Turning off an LDO using a mosfet - Page 1
Forum section: electronics / projects

author: OM222O
author_rank: Frequent Contributor
timestamp: February 13, 2019, 01:01:24 pm
content: I'm trying to use a DPDT latching switch to turn on/off my entire project.They are usually rated for something like 10mA but I've had no issues using them to connect/disconnect 1A! but that's still not good practice.I thought about using the switch to drive the gate of the mosfet in order to turn on the LDO, this way the switch is only a signal and carries no load.I'm already using the dual N channel fet (DMN3018SSS-13) multiple times on other parts of the board. unfortunately I can't do high side switching with a N channel fet, but it would be nice to reuse the same part. I was wondering what would happen if I were to disconnect the GND pin of the LDO using that fet? would that damage something or have some weird side effects? Would it be better to just use a P channel fet to disconnect the battery from LDO (high side switching)?

author: nsrmagazin
author_rank: Regular Contributor
timestamp: February 13, 2019, 01:18:17 pm
content: This "10mA" is not likely for a relay. You must be mistaken. Its best to use the relay or change it with a high power relay.Something from here maybe:http://www.cfcbazar.com/page/3/?s=relay

author: LapTop006
author_rank: Unknown
timestamp: February 13, 2019, 01:43:30 pm
content: You can probably also find an equivalent regulator with an enable pin.

author: OM222O
author_rank: Frequent Contributor
timestamp: February 13, 2019, 02:07:05 pm
content: I didn't say I'm using a relay. It's a small 8x8mm DPDT latching push button!I also don't want to swap out the LDO as it's bought in quantities for other projects.If switching the ground pin isn't a viable option, the best solution for me is a cheap P channel fet rated for about 2-3A.

author: mariush
author_rank: Super Contributor
timestamp: February 13, 2019, 02:26:04 pm
content: Ask yourself if the savings you get from using that LDO because you bought it in volume for other projects, are more than the cost of adding an additional part and the circuit board area and potentially an extra resistor to discharge the mosfet.LDOs are cheap, can be as cheap as a mosfet.

author: OM222O
author_rank: Frequent Contributor
timestamp: February 13, 2019, 02:40:32 pm
content: I'm using the AZ1117IH-5.0TRG1 bought from mouser. I need the minimum 1A capability and it costs £0.085 in 1000 quantity!Finding another alternative would definitely cost a lot more than a mosfet with similar price. Again, the mosfets themselves are used so it would be great if I'm able to reuse them! If you know,please just answer the question of what effect will switching the ground pin of the LDO have. Thanks

author: mariush
author_rank: Super Contributor
timestamp: February 13, 2019, 03:30:28 pm
content: Here's an example, costs ~ 0.2$ in volume (yours is around 0.11$ in 1000pcs) : LDL212PU50RDigikey:https://www.digikey.com/product-detail/en/stmicroelectronics/LDL212PU50R/497-16897-1-ND/6230213Mouser:https://eu.mouser.com/ProductDetail/STMicroelectronics/LDL212PU50RHas 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.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 13, 2019, 03:45:19 pm
content: Have a look at this thread:https://www.eevblog.com/forum/beginners/how-to-cut-power-to-arduino-after-a-set-time/

author: StillTrying
author_rank: Super Contributor
timestamp: February 13, 2019, 04:00:39 pm
content: Quote from: OM222O on February 13, 2019, 02:40:32 pmSwitching the LDO ground pin doesn't make much sense, and could be very bad! I presume you mean putting a switch in the whole -Ve side of the supply, you can usually put a switch in the -Ve side of the input supply if the input supply side is simple and fully floating from everything else.Keeping the whole ground line solid from input supply to output, and switching the +Ve is much safer than having a split ground each side of a -Ve switch.

author: Peabody
author_rank: Super Contributor
timestamp: February 13, 2019, 04:49:27 pm
content: I don't think there's any difference between using a P-channel to switch the positive rail and using an N-channel to switch the negative rail. If you look at the DW01 lipo protection IC, you'll see that it does all of its protection via the negative rail. But I agree that you don't want to switch just the LDO's ground. You want to switch the entire negative rail - i.e., between the negative battery terminal and ground. That would be my two cents.

author: OM222O
author_rank: Frequent Contributor
timestamp: February 13, 2019, 05:04:45 pm
content: Quote from: mariush on February 13, 2019, 03:30:28 pmThanks 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?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.

author: Kasper
author_rank: Frequent Contributor
timestamp: February 13, 2019, 05:16:10 pm
content: +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.

author: langwadt
author_rank: Super Contributor
timestamp: February 13, 2019, 05:17:27 pm
content: Quote from: OM222O on February 13, 2019, 05:04:45 pmthe battery is floating so it doesn't matter if you switch the low or high side

author: StillTrying
author_rank: Super Contributor
timestamp: February 13, 2019, 05:29:04 pm
content: 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.

author: mariush
author_rank: Super Contributor
timestamp: February 13, 2019, 05:49:49 pm
content: Quote from: OM222O on February 13, 2019, 05:04:45 pmI 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 5vIt 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

author: OM222O
author_rank: Frequent Contributor
timestamp: February 13, 2019, 06:15:11 pm
content: The application is a milliohm meter, so switching regulators could not be used. preferably I would use an external source, but it needs to be portable, so that was out of the question. 6AAAs also physically won't fit inside the box I have for this project. 9V battery was a decent compromise. Also for the voltage drop:Atmega 328P-AU is rated to work down to 1.8V! the thing I'm most worried about is the 24bit ADC which has a differential input voltage of maximum 2.048V. Even if the regulator drops to about 3 volt, I would be fineAnd at that point the MCU will kick in and display a warning or low battery symbol, telling you the 1A range is not usable anymore. That means you can still use the device to measure higher resistance values, or if you desperately need it, carry a separate 9V battery in the box, it can fit 2 9v batteries.Edit: I will create a new thread shortly, posting all the details about the project I have so far and asking for some feedback, feel free to comment about my newbie mistakes there

author: schmitt trigger
author_rank: Super Contributor
timestamp: February 13, 2019, 07:18:11 pm
content: "I was wondering what would happen if I were to disconnect the GND pin of the LDO using that fet? would that damage something or have some weird side effects? "The ground pin carries the quiescent current for the regulator's internal circuitry.Remove it and the regulator will cease to operate.

author: Krampmeier
author_rank: Regular Contributor
timestamp: February 13, 2019, 08:44:04 pm
content: If you want to use your N-channel FET, you can use a photodiode coupler like the TLP3905 to control it. It provides a floating gate voltage for the FET and just needs a few milliamps control current for the LED.