most examples use the diode method from your original link so that would be my suggestion. But you could always build it up and then disconnect one of the LEDs to test it before putting it into operation.

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Old 04-05-2018, 04:05 PM Thread Starter
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Gotcha. So in their example the BC337 is used to limit the current in string 1 to 445mA which is more than half the driver current but protects the string from the full 700mA. Since you're using 1.5A drivers you'll need to change the resistors from 1.5 ohms to 0.7 ohms (should limit to 1A if I understand their design). I tried simulating the circuit with their numbers and it didn't limit the current as they said it would, and most examples use the diode method from your original link so that would be my suggestion. But you could always build it up and then disconnect one of the LEDs to test it before putting it into operation.
By simulated, do you mean run through a computer simulation, or you built a prototype?

The reason I trust this design is that it is (or at least appears to be) an example from a Recom datasheet for one of their drivers. I figured it wouldn't be wrong. OTH I didn't find this exact circuit in their current literature; it may have been modified by the author from the original Recom graphic.

Regarding the 1.5Ohm resistors, why do you suggest .7Ohms? I could run the 1.5 resistors in parallel to achieve this resistance ( I bought more than enough), but what exactly is that accomplishing? If anything, I thought altering the 680ohm resistor would be required to alter the funtion of the BC337.

If I understand the function correctly, when a led in the first string goes out, both go dark (because of current mirror). If a led in the second string goes out, the remaining string is run at a reduced current (because the transistor is acting to reduce gain.. somehow). I didn't see a problem running my remaining string at 445mA, thus kept the values identical. Also, I figured it might scale somewhat with higher current, perhaps the resistors create a ratio of current that gets through. I know what the BC377 is (amplifier) but don't understand how it is working here, unless it can also attenuate.

Its odd to me that neither article explains the theory of operation or gives any math. If someone is an EE they already understand this information, if someone is not, they need to know how to adapt it to their own circuit.

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Old 04-05-2018, 04:16 PM
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By simulated, do you mean run through a computer simulation, or you built a prototype?
Computer simulation, there are various programs out there for this. A free version I've used with decent results is LTSpice if you're interested in trying it out.

Quote:
The reason I trust this design is that it is (or at least appears to be) an example from a Recom datasheet for one of their drivers. I figured it wouldn't be wrong. OTH I didn't find this exact circuit in their current literature; it may have been modified by the author from the original Recom graphic.
Yes I noticed it was removed from their recent datasheet as well and everything I found on current mirroring didn't use this limiting circuit. My guess is the diode circuit is more reliable since I found numerous people using that to protect the second string.

Quote:
Regarding the 1.5Ohm resistors, why do you suggest .7Ohms? I could run the 1.5 resistors in parallel to achieve this resistance ( I bought more than enough), but what exactly is that accomplishing? If anything, I thought altering the 680ohm resistor would be required to alter the funtion of the BC337.

If I understand the function correctly, when a led in the first string goes out, both go dark (because of current mirror). If a led in the second string goes out, the remaining string is run at a reduced current (because the transistor is acting to reduce gain.. somehow). I didn't see a problem running my remaining string at 445mA, thus kept the values identical. Also, I figured it might scale somewhat with higher current, perhaps the resistors create a ratio of current that gets through. I know what the BC377 is (amplifier) but don't understand how it is working here.
The math is in the link but not super obvious. The way the BC337 limits the current is that once the voltage across the 1.5 ohm resistor hits 0.7V the BC337 will start conducting and limit current in the main transistor. So it's just math to figure out what current causes 0.7V drop across the 1.5 ohm resistor:

I = V / R = 0.7V / 1.5R = 466mA

So in your case you want a current limit of ~1A so that it won't limit current when it's being split properly but will kick in if the second string opens. So:

R = V / I = 0.7V / 1A = 0.7 Ohm

Quote:
Originally Posted by ChrisX View Post
Its odd to me that neither article explains the theory of operation or gives any math. If someone is an EE they already understand this information, if someone is not, they need to know how to adapt it to their own circuit.
They do explain it, but without a background in EE it is hard to follow text vs diagrams.

Edit: So the problem with my simulation was that the current source was "Ideal" in that it had no voltage limit, so while the current wasn't being limited as expected the voltage across the whole circuit was 75V. So it does seem like this circuit should work as expected. Still not sure why every new example on current mirroring with protection uses the diode method from your original link though.

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Last edited by MrMan; 04-05-2018 at 04:35 PM. Reason: found the problem with the simulation
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Old 04-05-2018, 05:01 PM Thread Starter
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Originally Posted by MrMan View Post
Edit: So the problem with my simulation was that the current source was "Ideal" in that it had no voltage limit, so while the current wasn't being limited as expected the voltage across the whole circuit was 75V. So it does seem like this circuit should work as expected. Still not sure why every new example on current mirroring with protection uses the diode method from your original link though.
Thanks for explaining it. Glad you got it working in your simulation, I'll let you know how it works in practice.

I prefer not to use the other example because of the voltage loss. My cxa1304 leds use almost 10v at 750mA. I'm using a 36v supply with LDD-1500L drivers which are stated as 2-30v out, yet datasheet says output is -3v from in (so could be 33v). This organization looks like a perfect match, I'm not sure what would happen if it was voltage limited.

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Old 04-05-2018, 05:20 PM
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Originally Posted by ChrisX View Post
Thanks for explaining it. Glad you got it working in your simulation, I'll let you know how it works in practice.

I prefer not to use the other example because of the voltage loss. My cxa1304 leds use almost 10v at 750mA. I'm using a 36v supply with LDD-1500L drivers which are stated as 2-30v out, yet datasheet says output is -3v from in (so could be 33v). This organization looks like a perfect match, I'm not sure what would happen if it was voltage limited.
The max output for the driver is 30V so it will probably reduce the current until the forward voltage of the LEDs is 30V. BUT the current mirror circuit will have some voltage drop as well, across both the transistors and resistors. So considering how tight the driver voltage is with 3 LEDs I doubt you could use the current mirror circuit. You could do 2 LEDs and the current mirror though, but I'm sure that will increase the cost by a fair bit...

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Old 04-05-2018, 05:37 PM Thread Starter
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The max output for the driver is 30V so it will probably reduce the current until the forward voltage of the LEDs is 30V. BUT the current mirror circuit will have some voltage drop as well, across both the transistors and resistors. So considering how tight the driver voltage is with 3 LEDs I doubt you could use the current mirror circuit. You could do 2 LEDs and the current mirror though, but I'm sure that will increase the cost by a fair bit...
I'll have to check the output with multimeter. I bet there is some headroom. How much voltage is lost from the current mirror with your suggested resistors? Can you simulate what is lost?

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Old 04-05-2018, 06:41 PM
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Why not go with a 54v power supply or 48V supply adjusted to 53V (e.g LRS-350-48), 5x LDD-H 700mA drivers (\$25 vs \$16 for the 4 LDD-L) and not have to spend hours trying to get circuitry working that will end up costing you extra in components and effort?

I undervalue my free time at ~\$10 and there are still too many instances where coming up with extravagant ways to save money ends up with no net cost savings. It looks like you've been debating about this project for nearly 3 weeks, and a conservative 1 hour per day of pondering stacks up to \$200 in effort. Is it really worth it to save \$50 or so over that time period?
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Old 04-05-2018, 07:50 PM
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I'll have to check the output with multimeter. I bet there is some headroom. How much voltage is lost from the current mirror with your suggested resistors? Can you simulate what is lost?
This seems to be where it gets complicated. The voltage drop across transistor and resistor is 6.5V in my simulation. This can be adjusted by lowering the 680 ohm resistor. BUT when you lower the 680 ohm resistor you reduce the effectiveness of the protection provided by the BC337.

Conversely the diode protection circuit if you don't use a resistor on the base pullup the drop is less than 2V. You also need to add the drop across the additional transistor/diode which brings it up to just under 4V. So I think this circuit is actually better for both efficiency and protection. Or go with @SpringHalo and get a higher voltage supply and 700mA drivers...

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Old 04-05-2018, 08:18 PM Thread Starter
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Originally Posted by SpringHalo View Post
Why not go with a 54v power supply or 48V supply adjusted to 53V (e.g LRS-350-48), 5x LDD-H 700mA drivers (\$25 vs \$16 for the 4 LDD-L) and not have to spend hours trying to get circuitry working that will end up costing you extra in components and effort?

I undervalue my free time at ~\$10 and there are still too many instances where coming up with extravagant ways to save money ends up with no net cost savings. It looks like you've been debating about this project for nearly 3 weeks, and a conservative 1 hour per day of pondering stacks up to \$200 in effort. Is it really worth it to save \$50 or so over that time period?
This is a hobby. Part of the satisfaction comes from an efficient well designed system, hopefully better than what is available.

I purchased the LDD-L drivers last year, specifically for their analog dimming capability. This is a multi-channel build that I specifically wanted a
number of pots for independently adjusting channels. Eventually I will move to arduino based PWM dimming, but there is value in ability to switch to analog controls.

Additionally, having a 36V supply w/ 30v avail from driver is better sized for my supplemental color channels. I can fit 20x 3W leds on a LDD-1200L. (Incidentally, the epi colors come in 10x allotments.) If I went with a higher voltage supply and LDD-H, I would only be able to run 14x color leds per channel (which I don't think is enough, meaning extra drivers with unused capacity), and I wouldn't have enough power for all the channels.

4x 1500 Whites= 6 Amp
3x 1200 Colors = 3.6 Amp

Power supply has 9.7 Amps.

This is the best possible design that encompasses all my requirements. Sure I could buy another power supply, buy more LDD-H drivers just to avoid using a current mirror, but in the grand scheme of things, building a current mirror into each channel is negligible. I also learned some new things and find it interesting.

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Originally Posted by MrMan View Post
This seems to be where it gets complicated. The voltage drop across transistor and resistor is 6.5V in my simulation. This can be adjusted by lowering the 680 ohm resistor. BUT when you lower the 680 ohm resistor you reduce the effectiveness of the protection provided by the BC337.

Conversely the diode protection circuit if you don't use a resistor on the base pullup the drop is less than 2V. You also need to add the drop across the additional transistor/diode which brings it up to just under 4V.
Lol, this is not going to waste 4-6.5v. Thats 20% of the power. The resistors/transistors would go up in smoke if that was the case. Your assesment doesn't pass the "sniff test".

Worst case scenario if the current mirror pushes over the voltage limit... Each group of six cxa1304 are on the same heat sink so temperature should be shared and prevent thermal runaway. Testing might show they can be run without current mirror. However, I can still use the simplified current mirror with just two BD139. It wont protect the other leds if one burns out.

Alternatively, I could run them as a 2x3 (or 3x3) array instead of a 3x2. It may be that I'm running them at < 50%, so running them at 500mA would be no big deal.

A 3x3 array has built in protection. If one string goes out, the other two are still capable of handing the current.

There are a number of avenues for this to be a success.

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Last edited by Darkblade48; 04-06-2018 at 03:27 PM. Reason: Please use the edit function for back to back posts to keep threads cleaner
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Old 04-06-2018, 04:00 PM
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Lol, this is not going to waste 4-6.5v. Thats 20% of the power. The resistors/transistors would go up in smoke if that was the case. Your assesment doesn't pass the "sniff test".
Sniff test? "Lol, this is not going to waste 4-6.5v" Based on what?

You asked for EE help and I'm trying to provide that, no assessment here, just trying to provide numbers and explain the circuit as you asked. Of course adding a current mirror circuit is going to waste power, adding any extra circuit wastes power, figuring out if it's worth it is your decision.

Quote:
The voltage drop across resistor R3 compensates for the mismatching of LED voltage drops and holds both Q1 and Q2 in the linear region.
If the transistors are in the linear region they will have a voltage drop across them, they aren't being saturated to act as an on/off switch. As I mentioned before if you remove the base resistor, this does saturate the transistors and brings the voltage down to near zero, but you can't use the BC337 protection method.

As for "going up in smoke", 6V across the transistor with 0.75A is 4.5W and the sources you provided state that they transistors should be mounted to the same heatsink, a SOT-32 isn't going up in smoke with 4.5W if attached to a heatsink. But the resistor would have to go up from a 0.5W to a 1W for your increased driver size.

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Last edited by MrMan; 04-06-2018 at 04:01 PM. Reason: edit
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Old 04-06-2018, 04:24 PM Thread Starter
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Sniff test? "Lol, this is not going to waste 4-6.5v" Based on what?
Based on the fact that the circuit is worthless and wouldn't be published if it used that much. And certainly not without mentioning that it will rob 20+% of your power.

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Old 04-06-2018, 04:38 PM
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Based on the fact that the circuit is worthless and wouldn't be published if it used that much. And certainly not without mentioning that it will rob 20+% of your power.
Just because it's worthless to you doesn't mean it's worthless to everybody else. I agree that this design isn't as good as the other protection circuit, which is probably why Recom removed the design from their datasheet

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