Current Protection circuit for any PSU - Page 1
Forum section: electronics / projects

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 03, 2018, 06:43:52 pm
content: Hi all,Thinking of some primitive current protection, latch type (not a limiter) to any PSU , here my thinking below:a. I'm going to the right direction ?b. mosfet dissipation kinda unusually high, is it a simulator's thing ? ( load: R5 = 10 ohm)c. or I'm inventing wheels ?

author: Ian.M
author_rank: Super Contributor
timestamp: April 03, 2018, 07:31:09 pm
content: You cant use a N-channel MOSFET as a high side switch without a gate drive supply that is significantly above the main positive supply rail. As is, its acting as a source follower, with about 4.4V on its source, which with 9V on its gate, means Vgs is about 4.6V, pretty much what you'd expect to be needed for Vgs to get about 0.44A amp through it. (see fig.3 ofhttps://www.vishay.com/docs/91015/sihf510.pdf)If you rearrange your circuit to move the sense resistor to the high side of the load, after the MOSFET and connect the 12V supply negative terminal to the high side of the load, so it and the rest of your current sense and latching circuit swings up and down with the MOSFET source voltage as it switches, you'd be in with a chance. If you cant do that, you've got two choices - use a high side P-MOSFET, source to the supply, drain to the load, with a Zener clamp between its gate and source to prevent gate oxide breakdown due to over-voltage, or rearrange your existing circuit so its using the IRF510 MOSFET as a low side switch, i.e. with the load between the +40V supply and its drain, and its source connected direct to the low side current sense resistor.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 03, 2018, 10:07:24 pm
content: Thank you Ian, for an educated topic that force me to Chapter-3 of bibleWell, re-arange do work , but I'm guess it kinda bad .... cutting a ground. The whole schematic need redo with P-channel, hook all to positive , and mosfet with lowest resistance and switching voltage. OK, another education project I never did this, actually ...

author: Ian.M
author_rank: Super Contributor
timestamp: April 03, 2018, 10:25:28 pm
content: Try removing the Zener D1. Its clamping the IRF510 Vgs at 9V, which is a bit less than the 10v its Rds_on is quoted for, As its abs. max. Vgs is +/- 20V, it can easily stand the full 12V of your latch circuit supply, and any small improvement in its Rds_on is worth having to reduce that 6.55W dissipation.I agree a low side electronic fuse is less than ideal as it wont provide any protection against a short to ground. You'll probably have to go for a low Rds_on P-MOSFET if you cant float the fuse control circuit supply.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 03, 2018, 11:19:40 pm
content: Right, but 510 was for prototyping , for a real schematic need to find a better one. like 0.0X ohm.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: May 24, 2018, 07:31:34 pm
content: next iteration:R23 - to avoid simulator crashing, somehow it generated negative potential on load when OC off.U2 - curr. sense op-amp X20 0.5A == 5VS1, U5 - simulator OC off from a second channelU4 - accordingly to the second channelU3 -comparing predefined V and U2 output, triggering latchQ1, Q2 - latchQ4,Q5 - to shut the Q6 switchLED2 - indicatorQuestions where I'm not certain:Q4, Q5, Q6 - Have a feeling something not right, or elegant, something i did not correctly ...C1 - where Q1 - an idea to avoid false positive during turn-on process ...; seems doesn't work, and will it slow down? what else the better way ?and finally resistors for Q1 Q2 latch - im really fussy about nominals ....UPD, OK I know , i need to bump 7805 output, using 2 diodes, to avoid opamp ceiling of 5V, + vref - lm336-5