FET operation - Page 1
Forum section: electronics / beginners

author: veedub565
author_rank: Frequent Contributor
timestamp: April 15, 2022, 08:34:35 am
content: I've been reading a bit how FET's work. And I see they can operate as a switch effectively by applying voltage to the gate.I'm a bit confused with the following exampleGate = -25vDrain = +13vWhat should I expect to see on the source? All things being equal. I'm guessing it should be 0v and the FET is off, but I'm not sure.Second question, with 13v on the Drain what would I need on the Gate in order to turn the FET on

author: janoc
author_rank: Super Contributor
timestamp: April 15, 2022, 08:58:19 am
content: That is impossible to answer without knowing which specific FET it is and possibly seeing a circuit diagram.FETs aren't all made equal.

author: Ian.M
author_rank: Super Contributor
timestamp: April 15, 2022, 09:57:14 am
content: Also FETs generally aren't perfect, and pass a tiny leakage current when nominally OFF. Therefore the voltage you'd measure on the source would be highly dependent on the load on it, which if nothing else is connected to it would be the input resistance of your voltmeter. An ordinary 10Meg input DMM would read 10V per uA of leakage current so it wouldn't be unexpected to see as much as a couple of volts there, more if the FET is at a high temperature.

author: veedub565
author_rank: Frequent Contributor
timestamp: April 15, 2022, 05:39:49 pm
content: The part is a Siliconix J2086, I think that's a JFET. I've attached a couple of screenshots, one showing the FET in question (Q8 circled in red). The other showing what it's connected too (the negative input to a null detector/amplifier. Which happens to be the Drain of a J2317)The drain is connected to a 13v fixed reference voltageThe Gate voltage comes from the output of an LM339 and I measured it at -25v. (The LM339 has a +5v and -25V supply)I measured the Source voltage at 0vI perhaps ought to have posted this in the repairs section, as I am trying to fault find and repair an old Fluke instrument. It's a bit of a beginners question though, and I want to try and understand how the FET functions. Partly so I can check the operation is correct in this particular example, and partly because I need to learn these things for future reference.I have had a google, but feel free to point me in the direction of any useful information. It's probably better I learn about this, and understand why/what/how it works. Rather than somebody just giving me the answer to this particular example.

author: ledtester
author_rank: Super Contributor
timestamp: April 15, 2022, 06:38:10 pm
content: Quote from: veedub565 on April 15, 2022, 05:39:49 pmIs it a Fluke 5700A? And is the board the "A11A2 DAC Buffered Reference SIP PCA"?Q8 is listed as:QuoteFWIW, Marco Reps talks a bit about the A11 board in this teardown/repair video (at 18:18):https://youtu.be/qhr6h7wuqo0?t=18m18s

author: Gyro
author_rank: Super Contributor
timestamp: April 15, 2022, 06:51:45 pm
content: Ah, yes it is an N-channel JFET.JFETs are normally 'depletion mode' devices, where, if the gate voltage is zero relative to the Source, it conducts. The JFET is 'turned off' when the gate is taken negative (-25V in your case).JFETs are usually symmetrical too, and don't care if you exchange the Drain and Source - they don't have the parasitic Drain-Source diode that you find in MOSFETs.JFETS can of course be used as linear devices, but often as switching devices too - in your first schematic, it looks as if the marked part is being used to bypass the 'REF13 Filter', controlled by 'REF13 SEL'.I did a quick search and found a thread using the same part. It includes another schematic that you can study, and looks as if it shows the danger of driving the gate too far (I only did a brief scan)...https://www.eevblog.com/forum/repair/help-identifying-suitable-n-jfet-replacement-for-signal-switching-in-fluke-8505a/

author: veedub565
author_rank: Frequent Contributor
timestamp: April 15, 2022, 07:21:28 pm
content: Thanks for the explanation Chris, I was trying to work out how the -25v came into it. That makes sense now gate voltage zerorelativeto the Source means it's off. Looks like it puts +5v on the gate to select it, so then gate goes positive relative to the source and it turns on.Yes it's a Fluke 5700a I'm trying to fix, with a fault on the A11 card. I have a different thread running covering the overall repair. I just wanted some understanding of how a FET, or JFET in this case works. It's one of those things, I'm working through the circuit and.... bit of a beginners question how does a FET work ? It's a big project with a lot of learning.thanks for the help

author: Gyro
author_rank: Super Contributor
timestamp: April 15, 2022, 08:26:54 pm
content: You're almost there. You can't take the gate positive with respect to the Drain or Source - the J in JFET stands for Junction (i.e. a diode junction) this will be forward biassed if you take the gate more than a few hundred mV positive, pumping current into your high impedance signal. It's the degree of gate junction reverse bias that determines D-S cut-off current.This is a valid mode of operation though. JFETS are often used (with D and S shorted together) as very low leakage protection diodes. The maximum forward gate current will be specified in the datasheet.