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I want to design a voltage-auto selector for a battery powered board that I am designing.

First Question

Now I could just connect the LDO of my board to the battery output of the battery charger IC, and that would work perfectly fine. However for better optimisation, I would like to bypass the circuitry for that, if external power is connected. The first question is if it is worth trying this, or is the gain by doing so, so insignificant that it makes no sense to attempt it.

If it is worth it

In my research I came across an article that showed the circuit below as a reference design for such an application in mind.

enter image description here

You can find the whole article at "https://www.edn.com/use-op-amps-to-make-automatic-oring-power-selector/"

Now I absolutely love the idea of using op-amps and PFETs to reduce power waste. But is there another way to do something similar? Perhaps something that uses a comparator configuration? Otherwise I will just copy that circuit. In any case all sharing of thoughts and expertise would be much appriciated.

Below you can find the results of a small simulation I ran as requested by the accepted answer's author. The resistor symbolises the load attached.

enter image description here

JRE
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Emre Mutlu
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  • Why do you even need opamps? https://electronics.stackexchange.com/questions/525248/is-this-mosfet-upside-down/525249#525249 – DKNguyen Dec 16 '20 at 21:22
  • I dont if there is a better way :) – Emre Mutlu Dec 16 '20 at 21:23
  • look at page 2 and detailed info on page 5 of this [application note](https://www.microchip.com/wwwAppNotes/AppNotes.aspx?appnote=en533380) by microchip which considers using a simple load sharing method using a p-mos. – Tirdad Sadri Nejad Dec 16 '20 at 21:27
  • @DKNguyen I did a spice simulation of this circuit. That PMOS configuration is good for reverse voltage protection however it doesnt protect the battery if the Voltage on Drain(in this case 5V) is higher than on Source. – Emre Mutlu Dec 16 '20 at 21:28
  • @EmreMUTLU Which one are you looking at? I meant to point at the circuit in the question, not the answer. https://electronics.stackexchange.com/questions/525248/is-this-mosfet-upside-down/525249 – DKNguyen Dec 16 '20 at 21:31
  • @DKNguyen I see. I ran some simulations with mosfets configured in the diode configuration all resulted in worse results compared to a dedicated shottky diode. But I will run a simulation with that specific circuit too. – Emre Mutlu Dec 16 '20 at 21:35
  • What do you mean by worse performance? How does a Schottky give worse performance than a MOSFET when running off the battery? The mindset of the circuit I pointed to is that simplicity is more important than loss when running off external power so a diode is used for external power instead. – DKNguyen Dec 16 '20 at 21:36
  • @DKNguyen No I meant in the simulation using a Schottky diode connected to the battery yielded better performance compared to a mosfet configured as a diode. This is when no external load is connected. But maybe the circuit You sent should yield better performace, it isnt really configured as a diode. I didnt simulate it yet. – Emre Mutlu Dec 16 '20 at 21:42
  • @EmreMUTLU What kind of *specific* performance parameter are you talking about? I don't see how a Schottky could have a lower voltage drop than the MOSFET unless you were pulling so much current through the MOSFET that IR drop across it actually exceeded the Schottky's relatively fixed voltage drop. – DKNguyen Dec 16 '20 at 21:45
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/117361/discussion-between-emre-mutlu-and-dknguyen). – Emre Mutlu Dec 16 '20 at 21:46

1 Answers1

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I ran some simulations with mosfets configured in the diode configuration all resulted in worse results compared to a dedicated Schottky diode.

In your simulations, you are not using the MOSFETs properly. You are wiring them up such that the rest of the MOSFET is gone and only the crummy parasitic body diode remains. In this case, it would have a higher voltage drop than a Schottky diode. This isn't the goal.

The goal is to connect the PMOS in such a way that the source-drain actually conducts and flows through the low source-drain resistance. It's just that the current happens to flow in the opposite direction than usual through the source-drain. You don't actually want it to flow through the body diode. The body diode's conduction can be used to jump start the process but once everything starts the source-drain resistance will decrease and current will flow through that instead of the body diode.

Circuits like this are not drop-in replacements for regular diodes. You can't just put them anywhere you want in the circuit without thinking like you can with a real diode because the PMOS gate has to be connected to the correct potential (and sometimes additional circuitry is required for that).

You don't need opamps.

You can use this circuit which I will steal from another user's question (Is this MOSFET upside down?) on this website: enter image description here

With no external power (VBUS) the resistor is able to pull the PMOS gate to ground which turns it on and the PMOS conducts through the source-drain resistance rather than the parasitic body diode resulting in a very low voltage drop.

When there is external power, the external power overpowers the pull-down resistor and stops the PMOS from turning on, so the source-drain remains high resistance. However the PMOS body diode still tries to conduct, however, the Schottky diode has a lower voltage drop than the PMOS body diode so as long as the external supply voltage, VBUS, is at least as high as the battery voltage (and can even be a bit lower than that), the Schottky diode will conduct and cause the voltage at the PMOS source (the body diode cathode) to be higher than the PMOS drain (the body diode anode), thereby reverse biasing the PMOS body diode so it can't conduct.

DKNguyen
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