2
\$\begingroup\$

I am trying to light up 6 LEDs with the common garden light circuit (2v Solar panel and 1.2V battery). The problem is that I want to light up 2 yellow LEDs, 2 green LEDs, and 2 purple LEDs simultaneously. When I connect them to the QX5252 without any resistors only the yellow ones turn on. I added resistors to balance the current and it kind of worked. All the LEDs turn on but they turn off after like 1 minute. Then I disconnect and connect the battery and they turn on for another minute. Could you please let me know how I can make this project happen?

Here is the Schematic: Circuit

\$\endgroup\$
3
  • \$\begingroup\$ You need a much bigger battery and PV source 6 * 20mA * 1hr = 120mAh \$\endgroup\$ Dec 4, 2018 at 5:28
  • \$\begingroup\$ The battery is 1.2V and 1000 mah. When I use the QX5252 (LED driver), I can control the current that goes to the LEDs. Also, I want the LEDs to be on at night and the battery gets charged during the day with the 2V solar panel. My problem is the LEDs turning off after one minute even though the battery is fully charged. \$\endgroup\$
    – Miti
    Dec 4, 2018 at 21:21
  • \$\begingroup\$ I think your expectations are too high unless you learn how to supply all the necessary design details to analyze the fault including the design source and assumptions made. We dont spoon feed here very often \$\endgroup\$ Dec 5, 2018 at 0:40

3 Answers 3

1
\$\begingroup\$

You are expecting too much out of the QX5252, an NiMH battery, and a little solar cell.

You are missing a resistor.
And your inductor is too small.
You should not need more than a few mA if you use high luminous intensity LEDs.
Use only 2V red, yellow, and orange LEDs.
2 NiMH batteries in series may work better.
Pay attention to your LED forward voltages. Use current mirror transistors to balance rather than resistors: Current distribution in parallel LED strings

enter image description here


how can I make this project happen?

First off you need to use LEDs with a very high luminous intensity. This means you get more light with less current. Less current also gives you a lower forward voltage.

Yellow (or red) : Vishay TLCY5800
Green (or blue): Cree C503B
Purple: Lumex SSL-LX5093VC

The Cree 11,000 mcd blue LED will work much better than a 2,200 mcd purple.

Purple is very difficult to get luminosity. A 400 nm purple LED would need 773 times more photons than a 470 nm blue LED to achieve the same luminous intensity. A 12.4 µmols of photons are required to get 1 lux from a 400 nm LED. 470 nm blue needs 0.016 µmols to generate 1 lux.

So you would be better off mixing blue (preferably 450 nm) and red (preferably 660 nm deep red) to get purple.

I created this web page that shows the conversion factors from luminous lux, to radiometric watts, and quantum (number of photons) wavelength color conversion factors
The colors use the CIE factors to convert wavelength to RGB
I also made this one for Photopic Luminous Efficacy: Relative Sensitivity Curve for the C.I.E. Standard Observer



If you really want it to work, use an energy harvester.

Solar cell's voltage outputs have a large range of voltage from zero to their max. An energy harvester can work with a solar cell's output.

To make this work well with solar cells you need to use an energy harvester (like the TI bq25570) to charge a battery. The bq25570 charges an Li-ion battery. A higher wattage solar cell would help.

enter image description here


The battery is 1.2V and 1000 mah. When I use the QX5252 (LED driver), I can control the current that goes to the LEDs.

You will not likely get a 1000 mAh charge in a single sunny day.

What you are not saying is the measured wattage of your solar cell at the light levels you will be using. Solar cells need a lot of irradiance to produce usable wattage.

I have some 100 watt panels that will charge those batteries.

You need to measure the voltage and current while charging the battery on sunny and overcast days.

The proper battery charging with the solar panel will be my next step.

Once you get the battery charged consider using a real boost step up regulator. There are some that are made especially to work with a single alkaline or NiMH battery.

I would suggest you use a Low IQ Monolithic Power MP3416 with 0.86V to 5.5V output step-Up converter. This regulator uses a small 3.3mH inductor and is has good efficiency for your project even when the battery is nearly discharged at 1.1V.
enter image description here


enter image description here


Even with an AAA 80% efficient step up converter and 90% efficient resistors you should still get 60 hours with a fully charged AAA NiMH and over 120 hrs with an AA.

Discharge characteristics for an AAA NiMH.
enter image description here

With a voltage source, rather than current, you will not need current balancing. You can use a single resistor for each LED. Measure the forward voltage of each LED before calculating its resistor value. Tweak the voltage of the converter to just above the highest forward voltage for maximum efficiency.
If the LED has a 3.1V Vf and the supply is 3.3V the efficiency will be 94%. Set the supply voltage to 0.2V over the highest Vf. enter image description here
Source: Hobby Hour

You will need to run the purple at its highest allowable current. If you use a high intensity green you can run it at about 1 mA.

I am going to replace the yellow LEDs with white LEDs.

Very good choice.

For the white I recommend a warm/yellow white. The Cree C513A-MSS/MSN has 8200 mcd. This may seem low in comparison to the Cree 60,000 mcd green. But this LED has a view angle of 55° and the green has 15°.

60,000 mcd @ 15° = 3225 lumens
8,000 mcd @ 55° = 5680 lumens

Now, I am stuck in making the circuit work.

Get one LED working. Then keep adding an LED one at a time. I think using the white rater than yellow wikll give yu a much beter chance of getting the QX working. But I still highly recommend the MP3416 over the QX.

\$\endgroup\$
2
  • \$\begingroup\$ I really appreciate your answer. "Use only 2V red, yellow, and orange LEDs."--> I cannot eliminate the green and purple LEDs. I am going to replace the yellow LEDs with white LEDs. In this case, the Vf of all colors will be closer to each other(around 3.0 to 3.2 v). I am also going to try the current mirror circuit that you suggested. I will update you on the results. The proper battery charging with the solar panel will be my next step. Now, I am stuck in making the circuit work. \$\endgroup\$
    – Miti
    Dec 6, 2018 at 18:06
  • \$\begingroup\$ Added update to my answer. \$\endgroup\$ Dec 6, 2018 at 20:54
0
\$\begingroup\$

Looking at the QX5252 Datasheet the output inductance value affects the output current, the specific donor circuit board you are using is probably tuned to work with the current draw it needed.

Can you provide me with the specific circuit you are using and is it the small SIP or the DIP component you are using.

If it is the DIP the Photoresistor circuit may be getting in the way.

\$\endgroup\$
12
  • \$\begingroup\$ Thank you for your answer.I added the schematic diagram to my question. Please let me know if you have any other questions. I am using the small 4-pin QX5252F. \$\endgroup\$
    – Miti
    Dec 4, 2018 at 21:18
  • \$\begingroup\$ Looking at your schematic now, I would recommend an individual resistor per each LED. The problem here is the LED with the lower Forward drop is going to take all the current. As this, can you tell me what your Forward drop for each LED type is? \$\endgroup\$
    – robogeek78
    Dec 4, 2018 at 22:03
  • \$\begingroup\$ How many watts is the solar cell? What is the luminous intensity (mcd) of the LEDs? \$\endgroup\$ Dec 4, 2018 at 22:30
  • \$\begingroup\$ I will try individual resistor per LED and will update you. I added three resistors because I had three different colors. So I used one resistor for each color (2 LEDs). Here are the info on the LEDs: Yellow (VF:3.0-3.2v, Luminous:7k-9k), Green(VF:3.0-3.2v, Luminous:800-1000), and Purple(VF:3.0-3.4v, Luminous:200-300) \$\endgroup\$
    – Miti
    Dec 4, 2018 at 23:05
  • \$\begingroup\$ I really do not know how to choose the resistors because I have a 1.2v battery and the LED driver is using jould thief to turn on the LEDs so I don't exactly know how to add resistor to it. The data sheet says I do not need to use resistors, but it is based on using only one color LEDs. \$\endgroup\$
    – Miti
    Dec 4, 2018 at 23:09
0
\$\begingroup\$

I have many solar garden lights using the QX5252F controller IC. Many of them use colors-changing LEDs and a series Schottky diode then filter capacitor feeding them. A few have two same-color LEDs in parallel with no diode and no series resistor.

Pin LX is connected to the LEDs so that light on the solar panel turns off battery charging and the LEDs. 4.7uH is the lowest inductor I use and the resulting high current pulses make the LEDs very bright but then a AAA Ni-MH battery cell does not light the LEDs long enough at night then an AA cell is needed.

Maybe your QX5252X turns off the LEDs after 1 minute because it is overloaded and gets too hot. Try using a 10uH inductor.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.