How do I prove that a mosfet is running in the ohmic/Triode region.  - Page 1
Forum section: electronics / projects

author: finos
author_rank: Contributor
timestamp: June 20, 2023, 11:10:42 pm
content: Hello everyone!In the past year ive been designing and building a Electronic DC load using the knowledge that I gathered from the analog electronics classes at my university plus some help from my professors.The project is nearly complete (you can have a sneak peek, here(https://github.com/finos2/Electronic-DC-Load) , but I could do a more formal presentation when I get a bit more relaxed from assignments & exams )But there is a question that I was not able to answer using the theory I knew nor my professor was able to provide a answer.If we boil down my fancyPants schematic it boils down to this : (Op-Amp-With-MOSFET.png)Most other DIYers that have made similar projects state definitely that Q1 operates on the ohmic region . But (and here comes my question) the operation point depends (not only to the Vgs, a parameter that we can control by “riding another characteristic” ) but also heavly depends to the VDS as we see from the diagramhttps://www.electronics-tutorials.ws/wp-content/uploads/2018/05/transistor-tran37.gifVds in my case is the DUT . That is a unknown voltage . how then can I (or anyone else in this matter) say definitely the the mosfet is operating in a specific region ?Even though I designed it I have trouble understanding on paper (on the breadboard the circuit checks out… but when maths join the party just what I think I understand goes away..)

author: moffy
author_rank: Super Contributor
timestamp: June 20, 2023, 11:44:01 pm
content: Does it matter? As long as your electronic load behaves correctly it's of no consequence. Since you don't control Vds it could operate anywhere.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: June 20, 2023, 11:50:01 pm
content: That curve looks like a BJT curve. But nonetheless, I would think that if Id and Vds can be measured then RDS calculated then compared. If Rds is near or equal to Rdson then the device is saturated, otherwise you're in ohmic region. Also confused by Vds is the DUT. Is Q1 the DUT or the thing you're concerned about operating in the ohmic region.

author: jbb
author_rank: Super Contributor
timestamp: June 21, 2023, 12:46:44 am
content: Thinking about it…- in the fully-on region, Ids = Vds / Ron = f(Vds)- in the linear region, Ids = f(Vgs)So you need to choose a test which can prove Ids = f(Vgs)

author: bostonman
author_rank: Super Contributor
timestamp: June 21, 2023, 02:15:56 am
content: Think this is a good post to piggy back off of.If the MOSFET is dropping voltage across it with high current, is the power dissipation simply looking at the RDSon curve at a specific drain current, and calculating the power? I was looking at datasheets last night, but not certain I was looking at the correct graphs.A somewhat circuit that got brought up recently was 30V power from a battery, but the circuit can't see more than 20v. As the battery drops, the circuit needs to maintain 20v until the battery drops enough to turn off the circuit. Unfortunately the circuit draws 12A of current due to a low resistance load.

author: srb1954
author_rank: Super Contributor
timestamp: June 21, 2023, 03:18:16 am
content: Quote from: jbb on June 21, 2023, 12:46:44 amSimplest option would be to monitor the O/P of the op amp driving the MOSFET. If the MOSFET is still in the linear region the op amp O/P voltage will be still be controlled by the feedback and will remain somewhere within the normal operating range expected from an op amp. If the op amp O/P is saturated against its maximum positive level then the feedback loop has saturated, the MOSFET is turned fully on and is no longer in its linear region.This all assumes that the design O/P swing from the op amp is sufficient to drive a worst case MOSFET fully on.

author: SiliconWizard
author_rank: Super Contributor
timestamp: June 21, 2023, 05:14:56 am
content: Also, if it operates in its linear region, it will dissipate more power and thus will heat up significantly more.

author: Berni
author_rank: Super Contributor
timestamp: June 21, 2023, 05:32:37 am
content: Electronic loads will be operating to the right side of that graph! That's the so called linear region!The point of an electronic load is to electronically control the amount of load. Notice how the current on the right side of the graph is heavily dependent on the gate voltage? This means you can control it.Once you get too far to the left of that graph it means there is not enough voltage available for the electronic load to pull the desired amount of current for this transistor. No amount of gate voltage will make the desired amount of current flow, so your electronic load is no longer in control, so you get a smaller current that desired. This makes your electronic load near useless when operating in this region.The way to detect that you are in this region is to just monitor the gate voltage. Once the electronic load hits this condition the opamp will try to salvage the situation by applying more and more gate voltage until it runs out of supply voltage and clamps into its positive supply rail. You could connect a zenner diode and LED to that so that the LED turns on whenever the gate voltage becomes unreasonably high for the MOSFET to be in the linear operating region.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: June 21, 2023, 06:52:00 am
content: Quote from: Berni on June 21, 2023, 05:32:37 amWell Wikipedia says otherwise.QuoteBtw i have always found this name confusing, as there is not much "linear" about it. The equation is a quadratic, although in some definitions this is a linear equation.To me (and Bernie?) the saturation region looks more "linear".I guess they name it to be similar to BJTs where the linear region is close to a straight line.but otherwise i pretty much agree with Bernie.If fact my Electronic load mostly operates in the Saturation region, where Vgs controls the current pretty much independently of Vds.QuoteI think they are wrong.Thanks for posting the question. I would be interested to read more responses.

author: iMo
author_rank: Super Contributor
timestamp: June 21, 2023, 07:29:57 am
content: In the "linear region" ("constant resistance mode") Ids=Vds/Rds[Ids is a "linear" function of Vds, the Rds value depends on Vgs, for a specific Vgs is Rds "constant"]In the "saturation region" ("constant current mode") Ids=constant with changing Vds[Ids is "saturated" and does not react on changes of Vds, the Ids value depends on Vgs, for a specific Vgs is the Ids "constant"]Thus the "mode" and "region" matters here, imho.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: June 21, 2023, 12:19:53 pm
content: Had a look in Horowitz and Hill, found a note that saidQuotepage. 138 3rd edition.I always thought this but never saw it. I preferred Sedra and Smith.Though at a constant Vgs and a bias point within the Saturation region, I would argue that Vds/Ids is pretty close to a constant and so IMHO it should be the named linear region.Another example of confusing terminology is that the MOSFET Saturation region is the same as the what is known as the Active region.Whereas with a BJT the Saturation region is not the Active region.

author: shridmaster
author_rank: Contributor
timestamp: June 21, 2023, 01:21:20 pm
content: Woah woah woah, okay let's take a step back and define a few terms and actually properly follow the design process. We have terms being sloppily thrown around, and OP has an XY problem on their hands.----TERMINOLOGY----BJTs and MOSFETs both have similar-ish curves, and when they're on and properly function you have two major regions/operating modes of interest when you look at the I-V curve. In this case the "V" in this I-V curve is Vds or Vce, not Vgs or Vbe. There's the region on the left when Vds is low, and I is dependent on Vds. There's then the region on the right where Vds reach a certain point and I is relatively constant and independent of Vds.For BJTs, the left side is saturation, and right side is forward-active. For MOSFETs, the left side is triode, and right side is saturation. These are just terms, and are rightfully confusing. In order to a) unify the terms and lessen confusion and b) clarify them by *circuit purpose* rather than semiconductor physics.The two terms to instead use are OHMIC for the left side, and LINEAR for the right side. OHMIC is called OHMIC because the current through it changes with the voltage across it, like a resistor. LINEAR is called LINEAR because that is the region you need to be in to create amplifiers i.e linear circuits. Of course the I-Vgs curve of a transistor in isolation is not linear, but if you put the transistor in a CS amplifier, the *output voltage* is linear with the *input voltage*. The small signal gain remains constant, so inputting a sine wave outputs the same sine wave with a scalar gain. Linear.So we're going to use the terms "ohmic" and "linear", and it will universally apply to BJTs and MOSFETs. The transition point between these two is conveniently already called "saturation point", so we can keep using that.----XY PROBLEM----https://xyproblem.info/For the uninitiated, an XY Problem is when their real problem is X, but they don't know how to solve X, they (misguidedly) come up with Y as a solution, and ask how to solve Y even though Y doesn't actually solve X. What we need to do here is ignore Y and get to X to solve the actual problem.In this situation, OP is asking how to identify if something is in the Ohmic region and how to keep it ohmic. This is Y. We can ignore it. They're asking because they're trying to create an electronic load. This is X. Let's solve X instead.An electronic load is something that pulls a constant current regardless of the voltage across it. This is what a benchtop DC load does. If you wanted the current to change with the voltage, you would just simply use...a resistor. But we want to pull a programmable *current* instead.Now to the solution, if you look at the I-V curve of a transistor, the linear region gives you a constant current that is independent of the voltage across it. Indeed, read any microelectronics book or look at any microelectronics paper, this is how constant current sources are implemented using current mirrors. As long as the Vds (or Vce) is high enough that it stays in the linear region, the current through the transistor is purely a function of its gate/base voltage.So the solution to your problem is to simply measure the voltage across Vds, check if it's high enough, and match it to the datasheet. If it's above the saturation point, it's in linear region, and you have a constant programmable current source.

author: finos
author_rank: Contributor
timestamp: July 27, 2023, 11:55:34 pm
content: First and foremost, I would like to thank you all for taking the time to reply to my -kinda- rambling postI was AFK for this long cause Semester exams & life -happened-. There wasn’t enough time to fit everything into 24h .Now it’s time to get back and reply to everyone.Quote from: moffy on June 20, 2023, 11:44:01 pmThe whole question was born from discussions I had with colleges that had less “practical” experience (doing projects , breadboarding circuits etc.) but they knew the theory very well. I KNEW (Because this is what I was told in the classes) that I wanted the mosfet in the linear region cause there has the most power losses and disepeates more power as heat (exactly what we want on a DC load), plus in the ohmic region the Mosfet behaved like a Voltage controled resistor, this is what I was told. With the majority of other DIYers saying the same thing it generated kinda a confirmation bias on me. When I got asked for details on the circuit , I wasn’t able to “prove” in what region the fet was in , because of course I didn’t know the Vds .We don’t Technically need this info for the finished build , it doesn't affect the operation. But it kinda bugs me that I didn’t a) have answer to provide and b) effectively didn’t understand my circuit .Quote from: Ground_Loop on June 20, 2023, 11:50:01 pmAbout the curve , I think I attached to the post the correct one , see thishttps://www.electronics-tutorials.ws/transistor/tran_6.html, A bjt`s one is the “opposite”https://www.electronics-tutorials.ws/transistor/tran_2.htmlQuote from: jbb on June 21, 2023, 12:46:44 amQuote from: Berni on June 21, 2023, 05:32:37 amHmm , If I understand this correctly, you are saying that the circuit is opperaing at the saturation region ! Interesting … We were told in the electronics classes that the ohmic region was the more suitable one because there the fet would operate like a voltage controlled resistor , in the saturation the Channel would be fully open and we wont be able to control the current.In addition , Although the led circuit you mention could be useful , here is no need for such circuit because the mosfet I use has a max Vgs of +-20V and I supply the OpAmp with 12v . At best it receives 10V on the gate.Quote from: iMo on June 21, 2023, 07:29:57 amYou mean the “logic” modes (CV CC CP CR ) that I have implemented in my controler ?Interesting , but looking at the VI curve of a mosfet I cannot realy understand how this could be. Can you please explain it further?Quote from: shridmaster on June 21, 2023, 01:21:20 pmWoW That’s a very interesting way of thinking about a problem of approaching it .But I have only one objection to it though, I didn’t come here asking HOW to design a load, but more like the load schematic works in the theoretical standpoint.I hope that I replied to everyone, it’s getting late here for me, almost 3 Am better wrap this up not to wake anyone up.I’m looking forward to your replies!

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 28, 2023, 01:01:46 am
content: Quote from: finos on July 27, 2023, 11:55:34 pmYou have feedback around the MOSFET, so where it may sit on the transfer curve does not matter.Quote from: finos on July 27, 2023, 11:55:34 pm?? of course you can control the current, on all parts of the MOSFET curve, within the RDS limits.Quote from: finos on July 27, 2023, 11:55:34 pmPower loss is (voltage across the part * currrent thru part), which again does not care which 'region' the FET is in.

author: boB
author_rank: Frequent Contributor
timestamp: July 28, 2023, 03:12:22 am
content: Don't forget to stay within the SOA limits !Although keeping the case at 25 degrees C or less would be necessary which is next to impossible to do ?Companies like TDI (Transistor Devices) used to make these programmable loads called "Dynaloads" that used bipolar power transistors which used Emitter transistors to handle paralleling of MANY parts. TO-3 parts in the units we used.There are special FETs that are made for linear operation but I haven't looked lately for those. Ones they use in like the old Hafler audio amplifiers from way back when. Probably MUCH better parts to use these days. But Lateral MOSFETs come to mind. Don't know if there is anything better ?https://www.diyaudio.com/is one place to look but those the audio power amps may be more just for sourcing that sinking if that makes a difference besides reactive loads.IF it is just milliwatts you want to load, then no problem. I usually need hundreds of watts at least.boB

author: Berni
author_rank: Super Contributor
timestamp: July 28, 2023, 05:51:12 am
content: Here is a more realistic IV curve of a actual real life MOSFET IPB110N20N3LF:Here you can see how the slope of the curve changes with gate voltage, that's the changing resistance of the mosfets channel.But notice how this region of "Ohmic operation" gets narrower and narrower towards smaller currents. So at currents that make sense for an electronics load (a few Amps) you will already hit the channel saturation region after you apply only >0.5V across the transistor. Also notice how close together the lines are in the left area, this means a change in the gate voltage has less control over the actual current flowing.Then the 10V line that shoots off into space is what so called "gate saturation" looks like (not to be confused with channel saturation). This is what we think of as a saturated transistor in BJTs. This is where the transistor is turned on as far as it can turn on, so it will pass the maximum possible current. Large power MOSFETs became very close to a dead short at this point, so the curve you are seeing is largely the resistance of the pins and bond wires. In this case at 0.5V you get 50A flowing, so about 10 miliOhms. This area is only useful for switching things on/off, not linear control.As said above be aware of the Safe Operating Area (SOA) since transistors optimized for switching operation might not be able to handle operating linearly (internal unequal current sharing causing localised hotspots, thermal runaway and destruction of the transistor). For this IPB110N20N3LF the SOA does have a DC area because this is a transistor designed for linear operation:If you go out of the area under this curve for longer than the time written on the curve, then your transistor will blow up.