MOSFET as a power supply output switch  - Page 1
Forum section: electronics / beginners

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 12:03:15 am
content: Hi!I'm building a bench power supply and i need some kind of output overcurrent protection. I want when a certain set current is exceeded the output to be turned off. The current detection circuitry is done, i can detect the current, but i'm still figuring out how to turn of the output. First i tough of simple relay, but they can have very high contact resistance and will drop too much voltage. Now i'm thinking about a way to probably use a N-channel MOSFET to turn of the high side of the power supply because there are some that are cheaper that relays and have much lower on resistance. I'm not very experienced with MOSFETs and i don't know will this be a simple circuit with one transistor and some resistors or are there some oscillation problem than need to be taken in consideration. The output of the power supply is 1.2V-26V 2.5A maximum. So i guess the MOSFET should be able to operate with a minimum voltage between the drain and the source of 1.2V. Will this be a problem? Everything is being monitored by a microcontroller, so when the current limit is reached it will send a signal to the MOSFET, or rather it will turn of the signal depending on the transistor type. Can you recommend some simple standard circuit and some common MOSFET with low on resistance (lets say <0.005ohm). Thanks!

author: Audioguru
author_rank: Super Contributor
timestamp: January 18, 2016, 12:41:38 am
content: If your power supply will have a voltage regulating circuit then you can simply turn off the output by turning off the lower currentinput to the output devices. The input might even use a lower voltage than the output voltage.You can also simply regulate the output current to the current setting when the sensorreduces the output voltageuntil the output current satisfies Ohm's Law.

author: KL27x
author_rank: Super Contributor
timestamp: January 18, 2016, 03:31:15 am
content: QuoteStop thinking so much and use a regular relay. Or an SSR. Or a PFET to switch the high side if reverse current isn't a problem.If you're designing pcb from scratch, maybe go crazy. But otherwise, keep it simple. You bench equipment is supposed to be used to fix things. Making things complicated for no reason is just going to make you spend time fixing your bench equipment. Your reason for not wanting to use a regular relay is not valid.

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 10:11:33 am
content: I'm also doing this as a kind of a short circuit protection.A simple relay is too slow for that.

author: KL27x
author_rank: Super Contributor
timestamp: January 18, 2016, 10:44:34 am
content: I'm curious your reason for switching the high side with an N-FET vs a P-FET? I'm sure I'm missing something.Ok, went back and I suppose it's because your output voltage goes down to only 1.2V. Ne'ermind, then.

author: tszaboo
author_rank: Super Contributor
timestamp: January 18, 2016, 11:22:28 am
content: Quote from: blueskull on January 18, 2016, 01:14:39 amThey slow!Quote from: KL27x on January 18, 2016, 10:44:34 amFor N channel high side switching, you need power supply +4V to drive the gate. For P, you dont, but you have more Rds.If you make a one off project, I suggest to use an isolated power supply, mosfet driverand a back to back N channel fet on the output. Practically a discrete solid state relay. This is much more robust, and doesnt blow up, if you try to charge a battery with it.

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 11:28:08 am
content: The other reason is so i can put voltage on the gate and enable the output some tens of milliseconds after i turned on the power supply so if there are any voltage spikes and overshoots yo not be present at the output. Whats wrong with using a simple resistor from the gate to ground and a resistor from the output of the op amp to the gate of the N channel mosfet? Is this a bad practice?

author: Zero999
author_rank: Super Contributor
timestamp: January 18, 2016, 03:14:55 pm
content: Generally MOSFETs drop more voltage than normal relays.A charge pump is slow to turn on but it should be easy to turn off quickly by connecting the gate to 0V.

author: ice595
author_rank: Regular Contributor
timestamp: January 18, 2016, 04:07:24 pm
content: why not just use a P-channel MOSFET?

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 04:28:18 pm
content: I was looking for a relays and they generally have contact resistance of about 30-100mohms. The IRL1404 N-channel mosfet for example has on resistance of a 0.0031ohms, so at max current it will have less voltage drop. Im not after less power dissipation, i need less voltage drop. The only problem is that it needs about 10V GS voltage to be turned fully on if i read the datasheet correctly and i have a secondary voltage of max 6-7volts. As i said i don't have much experience with mosfets and i'm not yet sure should i go with N channel or P channel one.

author: edavid
author_rank: Super Contributor
timestamp: January 18, 2016, 04:45:25 pm
content: Obviously your power supply already has a pass transistor... why don't you just turn that off? That is the way (almost) all commercial bench supplies work. Why add a redundant switch?

author: tszaboo
author_rank: Super Contributor
timestamp: January 18, 2016, 04:53:35 pm
content: Quote from: edavid on January 18, 2016, 04:45:25 pmBecause this way the filter cap is already charged to the voltage, and there isnt overshoot when you turn on the output. I actually never seen a power supply doing what you described.Worth mentioning, that you can just get an extra winding to your transformer, creating a floating ~12V to drive the FET (s).

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 05:07:11 pm
content: To be more specific it's a LM2676 regulator and it has on/off pin, but even if i turned off the regulator the output capacitors are still charged (about 800uF total) and that's no good. I will be adding one more secondary winding for about 6VAC but i don't think i will go for 12VAC, it's a 100VA transformer and there will be a lot of winding to do for 12V.

author: rx8pilot
author_rank: Super Contributor
timestamp: January 18, 2016, 05:24:18 pm
content: SI7145 Pfet is super low Rds(on).I use charge pumps all the time for this exact scenario. Slow turn on is desired and its easy to dump the gate fast.There are lots of integrated controllers for this too from Linear and TI if you dont want the challenge of discrete design. I use LT1910 all the time for this purpose. N-FETs I am using are as low as .4 mOhm. That is REALLY low.Sent from my SAMSUNG-SM-G890A using Tapatalk

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 05:42:32 pm
content: What are those 0.4mohm mosfets? They will be nice if they aren't 10 bucks a piece

author: edavid
author_rank: Super Contributor
timestamp: January 18, 2016, 07:25:40 pm
content: Quote from: NANDBlog on January 18, 2016, 04:53:35 pmOP wants this for current limit, not output control...

author: rx8pilot
author_rank: Super Contributor
timestamp: January 18, 2016, 07:30:20 pm
content: Sound like an efuse to me. Not current limit.Sent from my SAMSUNG-SM-G890A using Tapatalk

author: langwadt
author_rank: Super Contributor
timestamp: January 18, 2016, 07:32:12 pm
content: Quote from: illusive on January 18, 2016, 05:42:32 pm10 seconds on digikey:http://www.digikey.com/product-detail/en/IPT004N03LATMA1/IPT004N03LATMA1CT-ND/4571877

author: Zero999
author_rank: Super Contributor
timestamp: January 18, 2016, 07:41:46 pm
content: Quote from: illusive on January 18, 2016, 04:28:18 pmWhat relay did you find with such a high contact resistance?Take datasheets with a big pinch of salt. The contact resistance for relays is normally an order of magnitude or more lower than it says on the datasheet. It represents the absolute worst case scenario and is often specified at the lowest current.Be careful with MOSFETs. Their resistance increases with temperature and their current/power ratings are often extremely optimistic.

author: dadler
author_rank: Unknown
timestamp: January 18, 2016, 07:54:33 pm
content: A physical relay is likely not going to react fast enough to be an effective current limiter.Not so sure I am a fan of requiring the mcu to do current limiting. Prefer an all-analog approach that still works if your mcu hangs or resets.

author: rx8pilot
author_rank: Super Contributor
timestamp: January 18, 2016, 07:58:02 pm
content: for this, analog is king.Sent from my SAMSUNG-SM-G890A using Tapatalk

author: edavid
author_rank: Super Contributor
timestamp: January 18, 2016, 08:03:17 pm
content: Quote from: rx8pilot on January 18, 2016, 07:30:20 pmTrue, but after he builds his power supply, he'll find out that he really wanted current limit after allQuoteCongratulations, but maybe it's time to turn this off?

author: rx8pilot
author_rank: Super Contributor
timestamp: January 18, 2016, 08:38:25 pm
content: Probably true. Current limited power supplies are about as important as the air we breathe.Sent from my SAMSUNG-SM-G890A using Tapatalk

author: illusive
author_rank: Regular Contributor
timestamp: January 18, 2016, 10:12:03 pm
content: So are you saying that i definitely need current limiting or is this more like a sarcastic joke ?My power supply is a simple LM2676 switcher. As i sad it has high output capacitance. In my mind if i set some current limiting with the feedback loop and set the current limit to 1A and the output voltage to 10V and if i connect a 5ohm resistor the output voltage will not go down fast enough because of the large output capacitors. A resistor will handle it but if i accidently forget the output voltage at 20V and i limit the current to 20mA a led diode will go pop. So if i'm correct in my case a "circuit breaker" will be as good as current limiting, but it seems more simple to implement.

author: illusive
author_rank: Regular Contributor
timestamp: January 19, 2016, 05:33:26 pm
content: OK, i got into reading about this stuff and and now i think i know why N-channel mosfet is harder to implement for highside switching.So basically something like this is my best shot ?If i'm going with this design i just need to find a N mosfet with low Rds at about 5-6 Vgs.

author: tszaboo
author_rank: Super Contributor
timestamp: January 20, 2016, 01:55:30 pm
content: Once again, if you ever plan to charge batteries with your bench power supply, use two FET, source and gate connected together. But in any case, it will be safer that way. Also I would toss the discrete transistors in favor to a FET driver, because they are designed to do this.

author: KL27x
author_rank: Super Contributor
timestamp: January 21, 2016, 05:23:10 am
content: The problem with a PFET is his power supply goes down to 1.5V. To drive the PFET gate fully at that voltage, he'd need to add a negative rail.QuoteThe easiest thing to do is to use this secondary voltage to drive an NFET to switch the low side. That might not be desirable for various issues of convention/expectation, especially if your power supply is grounded. This would be a problem if you have your circuit plugged into a something else that is grounded.And if you ever plan to, say, charge batteries with this power supply, might I suggest you ALSO put a relay on the high side output. In fact, in both of my DIY power supplies, I use two relays. I put one between the ADC and the output and one between the ADC and the load. So I can use my PSUs as voltage meters to read the float voltage of the load with the power supply output disconnected and conversely I can change and verify the no load voltage of the power supply without physically disconnecting the load. This is not as handy as having current limiting, but it adds a lot of utility for charging batteries, safely. And the audible click of the relays is oh so satisfying.

author: illusive
author_rank: Regular Contributor
timestamp: January 22, 2016, 07:01:36 am
content: I needed those 6-7 volts to power up the microcontroller and stuff. Anyway, i decided to get a small secondary transformer with two output wingdings, one for the low power stuff and the other to drive the MOSFET with a separate ground.

author: EPTech
author_rank: Regular Contributor
timestamp: January 22, 2016, 07:57:27 am
content: Hi there,You could put the feedback of your voltage regulation behind the high side FET switch, so as close as possible to the load, as it is supposed to be. That will get rif of the voltage drop of any FET you choose. The problem then is when you switch off the FET, you get a sort of open loop and the voltage before the switch will ramp up.It should be possible to switch the feedback loop together with the FET. I feed back loop will be slow enough not to even notice the switch-over.

author: illusive
author_rank: Regular Contributor
timestamp: January 22, 2016, 02:28:50 pm
content: Yes that crossed my mind. But then really the feedback loop will get messy. I can turn off the regulator (with it's on/off pin) and the mosfet at the same time tho. I got a 3.1mohm mosfet and i'm not sure if it's really to compensate for its voltage drop.

author: illusive
author_rank: Regular Contributor
timestamp: January 28, 2016, 05:26:11 pm
content: I was looking into these photovoltaic mosfet drivers and they seem like the easiest way to drive the mosfet. But i can't find many of them, the only accessible to me is VOM1271. Can you recommend more of these IC's?