0
\$\begingroup\$

I am still a beginner but I already managed to design a circuit around a PSoC 4 device with some peripherals (5V powered sensors, 5V powered mini fan, 5V step-down regulator for the main power supply and 5V step-up regulator for the sensors and fan). The circuit will be powered with 5V DC main power supply which will be either a wall plug supply or a power bank with minimum 5V voltage but it does not matter I hope.

Now I would like to add a battery backup power in case the main power supply fails or will be removed, just to be able to put the chip in sleep mode and keep the real time clock running, which is provided by the chip.

So far the application description.

Now to ensure that my backup battery lasts a long time and really provides current for the sleep mode only, which will be some uA, I would like to cut it off the circuit in case the main power supply is available. I woud like to use a N-MOSFET as a low side switch for this, but I am not really sure where to attach this mosfet in my existing circuit. After some research I came up with the solution in the attached schematic image, and here are my questions:

1) Is this correct that way in general?

2) Do I need the N-mosfet switch at all? I found several statements which state, that the battery won't be used as long as it's voltage is lower than the voltage of the main power supply.

3) I heard in some tutorial about power sources, that it's in general not wise to share a common ground between grounded (my 5V wall plug power supply) and floating power sources. Is this realy a "problem" in my case?

4) Can I switch the possition of D7 and U4 or is it bad when Vout of U4 gets the battery voltage?

enter image description here

Noe: Vgate will be set high by the PSoC 4 device as soon a voltage drop is detected by a voltage comparator, to switch the mosfet on. The power for this critical time, between the main supply failure and switching on the battery backup, will be provided by a 220uF capacitor.

I will appreciate any hints on how to make a simple battery backup for a long battery life time.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Many would opt for a P-FET on the high side. You arrangement has its benefits too. How are you planning to charge your battery? \$\endgroup\$
    – winny
    Commented Apr 9, 2017 at 21:18
  • \$\begingroup\$ I do not plan to charge it at all. I plan to use a CR2032 cell to be a replaced like the CMOS battery in a PC. \$\endgroup\$
    – Artur Cichosz
    Commented Apr 9, 2017 at 21:22

1 Answer 1

Reset to default
2
\$\begingroup\$

The key to these circuits is that you have to find a way for your circuit to turn itself on provided there is no voltage applied from the external power source. Many very basic reverse battery protection circuits (that go beyond a series diode) can actually be used with litte to no modification for this purpose - I would suggest starting there. Take a look at figures 2 and 3 from his application note from TI:

http://www.ti.com/lit/an/slva139/slva139.pdf

\$\endgroup\$
4
  • \$\begingroup\$ Hi, thank you for you reply. I will have a look into this, but my case assumes that there is allways some voltage applied, either main power supply or the battery backup voltage or (I did not mention it, sorry) a decent electrolythic capacitor e.g. 220 uF which should provide the power for the voltage comparator to detect voltage drop and to switch the n-mosfet on. \$\endgroup\$
    – Artur Cichosz
    Commented Apr 9, 2017 at 21:13
  • \$\begingroup\$ The Psoc device is fully operational with a voltage range 1.7-5.5 V, so I will be allways able to put it in sleep mode and wakeup given that one of the three powerr sources above deliver power as expected. \$\endgroup\$
    – Artur Cichosz
    Commented Apr 9, 2017 at 21:16
  • \$\begingroup\$ Joren, after I while I took a second look at the resource from TI you linked. Why is there (figure 2/3) a diode in parallel with the FET? It is conductive in the same direction as the FET itself so what is the point here? \$\endgroup\$
    – Artur Cichosz
    Commented Sep 18, 2017 at 8:55
  • \$\begingroup\$ @ArturCichosz that is not a external device that has been added, but is there to illustrate the internal body diode of the device. It is there because of the way a mosfet is built up. I discussed it in this question: electronics.stackexchange.com/a/308885/34873 \$\endgroup\$
    – Joren Vaes
    Commented Sep 18, 2017 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.