Is this a cheaper option than this 2A load switch for my 50mA load? - Page 1
Forum section: electronics / beginners

author: jnz
author_rank: Frequent Contributor
timestamp: August 12, 2020, 11:42:12 pm
content: I have a 50mA IC on a 5V. I need to high side switch it because it has no low power mode, need to do this with my 3.3V micro.I'm looking at a load switch.https://www.ti.com/lit/ds/symlink/tps22810.pdf?HQS=TI-null-null-mousermode-df-pf-null-wwe&ts=1597273276213&ref_url=https%253A%252F%252Fwww.mouser.com%252Forhttps://www.diodes.com/assets/Datasheets/AP22908.pdforsomething like that.The issue is the circuit I want to cut power to is 2mA min to 50mA max. It seems even at the 10 cents I need the part to be at volume that a 1-2A switch is just huge overkill. I tried to lay this out myself with NPN PNP pair, but I couldn't get the quiescent down to anywhere near where these special parts get it.I feel like I'm missing some obvious better option than relatively cheap to buy but likely expensive to place and solder sub 1mm BGAs.

author: Ian.M
author_rank: Super Contributor
timestamp: August 13, 2020, 12:16:11 am
content: What about a complimentary dual MOSFET like Rohm'sVT6M1? Use the N-MOSFET to provide levels shifted gate drive to the P-MOSFET. You'll also need a pullup resistor for the P-MOSFET gate. Off state current is just the sum of the two MOSFETs' leakage current. On state quiescent current depends on the pullup resistor which also determines turn-off speed. If you can accept the slowish turn off a 330K resistor gives it could be as low as 15uA - negligible compared to your load current.

author: jnz
author_rank: Frequent Contributor
timestamp: August 13, 2020, 04:12:18 pm
content: Yea... a complementary package with builtin resistors is a good idea.That specific part is too large for this project, but I'm sure I can find something with a no lead and go from there. I'm in a cost / placement game now, it's 3-5 cents for a placement so placing a two parts that total 15 cents isn't as good as one part at 20 cents. It's a higher RDS on than a specific load switch like I was looking at, but 3ohm at 5V (.15V drop) is still acceptable for this IC.I always forget how to calculate the turn time on of a mosfet like that. Mind reminding me?Edit: That part size is excellent.

author: Ian.M
author_rank: Super Contributor
timestamp: August 13, 2020, 07:43:09 pm
content: Any MCU I/O or logic output is going to be able to supply several mA, and provided you have some idea of the total gate charge you can get an order of magnitude turn-on or turn-off time estimate for the N-MOS level shifter by dividing the total gate charge by the max. I/O pin sink/source current.For the P-MOS pass transistor, its not the turn-on time that's an issue, as the level shifter pulls its gate down *HARD* during turn-on. However the turn-off time is determined by how fast the gate pull-up can discharge the gate to below the gate threshold voltage. You need to know the total gate charge at your working Vgs to even get a ballpark estimate. Unfortunately the one I pointed you at has a crappy datasheet so lets consider Diodes Inc'sDMC21D1UDA, which has a proper gate charge graph. See my annotated Fig. 23 Gate Charge below.Consider the discharge in three phases:5V to the 1.6V plateauthe width of the 1.6v plateau1.6V to 0.4V (Vgth.min)For phase 1, calculate the equivalent capacitance from the slope of the graph: Q@8V - Q@1.6V = 0.51nC, and 8V-1.6 V = 6.4V so C=0.51nC/6.4V = 80pF.For phase 2, delta Q=0.41nC - 0.14nC = 0.27nCFor phase 3, C = 0.07nC/1.6V = 44pF.Now you *CAN* calculate the discharge time for each phase and add them up, if you enjoy solving exponential equations, or you could simply use my LTspice arbitrary MOSFET gate discharge time simulator and plug in the voltages, capacitances, and plateau charge you've read off the graph in the datasheet, and out pops the turn-off time, which for a DMC21D1UDA with a 330K gate pullup on its PMOS half is about 107us

author: jnz
author_rank: Frequent Contributor
timestamp: August 13, 2020, 08:20:39 pm
content: Quote from: Ian.M on August 13, 2020, 07:43:09 pmAh. Ok, I'll need to re-read that a few times, but its coulombs. I knew there had to be a time unit somewhere!As to this application, doesn't matter if it takes a full second (minus the time that would keep the transistor in the linear range). As long as it cuts or restores power within reason that would be fine.So... I'm embarrassed to say, I modeled what I think it should look like, and I can't tell but either my modeling site sucks or I'm dumb, those aren't exclusive!It appears to me that the PNP in the sim is never fully saturating and I don't know why! I reduced the load resistor like crazy, but no matter what I do I can't get load to cross about 7mA through the PNP. This sim site, everycircuit doesn't allow for changing the mosfet RDS which is strange to me as that's pretty important. Can you or anyone tell me what might be going on here?

author: Ian.M
author_rank: Super Contributor
timestamp: August 13, 2020, 08:38:06 pm
content: Don't say PNP, its *NOT* a BJT!Your N-MOSFET model must have too high a gate threshold voltage for 3.3V logic drive. As a rule of thumb, you need a N-MOSFET with a max. threshold voltage of less than half your logic supply.You can get rid of the 100R resistor between the NMOS drain and PMOS gate. The current is limited by the turn-on speed of the N-MOSFET which is controlled by its gate drive resistor.

author: jnz
author_rank: Frequent Contributor
timestamp: August 13, 2020, 11:35:39 pm
content: Quote from: Ian.M on August 13, 2020, 08:38:06 pmCrap. You're absolutely right. NOT A PNP. P-CHANNEL.I understand about the turn on speed and so the P-CH doesn't need a gate resistor... but EVERY example I could find of this has it and remember, it's built into the combo package I'd be looking at.I agree, that it seemed like turn on voltage was wrong, but VTO is one of the options it does have but both are 400mV and -400mV respectively. That seems like it should work. P-CH Gate shows showes +10-11mV. It's connected to ground via the N-CH plus a little that the resistor and NPN aren't draining through. So... Why is this not showing 50mA at a 100ohm load? I'm trying to figure out if it's me or the software that's bad and should feel bad.I have a setting for KP (but not rdsOn!), that if I adjusted to 30 mA/V^2 seems to fully saturate and I get 50mA through it with a 100ohm load resistor. I have no idea if that's a reasonable trans-conductance or not, surprise, not in any of the data sheets I've seen. I'm learning towards bad software.