N Channel MOSFET Vds safety margin for driving a relais - Page 1
Forum section: electronics / projects

author: nemail2
author_rank: Regular Contributor
timestamp: June 18, 2019, 05:17:56 am
content: Hithe title already tells everything... What safety margin for the N channel MOSFET's Vds should I use when driving a 12V standard G2R Omron relais (about 45mA coil current)? Currently I'm using a FET which has 100V Vds but I'd like to reduce my BOM and maybe use a FET with 20V or 30V Vds which I was using somewhere else in my design already...Are there any general rules of thumb?Thanks!edit: diode across the coil is already present...

author: Psi
author_rank: Super Contributor
timestamp: June 18, 2019, 05:28:03 am
content: 30V fet for 12V is totally fine.A 20V fet is getting a bit close but probably still fine. I would use 30V though.Using the automotive world as an example, all parts in a 12V car should survive 24V without any damage. So a 30V fet would be ok, but 20V fet would be too low.(Of course in an automotive setting there are lots of voltage spikes, so you would also have some sort of TVS diode to stop spikes over 30V getting to the fet, but that's a separate issue)For current you probably want to spec something that can handle 200mA continuous. Nice round number with good safety margin, lots of mosfets existing in the 200 - 300mA rangeUsually heat production is more of a limiting factor than a chips max current rating.You should ideally look at the On Resistance (RdsON) at your gate drive current and work out how many watts will be used up as heat from your 45mA.Then look at the datasheet specs for temp (θJC/JA) and calculate how hot it will get above ambient temp. Then consider if that is ok.I doubt you will have any issue for your application because 45mA is tiny, but for bigger power stuff you may notice a mosfet rated for say 100V and 60A but find that it cannot actually run at 60A without water cooling it!ie, The chip is ok at 60A but its up to you to remove the heat. verses at 61A where the chip may fail as the package is to small to get heat out fast enough no matter how much cooling you have.

author: Ian.M
author_rank: Super Contributor
timestamp: June 18, 2019, 07:02:09 am
content: It also has a lot to do with how you snub the back-EMF from the relay coil.A simple anti-parallel diode puts the lowest voltage stress on the driver, but significantly slows the flux decay in the relay core slowing down contact opening, which is undesirable if the contacts are in a high energy circuit.Add a resistor matching the coil DC resistance in series with the anti-parallel diode, and the flux decays as fast as it builds when initially energised, but the driver must withstand double the supply voltage.Then there are arrangements like snubbing into a Zener, or with a Zener to the gate circuit to provide a controlled rate of turnoff, or using an avalanche rated MOSFET without any other snubber, all with their own driver voltage rating requirements.

author: nemail2
author_rank: Regular Contributor
timestamp: June 18, 2019, 07:22:53 am
content: cool bananas!regarding RDS(on), power dissipation, package limits and so on I'm quite familiar how to choose an appropriate part. My concern was only regarding the voltage spikes from the coil when turning it off.currently I do have a diode across the coil and I didn't see any issues with that in my prototype (regarding slow contact opening).The relay carries 4A at max @ 16V (max) when contacts are closed.Anyway, if I'd put a resistor matching the DC coil resistance in series with the anti-parallel diode, would I then want to use a MOSFET with 48V Vds or higher (due to saying double for margin and double due to the parallel resistor) or still just 30V?Thanks!

author: Psi
author_rank: Super Contributor
timestamp: June 18, 2019, 07:49:50 am
content: Quote from: nemail2 on June 18, 2019, 07:22:53 amThe relay voltage spike is negative, so gets clamped by the diode to the diodes forward voltage.

author: magic
author_rank: Super Contributor
timestamp: June 18, 2019, 08:28:26 am
content: Quote from: nemail2 on June 18, 2019, 07:22:53 amNot how it works. The coil just wants to maintain constant current flow. If you drive it at 10mA, it will drive 10mA peak through itself and the snubber.It's a series circuti: 12V-Lcoil-Rcoil-Rsnub-Diode-12V.The FET is connected between Rcoil and Rsnub, so it only sees voltage drop across the snubber at 10mA or less current. So for Rsnub=1k, that's 10.7V peak boosted on top of the 12V supply. The hidden ideal Lcoil will adjust its output voltage to drive 10mA so losses at Rcoil are invisible to the FET but they do contribute to deenergizing Lcoil.

author: Ian.M
author_rank: Super Contributor
timestamp: June 18, 2019, 09:41:49 am
content: Whether or not you need a resistor + diode rather than a simple diode snubber depends on your application and the relay characteristics. If you are switching a large inductive load at close to the relay's contact rating, its likely to be beneficial. You may also need a RC snubber across the relay contacts. If you are switching a resistive low voltage load at a fraction of the relay's rating, its probably unnecessary.Consider the case of a diode + matched resistor snubber for your relay. Assuming the Rds_on of the MOSFET is negligible compared to the coil resistance, you get nearly the whole supply voltage (12V) across the relay coil when on. As it draws 45mA, the coil resistance must be about 267 ohms. Lets assume you use a 270R resistor in series with the diode. Neglecting stray capacitance, if the MOSFET shuts off quickly enough, that 45mA gets rapidly diverted through the diode and resistor, and will rapidly develop a voltage drop of 11.8V across them, (IR + diode Vf @ 45mA) driving the drain to 23.8V then decaying exponentially as the stored energy in the coil is used up.23.8V is near enough to twice the supply voltage, but you need some margin above that. e.g. if the supply is 10% high, it will peak at around 26.2V which is getting pretty close to the rating of a 30V MOSFET. If you have to cope with a max 16V supply expect it to peak at 31.7V which is above the rating of a 30V MOSFET. Of course you can trade off slower contact opening against peak voltage by decreasing the resistor, but as its the total loop resistance + the coil inductance that determines the LR time constant, even if you do away with the resistor its only a factor of two worse, so the benefits of having a lower value resistor are marginal.Quote from: Psi on June 18, 2019, 07:49:50 amPlease explain your reasoning. Negative with respect to what? As 'Magic' explained, during the back-EMF spike, the MOSFET end of the coil will swing above the supply rail, not below ground.

author: nemail2
author_rank: Regular Contributor
timestamp: June 19, 2019, 08:17:55 pm
content: Thanks, learned much (once again)!The relais switches the output of my Lab PSU on and off so the load may be everything from capacitive to inductive or resistive. The relais is rated 5 A at 30 VDC (resistive load) and 3 A at 30 VDC (inductive load). My maximum output voltage is 16V, maximum current is 4A.According to the Omron SOA diagram of the G2R-2 series, the relais is fine for up to 5A @ 20V inductive load, so I should be ok there.The supply of the relais is regulated so it should not be way off 12V.If I understand all this correctly, a 30V Vds capable mosfet + regular anti-parallel diode should do just fine, right?Thanks!

author: magic
author_rank: Super Contributor
timestamp: June 20, 2019, 06:58:24 am
content: Don't believe the Internet, breadboard it and check with a scopeYes, it will be okay for the transistor. You could even use a BJT now, I think they are marginally cheaper because all those cheap PCB "modules" use them to drive relays when necessary.