Design power supply for SIM800L Core Board

Key topics : Powering SIM800L Module, SIM800L Core Board, Linear Regulator, Buck Converter, SMPS, Arduino

Introduction

As it seems most of the issues related to SIM800L core board is due to inappropriate power supply. If you want refer this SIM800L hardware design where manufacturer clearly specifies the suitable power supply for SIM800L module. 

Refer the power supply specifications 

• Voltage : 3.4 V to 4.4 V (4.0 V recommended, Some of the latest versions may work directly with 5V power supply)
• Current Rating : 2 A (To cater the demand at transmit bursts)

If you provide this voltage and observe the current through the Ammeter you will hardly notice a current of 2 A but few milli Amps. It will be using 2A at transmit bursts for less than a milli second at a time. Therefore it is really important to provide a power supply which can provide 2A with a very minimum voltage drop.

Most of the time issues related to power supply is because of people using linear regulators which cannot supply adequate power as well as not the suitable voltage. (Even in Arduino boards 5V power supply cannot provide more than 500 mA and 3.3V Power supply cannot provide more than 100 mA, Also they are not in the range of required voltage level) So directly connecting SIM800L module with Arduino will never fulfill the requirements. 

We'll see what data sheet recommends to power SIM800L module,

Manufacturer recommends to have the following arrangement as much as closer to the Vin pin of the SIM800L module. 

SIM800L Power Input

If you observe closely you can see that some of those capacitors are already included in the core board. However importance of those components are that,

• Zenor diode will be removing any voltage sure if present protecting the sensitive electronics
• 33pF and 10pF capacitors will remove any high frequency interferences
• 100uF and 1uF are also useful to make the input power supply smooth as well as to help with high current bursts during transmits. Also it is much more effective to use Tantalum capacitors due to low ESR (Equivalent Series Resistance) 

Since we cannot work with any of the commonly available power supplies or Arduino power outputs we'll see what are the other options we can have.

1) If you are using a 5 V Power supply manufacturer recommends to use a LDO (Linear Drop-Out) regulator which would be a great and efficient solution in such case. Here in this case they've recommended MIC29302A which can supply up to 3A with a 1% tolerance. 

MIC29302 Power Supply for SIM800L

Lets see the output voltage it is designed,

2) Since this seems little bit complicated we can easily achieve the same performance by using a diode with a forward bias voltage drop of 0.7V and a rating of 2A. (Consider forward bias voltage when selecting the diode)

Diode Power Supply SIM800L

3) If you are using a power supply above 5V using an LDO or another type of linear regulator would be very much inefficient. So the best option becomes using a Buck Converter (Switch Mode Power Supply) For such applications LM2596 based buck converters would be the ideal solution which is commonly available as well as less expensive.

LM2596 Adjustable Buck Converter Schematic

This is the schematic of most commonly available. Below image shows a commonly available commercial solution of this circuit.

Commercially available LM2596 Converter Module

However if you refer the above datasheet of SIM800L module, manufacturer recommends a slightly different setup with addition of another inductor and couple of capacitors which would deplete the amount of noise this produces. Normally buck converters tends to have a little bit high amount of noises compared to linear regulators. But this works fine with SIM800L module. Even though it is better to add a 100uF capacitor and a 100nF capacitor closer to the SIM800L module when using this.

4) Since most of the Li-Ion Cells are of 3.7 V -v4.2 V and it is in the recommended range we can directly use a Li-Ion cell to power up a SIM800L Module. However be careful when using small batteries as they may not be able to provide the adequate voltage at 2A current burst due to internal resistance.