# Capacitors and Leds

Discussion in 'LEDs and Optoelectronics' started by Tigermoth, Mar 5, 2014.

1. ### Tigermoth

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Mar 5, 2014
Good Morning All.

I am a novice at this so please excuss the terminology if not correct. I am trying to devise a circuit to light seven small, white LEDS in a train model (Vf=3.0 V, 30 mA each). The power would be straight DC (from the track), maximum about 16 volts. I plan to run the LEDs at a much lower current (about 10 mA or less each). The LEDS would be connected in parallel, each with their own resistor. I am going to use a Schottky bridge rectifier (since I want the LEDS to come on regardless of input polarity and I want the lowest possible Vf) and a voltage regulator (LP2950, again for the low Vf)). I understand the voltage regulator requires an input and output capacitor to function properly. I want the LEDs to light at the lowest possible input voltage and would like them to stay lit for about 10 seconds after the power is interrupted. I realize this is going to require a super capacitor in the range of at least .33 F. I think I have three options:

1. Install the capacitor between the bridge and the voltage regulator which means, due to the higher voltage a super capacitor cannot be used (unless in series) so may therefore have to use a "regular" capacitor with lower capacity.
2. Use a 3.3 V voltage regulator and put the capacitor after the voltage regulator in which case a single super capacitor should work although the differential voltage between the capacitor and LEDs may not be suffiicent to provide enough power for the LEDs.
3. Use a 5 V voltage regulator and put the capacitor again after the voltage regulator, providing a higher voltage differential. This would require larger resistors for the LEDs.

Thanks for any help,

Stephen

2. ### duke37

5,364
772
Jan 9, 2011
I do not think you need a Schottky bridge, any diodes will do.
I do not see how a regulator would be a benefit.

Some rough calcs, I'v oiled my abacus.

7 leds at 10mA = 70mA
Each will have a resistor but calculate the lot in parallel.
R = 13/0.07 = 186
Time constant = 10 sec = R*C
C = 10/186 = 54µf

Try it with one led and 1200R and C = 10µF

3. ### Tigermoth

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Mar 5, 2014
Thank you for your quick response. What if I wanted to keep constant light intensity of the LEDs? I think thats why I included the voltage regulator.

Stephen

4. ### duke37

5,364
772
Jan 9, 2011
Use a constant current supply. This can be made with a constant voltage regulator, connected differently.
Set for 70mA and put a lowish value resistor (470) in series with each led to make sure the current is shared fairly.

Power will be lost in the regulator so you will need a bigger capacitor.

5,164
1,087
Dec 18, 2013
Your going to need a large 0.5F 6V capacitor and as Duke says a constant current discharge down to 3V at 70mA to give you about 20 Seconds. You are also going to have to wait a fair while for this to charge back up again before you can re-use it.

6. ### Stese

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Dec 27, 2013
Hi Tigermoth

I'm assuming here, that the layout you want to run this on is NOT DCC?

Are you making lighting for your stock, or direction lights for locos?

regards,

Steve

7. ### Tigermoth

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Mar 5, 2014
Yes, I am still running DC (O Scale) but will undoubtedly embrace DCC soon (once someone comes up with a true Canadian Pacific Steam sound, since 90 % of my engines are steam). Presently I am converting the lamps to LEDs and the particular diesel (GP9) I am using as a starter project has a total of 14 LEDs (2 headlights, 2 marker lights, 2 number board lights and one ground light per side for a total of 14 in the locomotive). I want all the front lights and the ground lights on regardless of direction and the rear lights on only in reverse. I know there are a few circuits discussed on the web for this however rather than purchase one I thought it would be interesting to learn a little about electronics at this point. The kind gentlement who have responded thus far have provided some helpful direction which I shall pursue however I'm sure I will have many additional questions and therefore hope those on this forum will have the patience to assist.

Stephen

8. ### Tigermoth

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Mar 5, 2014
Gentlemen. With respect to the capacitor, I assume this would be on the discharge side of the voltage regulator (used as a constant current source). Also, shall I assume I will have to use the LM317 in this application? Will the output voltage from the LM317 not exceed the 6V for the capacitor you suggest? Thanks again,

Stephen

9. ### Stese

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Dec 27, 2013
Hi,

You might find that waiting until you get to DCC, as a standard DC system could get damaged by a DCC setup. (If I recall correctly, the Hornby system runs at 16v AC,)

Just my 2cents!

If you do find a solution for this, I'd love to know what you do, as I've got interest in a HO railway that will never go DCC, so could be very useful to modify some stock.

Regards,

Steve

5,164
1,087
Dec 18, 2013
The capacitor would need to power the constant current source and will just roll off at some point as the capacitor discharges. The LM317 will have a voltage drop across it and will limit the amount of energy you can take from the capacitor even with a 6 volt cap. You will need a different kind of constant current source, a low drop version for LEDs. Have a look and see if you can work it out. If not give me a shout and I'll design one for you.

5
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Mar 5, 2014

Thanks,

Stephen

12. ### duke37

5,364
772
Jan 9, 2011
The capacitor should provide the power to the constant current supply.
I get different values to Arouse 1973.
Voltages
led 3V
resistor 0.5V
regulator reference 1.25
regulator drop out 2V
Total 6.75V

So, from 16V, you have about 9V to play with.
Discharge of a capacitor dv/dt = I/C
so, C = I *dt/dv = 0.07 * 10/9 = 0.078 = 78µF
I would go for 100µF for a trial.

Edit Oops
0.078 is 78,000 µF? I.ve just driven a long way, at least that is my excuse.

Last edited: Mar 9, 2014

5,164
1,087
Dec 18, 2013
For some reason I have 20 seconds in my head? Dukes right but I made my calcs on two 3V caps in series.

5,164
1,087
Dec 18, 2013
Here you go. You need to choose a logic level FET and depending on the super capacitors used set the balancing resistors R5 and R4 to 50 times the leakage current.

Choose R1 for your current, see if you can work it out. Just remember that if you are using a higher supply voltage then add more caps in series and another resistor across each cap.

You may have to increase the 10R if your power supply can't supply enough current for the caps initial charging current. But with this in mind it will take longer to charge. The capacitor will have to be fully charged to get the time constant you require.

Give it a go, it might need some tweaking

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15. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Hi Stephen. What is the minimum supply voltage?