Electronic – AC/DC 4W vs 2A GSM transmission bursts

cpugsmpower supply

I have a big problem with the power supply of a SIM900.
I designed my own board with a 220VAC-5VDC tracopower, an ARM9 CPU, and a GSM/GPRS SIM900 module.
The output of the tracopower is 5V – 800 mA (so 4W).
To supply my GSM module, I use a LDO 5V – 4.1V.
When SIM900 is transmitting its 2A burst, 4.1V GSM supply don’t have any voltage drop. But with the 5V supply, I go from 5V to 4.3V, so the CPU reset.

I know that I did a big mistake when I designed the board because I can not supply the CPU directly from the AC/DC. The best solution would have been to use a AC/DC 9V and then create a 5V for the CPU and a 4.1V (from the 9V source) for the GSM. But I can't change the AC/DC so I need to use this AC/DC 5V-800mA.

So do you have a solution to avoid a voltage drop on the 5V ?
For information, the 4.1V can drop to 3.2 (the Sim900 supports that).

This is my 5V power supply :
5V Power Supply Schematic

And this is the GSM power supply :
enter image description here

Many thanks for your help.

Best Answer

2A bursts are 577 µs every 4.6 ms

The power for those bursts is equivalent to about 1.25 watts\$^1\$ - if the 4.1V supply were 100% efficient at taking energy from the 5V traco then the power needed would be 1.028 watts but because you are using a linear voltage regulator to create the 4.1 volts then it's 1.25 watts from the traco.

The question is, from your 4 watt traco supply, is the remaining 2.75 watts enough to power the circuits that don't take the 2 A pulse. If it is then that's OK you don't necessarily need to change the traco and you can provide the "surge" power with a large capacitor, preferably directly on the power feeds to the SIM900.

How big? 2 amps at 4.1 volts for 0.577 ms is an energy of 4.73 milli joules so rearranging the well-know energy-capacitance-voltage formula gives you: -

\$C = 2\dfrac{energy}{V^2}= 2\dfrac{4.73 mJ}{16.81}\$ = 563 uF BUT this needs to but at least 5 times bigger so the voltage at the terminals doesn't droop - consider that it needs to be about 3000 uF and use a low ESR type.


\$^1\$ average power is 5 volts x 2 amps x 0.577 / 4.6 = 1.25 watts.