Li-on charging ICs

[MattHollands]

I have a system that runs on 3.3V and I want to run it off a single 3.7V Li-on cell. I am planning to use a 3.3V LDO such as the BU33SD5 to lower the battery voltage. However I also want to charge the battery in-circuit from USB. My current plan is to use a MCP73811 to charge the Li-po at 85mA. So I would connect the VDD (power in) pin to the VBUS pin on the USB port, and connect the battery AND system-load (ie the LDO) to the VBAT pin (const current/oltage out). The obvious downside of this is that when charging, if you use the system then you may be stealing some of the 85mA charging current.

USB -> MCP73811 -> battery in parallel with LDO -> system

As this is my first rechargeable battery project, I just want to be careful that I don't do anything stupid - so is this a reasonable solution?

Is there a more elegant way to do this? Perhaps some kind of integrated solution that separates the battery from the system load and generates the 3.3V either from the battery or the USB voltage (depending on the situation).

[Mr.B]

This is what I do...
Excuse the rush job of the schematic.

Edit:
Battery is isolated from load while charging and the USB 5v supplies the load as well.
When USB 5v is removed, uninterrupted power is supplied by the battery.

[MattHollands]

This is what I do...
Excuse the rush job of the schematic.

That's something like I was imagining actually! Think I might borrow it 

I have a question about the choice of FET. Normally for a P-channel FET datasheet I would expect to see the Vds/Id curves with Vds and Id negative (because p-channels normally conduct from source to drain). But the datasheet for this transistor has positive Vds and Id. is that because they're just ignoring the polarity? Or is it known that the relationship is symmetrical? ie flipping the polarity of Vds flips the polarity of Id (assuming Vgs stays constant) but keeps the same magnitude. Or maybe this transistor is designed to be connected "backwards"?

Link to datasheet:
http://www.vishay.com/docs/70833/70833.pdf

[vealmike]

All good, but you can't ever turn it off...

How about:
    a 2nd PFET back to back with the battery PFET. Couple the gates together.
    Add a 2nd diode with resistor with anode to the battery and cathode to the FETs gate.
    Put a switch to GND or GPIO where your 10K resistor is GNDed.

[MattHollands]

All good, but you can't ever turn it off...

How about:
    a 2nd PFET back to back with the battery PFET. Couple the gates together.
    Add a 2nd diode with resistor with anode to the battery and cathode to the FETs gate.
    Put a switch to GND or GPIO where your 10K resistor is GNDed.

Is this what you're suggesting?

If so, the main issue I see is that you cannot use the system when USB is plugged in as both transistors will be off.

My counter suggestion would be this: This is basically same as original but added an extra transistor which is held on or turned off by the switch or a GPIO.

[MattHollands]

Even better, add a diode and the switch can also be used as an input to the microcontroller

[vealmike]

Almost... Try running this in LTSpice.

If USB power is applied, M1 / M2 gate is always at 5V less a diode drop. Battery is always disconnected.

If USB power is not applied and cathode of D4 is pulled to GND, (as shown) cct is powered from battery.

If USB power is not applied and cathode of D4 is not pulled to GND, cct is off.

You can play games with switches / pushbuttons and GPIOs on D4 cathode to create a cct. that comes on when a button is pressed then stays on until SW drives the GPIO high.

[vealmike]

Even better, add a diode and the switch can also be used as an input to the microcontroller

Yep, I think that works. Elegant too.

Hmmn. Simulate it and look at the current through the discrete diode attached to 5V... I think you may see something interesting.

[krho]

I'd suggest using something from texas instrumets with integrated MOSFET e.g. BQ24079 Or any other from that collection. AFAIR there are some that also support USB PD 1.2

[MattHollands]

Even better, add a diode and the switch can also be used as an input to the microcontroller

Hmmn. Simulate it and look at the current through the discrete diode attached to 5V... I think you may see something interesting.

I'm not seeing any issues in my simulation. But what I can kind of see is that if the lower FET is off, then the top FET may be slightly on and some current may flow through the diode from the usb to the battery. Is this what you're hinting at?

[MattHollands]

Here is my simulation

[Edit] Wait, did it again and I'm getting massive current spikes through the diode.... Why? [/Edit]

[mikerj]

Here is my simulation

[Edit] Wait, did it again and I'm getting massive current spikes through the diode.... Why? [/Edit]

Because you added C1. When the MOSFET switches on the diode see's a discharged capacitor, virtually a short circuit.

[MattHollands]

Nope! I added the cap to get rid of the voltage spikes that came along with the current spikes. Here it is without cap.

But maybe there is some capacitance on the FETs.....

[ali_asadzadeh]

BQ24160 is a good candidate for your project.

[Koen]

Linear battery charger + dc/dc converter(s) in a single chip : http://www.linear.com/parametric/Battery_Charger_Plus_DC%7CDC#!1069_Linear|Switching|Shunt!chem_!vmax_!icharge_!bat_bat

[vealmike]

[Edit] Wait, did it again and I'm getting massive current spikes through the diode.... Why? [/Edit]

 
Thought you might. I think you'll learn more if you try and work it out for yourself. Sooo, where does the current come from and where does it go to?

There are three components attached to the anode of D1. Plot the current through each and compare with the current through D1.
The cathode is easier, there are only two components to look at.

[MattHollands]

Sooo, where does the current come from and where does it go to?

Current is flowing from USB to battery and back again. I guess this is due to the reverse recovery of the diode and the PFET not turning off fast enough? But I have no idea how to fix this

[vealmike]

 :-+Bingo!
The diode turns on faster than the fet switches off. This is largely due to gate charge on the FET.

The reason I wasn't sure  you'd see it is that it depends on the exact models you picked.

You can't easily fix it, but a resistor in series with the diode would limit the current through the diode without harming the cct. function.

[MattHollands]

:-+Bingo!
The diode turns on faster than the fet switches off. This is largely due to gate charge on the FET.

The reason I wasn't sure  you'd see it is that it depends on the exact models you picked.

You can't easily fix it, but a resistor in series with the diode would limit the current through the diode without harming the cct. function.

Alright sweet! [emoji846] thanks I'll stick in a 10R resistor!