[moved] RC circuit backup supply

[Dots]

Hi

I am working on a design involving SIM7100 communication chip (**broken link removed**). 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?

 

[Audioguru]

Re: RC circuit backup supply

Why not use a rechargeable battery that holds its voltage high until it is almost dead? A Lithium cell is 3.2V or 3.4V when dead and is 4.2V when fully charged.
A capacitor voltage begins steeply dropping immediately when it is discharging.

 

[Dots]

Re: RC circuit backup supply

A design requirement asks for a capacitor. I came across supercapacitors, but I'm not sure if they would do the trick since I didn't deal with them before. The thing is that the communication chip needs around 2 seconds to send a message to a server alarming of a main supply cut off. Wouldn't an RC circuit with a supercapacitor be able to deliver constant power for like 4 seconds?

 

[FvM]

Re: RC circuit backup supply

I don't understand the idea behind the model.

- A GSM module doesn't consume 2 A average current, instead e.g. 0.2 to 0.4 A depending on the transmitter power setting and data rate.
- Capacitor voltage drops during discharge. To source e.g. 0.3 A for 2 sec while the voltage drops from nominal 4.1 to minimal 3.3V, your capacitor must have 0.75 F capacitance, which refers rather to a supercap than a regular electrolytic capacitor.

 

[Dots]

Re: RC circuit backup supply

Yes, the module in question consumes that amount of current on average, but it consumes a peak of 2A every 4.615 ms according the datasheet. This means that the supercap has to provide that amount of current while dropping from 4V (peak) down to 3.4V (minimum required to power module).

As for the 2Ohm model resistor, I thought that it would work to set up an experiment where the load (module) can be modeled into a 2Ohm resistor that will draw continues 2 amps peak once the capacitor start to discharge, and I can observe for how much time the voltage on the capacitor will keep close to ~4V. Otherwise, how would I be able to test the circuit without using a modeled load?

 

[erikl]

I think you can easily calculate the size of the necessary capacitor. From fig. 7 , from a single period:

0.577ms * 2A = 1.154 mAs
4.038ms * 0.36A = 1.454 mAs ; 360mA from Table 34: GSM900 (3Rx, 2Tx)
_________________________
in 4.615ms 2.60768 mAs needed ---> 0.565A mean current/period

0,565A * 2s = 1.13As = I*t

Q = I*t = C*V

Typ. voltage = 3.8V (from Table 6 , p. 24)
Min. voltage = 3.4V ---> Difference ΔV = 0.4V

C = I*T / ΔV = 1.13As / 0.4V = 2.825 F

So you need a 5V super capacitor ≥ 3F

 

[Dots]

How did you get 0.565A current/period?

 

[erikl]

How did you get 0.565A current/period?

2.60768mAs / 4.615ms = 0.565A

 

[samson A.]

Your cir uit looks good. Only that trying to charge the capacitor from 4V supply through a diode may not be a too good idea. Increase the charging voltage to 4.3v so you don't have to use a too hefty capacitor

 

[erikl]

Only that trying to charge the capacitor from 4V supply through a diode may not be a too good idea. Increase the charging voltage to 4.3v so you don't have to use a too hefty capacitor

samson is right. I'd try and replace the silicon diode with a germanium diode, so you can load the cap to about 3.7 , perhaps 3.8V.

 

[Dots]

I am definitely going to compensate for the voltage drop over the diode.

The problem now is to find a value for R1 to control the time that the cap will take to fully charge. Things are difficult because the initial charging current is currently limited to 20mA, so R1 should ideally be 200 ohm at least. 5RC (time required for full charge) equals ~70 minutes. The GSM module has to be powered within seconds, not tens of minutes.

What do you suggest?

 

[samson A.]

What you want is feasible, I just don't think this simple rc will do. Since you can't wait for the charging cycle of the capacitor to complete, say you have a semiconductor spdt switch which of course must be able to handle the required current by the module. NC to the 4.3V and NO to the charging capacitor. COM to the module. At power up the switch connect the module directly to your 4V supply while the capacitor is charging. Another circuit monitors the voltage on the capacitor and when it reaches a threshold, the spdt switch switches.

Just a concept I think will solve the problem

 

[erikl]

I am definitely going to compensate for the voltage drop over the diode.

The problem now is to find a value for R1 to control the time that the cap will take to fully charge. Things are difficult because the initial charging current is currently limited to 20mA, so R1 should ideally be 200 ohm at least. 5RC (time required for full charge) equals ~70 minutes. The GSM module has to be powered within seconds, not tens of minutes.

What do you suggest?

As your charging current from your 4V source is so low, wouldn't it be a simpler solution to buy a USB 5V charger with 1A or 2A capability and load the storage capacitor via a (1A or 3A) silicon diode very fast to 4.3V ? I guess you can get cheap USB chargers everywhere in the world.