1
\$\begingroup\$

I need an application with back-up battery. Like UPS when power go down device will work with battery. There are chips with power path option but it is diffucult me to get components from outisde of my country (not impossible but diffucult and expensive so i stick components can be found here) and the one i got bq24072 it is very diffuclt to solder. So i came up with below idea with mcp73831 but i am not sure it will work. I will try it with dead-bug prototyping after i get opinions (i am afraid of blowing battery on my face). I am planning to use Li-Po battery

my idea of power path

MOSFET's gates can be driven with transistors. The battery symbol is only for showing where the battery is. Will that work or can anyone suggest better solution. or i shoul stick with a chip which has power-path feature?

\$\endgroup\$
6
  • \$\begingroup\$ I do not think your idea will work. When VIN has no source connected to it there will be a path for the battery to source current through the body diodes of the two FETs back to the VIN terminal. \$\endgroup\$
    – Michael Karas
    Jan 4, 2015 at 19:18
  • \$\begingroup\$ So, in this circuit, you intend to use the battery node as the power source to the rest of your device, correct? (Basically, as the MCP73831 datasheet shows on page 17) I'm not sure what you intend with that top FET or why you feel the need for the switches at all. Is this to implement an on/off switch? \$\endgroup\$
    – caveman
    Jan 5, 2015 at 1:54
  • \$\begingroup\$ What are you hoping 'FET?' should do? (yes, properly labeling your schematics would be a good idea). If it were a JFET (or maybe a depletion-mode MOSFET?) then are you trying for a current-source? \$\endgroup\$
    – brhans
    Jan 5, 2015 at 6:15
  • \$\begingroup\$ @MichaelKaras yes you're right i made changes according to that. \$\endgroup\$
    – erkan
    Jan 5, 2015 at 12:42
  • \$\begingroup\$ @caveman I intended to use battery node to supply rest of device when tere is no VIN. \$\endgroup\$
    – erkan
    Jan 5, 2015 at 12:43

1 Answer 1

Reset to default
1
\$\begingroup\$

Everything looks good except for the BSS138, which would be better replaced with a Schottky diode.

The diode will have a voltage drop of 0.3~04V when the power is on, but the BSS138 would have much higher loss since it needs about 2.5V on the Gate to turn on (its body diode is also facing the wrong direction, but that could be fixed by simply swapping the Source and Drain).

Using a transistor here is unnecessary because Vin must be higher than the battery's peak voltage anyway (in order to charge it up fully) so the small drop through a diode should not be a problem.

If you are worried about excessive voltage sag as the power goes down then you could also put a Schottky diode across the IRLML6402. The FET will turn on and connect the battery directly to the load once Vin drops more than 2V below the battery voltage, but before that its body diode could drop 0.6V or more.

\$\endgroup\$
1
  • \$\begingroup\$ yes you are right simple schottky will suffice instead of BSS138. I will try that as soon as i can (without battery and IC) then try it with them. \$\endgroup\$
    – erkan
    Jan 5, 2015 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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