Capacitors placement in my project - Page 1
Forum section: electronics / projects

author: cofochips
author_rank: Newbie
timestamp: April 06, 2025, 08:43:06 pm
content: Hello, I have issues in my project, Google, nor my friends could help me, so I hope someone will help me here. Have I correctly placed capacitors in my project? they are used to smoothen/stabilize voltage. Thank you for your help.

author: lombster
author_rank: Newbie
timestamp: April 06, 2025, 09:14:51 pm
content: Hello.I'm still fairly new to electronics, but I can try to help. I see you have a decoupling capacitors and then some additional capcaitors on your regulator. Would you mind explaining more what isn't working?Lombster

author: TimFox
author_rank: Super Contributor
timestamp: April 06, 2025, 09:42:22 pm
content: The problems that I have seen with inappropriate or missing bypass capacitors in the past usually involved oscillation.Have you looked at the regulated DC outputs for signs of oscillation or incorrect voltages?(Systems in oscillation often result in "mean" voltages seen on DC voltmeters being far from expected values.)Is the circuit around "ModeOut1" and "ModeOut2" correct?

author: cofochips
author_rank: Newbie
timestamp: April 06, 2025, 09:53:16 pm
content: i mean i dont know if its working, but I just want to place capacitors to my circuit for better voltage regulation - not spikes, I don't know the term for it sorry

author: cofochips
author_rank: Newbie
timestamp: April 06, 2025, 09:54:27 pm
content: i mean it should be, it sends output from esp, and depends what mode is selected, will be grounded

author: lombster
author_rank: Newbie
timestamp: April 06, 2025, 11:19:54 pm
content: This is a motion sensing camera of some sort?If I'm understanding right you're trying to make sure your circuit gets a constant voltage from the battery pack. The regulator should be able to do that for you. Putting a capacitor at the input is something the data sheet (https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf) recommends. A capacitor at the output will help smoothen noise and stabilize the output from the regulator, however, you'll need to perform some calculations I believe in order to get the right capacitor size.This type of capacitor is called an output filter capacitor (bulk capacitor??) by the way. Decoupling/bypass capacitor is the one placed near ICs typically.

author: TimFox
author_rank: Super Contributor
timestamp: April 06, 2025, 11:32:35 pm
content: For linear voltage regulators, the capacitor at the input is used to avoid oscillation, while the output capacitor improves the response to quick changes in the load current.The traditional 7805, etc., series originally specified minimum 0.33 uF at the input, and said the output capacitor was not necessary, but might improve load response.See data sheet.

author: lombster
author_rank: Newbie
timestamp: April 07, 2025, 02:46:22 am
content: TimFox,Are you sure this is a linear voltage regulator? It appears to be a switching regulator based on the datasheet I linked above. Although the datasheet does seem to differ with the symbol in the schematic in terms of pin count.

author: TimFox
author_rank: Super Contributor
timestamp: April 07, 2025, 03:06:55 am
content: I was using linear regulators as an example, since I'm more familiar with them.Past that, read the datasheet for the individual device: it should specify capacitor requirements.

author: cofochips
author_rank: Newbie
timestamp: April 07, 2025, 11:06:33 am
content: okay thank you for your help

author: PGPG
author_rank: Frequent Contributor
timestamp: April 07, 2025, 12:16:01 pm
content: Quote from: cofochips on April 06, 2025, 08:43:06 pmYour schematic is hard to read.Try to have + voltage being over GND - MT3608 should be rotated against X axis so pins 2 and 4 be below pins 1 and 3. Then their connection to GND will be just with GND symbols attached to them.The same with PIR.Quote from: TimFox on April 06, 2025, 09:42:22 pmI think it is correct but doubt is made by the way it is drawn. It is easy to not notice that these labels are used 3 times. If you notice only two of them you think something is wrong. This is the other point saying schematic is hard to read.The best would be if switch and pull-up resistors were drawn near IO14 and IO15 pins (with resistors connected to 3V3 supply symbol (and 3V3 pin also connected with 3V3 supply symbol). I assume that ESP.. 3V3 pin is the 3V3 output and not pin needing be supplied by 3V3 source.

author: wraper
author_rank: Unknown
timestamp: April 07, 2025, 12:43:09 pm
content: There are several issues with this circuit. First of all it's very inefficient to boost voltage to 5V for it to be reduced to 3.3V right away by AMS1117 linear regulator of ESP32-CAM. Use 3.3V input and buck-boost module like thishttps://www.aliexpress.us/item/1005004818090848.htmlinstead. This way you can throw out one of the batteries for the same working time on a charge. Or connect two batteries in series and use step-down regulator to 3.3V and get 2x working time. What appears to be LED battery charge indicator will not work.EDIT: I see this particular PIR sensor needs 5V power, if using that (I suggest using 3.3V sensor, or do a simple mod to make it work from 3.3V), use two batteries in series and a buck converter to 5V. Also power it from converter output directly, not though ESP32-CAM module. Circuit on PIR module output does not require series resistors, not to say two in series (PIR module has 1k internally). If you use LED with high forward voltage, it won't work. R(PIR)3 is connected incorrectly. It should be used as a pull-up to 3.3V, not connected in series with GPIO. But it's best to just throw out BJT and LED and connect PIR module out directly to ESP32-CAM. Use some free GPIO pin to drive LED if you want one.