Electronics > Projects, Designs, and Technical Stuff
Turning off an LDO using a mosfet
<< < (3/4) > >>
OM222O:

--- Quote from: mariush on February 13, 2019, 03:30:28 pm ---
Here's an example, costs ~ 0.2$ in volume (yours is around 0.11$ in 1000pcs) : LDL212PU50R

Digikey: https://www.digikey.com/product-detail/en/stmicroelectronics/LDL212PU50R/497-16897-1-ND/6230213
Mouser: https://eu.mouser.com/ProductDetail/STMicroelectronics/LDL212PU50R

Has enable pin, has adj/sense pin, Rated for 1.2A , short circuit current at 1.5A min, 2A max , 350mV / max 600mV dropout voltage vs 1.3v for 1117 ,  max 380uA quiescent current vs 4-6mA for 1117, 
works with 4.7uF capacitance, maybe less (datasheet isn't clear) while your 1117 needs min 10uF according to datasheet ... also not all 1117 are happy with very low esr capacitors on output but your model  says it likes low esr ceramic capacitors)

1117 in general are 0.8A max current parts. I do see it says minimum 1A current limit on the datasheet, but that's the point where the regulator would "trip" and not output higher... so when you say you need the minimum 1A capability what exactly do you mean? I don't think you can rely on the regulator to output 1A continuously to devices.



--- End quote ---

Thanks for the recommendation.
My application can peak at 1A for a few seconds while performing some tests, but then drops to basically nothing (<10mA), so I need the regulator to be able to tolerate those bursts. It's powered from a 9V battery, so the battery wouldn't even last that long if I drew 1A continuous! As for the capacitors ... how does 470uF sound?  :-DD That is again another part I buy in bulk, so I just use it for both input and output! They are low ESR tantalums (<400m\$\Omega\$) and this combination has proven so stable and robust that I've used it pretty much in all of my designs so far without a single issue! I'm really leaning towards using a P channel fet.
Edit: Yes , sorry, I meant switching off the negative terminal of the battery, not just the LDO on it's own.
Kasper:
+1 for nmos to switch battery negative. That is often done in LiPo protection circuits and in reverse polarity circuits (though reverse polarity circuit has different fet configuration)

Clearly label the net for negative battery to try to reduce the chance someone will accidentaly think battery negative = ground. And its good to include a test point for that net.
langwadt:

--- Quote from: OM222O on February 13, 2019, 05:04:45 pm ---
--- Quote from: mariush on February 13, 2019, 03:30:28 pm ---
Here's an example, costs ~ 0.2$ in volume (yours is around 0.11$ in 1000pcs) : LDL212PU50R

Digikey: https://www.digikey.com/product-detail/en/stmicroelectronics/LDL212PU50R/497-16897-1-ND/6230213
Mouser: https://eu.mouser.com/ProductDetail/STMicroelectronics/LDL212PU50R

Has enable pin, has adj/sense pin, Rated for 1.2A , short circuit current at 1.5A min, 2A max , 350mV / max 600mV dropout voltage vs 1.3v for 1117 ,  max 380uA quiescent current vs 4-6mA for 1117, 
works with 4.7uF capacitance, maybe less (datasheet isn't clear) while your 1117 needs min 10uF according to datasheet ... also not all 1117 are happy with very low esr capacitors on output but your model  says it likes low esr ceramic capacitors)

1117 in general are 0.8A max current parts. I do see it says minimum 1A current limit on the datasheet, but that's the point where the regulator would "trip" and not output higher... so when you say you need the minimum 1A capability what exactly do you mean? I don't think you can rely on the regulator to output 1A continuously to devices.



--- End quote ---

Thanks for the recommendation.
My application can peak at 1A for a few seconds while performing some tests, but then drops to basically nothing (<10mA), so I need the regulator to be able to tolerate those bursts. It's powered from a 9V battery, so the battery wouldn't even last that long if I drew 1A continuous! As for the capacitors ... how does 470uF sound?  :-DD That is again another part I buy in bulk, so I just use it for both input and output! They are low ESR tantalums (<400m\$\Omega\$) and this combination has proven so stable and robust that I've used it pretty much in all of my designs so far without a single issue! I'm really leaning towards using a P channel fet.
Edit: Yes , sorry, I meant switching off the negative terminal of the battery, not just the LDO on it's own.

--- End quote ---

the battery is floating so it doesn't matter if you switch the low or high side
StillTrying:
Yes, as long as the floating battery is only powering the LDO the switch can be in either +Ve or -Ve.

I'd probably just parallel up the DPDT switch's contacts, :) and leave it in the +Ve. If you're worried about the inrush charging current into the first big cap. put the switch between the big cap and the LDO.
mariush:

--- Quote from: OM222O on February 13, 2019, 05:04:45 pm ---
Thanks for the recommendation.
My application can peak at 1A for a few seconds while performing some tests, but then drops to basically nothing (<10mA), so I need the regulator to be able to tolerate those bursts. It's powered from a 9V battery, so the battery wouldn't even last that long if I drew 1A continuous! As for the capacitors ... how does 470uF sound?  :-DD That is again another part I buy in bulk, so I just use it for both input and output! They are low ESR tantalums (<400m\$\Omega\$) and this combination has proven so stable and robust that I've used it pretty much in all of my designs so far without a single issue! I'm really leaning towards using a P channel fet.
Edit: Yes , sorry, I meant switching off the negative terminal of the battery, not just the LDO on it's own.

--- End quote ---

I know I'm off topic and I apologize for that, but wow...

So you have a product that for most of its life uses less than 10mA, but you're using a linear regulator that's eating through a few mA just by functioning on its own. And then, you're using 9v batteries which are known for having low capacity and low output current, risking a reset from your device when you're pulling near 1A.
What happens when the battery discharges down to around 7v and your product pulls 1A for a few seconds? Don't you risk going below 5v with your output voltage considering the 1.3v voltage drop on the regulator itself?


You're pissing away probably half of that 9v battery's energy.
If as much battery life matters, it would make more sense to replace your 9v battery with 2 or 3  AAA or AA batteries and use a step up/boost regulator to get your 5v
It would be more expensive, probably around 1$ in all, but AAA and AA batteries are cheap (and can be rechargeable)... and they can certainly handle the 5w output (5v 1A) for a few seconds without stress.   

Here's an example, Richtek RT 4812 .. around 95% at 2.5v..3v->5v conversion and I think <1mA quiescent current : https://www.digikey.com/product-detail/en/richtek-usa-inc/RT4812GJ8F/1028-1512-1-ND/5640521
 
Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod