RC circuit backup supply

[Dot]

Hi

I am working on a design involving SIM7100 communication chip (datasheet: http://simcom.ee/documents/SIM7100E/SIM7100_Hardware_Design_V1.02.pdf). As mentioned on page 24, its supply is rated for 3.4-4.3V with maximum current consumption of 2A every 4.615ms. I have a main 4V supply designed to deliver 8.4W.

I want to design a backup power supply, based on an RC circuit, that can deliver the required power, for at least 2 seconds, whenever the main supply is interrupted. This RC circuit should be charged from the main supply, holds its charge until the main supply cuts off and it starts discharging to deliver uninterrupted power to the chip for at least 2s.

So far, I've come up with the simple circuit below with no particular values except modeling the load (the chip) as a 2ohm resistor (4v/2ohm= 2amp). The diode is supposed to act as an open switch whenever the main supply cuts off, so that current only flows from the capacitor to the load. How reliable is this circuit? any suggestions on how to choose capacitor, R1 and diode given the high amperage required by the load for at least 2 seconds?

[ovnr]

This is not, ah, going to be terribly convenient.

Your GSM module will indeed pull bursts of up to 2A. I can't be bothered finding out it's worst case average consumption; if it were to pull a constant 2A, you'd need a 6F (yes, farad) capacitor to hold >=3.4V for 2 seconds, if you started off at 4. But you will also have to contend with the ESR of the capacitor, because the module does not permit the voltage to droop below 3.4V. So if your cap has a 1 ohm ESR - not uncommon for supercaps - you sacrifice a volt right there, and you're dead in the water before you even began.

You may want to describe why you want a two-second backup supply, so people can give you better suggestions than "slap a big honkin' cap on it".

[Dot]

The backup supply is needed for the module to send a message to a server alarming of a main supply cut off. In order to successfully deliver this message, 2 seconds is needed at least, so I'm shooting for an RC backup that can provide constant 2A while discharging from 4V down to 3.4V (minimum required to power the module) and can last for more than 2 seconds.

Using supercaps isn't out of the question if it's required.

[mij59]

Incorporate the backup function in the 4V power supply,  the module needs a fairly stable power supply, less than 0.3V voltage drop while transmitting.

[Ian.M]

The trick with maintaining a supply for a short time after mains failure is to get the extra caoacitance before the regulator as far 'upstream' as possible so you can tolerate a large percentage swing in its voltage while still maintaining regulation at the output. Lets assume you are using a  SMPSU for your 4V supply, Lets further assume you are in a country with a mains supply of 220V.

Consider a 680uF 400V reservoir cap on the DC input bus of your SMPSU, charged to the peak output of the bridge rectifier, approximately 310V.   Your PSU is delivering its full rated output of 8.4W, the power cuts off and assuming 90% efficiency, that means over the next 2 seconds it needs 18.67J from the cap.  The initial stored energy (from E=(1/2)*C*(V^2) ) is 32.67J.  After 2 seconds that leaves   14J in the cap. Calculating the final voltage from the remaining energy, the final voltage is 203V.  It isn't hard at all to design or specify a SMPSU that can supply its rated output for a few seconds while its input voltage drops by a third.

Then all you need is an optocoupler (with anti-parallel diode across its LED) fed via a RC dropper from the mains input before the bridge rectifier so the MCU gets a 50Hz pulse that goes away during power failure to warn it early enough to do its thing before the 4V rail starts dropping >2 seconds later.

If you aren't using a mains input SMPSU that you can beef up the bridge rectifier and reservoir cap of, you will need to tell us a lot more about your actual  PSU.