Power selector circuit question

I am working on a data logger design. I need to incorporate two power sources, one internal non-rechargeable battery (9V) and one external power source (12V-24V range).

I've looked at some online resources and reference designs. Here is a simple two P-MOSFET OR design:

Power Multiplexer Prevents Cross-Current Conduction POWER DESIGN By Scot Lester, Applications Engineer, Portable Power, Texas Instruments, Dallas

I'd like to base my design on this one instead of the much more complicated design in later part of this reference. Since the internal battery is non-rechargeable and lower in voltage, I don't think this is safe for my situation. if Q2's gate is pulled low, then a higher main WILL charge the non-rechargeable battery and possibly cause fire.

What if I connect Q2's gate to main instead of pulling it low? This way since main (external) is always higher than internal battery (aux), the selector always prefers external, which is what I'd like to do, to preserve internal battery as much as I can. I can also put fuses inline with main and aux. Will this be a sensible solution? I imagine I use a 100Kohm resistor inline with gates on both MOSFETS and the main's MOSEFT is always pulled low.

One rare situation would be that the external battery is run down to below 12V (unlikely if I have a legit Lithium battery solar charger in between), the internal battery will charge the external battery. Will this happen?

Thank you!

Why not use a DPDT relay?

larryd: Why not use a DPDT relay?

Thanks. How would the logger actuate the relay when it needs to switch power, with the external power applied to coil so it would switch back to internal battery if it loses power? Any suggested parts I can look at? Maybe a solid state relay? I thought maybe single pole is enough since the grounds are shared?

Voltages can be monitored with A0 and A1.

Add a Vin capacitor for transfer time.

https://www.jameco.com/z/DS2Y-S-DC12V-NAIS-Aromat-Electromechanical-Relay-DPDT-2A-12-Volt-720-Ohm-Through-Hole_99311.html

A Shottky diode between "Main"(9V batt) and Q1's drain?

Hi All, Here is another messed-up schematic that was published! I assume that the 'main' terminal is + because of the polarity of Cout. Look at Q1: the shunt diode is forward biased! Q1 is drawn wrong. It is reversed drain to source! (Likewise for Q2.) I like Larryd's suggestion. Herb

Thanks larryd. I assume that your MCU controls which power source to draw power from with that pin and Q1. The diode must be the flyback diode since it is possibly a coil relay.

herbschwarz,

I don't think the schematic I posted from the reference is necessarily wrong. It is using the MOSFETs' body diodes as a low voltage drop alternative to regular diodes so the MOSEFTS are "backwards". This looks like a diode OR circuit but it's not working for my purpose. I'll definitely use a relay (with some caps) anyway. Thanks for providing your opinion.

“I assume that your MCU controls which power source to draw power from with that pin and Q1. The diode must be the flyback diode since it is possibly a coil relay.”

The left battery would initially power the cct.

The controller has the ability to transfer the relay contacts by turning the transistor on.

There would be a capacitor from Vin to GND. It supplies power during the contact transfer time (milli seconds).

If needed, A0 and A1 could monitor the two batteries to monitor their voltages.

larryd,

Thanks for clarifying. Could you also explain the double pole? Is that because it allows twice the current of a single pole?

Yes, the contact current capability would effectively double.

I think you will find DPDT relays similar to the one I linked ($2.95) are more easily found and many are less expensive than a SPDT version.

With a DPDT relay you do have the option of switching both the plus and negative side of the supplies.

Great! Thanks again larryd. I'll use the relays.

Updated:

Thanks for adding the capacitor in.

Say if I want to sense the battery voltage as an indication of its charge, is there any alternative to voltage dividers? The battery will be 9V and I can use say a 100kohm trim pot to read its voltage. On the other hand, 9V/100kohm=90uA constantly discharging the battery. Maybe I should use a MOSEFT to turn on the divider only when I need to sense the battery?

You could use a second DPDT to connect the A0 and A1 voltage dividers to the two batteries as needed.

You would connect the top of a divider to a N.O. open contact, the common terminal connects to battery plus.

Energizing the relay connects the battery to the voltage divider, read the A0 voltage then de-energize the relay.

During the relay de-energize state, the battery has no discharge current.

Sorry to bring this older thread back. Just thought that if I had to keep the mechanical relay connected, I would have to spend say 200mW to do so, not efficient enough. What if I just replace the bottom MOSFET with a diode like JCA79B said? Since the internal battery is rarely used, probably only very occasionally, the diode protects it from getting charged and in case it is needed, the diode drops a small voltage, until external battery is connected again.