[Solved] Resistor in constant current circuit

Thread Starter

Madelaire

Joined Mar 1, 2018
11
Hi
I am making some lighting for a project, using LEDs. There are going to be two sources of light:

  1. 13x 3W LEDs
  2. 2x 3W LEDs
I have acquired a LED-driver 0.7A with a minimum load of 25W and a maximum of 60W.
When connecting all 15 LEDs in series there is no problem. But, I would like to connect a Toggle Switch (ON-ON-ON) to give me three options.

a. Turn on all lights
b. Turn on only source 1
c. Turn on only source 2

Options (a-b) are no problem. Because they exceed the minimum load of the LED-driver. But option (c) does not.
To accommodate this I thought of placing a resistor in option (c), as shown in the circuit below. NB: I could not find the right switch, so only option (b-c) is illustrated. But it gets the point across.

Ikke-navngivet.png

I thought of using a 50W / 56 ohm Power Resistance. But I am not sure how to choose a resistance for a constant current circuit.

Any input will be appreciated!! :)
 

MrChips

Joined Oct 2, 2009
27,654
In theory, you do not need a series resistor because the supply is already a constant current source.
If you wish you can use a resistor to provide some additional load and some level of protection.

Use the 3W @ 700mA to calculate the voltage drop across each diode. 3W/700mA = 4V
Each diode represents a load of approx. 6Ω
For a 10Ω series resistor, the power dissipated is 5W.
Or you can use any number of 5Ω /5W resistors in series.
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
In theory, you do not need a series resistor because the supply is already a constant current source.
If you wish you can use a resistor to provide some additional load and some level of protection.

Use the 3W @ 700mA to calculate the voltage drop across each diode. 3W/700mA = 4V
Each diode represents a load of approx. 6Ω
For a 10Ω series resistor, the power dissipated is 5W.
Or you can use any number of 5Ω /5W resistors in series.
Yes since it is constant current there should be no problem. But the manufacture said that the minimum load of the Current source is 25W. And if I only place 2 diodes the load will be 6W. Doesn't that mean that I need a resistance to increase the load to a minimum of 25W?
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
I was thinking that the load had to be 25W. Therefore I as missing 25W-3W-3W=19W.
WIth a current of I=0.7A the voltage drop over the resistor had to be 19W/0.7A=27.1V.
In ohms that is R=27.1V/0.7A=38.7ohm.
Then to be safe i increased to 56ohm
 

MrChips

Joined Oct 2, 2009
27,654
Sorry, I missed the part about minimum load of 25W and maximum of 60W. This means that the supply has a voltage range from 35V to 85V.
A 56Ω/50W resistor will get extremely hot!
Have you ever touched a 25W soldering iron or a 40W lightbulb?

Consider building a 50Ω dummy load using five 10Ω/10W resistor mounted on a heat sink.
 

mcgyvr

Joined Oct 15, 2009
5,394
It makes zero sense to me to make a LED fixture then to need to dump all that heat (waste all that energy) to make it work the way you intend..

What is the reason for the switch? Lower light levels? if so have you looked into a dimmable driver or using PWM to achieve dimming instead?
Adding another driver for the smaller load?
or many other more efficient options..
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
Sorry, I missed the part about minimum load of 25W and maximum of 60W. This means that the supply has a voltage range from 35V to 85V.
A 56Ω/50W resistor will get extremely hot!
Have you ever touched a 25W soldering iron or a 40W lightbulb?

Consider building a 50Ω dummy load using five 10Ω/10W resistor mounted on a heat sink.
May be a dumb question, but why not a 50ohm load with a heatsink?
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
It makes zero sense to me to make a LED fixture then to need to dump all that heat (waste all that energy) to make it work the way you intend..

What is the reason for the switch? Lower light levels? if so have you looked into a dimmable driver or using PWM to achieve dimming instead?
Adding another driver for the smaller load?
or many other more efficient options..
You are completely correct. And it is a feature that almost never is going to be used.
I am making a small aquaponic system. To this I need lighting for the plants and the fish. They are going to be active at the same time. But if the water in the tank has to be changed or fix something with the plants I need to be able to disconnect one of the sections.
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
Possible, but is it practical? How much space have you got for a 19W (at least) heatsink? What materials (flammable, non-flammable) will be adjacent/surrounding the heatsink?
Okay :)
I have come to understand that I should drop it. And just live with not being able to disconnect a section.
 

mcgyvr

Joined Oct 15, 2009
5,394
Okay :)
I have come to understand that I should drop it. And just live with not being able to disconnect a section.
Not necessarily.. But only you know the specifics/design parameters around which you are making that decision off of..

I'm not clear why you need to split them nor why you need to do so unevenly..
With a constant current driver you can certainly have a range of LEDs on it..
Provided that the smallest group still allows the min output voltage spec to be met and that the largest group still allows the max output voltage to spec to not be exceeded..

"IF" the DC output range of that driver is 35V to 85V and assuming each LED has a 3V forward voltage then you could have as few as 12 LEDs or as many as 28 LEDs and it would work just fine..
Its just once it gets below 12 or above 28 it will not be able to regulate the current properly.. But inside that range all is fine..
 

ebp

Joined Feb 8, 2018
2,332
Are you certain the DC output is galvanically isolated from the AC line? The load requirement strikes me as odd, and I'm a bit suspicious the driver is essentially a buck type non-isolated converter that is not isolated.

Can you show us some pics or a link to data, etc?
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
Are you certain the DC output is galvanically isolated from the AC line? The load requirement strikes me as odd, and I'm a bit suspicious the driver is essentially a buck type non-isolated converter that is not isolated.

Can you show us some pics or a link to data, etc?
I tired finding the correct data sheet, but I cannot.
It is a Mean Well LPC-60-700 (60W and 700mA).
Beneath is a link for the data sheet LPC-60-1050 / 1400 / 1750. It should tell you the necessary information.
http://www.meanwell.com/productPdf.aspx?i=248
 

ebp

Joined Feb 8, 2018
2,332
Mouser does not list an 700 mA CC supplies from Mean Well. It is a reputable company, so I would be surprised if the supply is not fully isolated and has safety agency approvals.
 

Thread Starter

Madelaire

Joined Mar 1, 2018
11
Mouser does not list an 700 mA CC supplies from Mean Well. It is a reputable company, so I would be surprised if the supply is not fully isolated and has safety agency approvals.
That might be, but that is never the less what I got. I have attached a picture of if. Please correct me if I am wrong.
But your point was that you do Not think there is a minimum load requirement on the power supply?
 

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mcgyvr

Joined Oct 15, 2009
5,394
There is absolutely a lower limit to them and when you drop lower than it they tend to flash on/off as it doesn't regulate properly anymore..
Easiest solution is to purchase a different driver with dimming option..

Or just try seeing if it will work with only 3 or 4 leds. (I suspect it should). Then you can wire it up so that you have your full 15 LEds on one setting and 3 or 4 on the other setting
 

ebp

Joined Feb 8, 2018
2,332
No, my point was with regard to safety. A lot of hobbyists buy LED drivers from Chinese vendors. Some of them are unsafe. Some are safe for the original purpose which was built entirely inside a "light bulb" where isolation is not an issue. I was concerned that your supply might have been intended to be built inside something that removed risk of shock and hence did not require isolation of the DC output from the AC line.

It is not uncommon for a constant current supply to have a minimum output voltage requirement because internal control power comes from the same transformer and tracks in proportion to the output voltage.

I don't know what to think here. From you photo I would conclude the supply an ordinary 78 VDC constant voltage supply, not a constant current supply, yet the datasheet, which does not include that model, says constant current for the series. There are no safety agency markings it.
 
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