Efficient current through 30 leds

[AmateurDesign]

Hey there, I've gotten some good advice on here before and was hoping for some help once again.

I am working with a uni group to develop a circuit that powers 30 leds. The arrangement of the leds is up to me.
Ideally, I want it to be low voltage, so that the device can be powered by batteries. But it also needs to power the leds for at least 60, possibly 90, minutes.
The leds that I am using are the ThorLab 830L:

Thorlabs - LED830L 830 nm LED with a Glass Lens, 22 mW, TO-18

Thorlabs specializes in the building blocks for laser and fiber optic systems. From optomechanical components to telecom test instrumentation, Thorlabs' extensive manufacturing capabilities allow us to ship high quality, well priced components and devices for next-day delivery. Optomechanics...

www.thorlabs.com

They each have a forward voltage of 1.6V, so if I do all 30 in series, that's far too high of a voltage than I want.
Here's the catch though...the leds are designed to emit a specific wavelength light (830nm) at 50mA current. And the more resistors in the circuit, the more variation that I have between led current values.

Lastly, in addition to the sensitive current parameter, there's also the issue of intensity. I need each of these leds to be at the same intensity or at least as close as possible.


I considered having six parallel branches containing 5 leds in series with a single resistor. But 5 leds in series is already 8V, disregarding the voltage across the resistor. I'm not sure if putting all 5 branches in parallel will simply add up my voltage requirements...meaning 8V x 6 branches = 48V total.
Any help would be greatly appreciated! Any additional info I can provide I'll do so as promptly as possible.

 

[Pommie]

Putting six strings in parallel will increase the current to 300mA and keep the voltage at 8V but you will need a resistor in each string.

Mike.
Edit your power requirement (assuming a 12V supply) is 12*0.3 = 3.6W. A 2Ah 18650 cell holds about 8W so could power it for ~2 hours. You'd need a boost converter and an 80Ω (1/4W) resistor in each string.
Edit2, this setup will give you 50mA in ALL LEDs so they should all be the same brightness.

 

[For The Popcorn]

Your 8 volt LED strings in parallel will need only 8 volts across them.... Mike beat me to it.

Rather than resistors for each string, a constant current driver circuit may be a better bet, and provide current adjustment for each string. There are many ways to achieve this, from dedicated LED driver chips to simple transistor circuits.

 

[Pommie]

I think a boost converter and series resistor in each string will be the simplest approach. See my edit above.

Mike.

 

[Pommie]

Or use one of these and a power bank. (Plus the resistors of course)

Mike.

 

[Pommie]

Wow, those LEDs are very expensive.

Mike.

 

[AmateurDesign]

Putting six strings in parallel will increase the current to 300mA and keep the voltage at 8V but you will need a resistor in each string.

Mike.
Edit your power requirement (assuming a 12V supply) is 12*0.3 = 3.6W. A 2Ah 18650 cell holds about 8W so could power it for ~2 hours. You'd need a boost converter and an 80Ω (1/4W) resistor in each string.
Edit2, this setup will give you 50mA in ALL LEDs so they should all be the same brightness.

I'm very much looking forward to fully understanding everything here one day, but I'd love some clarification/verification please and thank you. lol

I'm going to include a picture of my work, with letters to serve as a guide.
A) I know that current through a series circuit stays the same and, since I need 50mA and am assuming a 12V source, I can calculate my resistor requirement for a single branch with 5 leds. I get: 80Ω

B) This was work from a previous page, but verifies that my calculations are correct.

C) It seems pretty obvious (now that I think about it) that the current needs would just add up.

D) P=IV gives me the wattage requirements for the circuit.
And for the source, if we use this battery as an example:

Samsung 25R 18650 2500mAh 20A Battery

Get the Samsung 25R 18650 2500mAh 20A Battery, the most widely sold and respected 18650 cell on the market. With a high capacity of 2500mAh and a maximum continuous discharge rating of 20A, this battery is perfect for power packs, power banks, power tools, and more.

www.18650batterystore.com

It has an I=2.5A, V=3.6V nominal. If I do P=IV = 20(3.6) = 9W

Then I can simply divide that value by my circuit wattage requirements to find around how long in terms of hours?
9/3.6 =2.5 hrs (ideal)
I thought watts were in terms of joules/sec? But we can treat these in terms of hours?


I'm going to have to watch a video and do a bit of research into the boost converter before I can respond to that. I've got to get some homework done and have class right now. But once that's done, I'll post once more.

Thank you Mike and Popcorn, you two rock!

 

[Pommie]

Yes, all correct.
Batteries are rated in Ah - your battery is 2.5Ah (not 2.5A) meaning it can supply 2.5A for 1 hour or 0.25A for 10 hours or 1.25A for 2 hours. So to say it contains 9W is wrong (sorry, I started that) and it should be Wh. This is why we can use the number to calculate the time the battery will last.

Calculating your series resistor was also easy. I know that 50mA is one twentieth of an amp so to drop 4V requires 4*20Ω.

After seeing this thread I googled 830nm and found some very interesting articles. This being the most informative,

Adjunctive 830 nm light-emitting diode therapy can improve the results following aesthetic procedures

Background: Aggressive, or even minimally aggressive, aesthetic interventions are almost inevitably followed by such events as discomfort, erythema, edema and hematoma formation which could lengthen patient downtime and represent a major problem to the ...

www.ncbi.nlm.nih.gov

I also checked the price of 830nm LEDs and found these. Less than ¼ the price and twice the output. This is the Australian Farnell site.

Mike.
Edit, The boost converter you can think of as watts out = watts in * efficiency (normally ~95%).

 

[crutschow]

Below is the LTspice simulation of a simple two-transistor constant-current circuit that will operate with an LED voltage (yellow trace) to within <1V of the battery voltage (red trace) with the LED current (green trace) staying quite constant above that:
That allows good use of the battery voltage for better efficiency, and minimizes change in LED brightness with battery voltage.

Note that the LEDs I used in the simulation have about a 1.37V drop @ 50mA rather than the 1.6V drop of your LEDs.
For ones with 1.6V drop you should be able to put 6 LEDs in series, giving 5 strings, to operate below 11V battery voltage.

 

[AmateurDesign]

Yes, all correct.
Batteries are rated in Ah - your battery is 2.5Ah (not 2.5A) meaning it can supply 2.5A for 1 hour or 0.25A for 10 hours or 1.25A for 1/2 an hour. So to sat it contains 9W is wrong (sorry, I started that) and it should be Wh. This is why we can use the number to calculate the time the battery will last.

Calculating your series resistor was also easy. I know that 50mA is one twentieth of an amp so to drop 4V requires 4*20Ω.

After seeing this thread I googled 830nm and found some very interesting articles. This being the most informative,

Adjunctive 830 nm light-emitting diode therapy can improve the results following aesthetic procedures

Background: Aggressive, or even minimally aggressive, aesthetic interventions are almost inevitably followed by such events as discomfort, erythema, edema and hematoma formation which could lengthen patient downtime and represent a major problem to the ...

www.ncbi.nlm.nih.gov

I also checked the price of 830nm LEDs and found these. Less than ¼ the price and twice the output. This is the Australian Farnell site.

Mike.
Edit, The boost converter you can think of as watts out = watts in * efficiency (normally ~95%).

Gotcha! 2.5Ah makes sense, thank you.
Also, thank you for the led recommendation. I was brought onto this team and they had already gone with those really pricey leds. Not exactly sure why, but didn't come out of my pocket, so I don't fret too much.

So the boost converter just boosts voltage by reducing current? Since I only need 300mA, then I could afford to lose some current for the sake of voltage.

Basically, I need to arrange these leds on a breadboard and try to get a working circuit with all 30 leds. If I have several strings of leds connected in parallel and put current through it, is there a risk of intensity variation due to the current flowing through each string at a slightly different time? This is something that I was unsure about.

 

[AmateurDesign]

Below is the LTspice simulation of a simple two-transistor constant-current circuit that will operate with an LED voltage (yellow trace) to within <1V of the battery voltage (red trace) with the LED current (green trace) staying quite constant above that:
That allows good use of the battery voltage for better efficiency, and minimizes change in LED brightness with battery voltage.

Note that the LEDs I used in the simulation have about a 1.37V drop @ 50mA rather than the 1.6V drop of your LEDs.
For ones with 1.6V drop you should be able to put 6 LEDs in series, giving 5 strings, to operate below 11V battery voltage.

Hey Crutschow, thank you for your response! I had been told that a transistor might be better at controlling the current through the leds, so that there's no variation in intensity. I do have a microcontroller that can simply issue the command to flip the transistor.

I've used Multisim in a previous class and have wanted to get into LTspice, just because it seems to be popular.
Your circuit is a bit above my head, though I did already watch a couple of videos on something like yours.
It might be exactly what I need, I'm not sure just yet.

I noticed that there's no resistor immediately before the leds. Is that because the resistors in your constant-current circuit make up for that? Would the 5 strings just be put in parallel at the OUT point?

Lastly, do you have any recommendations for me to better understand what you've shared?
Thank you again, I appreciate your time and energy!

 

[crutschow]

I do have a microcontroller that can simply issue the command to flip the transistor.

What do you mean "flip the transistor"?

noticed that there's no resistor immediately before the leds. Is that because the resistors in your constant-current circuit make up for that?

Yes, the value of R3 determines the LED current, as the equation next to it shows.

Would the 5 strings just be put in parallel at the OUT point?

No, each string needs its own current limiter circuit to avoid one string taking more current than the others.

do you have any recommendations for me to better understand what you've shared?

If you understand how BJTs work, then understanding its operation should not be difficult.
R1 provides base current to Q2 to turn it on (current from emitter to base turns on a PNP transistor).
When the output current through R3 generates sufficient voltage across it to turn on Q3 (about 0.65V) then it starts to conduct and raises the voltage on R1 until Q2 starts to turn off, thus limiting the current to that value.

The circuit could also be built with NPN transistors, but the circuit would be put in series with the ground side of the LEDs instead of the high side.

Make sense?

If you simulate the circuit, then you can look at all the node voltages and currents to help understand how it works.
Might be a good time to try that with LTspice, since it's a simple circuit to start with.
I can post a copy of my .asc simulation file.

 

[danadak]

This uses a 3 Terminal adjustable regulator for a constant current source.
Has advantage it is both internal short circuit and thermal protected.

The current is Iled = 1.2V / Rout

Note I used regular diodes as I did not have a spice model, and modified their threshold to 1.6V.


Regards, Dana.

 

[danadak]

Note you can get LM317HV versions that can handle 60V input -

https://www.ti.com/product/LM317HV

https://www.ti.com/lit/ds/symlink/lm317hv.pdf?ts=1650464501646&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM317HV%253Futm_source%253Dgoogle%2526utm_medium%253Dcpc%2526utm_campaign%253Dapp-null-null-GPN_EN-cpc-pf-google-wwe%2526utm_content%253DLM317HV%2526ds_k%253DLM317HV%2526DCM%253Dyes%2526gclid%253DEAIaIQobChMIrc_fr-ui9wIVUrTVCh10QALgEAAYASAAEgJndPD_BwE%2526gclsrc%253Daw.ds

Regards, Dana.

 

[crutschow]

This uses a 3 Terminal adjustable regulator for a constant current source.
Has advantage it is both internal short circuit and thermal protected.

The downside for battery operation is that its dropout voltage is about 3V, whereas the circuit I posted requires less than a volt.

 

[danadak]

Of course depends on # LEDs / string and what battery V he plans on using.

Otherwise use an LDO....dropouts << 1V, plus you get short circuit and thermal protection.


Regards, Dana.

 

[danadak]

Sim, you do get the low dropout at the expense of circuit protection -

Regards, Dana.

 

[Pommie]

The op stated he wants to use one 18650 battery.

Mike.

 

[danadak]

I focused on 5 leds / string., constant current (same in each LED).

So I guess the ideal would be a const current boost converter to
help in LED to LED brightness variation reduction ? Still with
some # LEDs in series. or just live with variation due to T and V
and Die the V drive R + LED approach results in ?


Regards, Dana.

 

[Pommie]

Due to the OPs lack of experience (but a fast learner), a simple approach would be best. Hence why I suggested a (voltage) boost converter and series resistors.

Mike.