Software poweroff circuit, which turns on in intial power up - Page 1
Forum section: electronics / projects

author: rsd
author_rank: Newbie
timestamp: November 09, 2019, 10:35:31 am
content: Welcome to the thread with the bad title, if you got something better please tell me.I am looking for a circuit that is capable of cutting the power supply and keeping it off. The signal for cutoff comes from a microcontroller. Normally I would use a power Mosfet as a switch and a button to enable the circuit initially, while the button is pressed the microcontroller is booted and takes control of the MOSFET gate circuit. But this time I would like the same, but with an additional circuit trick to enable the MOSFET switch on an initial power-up.I attached a circuit that cloud work. C1 should keep the P-Mosfet gate long enough low to enable the microcontroller to pull the gate on its own.I am curious if there is a better option? Maybe this could be integrated into an N-Mos Reverse Polarity Protection.Thanks for your time!

author: Gandalf_Sr
author_rank: Super Contributor
timestamp: November 09, 2019, 02:19:20 pm
content: I've done a couple of these recently, tell me what your Vcc voltage is and I'll post a circuit for you. More questions...How much current is your system pulling?Is there a regulator in circuit that has an enable pin?

author: rsd
author_rank: Newbie
timestamp: November 09, 2019, 03:27:13 pm
content: The DC Input is between 10V and 30V with Currents up to 5A.There are Regulators with Enable but not all supplied from the DC Rail have enabled, one Load is connected directly, so working with the enable pin is not an option.I am curious about your solution

author: Gandalf_Sr
author_rank: Super Contributor
timestamp: November 09, 2019, 03:32:49 pm
content: Quote from: rsd on November 09, 2019, 03:27:13 pmSo is this right...1. You want to use that 10V to 30V input voltage to switch on power and then have the power held on by a microcontroller signal?2. If 1 above is true, what voltage is the microcontroller running at?3. If the on switch is not pressed and the microcontroller wants to shut off power, it changes the 'hold on' signal and the input voltage is disconnected completely, is that right?

author: Gandalf_Sr
author_rank: Super Contributor
timestamp: November 09, 2019, 04:15:24 pm
content: Here's something to chew on...It's similar to your circuit but you have an issue in that the max source-gate (s-g) voltage of most FETs is 20V and, if your input voltage is 30 V, then you would have 30V across the main switch transistor's s-g (Q801).My circuit below (taken from a proven design, built and tested) solves this problem by adding R43 so that, if Q7 or the switch are 'on', the input voltage is split between R43 and R41.So, for Vin = 30 V, the s-g voltage on Q3 would only be 15V (C77 protects against voltage spikes). My design is for a 12.6 V Vin but the DMP6023LE is rated for 60V and 7A so it should work for your design but you can use any PFET. If you use my circuit, I'd increase R43 to say 75 k because you just don't need 15 V across Q3's s-g to switch it on. The voltage that comes in from the right is the 'hold on' signal and is +3.3V in my design but, as long as you don't exceed the s-g limits for Q7, you could run it at a different voltage. Note that, if the micro is running at 1.8V then you need to make sure that the gate voltage for Q7 is low enough that the micro hi signal level can turn it on.I call this a 'suicide' circuit as it kills power to itself.PS the top side of the switch is connected to the junction of Q7 pin 3 and R43 pin 1, I added it in Paint and forgot the dot. In my design, Q7 is a 60V FET (d-s) which is needed as, in the 'off' state, this circuit puts the full input voltage across Q7's d-s. Make sure that the lower NFET you use can handle the full input voltage and leave a bit of headroom.

author: SiliconWizard
author_rank: Super Contributor
timestamp: November 09, 2019, 07:05:15 pm
content: @rsd: you're assuming that the capacitor will be discharged enough at power-up so that the transistor will conduct. That may not be always the case, so that's not very robust.

author: Gandalf_Sr
author_rank: Super Contributor
timestamp: November 10, 2019, 11:09:50 am
content: Quote from: SiliconWizard on November 09, 2019, 07:05:15 pmIn fact this is a huge issue because - I think - that on power application, RSD does not want the circuit to the right of Q801 to get power until the switch is pressed; in reality, there will be a brief period of power fed through Q801 as the circuit powers up...As power comes on, C801 will be at zero volts which will mean that the gate of Q801 will see ground and Q801 will be turned on for approx C801 x 33,000 seconds, the time for C801 to charge through the 33k resistor R801.My circuit has the opposite effect as there's a capacitor in parallel with a resistor across source-gate of Q3 ensuring that it is off at power up. In my circuit, after the switch is pressed, there will be a small time delay for Q3 to switch on as C77 is charged up through R43. If C77 is 1 uF and R43 is 75k that time will be approx 0.000001 x 75,000 = 75 mS. I am also assuming that whatever circuit is powered up by the main NFET switching on causes the microcontroller to start and apply a high state (+3.3V) to the 'hold on' line.If C77 seems high at 1 uF, it's there to provide a soft start for the power up rather than hit the rest of the circuit which may well have large capacitors on the input and cause a sudden inrush of current. The slow turn on of Q3 means that it briefly acts as a current shunt but the timing has to be set carefully to ensure that Q3 doesn't get damaged by the inrush dissipation. Made to correct multiple typos (on reference designators) and add detail about slow start.

author: rsd
author_rank: Newbie
timestamp: January 06, 2021, 05:28:51 pm
content: Thanks for your input! I know its late but at the time it did not fully understand your concerns. Today i now it betterI actually used my circuit but did not connect the output side of the transistor with the power rail directly. Instead it was connected to the enable of the buckboost converter, which seems to be slow enough to not fully start with the initial "power-through". So i was kina luckyEDIT:I did not measure it, the startup was an assumption, but maybe wrong again, since the time should be enouth to turn on. Maybe it just was not long enouth for the microcontroller to pull the enable.