Turn on 88 leds in two modes

Thread Starter

amelkhalifa

Joined Jul 3, 2019
7
Hello,
I want to turn on 88 leds in two modes DIM and BRIGHT.
The mode DIM is specified by low voltage (25VAC to 35VAC) and the current consumption must be between 5mA and 15 mA.
the mode Bright is specified by hight voltage (98VAC to 132VAC) and the current consumption must be between 5mA and 50 mA.
I should use passive components so I try to calculate the adequate current limiting resistor but it does'nt work when the voltage increases the current also increases and becomes out of specification.
can anyone help me to do this with passive components.
thank you,
 

LesJones

Joined Jan 8, 2017
4,174
Your schematic shows 3 strings of 29 LEDs in parallel. Each string has an effective series resistance of 4 times the value of the resistors used. (This is assuming that all the resistors are the same value.) Even if the LEDs are red ones (These have about the lowest forward voltage of about 1.8 volts except for IR LEDs.) So 29 would require at least 52.2 volts. So they would not work with 25 volts AC into the bridge rectifier as even the peak voltage of the waveform would only be 25 x 1.414 = 35.35 volts.

Les.
 

Thread Starter

amelkhalifa

Joined Jul 3, 2019
7
Thank you for your response.
The leds are white and the forword voltage is 3.2V.
I m not the designer of this schematic, they give me this schematic to correct it with passive components so I change the number of leds in evry chain and I calculate the adequate current limiting resistor but it does'nt work when the voltage increases the current also increases and becomes out of specification.
 

ebeowulf17

Joined Aug 12, 2014
3,307
If you don't want the current to scale up way faster than the voltage, you need to make more of the voltage drop be across the resistors, and less across the LEDs. Unfortunately, this is inefficient and will run hot.

4 LEDs in series per string
22 strings in parallel
12.8Vf per string
12.2V across resistors at 25V
2440ohms per string
5mA per string

119.2V across resistors at 132V
48.9mA per string

I can meet voltage and current requirements, assuming that was current through each LED, not total current through the device. I'm afraid dissipation and efficiency figures will be bad compared to an optimized single-voltage design.

At 25V, dissipation in resistors is 61mW per string. At 132V, it's 5.83W per string (128W for all 22 strings!!!)

I don't actually think the scheme above is a good idea, but it does meet the design requirements as described. I think this two voltage concept is a bad idea. However, if you really need it to work, you should probably consider active components to provide more control.
 

ebeowulf17

Joined Aug 12, 2014
3,307
On second thought, I wonder if this is homework and I shouldn't have given numbers. That would explain the apparently unrealistic requirements.

If moderators want to hide my previous post, that's cool.
 

Thread Starter

amelkhalifa

Joined Jul 3, 2019
7
On second thought, I wonder if this is homework and I shouldn't have given numbers. That would explain the apparently unrealistic requirements.

If moderators want to hide my previous post, that's cool.
thank you for your response, it is not a homework, it a given project, I want to change old project with fluorescent lamp to new one with LEDs
 

MisterBill2

Joined Jan 23, 2018
18,176
The fundamental concept of the two voltages is in error for LED devices, because LEDs are DIODES and the voltage to current relationship is very non-linear. If you stay with the 3 strings of 29 LEDS in series then you will not get much light at all below a forward voltage of 90 volts. That would be about 3.03 volts across each LED device. (Illuminated but not very bright)When you go to 92.8 volts you will have 3.2 volts across each LED and they will be quite bright. You will also have a lot more current.
Please note that my numbers do not include any voltage drops across resistors, which you will need to have at least 2 resistors to allow stable operation. And your dim/ bright voltages will not be so very far apart.
 

Thread Starter

amelkhalifa

Joined Jul 3, 2019
7
The fundamental concept of the two voltages is in error for LED devices, because LEDs are DIODES and the voltage to current relationship is very non-linear. If you stay with the 3 strings of 29 LEDS in series then you will not get much light at all below a forward voltage of 90 volts. That would be about 3.03 volts across each LED device. (Illuminated but not very bright)When you go to 92.8 volts you will have 3.2 volts across each LED and they will be quite bright. You will also have a lot more current.
Please note that my numbers do not include any voltage drops across resistors, which you will need to have at least 2 resistors to allow stable operation. And your dim/ bright voltages will not be so very far apart.
thank you for your response, according to you, how can I do this, what is the best solution? I thought about voltage amplifier when I have 25V to 90V and I minimize the number of leds in evry chain. but I don't know how to do this! how can I amplifier the voltage when I have 25V to 35V as input??
 

MisterBill2

Joined Jan 23, 2018
18,176
thank you for your response, according to you, how can I do this, what is the best solution? I thought about voltage amplifier when I have 25V to 90V and I minimize the number of leds in evry chain. but I don't know how to do this! how can I amplifier the voltage when I have 25V to 35V as input??
The method would be simpler than an actual amplifier, what is required is a regulated power supply with two voltage set-points. Other folks who participate here should be able to provide links to suitable designs for such a supply. AND I have a caution, which is that given the supply voltages described there would be a temptation to use a non-isolated power supply, which I do not recommend. A transformer isolated power supply is a much better choice.
 

AnalogKid

Joined Aug 1, 2013
10,987
An LED replacement bulb for a 48 inch fixture is $7 (US). There is nothing you can build for anything close to that price.

ak
 

MisterBill2

Joined Jan 23, 2018
18,176
An LED replacement bulb for a 48 inch fixture is $7 (US). There is nothing you can build for anything close to that price.

ak
AK, please know that I recently had to replace 26 of those LED 48 inch tubes because groups of the LEDs were failed. The failure starts wit one group of 24 flashing and eventually all 4 groups flash for a few hours and then fail. I have a post about that problem but no responses other that from the folks who always want a schematic of the circuit. But it appears to be the actual LEDs, since there are 4 strings of 24 in parallel, and the failure is one string at a time. So the replacement cost and effort soon wipes out the savings with that product.
 

AnalogKid

Joined Aug 1, 2013
10,987
My rebuilt kitchen / bath / laundry has about 20 PAR 30 LED floods. It will be interesting to see how they hold up compared to your and other experiences. Given that white LEDs are in some very high-reliability applications such as street lights and auto headlamps, it's surprising that the tube failure pattern points to individual LEDs rather than the regulator.

My basement has nine 4-tube fluorescent fixtures. I'm on my next-to-last case of used-but-good 4-foot tubes harvested from my workplace when they replaced all ballasts and tubes with high-efficiency components. Once they're gone, I'll join the LED club. I wonder if anyone has tracked $50 vs. $10 tube reliability.

ak
 

DickCappels

Joined Aug 21, 2008
10,152
The regulators are great places to look for cost savings, but they can also drive more than an appropriate amount of current to get more lumens per dollar. It might come at the expense of lumens per watt but there is big money to be saved when using the cheaper LEDs or fewere LEDs driven at higher current.

I have received datasheets from some Chinese manufactures and many of these datasheets have the names of other manufacturers. I asked why they were sending out datasheets with the wrong company name and the answer came back: When the company gets into trouble they merely change the name and continue with business under that name, in these cases they had not yet got around to printeing new datasheets.
 
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