Bypass serial chip for better battery life on Wemos D1 Mini Pro - Page 1
Forum section: electronics / beginners

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 20, 2022, 09:41:05 am
content: In deep sleep mode i get around 200uA (mico ampere) consumption. i power the Wemos via HT7333 regulator to the 3.3V pin with an 1000uF and an 100nF capacitor.The unit works fine, but i want it to consume less current in deep sleep.i have read that i can cut the copper area between 3.3V input and the serial chip for bypassing the built in regulator, but where di i cut the copper?Here is too images taken of my controller.

author: ledtester
author_rank: Super Contributor
timestamp: February 20, 2022, 07:38:00 pm
content: Apparently the WEMOS D1 Mini Pro uses a SiLabs CP2104 for the USB UART.According to the datasheet:https://www.silabs.com/documents/public/data-sheets/cp2104.pdfpin 6 is Vdd.However, you should check the datasheet to see if there are any other issues to consider. For instance, there are other power inputs such as pin 5 and pin 7.I'd also look at pulling /RST (pin 9) low.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 22, 2022, 04:12:31 pm
content: On side 18 in datasheet i can read that i just can make and bridge between VDD and REGIN pin for bypassing regulator. Have i understand that correctly?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 23, 2022, 02:51:32 pm
content: Someone?

author: Peabody
author_rank: Super Contributor
timestamp: February 23, 2022, 03:26:13 pm
content: It's not clear how you are powering the 2104. Typically it would be powered from USB, and would only sink current from USB when it is connected. The alternate circuit would power the 2104 from your 3.3V supply at Vdd, and the built-in regulator is already bypassed. Do you have a schematic you could post?The question is whether bypassing the regulator actually reduces the current the 2104 draws. I don't know the answer to that. Table 5 shows a maximum bias current of 120uA, but if that's a resistor divider feedback, it will still be sinking current whether the regulator is active or not.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 23, 2022, 05:39:14 pm
content: Quote from: Peabody on February 23, 2022, 03:26:13 pmMaybe it is better to just use an ESP 12F without the circuitry and make my own 3.3V regulator with HT7333 etc.?

author: MLXXXp
author_rank: Frequent Contributor
timestamp: February 23, 2022, 08:58:21 pm
content: A Wemos D1 Mini Pro with a CP2104 USB to serial converter does not appear to be an official Wemos product. The real D1 Mini Pro seems to have a CH340G serial chip.I can't find any schematics for the one like you have with a CP2104. If you can provide a schematic for your board, we may be able to tell you how to remove its power drain.

author: Peabody
author_rank: Super Contributor
timestamp: February 23, 2022, 10:16:44 pm
content: I think it's an earlier version of the D1 Mini Pro. I think the attached v1.1.0 schematic is the one. I actually have one of these with the 2104, and it's definitely genuine Lolin (I bought it directly from their Aliexpress store). But it's labeled v1.0.0.So it appears Vdd is directly powered from the 3.3V supply, which is also connected to REGIN. So the regulator is already bypassed. And the 3.3V regulator on the module is also bypassed if the module is powered externally. But both of those regulators can sink current if they have resistor divider feedback circuits on their outputs. You might actually draw less current if you use the module's regulator - a ME6211, which is a pretty good regulator - instead of the HT7333.If you cut the 3.3V supply to the 2104, you also have to be sure that any ESP I/O pins connected to the 2104, such as Rx, Tx, DTR and RTS, are brought low during sleep. That's because those pins on the 2104 will sink current when its power is off. I don't know what happens to the ESP's I/O pins during deep sleep. They need to be low or floating.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 24, 2022, 01:54:03 am
content: Okay, just don't know...I just want to reduce the power consumption. For now it draws around 200uA in deep sleep.

author: Peabody
author_rank: Super Contributor
timestamp: February 24, 2022, 03:56:28 pm
content: Andreas Spiess' Youtube video #324 shows the trace cutting on a D1 Mini, but I don't understand what he is doing, or how the USB chip can work after the trace is cut. And I don't know what the equivalent procedure would be on the Pro.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 24, 2022, 05:03:01 pm
content: Quote from: Peabody on February 24, 2022, 03:56:28 pmI have seen that video.But i do not have the board he using.

author: Peabody
author_rank: Super Contributor
timestamp: February 24, 2022, 06:46:01 pm
content: In the comments below his video, there are several people who report that his fix didn't work for them. Apparently there are several versions of the D1 Mini, with different layouts, so cutting the trace affected different portions of the circuit. I don't know how you would find what to cut on your board, if in fact anything would work. I guess in theory the idea is to have the UART chip disconnected from the 3.3V pin, but still connected to the regulator output so it would work when USB is plugged in. But there's no guarantee the board is laid out in a way that would permit that.

author: Peabody
author_rank: Super Contributor
timestamp: February 24, 2022, 08:49:33 pm
content: Let's back up and look at this again. First, the schematic I posted earlier is not correct. Looking at your pictures, it's clear that pin 8 (VBUS) of the 2104 is connected to the USB power line, and pin 7 (REGIN) is directly connected to VBUS. However, it's not clear that pin 6 (VDD) is connected to anything except the capacitors. It is not connected to the module's 3v3 pin. So it appears the 2104 draws current only when USB is connected. If that's true, it would not account for your sleep current.However, on my copy of this module, pin 5 (VIO) is directly connected to the 3v3 pin of the module. So you have a situation where the 2104 is powered down, but 3.3V is being applied to VIO. I don't know if the datasheet tells us what happens in that circumstance. In addition, there is the issue of the ESP pins which are connected to the 2104, but I think you can control the state of those pins during deep sleep in software (but it's complicated).If VIO is a problem, one possibility would be to cut its connection to 3v3, and connect it instead to VDD. But that's just a guess. I'll try to measure current through VIO when 3.3V is applied and everything else is off.

author: Peabody
author_rank: Super Contributor
timestamp: February 24, 2022, 11:57:00 pm
content: I wasn't able to make any measurements on my module that would be useful. Applying power to VIO produces lots of current flow, but that may be the protection diodes kicking in.I did test a mailbox notifier sketch that ends in deep sleep, and measured the current at 8.25mA when powered at the 3.3V pin, which of course is much worse than you are getting. I don't know why.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: February 27, 2022, 09:40:53 am
content: Thanks for all answersI have ordered ESP 12F now, without surrounding electronic and i think its easier to use this regards to low currents