voltage divider not working when load is applied - Page 1
Forum section: electronics / beginners

author: abdullahseba
author_rank: Frequent Contributor
timestamp: May 05, 2018, 10:44:44 pm
content: HiI have a wemos D1 mini pro that I want to power with a 3.7v LiPo battery and a TP4056 module.I have a 91k and a 750k resistor acting as voltage dividers on the output of the module to reduce the voltage from 3.7v to 3.3v.Measuring the output of the divider directly does give 3.35v as expected. However if I connect the D1 mini to it, all I get is 0.6v and the wemos does not power on.Nothing gets hot and the charging module shows the battery as fully charged. The D1 mini works fine with USB power.Here is a schematic of the D1 mini:https://wiki.wemos.cc/_media/products:d1:sch_d1_mini_pro_v1.1.0.pdfI have the voltage divider output connected to the 3.3v and GND pin. There is no diode, so it should work.I have the TP4056 module's input connected to the 5v pin of the D1 mini. This lets me charge the battery when the D1 is plugged in and works fine. I have tried disconnecting it though to make sure that was not the issue but I still get 0.6v.Anyone have any ideas?Thanks in advance.

author: Karlo_Moharic
author_rank: Regular Contributor
timestamp: May 05, 2018, 10:52:53 pm
content: You are putting to much load on your voltage divider (it uses too much current). You need voltage regulator.Something like thishttps://www.ebay.com/itm/5PCS-5V-3-3V-DC-DC-Step-Down-Power-Supply-Buck-Module-AMS1117-LDO-800MA/201662468680?hash=item2ef4051248:g:Ga0AAOSwMgdXzhOz

author: abdullahseba
author_rank: Frequent Contributor
timestamp: May 05, 2018, 11:30:53 pm
content: Quote from: Karlo_Moharic on May 05, 2018, 10:52:53 pmDamn, The problem is its only 3.7 to 3.3V.I'll try this thinghttps://www.amazon.co.uk/KKmoon-Converter-Module-Battery-Development/dp/B075FSL1LW

author: IanB
author_rank: Super Contributor
timestamp: May 05, 2018, 11:37:13 pm
content: Also a lithium ion battery does not produce 3.7 V; the voltage varies with the state of charge and the load between 4.2 V and 3.3 V or so.

author: PA4TIM
author_rank: Super Contributor
timestamp: May 05, 2018, 11:37:37 pm
content: Are you trying to feed the D1 via the voltage divider ? Without a load there is 5uA going to the divider.If you measure 0,6V there is a 3,1 voltage drop over the 91k there is 34 uA running to the resistor. 800nA for the 750k resistor and 33uA for the D1. That is not enough.Use a low drop 3V3 voltage regulator

author: Brumby
author_rank: Unknown
timestamp: May 06, 2018, 12:20:14 am
content: The very first question is - how much current are you going to need? This is the question you did not ask, nor did you include in your voltage divider.I agree on the low dropout regulator - and if your current needs do not exceed 100mA, then I suggest checking out this page:http://www.ti.com/product/LP2950-N

author: langwadt
author_rank: Super Contributor
timestamp: May 06, 2018, 01:48:17 am
content: Quote from: abdullahseba on May 05, 2018, 10:44:44 pmremove F1 on the D1 mini and connect the output of the charger to +5V on the D1 mini

author: Jeff1946
author_rank: Regular Contributor
timestamp: May 07, 2018, 09:03:54 pm
content: Look at specs on a Schotty diode. If you put it in series with the power supply, it should workhttps://www.mouser.com/datasheet/2/308/MBR150-D-106444.pdfA quick look at an MBR150 shows it starts conducting at 0.3 V and at 0.1A is dropping about 0.4 volts.

author: Naguissa
author_rank: Regular Contributor
timestamp: May 07, 2018, 09:16:53 pm
content: If it's nothing serious, you can use a diode in between.4.2 - 0.6 = 3.6v3.6 - 0.6 = 3vESP specs are 3-3.6v (i think) but they work well between 2.8v and 3.9vEnviado desde mi Jolla mediante Tapatalk