Charging 3.7 lipo Battery and using it the same time.

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Dumken

Joined Oct 7, 2014
31
Hi. From the diagram shown here, I am trying to provide an alternative power supply for my esp-12E using lipo 3.7V battery. But i noticed that after adding the 3.3V regulator to the design my voltage comes down to 2.8 which is not enough supply for my esp12. again i know that the circuit is not a good practice for the lipo battery. Can anyone help me out with a standard way of doing this so that i van use my esp8266 while charging the lipo battery as it is in the mobile phone. Thank you.
 

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jayanthd

Joined Jul 4, 2015
944
Hi. From the diagram shown here, I am trying to provide an alternative power supply for my esp-12E using lipo 3.7V battery. But i noticed that after adding the 3.3V regulator to the design my voltage comes down to 2.8 which is not enough supply for my esp12. again i know that the circuit is not a good practice for the lipo battery. Can anyone help me out with a standard way of doing this so that i van use my esp8266 while charging the lipo battery as it is in the mobile phone. Thank you.
Which 3.3V regulator are you using ?

Have you tried Microchip MCP1802T-33 ? It is specifically made for battery power devices.

What is your current requirement ? If less than 300 mA then use the regulator which I mentioned.
 

ArakelTheDragon

Joined Nov 18, 2016
1,353
Aha ok, I thought he wants to charge it and use at the same time with 3.3V. Its not the best option to use a regulator. There are much easier ways. Even with 1 resistor it will work.
 

jayanthd

Joined Jul 4, 2015
944
Aha ok, I thought he wants to charge it and use at the same time with 3.3V. Its not the best option to use a regulator. There are much easier ways. Even with 1 resistor it will work.
Yes, He also wants to charge the battery but that is a different circuit and not the question of this thread.
 

ArakelTheDragon

Joined Nov 18, 2016
1,353
The easiest way with a regulator is to regulate the voltage with a maximum current that can change based on the load. The regulator will be before the battery to charge it with a stable voltage and varying current and after the batter you put 1 resistor and 1 zener in order to limit the current further and stabilize the voltage. That way there will always be charging current for the battery and working current for the rest of the circuit. Or you could just put a zenere after the battery and use the rest of the circuit as the resistor, but the current will be limited only by the regulator and when the circuit after the zener is off, the whole current will go to the battery.
 

jayanthd

Joined Jul 4, 2015
944
The easiest way with a regulator is to regulate the voltage with a maximum current that can change based on the load. The regulator will be before the battery to charge it with a stable voltage and varying current and after the batter you put 1 resistor and 1 zener in order to limit the current further and stabilize the voltage. That way there will always be charging current for the battery and working current for the rest of the circuit. Or you could just put a zenere after the battery and use the rest of the circuit as the resistor, but the current will be limited only by the regulator and when the circuit after the zener is off, the whole current will go to the battery.
If you use 3.3V zener then when 3.7V Li-Po battery is charging the battery terminal voltage will be 4.3V and this will make the zener to conduct in breakdown region and battery will not charge.
 

ArakelTheDragon

Joined Nov 18, 2016
1,353
Sorry, I just put the elements for a quick example, I didnt mean to connect it in reverse :D. This is what happens when you do 50 times at the same time :D (in modern day world called "multitasking").
 

jayanthd

Joined Jul 4, 2015
944
Sorry, I just put the elements for a quick example, I didnt mean to connect it in reverse :D. This is what happens when you do 50 times at the same time :D (in modern day world called "multitasking").
According to my understanding of zener diode working if you take 3.3V zener diode as example and then apply 4.3V or even 3.7V accross it then the breakdown occurs at 3.3V and zener starts conducting heavily. All the current flows through zener but voltage across zener is maintained at 3.3V. So, no current will be available for ESP8266.

Am I correct ? Maybe I have forgotten the basics.
 

jayanthd

Joined Jul 4, 2015
944
According to my understanding of zener diode working if you take 3.3V zener diode as example and then apply 4.3V or even 3.7V accross it then the breakdown occurs at 3.3V and zener starts conducting heavily. All the current flows through zener but voltage across zener is maintained at 3.3V. So, no current will be available for ESP8266.

Am I correct ? Maybe I have forgotten the basics.
See this video which I made but zener model is not working. Proteus bug. Anything above 3.3V should be clamped at 3.3V and load current will be zero.
 

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ArakelTheDragon

Joined Nov 18, 2016
1,353
The current will be divided according to Kirhov's law of currents. In 1 node all currents summed equal zero. To limit the current flowing through the diode you can put a resistor after it. The diode will consume "0.6V", so the resistor should be "(3.3V-0.6V)/diode current". The current depends on the diode. The rest of the current should go towards the load. I will simulate it on proteus and show you a graph.
 

jayanthd

Joined Jul 4, 2015
944
The current will be divided according to Kirhov's law of currents. In 1 node all currents summed equal zero. To limit the current flowing through the diode you can put a resistor after it. The diode will consume "0.6V", so the resistor should be "(3.3V-0.6V)/diode current". The current depends on the diode. The rest of the current should go towards the load. I will simulate it on proteus and show you a graph.
Ok.

I know I have to put a series resistor in zener circuit to limit the zener current. I was just testing the zener diode working in Proteus but using a 3.3V model it doesn't clamp the o/p voltage at 3.3V if i/p is > 3.3V.
 
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