1
\$\begingroup\$

For a hobby project I'm trying to power a PIC microcontroller and 4 RGB leds from two AA batteries. I have both the PIC and the circuit shown below being powered by the batteries and the PIC communicates with the PCA9635(for brightness control) through I2C. Since the blue and green leds have a forward voltage of 2.8V and I wanted to ensure that they could stay on as the battery discharges, I added a boost converter (as shown), which is capable of 300ma at 3.3V. The PCA9635 is capable of sinking 25 ma on each input pin and since there are 12 total leds (RGB for each), each rated at 20ma, the total current consumed should be less than the 300ma limit of the boost converter.

Schematic

On fresh batteries the circuit works. However, with slightly used batteries, when I begin to turn on LEDs the voltage of the batteries drops sharply causing the PIC to brown out/reset. It seems to me that two AA batteries should be able to source the amount of current required for this. So my questions are:

Is it possible to power everything both without causing a brown out and without using an alternate/extra power source?

What is causing the sharp voltage drop?

Is there a better way to power the 4 RGB leds than using a boost converter?

EDIT: The actual leds being used are these: http://www.dialight.com/Product/Details/445022/5988710307F.

\$\endgroup\$
  • \$\begingroup\$ I wonder what the boost's effeciency at 3v vs 1.5v input is. You might have better performace with two in parallel instead of series. \$\endgroup\$ – Passerby May 20 '14 at 13:51
  • \$\begingroup\$ I too,suspect the boost circuit. Why not go straight from 3v? \$\endgroup\$ – EkriirkE May 20 '14 at 18:36
1
\$\begingroup\$

(1) Probable issue is attempt to massively over drive LEDs - see below.
Series LED resistors will be needed.

(2) Boost converter MAY be not working properly - see below for testing method.

LED datasheet here
TPS61201 boost converter datasheet here

Two x AA alkaline with provide a voltage between 3.2V and about 2V.
A larger value of C1 on Vin will do no harm and will help very low voltage/bad battery startup.

The TPS61201 boost converter will happily start and run on this voltage range.

The LEDs are NOT rated at 20 mA continuous - see data sheet. LEDS are rated at

  • 10 mA continuous

  • 20mA peak at 10% duty cycle at 0.1 ms pulse width !!!.

Also thermal limitations of IC must be observed.
Running LEDs directly off IC pins risks LED damage and possibly IC malfunction.

What is design requirement?:

Say 10 mA/LED and all on.
12 LEDS x 10 mA = 120 mA.
120 MA x 3V3 =~ 400 mW.
Efficiency at 120 mA out and 2V Vin ~= 75% - see fig4 from datasheet below.
So 400 mW/75% = 560 mW into converter.
At 2V Vin Iin = 560 mW/2V = 280 mA Iin.
This is well within IC capability.

So - IC is capable of providing requerement IF LEDs are correctly driven.
Problem may be excess LED drive OR bad converter components.

Test: Provide a 120 mA resistor load to converter.
R = V/I = 3.3 V / 120 mA = 27 Ohm.
Will converter supply 3V3 to 27 Ohm with 2V supply including startup?
Use lower R's for higher load current if desired.

If converter will not support desired load current then an inadequately rated inductor is the most likely problem.

Your Cout = 22 uF = 2 x data sheet value - should not be a problem. Sometimes high Cout can cause startup problems but 22 uF should be fine.

Most likely problem is massively high LED currents.
Add series resistors to set currents to 10 m max. Note that Vf LED varies with colour - see datasheet.

enter image description here


ADDED:

New information:
LEDs are not as shown on diagram.
Assume max per LED current is 20 mA.

Actual LEDs are Dialight 5988710307F.
These LEDs are rated at 20 mA ABS MAX so you could run them at 20 mA, perhaps.
[Are you feeling lucky, punk?]
If so then double figures I supplied above to 1
20 mA x 12 = 240 mA.

This is still on the curve in Fig 4 above with 1.8V Vin so the converter can handle it.

TRY MY RESISTIVE LOAD TEST with R to suit real load.
If this passes OK then converter is OK.
If this fails then fix it first - chasing driver problems when the power supply is failing is liable to be unproductive :-).


ADDED:

A deleted answer suggested that rising battery impedance would mean that AA batteries could not be used in this application and that C or D cells were needed.
IMHO this is not true.
While AA cells will probably not reach full discharge potential due to falling current capability, they should work reasonably well.

The converter has a performance curve in fig 4 of the data sheet which shows operation at 1.8V = 0.9V/cell. As long as the batteries will provide the required load current at this voltage OR HIGHER the system will work OK.

At I_LED = 10 mA and all segments lit Ibattery at Vbattery=2V will be ABOUT 280 mA (see above) and at I_LED = 20 mA I_Battery at Vbattery = 2V will be about 550 mA (efficiency slightly HIGHER at higher load - see graph).
IF the battery is capable of providing about 500 mA at 2V+ then it will work. This is getting extreme for AA Alkaline but the battery will provide more power than this for much of its discharge life.
My simple resistor loading test will show whether the converter is OK at any given battery state.
Note that input capacitor matter muchly when batteries are near end of life. A large capacitor greatly reduces battery effective impedance on current peaks. Mean ESR is not altered but failures usually occur when current peaks occur during the boost cycle.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. I should have mentioned that the actual leds that I'm using are these: dialight.com/Product/Details/445022/5988710307F. I'll try the test from your post. \$\endgroup\$ – James May 20 '14 at 17:27
  • 1
    \$\begingroup\$ @James Yes, you waste people's time with incomplete information - and do so evenmore so with WRONG information. LED part numbers on your diagram are assumed to be the parts used in the absence of information to the contrary. New specd LEDs are rated at 20 mA ABS MAX so you could run them at 20 mA, perhaps. [Are you feeling lucky, punk?]If so then double figures I supplied to 240 mA. This is still on the curve in Fig4 with 1.8V Vin so the converter can handle it. TRY MY RESISTIVE LOAD TEST with R to suit real load. If this passes OK then converter is OK. If this fails then fix it first. \$\endgroup\$ – Russell McMahon May 21 '14 at 3:41
  • \$\begingroup\$ @RussellMcMahon The merit of the "deleted answer" as worthy of response fell to zero with the use of the term "Tony's law": That poster is well known for limited knowledge and unlimited bovine excreta. \$\endgroup\$ – Anindo Ghosh May 21 '14 at 4:20
  • \$\begingroup\$ @AnindoGhosh - Yes - I was aware of who posted it and past history. Some of his material is quite good. Other "not so". Sorting one from t'other can be 'a bit of an ask' for beginners. The issue of battery impedance had also been raised by Andy and I thought it (may be) worthwhile to address what the root issue is - which is, will the battery stably source at least enough current to support the system at a chosen voltage. In this case I'd expect my original 10 mA / LED result to work to quite low battery SOCs and 20 mA/LED to be somewhat more challenging. ... \$\endgroup\$ – Russell McMahon May 21 '14 at 7:30
  • \$\begingroup\$ ... Long ago (2001 I think) I designed a power supply for use in a Taiwanese made exercise machine. The aim was to allow 4 x C cells to supply 5V & 3V3 rails including when a motor actuator was changing the load level. 4 x C Alkaline get to well below 5V withe still some energy left and their systems were dropping out when the user changed settings but working if left alone. I added a boost converter based on a 74C14 Schmitt trigger inverter pkg. Worked beautifully. Tiwanese factory engineer went and got his used battery box and tried many combinations. If it would drive the ... \$\endgroup\$ – Russell McMahon May 21 '14 at 7:34
4
\$\begingroup\$

Your LEDs need current-limiting resistors. This does not appear to be a constant-current LED driver IC.

See page 23 for an example circuit: http://www.nxp.com/documents/data_sheet/PCA9635.pdf

\$\endgroup\$
  • \$\begingroup\$ I think you're right, but the NXP says: LEDs can be directly connected to the LED output (up to 25 mA, 5.5 V) ... Thus I would presume they simply rely on the internal resistance of the driver to limit current to 25mA at 5V. \$\endgroup\$ – Dzarda May 20 '14 at 11:08
  • \$\begingroup\$ Yeah, I saw that, but I think it's VERY poor wording for "no output transistor is needed, up to 25mA" If you look at Table 14, you can see that "LED current limiting R" is required for any type of direct connection to the LEDn pins. \$\endgroup\$ – Daniel May 20 '14 at 11:16
  • \$\begingroup\$ Yep, I'd back you on that. \$\endgroup\$ – Dzarda May 20 '14 at 11:19
  • \$\begingroup\$ I thought exactly what Dzarda posted in the first comment. I also thought that if the IC pin is only capable of sinking 25ma wouldn't that be the limit? Or is that telling you to limit the current to 25ma yourself with outside resistors? \$\endgroup\$ – James May 20 '14 at 17:33
  • \$\begingroup\$ That just means "If you need more than 25mA, use an external transistor" I'm a little surprised you didn't damage the chip. \$\endgroup\$ – Daniel May 20 '14 at 21:31
2
\$\begingroup\$

Take a look at this: -

enter image description here

The internal resistance of your battery will likely cause the voltage to droop soon after power is applied and you have two in series, doubling the effect.

If you are taking 500mA (in order to provide 3.3V at 300mA on the booster output) the battery voltage is probably going to droop > 0.2V and this will get worse as the battery becomes more discharged.

Also take a look at this -it's where I got the document from - go to the link at the bottom of this wiki page.

\$\endgroup\$
  • \$\begingroup\$ Typical AA Alkaline batteries will have no problem providing the load he SHOULD be drawing across the battery voltage range. Overcurrent load is likely main issue. \$\endgroup\$ – Russell McMahon May 20 '14 at 13:48

protected by Community May 20 '14 at 16:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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