AP3019A low efficiency at low input voltage? - Page 1
Forum section: electronics / projects

author: ifonlyeverything
author_rank: Regular Contributor
timestamp: November 26, 2023, 01:58:49 pm
content: I have some AP3019A boost drivers on a string of RGB LEDs, schematic is below and all balancing resistors on the LEDs are 100R. At 9V and 5V input voltages I'm getting the rated 80% efficiency on the R, G, and B channels of the LEDs. When I drop it down to 3.7V input voltage (to replicate the nominal voltage of a lipo cell) the red channel's efficiency stays at about 80% while the G and B channels dump to 60% efficiency and the driver ICs get very hot.Granted my LEDs are drawing 40 milliamps from the boost driver rather than the 20 milliamps from the data sheet, but they suggest that 80% efficiency should be possible for a 3.7V input voltage. Am I doing something wrong or is the datasheet misleading? I've attached a layout of one of the driver sections, they are laid out identically. PCB is two layers with a solid ground pour on the same side as the drivers, the other layer has no pours and is for routing the LEDs.

author: mtwieg
author_rank: Frequent Contributor
timestamp: November 26, 2023, 02:34:56 pm
content: A bit more info would help. Can't even tell whether the chip will even be operating within its stated limits. What duty cycle and output voltage do you see for each channel? Are the boost circuits operating in CCM or DCM? Is the inductor part number on that schematic the part you're actually using?

author: ifonlyeverything
author_rank: Regular Contributor
timestamp: November 26, 2023, 02:42:29 pm
content: Quote from: mtwieg on November 26, 2023, 02:34:56 pmOutput voltages are generally the same across varying input voltages, +/- a few hundred millivolts.Red: 13.6VBlue: 20.2VGreen: 20.8VYes, the inductor part # in my schematic is what is on the physical PCB: Sunlord SWPA5020S220MThttps://datasheet.lcsc.com/lcsc/2310251551_Sunlord-SWPA5020S220MT_C96915.pdfpage 10/20 of the PDFI'll have to dig my scope back out, I can't remember the duty cycle % off the top of my head. Is there a way to determine CCM/DCM without having access to a high bandwidth current probe? I only have a Hantek CC-65 (20 kHz).

author: ifonlyeverything
author_rank: Regular Contributor
timestamp: November 26, 2023, 03:08:29 pm
content: U2 pin 1 (green channel switch node) @ ~3.7V input voltage:

author: frojac
author_rank: Newbie
timestamp: September 29, 2025, 07:16:39 pm
content: Sorry for gravedigging, looking at these ICs myself, and came over this post, the devil is in the detailsI'm looking to make a circuit for myself, and noticed your circuit has a flaw compared to the datasheet, increasing the boost voltage unnecessarily by 4 volts (40mA*100ohm and 0,16W powerloss), in your LED resistors, making the circuit work hard to boost the voltage 4V more than it needs.They are not needed, as the IC limits the current to 40mA (that's why it was designed in the first place).Anyway, looking for tips on good efficiency small IC to run some LED with a cell of lithium, Looked at MIC2289, AP3019, MIC2287 so far.If you came across a better IC, please shareAll the best,Frode.

author: jwet
author_rank: Frequent Contributor
timestamp: September 30, 2025, 12:26:03 pm
content: 1. These parts are current controllers. The balancing resistors are just burning power and are not needed. You're dropping an extra 4v through the 100 ohms. Adjust the low value R's to set the R,G and B currents or use the CTRL input.2. You're doubling the power output by running at 40 mA. This is like doubling the number of LED's and the part is not designed for it. These parts are designed to run at 20 mA. The schottky drop, etc will suffer and dissipate power. The thermals aren't great in these tiny packages- 260 degrees C/W. 18V x 40 mA is 720 mW. At 80% eff, part dissipates about 150 mW making a 40 deg C rise which lowers efficiency, increases temp, and you auger in. Thermal runaway.3. These things have a little switch internally- about 1.5 ohm/ish (calc from vsat at 300 mA). with 6 led's, you're boosting to about 20v depending on color. Use more of them if you want more light.

author: Zero999
author_rank: Super Contributor
timestamp: September 30, 2025, 12:42:33 pm
content: Generally speaking, the eficiency of a boost does drop at lower input voltages, because the duty cycle and peak current through the inductor increases, leading to greater I2R losses. If you need to operate at a large voltage difference, then use a transformer for the inductor. It can be an autotransformer, which will be smaller and should be more efficient.