Review my power supply circuit

[DTJ]

Anyone interested in a quick review of my power supply circuit?

The application is a hand held battery operated device. I need soft power on / off via a button.

I could use a switcher to get the most out of the battery but this thing is only low power so I'm happy to keep it stick with a LD linear regulator.
I'm not using a 78L05, I just stuck that in there for the purpose of the review.
D2 will also probably be deleted.

I'm not sure that I need D3.

[Richard Head]

Looks good.
What's the purpose of D3?

[Kalvin]

You may want to flip Q1.

[Gyro]

You're not making very efficient use of your Alkaline battery discharge curve. The fully discharged point is 4.8V. Having D1 in series with the supply immediately wastes quite a bit.

An LP2950-50L makes a good LDO substitute for a 78L05 (needs a bit more output capacitance for stability). I'm not sure if it's actually worth going to a switcher or not. With the substituted LDO and D1 removed you'd get down to 5.something quite small, which is about 95% of the battery capacity.

[Rick60]

yes swap D<->S on Q1 otherwise the body diode will be in forward conduction ,and the circuit will never switch off

Be wary of internal pull up resistors on micro ports ( some micro's have them  enabled by default ),so when you power down if the micro has brown out , it will reset  ,set  the ports back to  input , the pull up could now  provide just enough voltage to turn Q2 back on ,so your circuit re-powers itself, simple fix put a 1-10K to ground on pwron   

[DTJ]

You're not making very efficient use of your Alkaline battery discharge curve. The fully discharged point is 4.8V. Having D1 in series with the supply immediately wastes quite a bit.

An LP2950-50L makes a good LDO substitute for a 78L05 (needs a bit more output capacitance for stability). I'm not sure if it's actually worth going to a switcher or not. With the substituted LDO and D1 removed you'd get down to 5.something quite small, which is about 95% of the battery capacity.

Yes, I've actually used an LP2951 in the test circuit I've lashed up. I was trying to utilise the shutdown pin to control the power on/off but the battery drain when off was a bit too high.
D1 was initially placed to protect against reverse battery connection, now that I've added the high side FET I can probably delete it.
The micro needs 5V so I guess with an LDO I could run the battery down to 5.5 to 6V without the series diode.

You may want to flip Q1.

Yep, thanks D<=>S are arse about.

Looks good.
What's the purpose of D3?

D3? I'm not sure! I think it can be deleted. The circuit has evolved a bit and it now serves no purpose.



Changes so far!

Flip the BS250 S/D reversed.
Delete series diode D1.
Delete protection Zener D2.
Use an LDO - run battery down to ~6V.

Put a 1-10K to ground on pwron

I might see if I can find a SOT23 dual diode to replace the two 1n4148 diodes.

[DTJ]

yes swap D<->S on Q1 otherwise the body diode will be in forward conduction ,and the circuit will never switch off

Be wary of internal pull up resistors on micro ports ( some micro's have them  enabled by default ),so when you power down if the micro has brown out , it will reset  ,set  the ports back to  input , the pull up could now  provide just enough voltage to turn Q2 back on ,so your circuit re-powers itself, simple fix put a 1-10K to ground on pwron

Good point on the pull down, good to start and finish in a known state. Thanks.

[Kleinstein]

If the µC can go to a very low power mode, one could use a low standby current regulator like MCP1702 and get away without turning the regulator off. The 78L05 needs quite some current and higher voltage than an LDO -  so not a good choice for battery operation.

It's also a question if you really need 5 V - many µCs also run on less and need less current this way.

[DTJ]

If the µC can go to a very low power mode, one could use a low standby current regulator like MCP1702 and get away without turning the regulator off. The 78L05 needs quite some current and higher voltage than an LDO -  so not a good choice for battery operation.

It's also a question if you really need 5 V - many µCs also run on less and need less current this way.

I would have used the sleep mode but the other circuitry around the uC draws too much quiescent current and would need a supply that could be turned off.
The Vreg has been changed to a LDO type form microchip.

I'd normally run at 3 or 3V3 but the peripheral parts in this project all require 5V so I've just gone for 5V across the board.

Cheers.

[DTJ]

Ok - here's the revised circuit. Parts are ordered, I'll see how it performs.