Resistors for LEDs

[Colin HeseltineParticipant @colinheseltine48622]

After some help with resistors used with Led’s

i have led’s with forward voltage between 2.1( current 30ma) and 3.1 volt (current 20ma)

power supplied by 7.4v battery

i think I need resistors of 176ohm (nearest 180 ohm) & 215 ohm (nearest 220ohm)

Can someone please confirm this is correct before I blow anything up.

Many thanks

Colin

[Grindstone CowboyParticipant @grindstonecowboy]

Yep, sounds fine to me.

Rob

 

[John HaineParticipant @johnhaine32865]

Take the 2.1V one.  Supply is 7.4V so resistor has to drop 5.3V with 30mA flowing, so its value is 5.3V/0.03A = 177 ohms.  Similarly for the other.  Just Ohm’s law.

[Robert Atkinson 2Participant @robertatkinson2]

Yes, your sums are correct.

You have not considered thr power rating. For the first this is the voltage across the resistor times the current Nominally (7.4-2.1) * 0.03 = 0.16 W. This indicates a 1/4 W rated resistor is adequate. A 180 Ohm 1/4 W resistor is not OK for a short across a 7.4 V battery (P =Vsquared/R) =0.3 W

Note that unless this is specifically for illumination or use in sunlight you may find the LEDs too bright. You may be able to use 50% of the rated current. Reducing current increases the reliability of the LED.

Robert.

[Colin HeseltineParticipant @colinheseltine48622]

These are for the lights on a BR08 shunter. I have 3D printed it as since my surgery 3 months ago and subsequent 2 weeks in hospital with infection I am not allowed near workshop.
the lights will be quite shrouded

 

Am I correct in thinking that if I double the resistor value, eg by putting two in series this will dim the led.

Thanks for the rapid replies.

Colin

[not done it yetParticipant @notdoneityet]

The value of resistors in series is the sum of the individual values.  Then you can apply Ohm’s Law for the new current and use Power = current squared*Resistance .  [Alternatively rearrange the algebra to find the power (Wattage).]

For resistors in parallel the reciprocal of the total resistance is the sum of the reciprocals of the individual resistors.

[Colin Heseltine]

On Colin Heseltine Said:

These are for the lights on a BR08 shunter. I have 3D printed it as since my surgery 3 months ago and subsequent 2 weeks in hospital with infection I am not allowed near workshop…

Poor you! Get well soon.

 

Am I correct in thinking that if I double the resistor value, eg by putting two in series this will dim the led.

Thanks for the rapid replies.

Colin

Yes, and note that Robert suggested increasing the calculated resistance because that runs the LED at full brilliance, which may be too bright.

Brightness depends on the age and colour of the LED.   LEDs have improved enormously since they first hit the streets, and the latest are likely to be too bright for comfort!  If so, increase the value of the resistor.

I tend to use 330Ω for older LEDs (my junkbox is full of them),  and 1kΩ for the bright new ones.  Although too low a resistance shortens their lives, LEDs aren’t fussy – almost any resistor up to 2k2 will light one up.

If the shunter is battery powered, might be worth dimming the LEDs by pulsing their power input.   Dropper resistors work by converting unwanted current into useless heat, not good.   More efficient to chop the power up, so that the LED is dimmed by only being ‘on’ for part of the time.   The human eye won’t detect flicker provided the chop rate is faster than about 40Hz.  This table is for 200Hz:

 

The NE555 chip was/is popular for this, there are some dedicated chips, and the functionality is trivial for a microcontroller.   The cost is extra complexity.   Dropper resistors may be inefficient, but they’re nice and simple.

Dave

[MacolmParticipant @macolm]

Er no, not unless there is a circuit with an inductor rather than a resistive dropper. Pulsating a resistive LED dropper will make the efficiency slightly worse.

[Macolm]

On Macolm Said:

Er no, …

I got as far as typing “Er no, but no” when it hit me Macolm is on to something!   Gut feel is my PWM suggestion is misguided, no use to Colin.   Belly hurting too much to get my head round why not!  Yet  another problem on my ‘To Do’ list…

🙁

Dave

[Robert Atkinson 2Participant @robertatkinson2]

Sorry you are ill SoD. Hope you get well soon.

Actually using pulse drive on LEDs is very effective.  While Macolm is correct and an inductor is normally used to smooth the current in a PWM supply this is not the best approach for LEDs.
It’s a physiological thing. The human eye responds to peak intensity not average. So a LED driven with pulses looks brrighter than the same energy used for constant illumination. The trick is to use something like 80% of the peak current rating and appropriate duty cycle. A couple of random datasheets for Red LEDs rated at 20mA nominal shows 100mA peak and 10% to 30% duty cycle.
https://www.farnell.com/datasheets/2925628.pdf
https://4donline.ihs.com/images/VipMasterIC/IC/AVGO/AVGO-S-A0004682023/AVGO-S-A0004682023-1.pdf?hkey=6D3A4C79FDBF58556ACFDE234799DDF0
So if we use 80mA and 15% duty cycle the average current drops to 12mA, The total power drops to 0.089W (down from 0.148W)  and the average power “wasted” in the resistor (now 68 ohm) drops to 0.062W
All this and the LED looks brighter.

Robert.

[duncan webster 1Participant @duncanwebster1]

Another way of reducing consumption is to put 2 leds in series, thus the same current lights both, you then need a smaller resistor of course. I find you can run them well below max without noticeable reduction in brightness, just experiment. They last a lot longer then as well.

[MacolmParticipant @macolm]

An inductor used as a simple buck regulator where the LED replaces the capacitor, and with the feedback to control the current will deliver pretty good DC and stable brightness to the LED. For many applications that is a very good solution. Granted pulsed LED operation may provide a slight boost to subjected brightness, but it also risks strobe effects which can prove a nuisance. Both will also need supply filtering.

[Robert Atkinson 2Participant @robertatkinson2]

Duncan’s suggetion of LEDs in series is a good arrangement when using and existing powersupply that is well above the LED forward voltage. Ideal for vehicle models where lights come in pairs.

Macolm’s buck regulator is overly complex. In particular you don’t have to have an inductor. Another problem is most smitchmode regulator IC’s are not designed for constant current operation. This does not mean you can’t arrange a circuit for constant current but it can be more complex. I have actually done just this, many years ago using a LM2575T-ADJ. By coincidence I dug out my prototype this month because I was adding a LED illuminator to my H&W Autocollimator. I tried both with and without inductor. Without appears brighter because of higher, pulsed current but needs a extra capacitor in the feedback loop. The LM2575T-ADJ has a 1.25V reference and a feedback pin. To use for constant current you have to put the current sense resistor in the cathode circuit of the LED. A 10 Ohm resistor gives 125mA.
This is only really suitable for high current / power LEDs. The quiesent current drawn by the IC would be significant comepared to the output current for a small LED. My original application for the circuit used 6 3W LEDs in series on a 24V supply.

For the OP’s application two LEDs in series with a single resistor is optimal.

Robert.

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