Downsizing the main electrolytic on a tiny module - Page 1
Forum section: electronics / projects

author: microvia
author_rank: Regular Contributor
timestamp: August 23, 2025, 08:46:21 am
content: Hi,I designed some small modules which use a PIC12F or a PIC16F along with a few other components, of which an FM receiver chip.As it can be powered in AC or DC, I use a half-wave rectifier circuit along with a Microchip MIC5211 dual voltage LDO powering the PIC at 3.3V.The 3.3V part sinks around 30mA and the 5V part around 40mA when used.I chose a low ESR (13 mOhms) 1000uF electrolytic is preceeded by a shottky and a 5 ohms resistor.As the electrolytic is the second expensive component there, and supposing that I overthaught its choice, could I safely reduce its value ? If so, up to what level and under what acceptable ESR ?Suppose for example that the actual electrolytic costs $0.7 per 100pcs and that a more common type can replace it at $0.25 per 100pcs. Could I replace it by something else while staying withint thouse cost limits ? Are there specific ICs designed to filter half-wave at a high frequency so as to require only one or to small MLCC capacitors ?Thanks.

author: wraper
author_rank: Unknown
timestamp: August 23, 2025, 09:16:19 am
content: If you can use a 5 ohm series resistor, you certainly do not need low ESR (increases the possible minimum size and price) in this design.

author: mariush
author_rank: Super Contributor
timestamp: August 23, 2025, 09:27:49 am
content: You don't need very low ESR capacitors to smooth out voltage after rectification. You could literally use a 5-10 cent electrolytic and you'd be fine. At the current levels you have even the shittiest would last years. But polymer capacitors are like 10-25 cents even if you order a single one, and they'd last forever.You can estimate how much capacitance you need using the formula (assuming full wave rectification):Capacitance (in Farads) = Current / [ 2 x AC Frequency x (Vdc peak - Vdc min desired) ]where Current is the maximum current you estimate to need, AC Frequency is 50Hz in 230v countries or 60 Hz in 120v countries, Vdc peak is the voltage after rectification and Vdc min is the minimum desired voltageMIC5211 has a maximum dropout voltage of 0.25v typical, maximum 0.5v according to the datasheet, page 3:https://www.microchip.com/content/dam/mchp/documents/OTH/ProductDocuments/DataSheets/mic5211.pdfSo if you want 5v output, you want minimum 5.5vThe Vdc peak is calculated with formula Vdc peak = sqrt(2) x Vac - 2 x voltage drop on diode in rectifierThe voltage drop on rectifier can be as little as 0.3v if you're using very low voltage drop schottky diodes (ex SS14 diodes), around 0.5v if you're using 1n5817-1n5819 ... or as much as 0.7v if you're using plain diodes like 1n4007 .. 1n4007If you power with 6v AC , then the peak DC voltage is Vdc peak = 1.414 x 6 - 2 x ~0.5v = 7.5vYou say you need ~40mA on 5v, ~30mA on 3.3v , so 70mA in total - linear regulators just throw the difference in voltage as heat, so it's not like with buck regulators where the conversion could be up to 95% or so efficient. Let's say you aim for maximum 100mA power consumption or 0.1A , so let's put the numbers in formulaC = 0.1A / [ 2 x 50 Hz (to be conservative) x (7.5v - 5.5v) ] = 0.1 / 100 x 2 = 0.1/200 = 0.0005 Farads or 500uF (https://www.wolframalpha.com/input?i=0.0005F+to+uF)I would use a 470uF capacitor rated for 16v (use rated for 25v or even 35v if you may use a transformer that outputs higher AC voltage, but be careful as the linear regulator you chose supports maximum 16v dc input), and I'd accept that I may get a bit less than 5.5v if the current gets close to 100mA.You can get polymer capacitors that are super tiny, for example 6.3x 10 470uF 25v :https://www.lcsc.com/product-detail/C42377047.htmlor a 560uf/16v in 5.5mm x 10 :https://www.lcsc.com/product-detail/C42460431.htmlYou could also have a cutout in the pcb and put the capacitor horizontally inside the cutout, and then it may sit less than 5mm above the board (the standard pcb height is 1.6mm) and it probably doesn't sit flat on the bottom of a case)keep in mind that most ac transformers will idle at a higher voltage, usually up to 10-15% higher ... so your 6v AC transformer could idle at 6.5-7.0v AC, and after you rectify you could have higher dc voltagealso your mains voltage is not constant, could be 210-245v ac in some european countries, depending on time of day, so the output of a classic transformer will also vary, so you may get less or more dc voltage after rectificationso don't use lowest voltage rating (ex with my 6v ac transformer math, knowing the peak is 7.5v dc after rectification, you could say 10v rated capacitor is plenty, but no, not safe)

author: microvia
author_rank: Regular Contributor
timestamp: August 23, 2025, 11:56:33 am
content: Quote from: wraper on August 23, 2025, 09:16:19 amThis is so obvious that I didn't think about it. Perhaps "because of the shottky between them" ?...At the same time I used this resistor to reduce the amount of internal heat in the electrolytic while charging, and also the shottky after seeing several BAT43 fail sometime ago when I used 2200uF caps.

author: microvia
author_rank: Regular Contributor
timestamp: August 23, 2025, 01:53:21 pm
content: Thanks a lot for your detailed answer.I'll probably opt for a 560uF or a 680uF depending on the tolerances.In any case I can only use SMD devices as my 4L PCB has internal tracks passing under it.