I've found a lot of recommendations to do not use a low side switch when you want to turn off some complicated load (with IOs). In my case, I'm designing an undervoltage protection for MCU but it'll turn off all things on this power line (MCU, radio transceiver, i2c sensor). The decision-maker, won't be the MCU itself, but a separate voltage monitor (NCP300LSN18T1G).
What disadvantages I'll get using a low side switch instead of a high side in this case?
strange_v:
What disadvantages I'll get using a low side switch instead of a high side in this case?
The risk that some grounded parts of the system you are low-sided switching - such as a radio transceiver which may have an external antenna - will at some time or in some way be connected to the ground of your power supply. Or worse, to another power source.
Logic level PFETs are available!
@Paul__B, it shouldn't be connected to the ground of the power supply in any way (at least intentionally).
The main advantage of the low side switch for me is tiny power consumption.
In the case of using PFET, I'll lose power on R8 and through R3 - Q1. I'll try to change Q1 to N MOSFET and use high value for R8, but I'm not sure how high I can go. Ideally, I would like to see power consumption less than 1uA.
UPD: peak power consumption of the load is 50mA (~6ms).
Low side switching is usually better than high side. The problem with connections between powered and unpowered part is the same. If low side is more convenient for you there is probably no good reason to use high side.
Two different opinions for now. Seems I have to test it in the real world.
strange_v:
In the case of using PFET, I'll lose power on R8 and through R3 - Q1. I'll try to change Q1 to N MOSFET and use high value for R8, but I'm not sure how high I can go. Ideally, I would like to see power consumption less than 1uA.
Then use a CMOS logic gate. Not so expensive (and actually, you can get single gates; they come in a 4 or 5 pin SMD).
Paul__B:
Then use a CMOS logic gate. Not so expensive (and actually, you can get single gates; they come in a 4 or 5 pin SMD).
Could you show a particular example? Nothing that I found is really low-power, e.g. LTC1981 - 20uA. In my case, MCU (in a deep sleep) + sensor consumes ~5uA. One more limitation is voltage, the circuit should work till at least 1.5V (or keep the load off below min voltage).
Based on the output pulse shown on page 8 of the datasheet for the NCP300, and the schematic shown in the attached photo, there appears to be a discrepancy between the hand drawn schematic in your OP that shows
Vin shorted to Vcc and the attached photo that shows Vcc switched by the NCP300 controlled mosfet . Can
you clarify that ?
What is the circuit supposed to do ?
It appears to briefly switch power to the "MCU" during the high portion of the NCP300 output pulse, after which
the MCU power is switched off. Is that what you are intending to do ?
In the case of using PFET, I'll lose power on R8 and through R3 - Q1. I'll try to change Q1 to N MOSFET and use high value for R8, but I'm not sure how high I can go. Ideally, I would like to see power consumption less than 1uA.
I don't understand why R8 or Q1 are needed in the high side version. It appears they make versions of this chip with either active high or active low reset. Why not connect directly to the gate of the PFET using the version that brings the gate low when power is good (I guess that would be the active high reset version). With the totem pole output it should work just like the original low side version, except upside down.
If you are trying to reset the MCU by cycling power
your logic is reversed. The schematic you posted
will turn ON the MCU briefly instead of turning it OFF briefly. Why can't reset the MCU directly ?
raschemmel, on page 8 of the datasheet, we see 3 charts with the synchronized x-axis. The first chart shows the input voltage, 2nd, and 3rd reaction of the output pin to the on the input. At least it is how I see this.
If we scroll to examples, Figure 26 (page 17), we will see a typical reset circuit, how in hell it will work in short pulse mode? But okay, they haven't stated that this is active low IC. Figure 27, Active Low, it definitely outputs low when voltage < 2.7V and continuous high when > 2.7V. The same logic behind the next schematic, Figure 28.
About shorted Vin and Vcc. I've posted two different schematics, one with a common ground and high switch, another one (hand-drawn) with a low side switch, and yes, in that case, Vin and Vcc are connected.
The circuit should keep MCU, radio, and sensors off (disconnected from the power line) when voltage < 1.8V and consume as little as possible.
ShermanP, availability issue. I'm able to source only active low IC, already ordered 
raschemmel:
Why don't you drive the mosfet directly ?
What will drive it?
An example, a supercapacitor is discharged and a solar panel tries to charge it. During this process, MCU should be powered off because when the voltage reaches ~1.6V it tries to start and go into the continuous reboot with power consumption up to 3mA. But solar panel produces 1mA only, so it'll never start.
What can drive the MOSFET in this case?
Yes I realized that which is why I deleted that post. I understand why you are using the NCP300. It's a voltage monitor. I think your base resistor is way too high but the NCP300 has an output current of 70mA and the transistor has an hfe of at least 500 so maybe it will work with a 1.5M resistor but I'm curious how you chose so a high value for a base resistor ?
Haven't noticed that the post has gone.
Unfortunately, it's mostly guessing. I placed the highest theoretical value, but yes, probably it is too high. Also, I can replace Q1 with a MOSFET (and set some small R3 value) to reduce power consumption, but even with this, a single NFET will be a winner...
Unfortunately, it's mostly guessing
Why ?
The transistor has a DC gain (hfe) of at least 500 and the PMC160IP gate is not going to draw much current so it
might actually work with a 1.5M base resistor because almost any base current is going to turn it on enough to
ground the p-channel gate.
Even if you didn't know how to calculate or estimate it you knew enough to post here so you could just add that
question to your post.
What's the MCU current ?
~5uA in a deep seep
~50mA during the transmission (I have a 2200uF cap near the radio to eliminate voltage drop)
It is for all components: MCU, radio, sensor.
~5uA in a deep seep
~50mA during the transmission (I have a 2200uF cap near the radio to eliminate voltage drop)
Oh yeah, you're more than covered with a 1.2A mosfet.
FYI,
The gate of a MOSFET is composed of a silicon oxide layer. Since the gate is insulated from the source, an application of a DC voltage to the gate terminal does not theoretically cause a current to flow in the gate, except in transient periods during which the gate is charged and discharged.
In plain English that means that the current the transistor passes is essentially negligible and the only purpose is
serves is to apply 0V to the mosfet gate.
OMG why do you all force OP to the high side switching? When the whole system is powered off by the switch it does not matter which side is switched. And all other things being equal low side switching is better. Since for OP low side switch is more convenient it would be foolish trying to use high side switching.
Smajdalf, one thing I can imagine could cause issues with the low side switching is connecting a programmer. But typically, for this purpose, I disconnect a supercapacitor and connect the programmer.
I've played a bit with the previous schematic, really close to this one (but U1 is a different IC, V2 is embedded into U1, R4 is pull-up)
But I'll try both schematics, low side switching, in this case, looks too good to ignore (from the power consumption point of view). Probably I'll order two different PCBs to have options.
UPD: raschemmel, haven't noticed your update again
raschemmel:
FYI,In plain English that means that the current the transistor passes is essentially negligible and the only purpose is
serves is to apply 0V to the mosfet gate.
Probably, you should also quote here how BJT transistor works, because we discuss the base resistor for BJT.
As a small current flowing into the base terminal controls a much larger collector current forming the basis of transistor action.
Summarizing: I'll lose power through Base - Emitter of the transistor, and then through R8 and Collector - Emitter of the transistor, and yes MOSFET's gate won't consume almost anything (except during switching), but do you see all other losses here?
UPD2:
Paul__B:
Then use a CMOS logic gate. Not so expensive (and actually, you can get single gates; they come in a 4 or 5 pin SMD).
Seems I've found a nice candidate TPS22917 Ultra-Low Leakage Load Switch. Voltage range: 1V to 5.5V, power consumption 0.5uA.
"UPD: raschemmel, haven't noticed your update again"
You're good. You don't need my help.
PS-I don't see why R8 is needed.
(evidently I'm not the only one)
