Hello! I’m working at a circuit, which can use either 5V from USB or 9V from external battery. I implemented a jumper selector through which the user can choose where the voltage comes from.
The problem is that I would have 2 different 5V signals: one from USB and one from a voltage regulator from the 9V signal. How can I design a circuit that automatically switches between those 2 whenever one is active and the other one is not.
Remember that the 2 power supplies can never be activated at the same time.
Thank you!
Sorry, but circuit design is outside the scope of this forum, which is for questions relating to KiCad. Even though you may be using KiCad to draw the schematic, the problem would not change if you were using another ECAD suite.
Try another more suitable forum, pehaps one of those under EEVblog.
This topic will be closed after a grace period for others to suggest better places to find answers.
The pensionable pussy is quite correct of course and EEVblog is a good start but any site or forum that deals with development boards and the like will be able to help. You basically will need a P Channel MOSFET and you will be switching the ‘high side’ of one supply when the other one isn’t present. ‘Arduino battery backup’ would be one search term to get you going
Of course when you have your design and find a problem relating to its PCB and Kicad you can come straight back for help.
If you can tolerate a few hundred millivolt drop then diode-OR with a couple of schottkys. This is perfectly fine if you are feeding a 3.3V LDO as there is plenty of headroom. Fancy ways to do this with pass mosfets and comparators are overkill unless the drop is an issue.
I was thinking more along the lines of this, that if the USB fails or is switched off the power is switched to the lithium battery, but both cannot be on at once
Also…see the attached if you want to “roll your own.”
This will select the higher of the two inputs. One input being shut down should be OK. This design is slow (it uses compensated op amps with negative feedback) but DC-wise it will work as well as you are willing to “pay for.” In other words, the forward voltage drop is limited by the input offset voltage of your op amp and the “on” resistance of your FETs. If you want to buy precision op amps and big FETs with low “ON” resistance, the voltage drop can get below 1 mV. But even with cheap parts, a 15 - 20 mV voltage drop is reasonable.
Oh yeah ! I just grabbed it to illustrate my point, should have looked a bit closer but we’re not supposed to do it anyway I’m sure the OP can fix it with a diode, Thanks ! Its late and thats my excuse and I’m sticking to it
