Input protection of DC electronic load - Page 1
Forum section: electronics / projects

author: MikeEagle95
author_rank: Newbie
timestamp: January 06, 2018, 10:33:22 pm
content: HiI'm designing programmable dc electronic load based on MOSFET-N, and now I'm thinking about input protection, namely reverse polarity and overvoltage protecion circuit. I've been thinking about using relay at first, but its relatively long time between switch disqualifies it as a reverse polarity protection.I could use another MOSFET, but my load is going to be capable of 5A so I will need to use another high power mosfet and I don't like this idea.Can you suggest me what could I use to protect my device. All of my ideas take some significant resistance into path, but in case of electronic high power load that is not a right option.Thank you in advance!

author: jbb
author_rank: Super Contributor
timestamp: January 06, 2018, 10:57:26 pm
content: You’re right that a relay is too slow. They take some ms to start opening and are infamous for arcing when trying to switch off DC currents (when supply is > 40 V DC or so).For reverse V protection, how about a basic series diode? It’ll add about 1V to the minimum voltage limit, but that might be OK. You could also do a second MOSFET with appropriate driver for reverse protection.For over V protection, your best bet is to use a main MOSFET with lots of voltage headroom (say 3x your rated load voltage) and a big TVS clamp set to say 1.5x rated voltage to clamp spikes. Finally, a fuse is a nice idea for a high-quality system.

author: MikeEagle95
author_rank: Newbie
timestamp: January 07, 2018, 09:30:45 am
content: Thanks for ideas.When it comes to diode as a reverse protection, even if I use schottky with low voltage drop, I could not measure a full range of lets say 30V/5A power supply, because the voltage will be lower due to diode.You are right about chosing MOSFET with higher voltage rating.I was thinking about using fuse, but it will be another component that brings resistance into path(another voltage drop)

author: max_torque
author_rank: Super Contributor
timestamp: January 07, 2018, 01:13:10 pm
content: IS a rely really too slow?If you add some "smarts" ie a processor, you can check the input is correctly polarised BEFORE pulling the relay in, and the mostly likely time to accidentally reverse connect a DC load is when it is connected up, rather than once it has been connected up?Use the uC to measure the polarity and voltage applied to the input terminals, and display a "DANGER WILL ROBINSON" type message to the user if they are applying the voltage backwards or in excess etc?

author: Cerebus
author_rank: Super Contributor
timestamp: January 07, 2018, 02:06:49 pm
content: Yes, relays are too slow. A part can die from overvoltage in fractional milliseconds - the only relays that open in even millisecond scale times are signal reed relays and anything that can take some current is 10s of milliseconds minimum.The best general purpose protection for overvoltage is a crowbar style voltage clamp (e.g. SCR and trigger circuit) combined with a fuse to protect the connected power source from the short circuit a crowbar would present it with. I'd also add a gas discharge tube or TVS for ESD overvoltage protection.As far as reverse protection, little beats a beefy reverse biased Schottky diode slapped directly across the input (i.e. acting as a voltage clamp at around -0.5V). This has two problems however: reverse leakage current (which may or may not be significant for the OPs purposes), the violent near short circuit load that this presents to any supply misconnected to the load in reverse. Again, as above, a fuse helps to minimise the possible damage.There's a trade off to be made in how well you protect your electronic load from the supplies it is connected to, and how well you protect connected supplies from the electronic load. You will not be able to do both perfectly, so you have to choose which you're more prepared to potentially damage - your supply or your load. There's something to be said for compromising on the load side, and designing the load so that only the front end components are vulnerable (i.e. lots of protection just 'inside' the load transistors) but are designed to be easily and cheaply replaced.

author: Zero999
author_rank: Super Contributor
timestamp: January 07, 2018, 02:38:06 pm
content: Quote from: MikeEagle95 on January 07, 2018, 09:30:45 amThen measure the voltage before the diode.

author: Yansi
author_rank: Super Contributor
timestamp: January 07, 2018, 02:42:17 pm
content: You have substrate diodes in the N-FETs.

author: Cerebus
author_rank: Super Contributor
timestamp: January 07, 2018, 02:55:35 pm
content: Quote from: Yansi on January 07, 2018, 02:42:17 pmParasiticsubstrate diodes. It's well not to rely on these unless the manufacturer characterises and tests them. It's quite possible that a part that has 20A forward rating has a parasitic diode that can only handle 2A thanks to a single bonding wire or narrow metallization between the source and the substrate. I have yet to see a datasheet that includes good data on their characteristics - not to say that there are none, but I haven't ever seen one. Without data you're just whistling in the dark and hoping that some unspecified diode is up to the job.

author: Yansi
author_rank: Super Contributor
timestamp: January 07, 2018, 03:00:11 pm
content: Not true at all. The diode almost always can pass some heavy current and is well characterized, as the parameteres of the diode are important for most switching designs!See this. First mosfet I pulled: I think the body diode is pretty well characterized. At 50A continuous currenthttps://www.vishay.com/docs/91291/91291.pdfIt does not matter whichever mosfet will you pull, the diode is always well characterized and can pass heavy current. Typically the same, as the mosfet itself.http://www.vishay.com/docs/91054/91054.pdfhttp://www.irf.com/product-info/datasheets/data/irf3205.pdfhttp://www.mouser.com/ds/2/389/en.CD00002970-1219767.pdfhttps://www.infineon.com/dgdl/Infineon-IPW90R1K0C3-DS-v01_00-en.pdf?fileId=db3a30431b3e89eb011b8da2ee4e1071https://www.vishay.com/docs/91084/sihf9630.pdfhttp://www.st.com/content/ccc/resource/technical/document/datasheet/b4/15/c5/14/18/c1/4f/1c/CD00003430.pdf/files/CD00003430.pdf/jcr:content/translations/en.CD00003430.pdfhttps://datasheet.lcsc.com/szlcsc/IRFB4227PBF_C2652.pdfAnd btw, the diode is not parasitic, it is there (due to the nature of the mosfet construction of the mosfet of course) and the part is being typically designed so the diode will behave well in switching designs, where the diode does matter a lot.So I think you are the one whistling the nonsense.

author: Cerebus
author_rank: Super Contributor
timestamp: January 07, 2018, 03:46:19 pm
content: Quote from: Yansi on January 07, 2018, 03:00:11 pmLike I said, I didn't say they didn't exist just that I'd never seen one properly characterised. And "whistling in the dark" has nothing to do with nonsense, it's an idiom implying that one is trusting everything to go right relying on nothing more than wishful thinking.Anyway, that 50A is at a Vfof 2.5V, not what one wants to see in a protection circuit (and you're going to see more voltage drop across the sense resistor(s) in series with that). First 50A Schottky diode datasheet I pulled (DSA50C100QB ) Vfat 50A is 1.07V. I'd be much more comfortable with the latter for reverse voltage protection clamping.Also it's classic datasheet nonsense. it's rated 50A @ 25C junction temperature. If Vfis 2.5V and it's at 50A continuous, then that's 125W dissipated in a part with a junction to case RthJCof 1C/W and a case to sink RthCSof 0.5 C/W. That's 187.5C junction temperatureriseon a part with a specified max junction temperature of 175C (At least it would lower the Vf.). That'swell characterisedas acontinuousrating is it? Or is it just a little bit misleading?Granted, derate the part appropriately and you might just get away with it, but it's a poor substitute for a properly specified Schottky clamp right across the rails. Alternatively, be a cheapskate and rely on the substrate diode, with its high Vfin series with a sense resistor as a reverse voltage protection clamp and hope that -(2.5V + (50A * Rsense)) isn't enough to scrag the rest of your circuit.

author: Yansi
author_rank: Super Contributor
timestamp: January 07, 2018, 04:12:50 pm
content: QuoteThat is not a nonsense, that is you not getting the grasp of how to calculate the derating for elevated temperatures. Do not blame the datasheet.I do not see a reason why 2.5V fwd drop at 50A is an issue. We do not know shit about the OP's designs regarding what negative voltage it can tolerate or what parts and how many does it use (but it seems you do), but I can tell you adding non-required expensive parts (DSA50C100QB) to the BOM and comparing a schottky diode with a price of 4 times or more of those IRFZ mosfets is a bullshit that won't stand much.And as you seem to know everything about the OP's design and as far as I can't stand discussing with someone so arogant like you, I am leaving this thread.

author: Cerebus
author_rank: Super Contributor
timestamp: January 07, 2018, 04:41:02 pm
content: Quote from: Yansi on January 07, 2018, 04:12:50 pmIf anyone is wondering why Yansi is having such a hissy fit with me I'm guessing it's because earlier I told him off (very mildy, and very indirectly) for being rather rude to someonehere. I hadn't make the connection myself until I just happened to re-read the other thread from the top because I was sure that I was repeating myself there (I was).If he really has stomped off perhaps we could get back to trying to help the OP. I'd be particularly interested in any criticisms of my suggestions (or plaudits for that matter).

author: duak
author_rank: Super Contributor
timestamp: January 07, 2018, 06:55:00 pm
content: Yansi notes that power FETs have intrinsic body diodes - I'm not aware of any that don't. Unless I've missed something, these will also make any series FET useless as a reverse protection switch, ie. a reverse diode is still needed.Food for thought: I've been toying with the idea of designing a load that can draw current to an internal negative supply of about a volt or so rather than to 0 V. The load would then become an Quadrant and a Half where its input port voltage could go slightly negative. This would compensate for the voltage drop across the sense resistor, pass transistors and a reverse protection diode and allow for essentially zero resistance. The negative supply could be based on one of the low voltage mulitphase processor converters. Five amps would be no sweat.I'd also been toying with the idea of a switching based load that actually converts the energy to be dissipated back into AC and back onto the line. It started when I was experimenting with a 4 quadrant PWM servo drive to synthesize a complex impedance. It was no problem at DC but 500 Hz was a pain. When the power appled to the drive's output exceeded what was needed by the drive, the drive's DC bus voltage started to rise indicating that it was converting excess power. I had a shunt regulator that dissipated this to prevent the bus voltage from going too high. Except for the switching noise from the servo presented to the source, it would make a jim-dandy load.Cheers,

author: Siwastaja
author_rank: Super Contributor
timestamp: January 07, 2018, 10:08:44 pm
content: Quote from: Cerebus on January 07, 2018, 04:41:02 pmThanks for this info, but he just happens to be right this time. I was wondering what world do you live in, becauseIdon't remember seeing any modern MOSFET (designed for switching) datasheetwithoutproper specification of the body diode, andwithoutthe diode being rated in the same order of magnitude of max forward current than the FET itself.Sure, what can be considered "proper" is not easy to define. The specs are far from perfect and often require manual extrapolation and some guesswork regarding SOA, but IME that's almost equally true with the MOSFET itself! We all want better specs and better diodes, but to me it looks the body diodes arefarfrom being unspecified, or unusable.I really don't remembereverseeing an unspecified body diode in a modern switching MOSFET datasheet, or one only usable at about 1/10 of the Id, like you explained earlier.The body diode is often very important in a switching converter, hence the manufacturers do the basic specifications.

author: MikeEagle95
author_rank: Newbie
timestamp: January 08, 2018, 12:31:53 pm
content: Well, due to what all of you said I realize that bringing some resistance into path is necessary.I haven't said yet, that my load needs to handle 5A,50V,100WI was thinking of using 6.3A fuse and 1 way transil lets say for around 60V stand-off. If the voltage will be higher than Vbr, than fuse blows up.I've never used a GDT. How could it help with protection?

author: tjhoe16
author_rank: Newbie
timestamp: January 10, 2018, 02:52:23 pm
content: I would use another power MOSFET is series with your load but with source/drain terminals swapped around so the body diode is in the blocking direction for reverse polarity. You can use the load voltage to turn on the FET and then all you have is the Rds,on*Iload drop. Of course, you'll need a few volts to get it to turn on but until then you have the body conducting the load current. Just make sure that the FET is off when nothing is connected and if you have a reverse polarity applied. Also, make sure you limit the Vgate-source voltage.The trouble with diodes is the power loss and associated temperature rise. Even 5A through a diode can be interesting to manage.I use back-to-back MOSFETs all the time for reverse polarity and over-voltage protection on the high-side of vehicle power supplies. This technique is very effective and I've got production designs conducting more than 20A with PFETs. Linear technology has an IC for driving NFETs on the high-side for exactly this purpose as well. I can link in the part if you're interested.

author: langwadt
author_rank: Super Contributor
timestamp: January 10, 2018, 04:04:31 pm
content: Quote from: duak on January 07, 2018, 06:55:00 pmthe diode can be the body diode, just use the fet in "reverse"

author: duak
author_rank: Super Contributor
timestamp: January 10, 2018, 06:42:00 pm
content: I agree with the previous two posts; a reversed series FET should work - I was thinking of standard operation. I had understood that third quadrant (reversed) FET operation was possible but I'd never seen any characteristic curves in the data sheets showing this and never knowingly used it. I found a few articles that indicate that the characteristics are the same as in the first quadrant except that they are in parallel with the intrinsic body diode and so the maximum Vds would be limited to the diode's forward voltage drop.Cheers,

author: MikeEagle95
author_rank: Newbie
timestamp: January 10, 2018, 10:27:10 pm
content: Quote from: tjhoe16 on January 10, 2018, 02:52:23 pmWell, I did some simulation in LTSpice, and I could use P-channel Mosfet with its drain connected to the positive input terminal of load.From what I saw, I could connect its gate to the +5V positive rail that I use in my project to supply other stuff.Correct me if I'm wrong, but If my load is capable of max.50V, so I should choose P-MOSFET capable of ~100V and a bi-directional transil for 50V stand off with fuse should do the business?

author: tjhoe16
author_rank: Newbie
timestamp: January 10, 2018, 11:37:59 pm
content: Quote from: MikeEagle95 on January 10, 2018, 10:27:10 pmYou'll want to use an NCH MOSFET if you're working on the low side (I'm not sure exactly what topology you're using). My example using PFETs is because I'm working on the high-side, I'll leave it up to you to figure out why. I would put this either right after your main MOSFET or even after your sense resistor (if you're doing differential measurement). To turn on the reversed FET you need to have a gate-source voltage that exceeds the threshold voltage of that FET. Placing the FET on top of your other MOSFET will force the source terminal to whatever your load voltage is. This would be difficult to control across your load voltage. Lets assume you place it below your sense resistor, now the source terminal will always be a diode drop above ground (until you turn it on where it would be even closer to ground). The next thing is that you only want to turn the FET on when you have a positive voltage on the load terminal otherwise you won't have reverse polarity protection. The simple solution here is to tie the gate to the load voltage through a resistor and then limit the gate voltage with a zener diode (I usually choose 10V as most MOSFETs specify their drain-source resistance at this voltage). This simple circuit has the benefit of turning off the FET in the reverse polarity situation and rather quickly because it would just see the zener diode (forward biased) charging the gate to a diode drop above the source.For overvoltage protection, a transorb *could* work but it really depends on what you're trying to protect againt. Do you have some specific transient you're trying to guard against? Or are you trying to protect against a DC overvoltage event? I personally don't like the idea of having to replace a fuse in everytime there is an overvoltage event. I do overvoltage protection on vehicle power supplies against some standard like SAE J1113-11 but in that case I'm designing to protect against specific transients. If you do end up using a transorb, keep in mind that during conduction, the voltage across the transorb will be much larger than the breakdown voltage and you need to take that into account when you select the rest of the components in your system.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: January 11, 2018, 12:49:37 am
content: Many cheap commercial electronic loads claim 'reverse polarity protection' as a feature but in reality they just have an Max Eload current rated fuse to protect the Eload and maybe a led to inform you your DUT is being shorted, they do not actually prevent the S/C and offer no protection to the DUT. If it's a requirement to protect the DUT from possible S/C then here's a few methods of reverse polarity protection :-Schottky diode .Advantages :- Very simple , fastDisadvantages :- extra Voltage drop (1V or so at high current) this limit's your eloadcurrent when working with Dut's of low input voltages (e.g working with 1.2V or lower batteries). At high current ratings the power dissipation means it must be fairly large and just as expensive as the alternative low rds mosfet.A reversed drain/source Nmos used as a Low Vf diode.Advantages :- Very little forward voltage drop and power dissipation so something like a to220 case will be sufficient. Also Fast.disadvantages :- extra circuitry to drive it.Here's a very rough schematic of a one way to drive a reverse polarity protection Nmos.Below is how I did it in an old diy eload project, It was a bit trickier than the above since it was a multiple output range Eload and one reverse mosfet was used to provide RP protection to all the output's.Because of the way I chose to implement multiple ranges In that version it made it's placement less than ideal.A TVS diode was used on the input's to provide transient over-voltage protection.https://www.eevblog.com/forum/projects/fully-portable-150w12a-e-loadcurrent-reg-with-multiple-ranges/msg233642/#msg233642.But what also about forward S/C protection. ?If C.C load intended main use is to work with batteries then it should really also have some form of over current protection internally (fire hazard : especially lithium based batteries if they are shorted and no ones present) (if no internal protection then I would always use a suitably rated inline fuse in the test lead or fitted on the battery cradle that's used when working with batteries). Main Mosfets going S/C are certainly not unknown in electronic load's.If eload is only low current device (<5A) then an internal similar rated fast fuse 'may' suffice for thisbut if eload is rated higher (say 10A) then a similar rated internal fuse is just too high to be useful to protect batteries from overheating in event of fault. So some sort of programmable over current protection would be a good idea which can be set to trip at some slightly higher value than the current Eload setting. A mechanical relay (a few 10's of mS trip time) or ss relay (< 10 uS trip time) could be used for this and since they are bidirectional a single device could provide both forward over current and reverse polarity protection. A mechanical (S.P.N.O) relay with typically a few 10's of mS opening time isn't going to offer Efuse trip speed's or micro Second RP protection but it's plenty fast enough to protect batteries, wires, pcb traces and larger components in DUT's. You can buy 'efuse' and high side 'load switches' which may offer similar features but at typical eload current ratings it's going to be a lot cheaper and more rewarding to roll your own from a relay/mosfet's and a driver/comparator If you enjoy building/designing your own eload.To use that relay also for RP Protection as a previous poster suggested you would use the uC to test the input's of the eload for presence of a reverse polarity voltage BEFORE you enabled the relay so prevent 90% of miss-connections right there, then you can perhaps pole the input's at say 1mS intervals during use to test if RP present.Regards

author: tjhoe16
author_rank: Newbie
timestamp: January 11, 2018, 02:00:47 am
content: Just for reference, here's what I was suggesting.Note, it's important for a fast turn-off for M2 during reverse polarity. Too slow and you'll exceed the SOA of M1 and M2. In this case, when M2 is on, there is a current path through D1 to charge the gate and quickly turn it off.Obviously M2 won't be on until the load voltage exceeds at least the threshold voltage (during normal operation) so if you're expecting to have high currents at low load voltages you might need to consider some added complexity. The body diode will still pass the current though so it won't cease to function, you'll just have to deal with the voltage drop until you have enough load voltage.

author: MikeEagle95
author_rank: Newbie
timestamp: January 11, 2018, 06:21:15 pm
content: If I use another n-mosfet after current sense resistor, how will error-opamp be able to drive the main n-mosfet.The whole idea behind this type of project is that the error-amplifier tries to do whatever it needs on its output to keep both its input at the same potential.For example:I want to have 3A.So using 0.1Ohm resistor I cannot put 0.3V on the error-opamp input because after resistor there will be no 0V due to another n-mosfet.I would have to first measure the voltage drop across resistor using current sense opamp, and then calculate the necessary voltage on the input pin of error-opamp.

author: tjhoe16
author_rank: Newbie
timestamp: January 11, 2018, 06:41:42 pm
content: You can use a differential amp for the current sense OR just swap the current sense with the reverse polarity circuit. You'll just have to have a higher voltage on the load (few hundred millivolts in your case) to turn on the FET.

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 15, 2018, 10:31:07 am
content: in my circuithttps://www.eevblog.com/forum/projects/check-my-constant-current-load-(with-mcu-control)-2-days-worth-of-initial-design/msg1191821/#msg1191821,i used voltage sense (through a schottky diode D2) to close the relays (concentrate on the top right of the schematics). when there are voltage registered by vpdiv (indicating forward voltage), mcu will close the relay SR2 and SR3 during energization of the constant current loading. no need protection diodes in series, that will take some power and voltage drop in the line during operation. (you can remove SR1 if you want to save cost, it will be always closed during operation, thats only for sense circuit protection in case of really nasty over voltage occurred iirc)