How does this power mosfet work in a constant current load? - Page 1
Forum section: electronics / general

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 04, 2010, 01:36:17 pm
content: Hi,The electronic load shown in attached image is constructed using a power mosfet Q1 (IRF540). Let's say the opamp MAX480 is powered by a 9V battery, thus the maximum output voltage of the opamp should be no more than 9V. If the load current is 10A, the voltage drop across the R10 resistor is 0.1x10=1.0V. If it's a 12V load, then the Vds=12-(0.1x10)=11V. To conduct current, the Vgs must be larger then Vth of the power mosfet. In this case, the Vds>(Vgs-Vth) and the power mosfet should be operating in "saturation" region. However, the power mosfet is said to act like a voltage controlled variable resistor (in "linear" region) in several articles. Which one is correct?Thanks.

author: scrat
author_rank: Frequent Contributor
timestamp: November 04, 2010, 01:50:11 pm
content: The MOSFET acts like a voltage controlled current source, in the case you described. It can't be in triode region, your guess is correct. However, since the voltage is almost constant (from 11 to 12 V, since resistor is very small) the result is that the V/I ratio is variable with current, acting like a voltage controlled resistor.

author: jimmc
author_rank: Frequent Contributor
timestamp: November 04, 2010, 05:20:33 pm
content: I don't think that that is meant to be taken literally, it merely indicates that the current varies with gate voltage in a similar manner to a voltage controlled resistor. The analogy does not extend as to how the current varies with output voltage.To be honest it doesn't matter what region the MOSFET is working in, the current is stabilised by negative feedback taken from the source resistor.The output voltage of the Op-Amp (and hence the MOSFET's gate voltage) will rise until the current through the 0.1 ohm resistor produces a voltage equal to that on the wiper of the pot (R8).The Op-Amp has a gain of over 100,000 so the voltagechangeacross the 0.1ohm resistor required adjust the gate voltage to compensate for the MOSFETs characteristics is negligable.As long as the MOSFET can conduct sufficient current its characteristics do not matter.Jim

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 04, 2010, 05:29:09 pm
content: I thought, by understanding how the circuit works, it will help to select a suitable power MOSFET for my application. I'm not sure if every power MOSFET (with the right voltage and current ratings) can be used to build the similar electronic load. Some articles said "logic level" power MOSFET is better. Is that true?

author: scrat
author_rank: Frequent Contributor
timestamp: November 04, 2010, 06:23:42 pm
content: Jim is right, in the sense that it is strictly a current source, controlled to reach a reference.If your opamp was required to be operated into saturation (as in the case of switching), then you would need a voltage much above 5V to achieve a low resistance with a normal MOS.In this case you won't need any logic level MOSFET, because you don't need to put it in saturation and (as far as you wrote) you don't mean to operate the opamp at 5V or so.

author: blackdog
author_rank: Frequent Contributor
timestamp: November 04, 2010, 06:27:01 pm
content: Hi,If you look at the datasheet of the IRF540 you will see that about 5.5V is necessary for 10A current.The output of your opamp wil be 5.5 + 10 x 0.1 is about 6.5 Volt.And you wil KILL the IRF540 because PD = 100Watts if you don't keep it perfectly cooled @ 25CTipPlace a power Diode in series with the drain, for in case you switch the + and de -Use 3 or 4 IRF540 and a good heatsink.Kind regarts,Bram

author: tecman
author_rank: Frequent Contributor
timestamp: November 04, 2010, 10:12:54 pm
content: A large R and perhaps a small C may be needed from the op-amp output to the "-" input for stability. The open loop gain and GBW product of the amp is very high and you may have stability issues in the configuration shown. Instability will likely be high frequency oscillation at the gate.paul

author: Time
author_rank: Frequent Contributor
timestamp: November 04, 2010, 11:02:44 pm
content: IXYS makes good stuff. I have done some amazing things with their switches at high voltages at my previous job.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 04:14:43 am
content: Quote from: blackdog on November 04, 2010, 06:27:01 pmHi,Thanks for your explanation. By the way, would you please share which IRF540 datasheet that you referred to in your reply above? I wish to learn how to read the output characteristics of the IRF540 and apply it to the electronic load. It looks to me that many manufacturers produce the IRF540 and their characteristics are different from one manufacturer to others.Thanks for the protection idea as well as the idea of paralleling few IRF540 and the heatsink.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 04:27:13 am
content: Quote from: jimmc on November 04, 2010, 05:20:33 pmHi,I'm confusing because what I study in the class is apparently different from what the circuit does. By knowing which region the mosfet works in will help correcting my misunderstandings.Due to my poor background of power mosfet, I do not quite understand the underlined sentence above. Would you please advise which variable, due to the high gain of the opamp, is negligible? Also, what are the mosfer characteristics to be compensated?Thanks.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 04:42:44 am
content: Quote from: scrat on November 04, 2010, 06:23:42 pmHi,Actually I do not know what will be maximum Vgs the circuit needs to operate correctly (e.g. for a 12V,10A load). Initially, I plan to use a single supply (up to 5V) opamp. May be now I should consider to use an opamp which can provide higher output voltage.As you mentioned earlier, a 12V load with 10A will drive the power mosfet into its saturation region. If we refer to the datasheet of IRF540 (attached in my first post in this thread), can I say that the required Vgs to achieve this is some values between 4.5V and 5V?

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 05:20:27 am
content: Quote from: AcHmed99 on November 04, 2010, 10:48:45 pmHi,Thank you very much for your reply. I learn one thing today: IXYS Linear Power Mosfet. I also found many useful literature at the IXYS website.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 07:11:02 am
content: According to Wikipedia:http://en.wikipedia.org/wiki/MOSFET... the "ohmic" region is also known as "triode" or "linear" region, whilst the "saturated" region is also known as "active" region.According to this IXYS article:http://ixdev.ixys.com/images/technical_support/Application%20Notes%20By%20Topic/Power%20mosfets/IXAN0068.pdf...the output characteristic of an N-Channel power mosfet can be divided into "ohmic", "current-saturated", and "cut-off" regions. The article states:[-a-] Applications likeelectronic loads, linear regulators or Class A amplifiers operate in thelinear region of the Power MOSFET, which requires high power dissipation capability and extended Forward Bias Safe Operating Area (FBSOA) characteristics. Such mode of operation differs from the usual way of using Power MOSFET, in which it functions like an “on-off switch” in switched-mode applications.[-b-] In thelinear-mode of operation, the device operates in the"Current-Saturated" regionwhere the drain current (Ids) is a function of the gate-source voltage (Vgs).Thus, it looks to me that it is the definition of "linear" region that causes the confusions here. Which is the right definition for "linear" region/mode, the Wikipedia article or the IXYS article?Thanks

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 05, 2010, 07:14:15 am
content: Wish to confirm... is it correct to say that in switching mode application, e.g. SMPS, power mosfet mainly operates in the "ohmic" and "cut-off" region? Thanks.

author: scrat
author_rank: Frequent Contributor
timestamp: November 05, 2010, 07:39:43 am
content: Quote from: onemilimeter on November 05, 2010, 07:14:15 amYes. It is made "jump" from one to the other, because in those regions power dissipation is relatively low.Quote from: onemilimeter on November 05, 2010, 04:42:44 am

author: blackdog
author_rank: Frequent Contributor
timestamp: November 05, 2010, 08:30:58 am
content: HiHere are the datasheets from different brands...www.bramcam.nl/IRF540-Fairchild.pdfwww.bramcam.nl/IRF540-Internationl-Rectifier.pdfwww.bramcam.nl/IRF540-Philips.PDFwww.bramcam.nl/IRF540-SGS.pdfwww.bramcam.nl/IRF540-ST.pdfHappy readingBram

author: scrat
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timestamp: November 05, 2010, 11:54:23 am
content: Sorry for the mess in my above post! I hope it is understandable.

author: onemilimeter
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timestamp: November 05, 2010, 12:25:19 pm
content: Quote from: scrat on November 05, 2010, 11:54:23 amI can understand the description in your post. Thanks.

author: blackdog
author_rank: Frequent Contributor
timestamp: November 05, 2010, 03:20:08 pm
content: HiYour schematic is a voltage controled currend source ("voltage controled resistor")If you stay within the specs of the opamp and the IRF540 you don't have to think about the Vgs.The IRF540 in the control loop of the opamp.The IRF 540 is NOT saturated! at 10Amp'sOK, think with me...Your powersupply is adjusted so dat it gives 12V @ max 10AmpsSet your electronic load to 10Amps10Amp and a 0.1 Ohm resistor = 1VOn the "+" input of your opamp = 1VThe opamp ouput wil go up driving the gate until its "-" input wil also be 1V ( its the nature of opams to have the same voltage on both inputs)The voltage the IRF540 needs for 10Amps you wil find in the data sheets ( about 5.5V) +- 10%The gate voltage against ground level wil be 6.5V @ 10Amps +-10%OK, there is 10 amps comming out of your powersupply.The total load (without your connection wires) = U / I = 12 / 10 = 1,2 OhmThe "R" from the IRF540 = 1.2 - 0.1 Ohm drain resistor = 1.1 OhmThe IRF540 is disipating 11V x 10 Amps = 110Watts !!!! (It can boil your thea water)A saturated IRF540 wil have a resistance of < 0.05 Ohm @ Vgs of > 10VI hope my 2 cent helpsKind regarts,Bram

author: Time
author_rank: Frequent Contributor
timestamp: November 05, 2010, 03:32:58 pm
content: Are you aware this is similar to the current source Dave describes in a blog?

author: scrat
author_rank: Frequent Contributor
timestamp: November 05, 2010, 04:40:38 pm
content: Quote from: Time on November 05, 2010, 03:32:58 pmIt's the same thing, in fact.@blackdog: there are two things about which I don't agree with you.- The MOSFET's active region (far from the two axes) is called saturation, while the region corresponding to the BJT's saturation (near the Id axis) is called triode or linear region. So this MOSFET IS in saturation.- In my previous post, I was wrong because I considered linear coordinates instead of logarithmic.However, finding the right answer is even easier than I said above.If you look at Fig.1 of the datasheet and find the Vds-Id point in the axes (Id = 10^1A, Vds=1.1e^1V), you will find that it's in the middle between the 4.5V and the 5 V characteristic. So Vgs must be in that range.

author: jimmc
author_rank: Frequent Contributor
timestamp: November 05, 2010, 05:20:04 pm
content: Quote from: onemilimeter on November 05, 2010, 04:27:13 amImagine that you MOSFETs, for a drain current (Id) of 10A at a given drain voltage, one requires a Vgs of 3v and the other a Vgs of 5v.Plug the first MOSFET into the circuit and adjust the pot for 10A, the Op-Amp output will be Vgs +(0.1 ohm * Id) or 4v.Change to the second MOSFET without changing the pot setting.Initially the drain current will be lower leading to a reduced voltage across the 0.1 ohm resistor, this will cause the input voltage to the Op-Amp to rise (ie difference between the voltage set by the pot and the voltage across the 0.1ohm resistor).Thus the output of the Op-Amp will rise until equilibrium is reached.For the Op-Amp output to rise by 2v (from 4 to 6v) the voltage across the 0.1 ohm resistor has to change by 2v/gain or 20uV for an Op-Amp gain of 100,000.Through the action of negative feedback the Op-Amp has reduced the effect of a change in Vgs of 2v to a change in Id of 20uV/0.1ohm = 200uA.Hope this helps.JimpsDave did an episode on a low power version herehttp://www.eevblog.com/2010/08/01/eevblog-102-diy-constant-current-dummy-load-for-power-supply-and-battery-testing/

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 07, 2010, 01:53:49 am
content: Quote from: scrat on November 05, 2010, 04:40:38 pmHi...blackdog has been very helpful. I still could not figure out how blackdog came to the Vgs=5.5V for 12V/10A load. I agree with scrat that the Vgs should be within 4.5V and 5V. I have checked all the datasheets given by blackdoghereand none of them shows that the Vgs is near to 5.5V.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 07, 2010, 02:02:43 am
content: Quote from: blackdog on November 05, 2010, 03:20:08 pmHi...I really appreciate for the detailed explanation.Like scrat pointed out that (1) the mosfet should works in "current saturated" region, (2) the Vgs is between 4.5V and 5.0V for a load of 12V/10A. Would you please kindly advise how did you come to the conclusions which were slightly different from what scrat and I would have expected?Thanks.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 07, 2010, 02:03:21 am
content: Quote from: Time on November 05, 2010, 03:32:58 pmHi...Yes. I'm aware of the Dave's video. It's a great video!

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 07, 2010, 02:06:58 am
content: Quote from: jimmc on November 05, 2010, 05:20:04 pmHi...Great explanation! I love reading it... Thanks a lot!