What is this JFET and what's it's purpose ? - Page 1
Forum section: electronics / repair

author: MathWizard
author_rank: Super Contributor
timestamp: June 14, 2023, 09:19:08 pm
content: In this PSU I'm working on, in the PWM gate drive, I found this sot23 thing, that seemed to be an N-ch JFET. It's marked NAN, searching around for NAN sot23, IDK what that is. The 1st 2 things that show up are in fact JFET listings, but they don't match the pinout.https://www.google.com/search?client=firefox-b-d&q=NAN+sot23Out of circuit, in a device tester, that I don't think makes a negative rail, it said it was an N-ch, with Idss=2.7mA and Vgsoff=63.8V. So IDK what it does to get that. But it said the pinout, with the middle pin pointing away from me, is from left to right, G-D-S.On a breadboard, it seemed to turn on around -1.6Vgs, and had no gate current, and was having around 10-15mA Id for 10-20Vdd from a 1k resistor. But then between my DMM changing ranges, possibly bad banana plug connections, stopping all current, and me turning up the Vdd to ~24V, I seemed to have zapped it, and now some current flows into the gate. Sometimes the tester sill says the same info tho.The circuit is below, as the way it should be wired, if in fact the tester is right. But it is setup up more like a MOSFET, with the source on GND, the drain sinking the primary side L1 inductor current, along with a parallel SMD diode marked D1. And the gate is connected to the PWM, but like a MOSFET, taking it higher than the source.In the sim, with a JFET with Vgsoff=-1.8V, the circuits works pretty much the same if I swap D and S, thats typical I know. And not much worse, if I GND the gate, put PWM on the drain, and inductor on the source. I guess because that diode is there too. A BJT works almost exactly the same too, almost the same current in the main switching inductor L4.Another thing tho, with the JFET, (and a zero output resistance PWM source), I see the gate current is ~0.73A peak. So with an on time of 10us, that's 7.3uC, and with Vgs=1.6Vp, that would be 4.6uF if it was a normal capacitor. So how is that gate current actually so high? And if typical gate source capacitance is a few pF, how is the current so high for so long ?? Vdg is also about 1.8Vp. Is this the Miller effect, is it really that big ? ?Also the diode current from the gate drive inductor L1 seems crazy at an average ~1.43A . I wonder what it is IRL. But thats basically the circuit (IDK the C4 and C15 values, or the exact diodes/FETs)

author: fzabkar
author_rank: Super Contributor
timestamp: June 14, 2023, 09:59:48 pm
content: The circuit doesn't make any sense to me. For example, I would think that D1 should charge a capacitor rather than going to ground, otherwise where does the inductor's energy go when the FET switches off?

author: MathWizard
author_rank: Super Contributor
timestamp: June 15, 2023, 01:15:23 am
content: Thats the right way tho, for both on either side of L1, and I took out the transformer to get the inductance. Other zeners are marked Z, these are just marked D. And both labels on the package are D1, so something like a 1N4001, and their Vf reads around right. And both bases and both emitters are tied together.Ok so the when the NPN is on, current flows through L1 into D1, and when the base drive goes low, to keep the current flowing the same way, the inductor creates an opposite polarity voltage across itself, that gets clamped by D5 and the PNP, which is still reversed biased and doesn't operate in normal mode. And current keeps flowing through L1 in the same way.Yeah IDK if the gate drive inductors are a regular transformer, just transfer energy without storing it, but the B or H field is growing and shrinking as the current goes up and down.I need to learn some more equations for all that, like magnetizing current.

author: fzabkar
author_rank: Super Contributor
timestamp: June 15, 2023, 01:36:51 am
content: I understand the point of D5, but not D1. As I see it, the voltage on D1 switches between 0V and 0.6V. What's the point of that? "NAN" may as well not be present.Where is the PWM signal coming from? Maybe the datasheet for the source IC has an application circuit?

author: jwet
author_rank: Frequent Contributor
timestamp: June 15, 2023, 04:14:54 am
content: SOT marking is a mess- I don't think NAN is a JFET- it just can't be and make sense.I guess this is a circuit fragment that you simulating, its not clear what's input and what's output, etc. A little annotation could help.Looking at what's there- the NAN device must be a medium resistance N channel MOSFET like a 2N7000. My clues are the polarity of the gate drive and its size, its not going to be a very low resistance device. I don't know what the PWM (is it PS controller?) is either but all the diodes and R's that follow it are designed to direct the source and sink currents for the bipolar drive and the FET drive. The two bipolars are a low impedance emitter follower driver (10 ohms?). They are driving against that inductor that has that D1 looking like a free wheeling diode with a N channel across it- a little like a sync rectifier. Need a few more details.

author: MathWizard
author_rank: Super Contributor
timestamp: June 16, 2023, 10:31:34 am
content: OK of course the L1 inductor is storing energy, that's why the current keeping flowing but decreases after the PWM shut off. I guess a non-saturated transformer current would shut off right away.The PWM input is directly from the common PFC/PWM controller CM6802. In the datasheet it says the chip output drive can be up to 1.5uJ of energy / cycle,from either PFC or PWM output. And the max PFC or PWM output sink/source current can be 1A. But IDK what the output impedance would be. I need to power up the PSU and see what the Vaux is, that powers the CM6802 Vcc curcuit.https://www.techpowerup.com/review/corsair-rm850/images/CM6802.pdfWhen I tested the NAN, at 1st, I had gate-source shorted, and a 1k drain resistor. So a mosfet shouldn't have worked, unless I damaged it with the hotair gun, or its upside down. I see in the sim, that an upside down Nch 2N7002, with drain-gate on gnd, and 1k on source to 12V, passes 11mA to GND no problem.With the NAN, I hooked the ch2 positive output of my bench PSU to the ch1 negative, and put the source on ch1 negative terminal, the the gate on the ch2 negative terminal, it it really was shutoff until ~-1.8V approaching 0V . Now with damage, around 3mA enters at -2V.My cheap tester should easily identify a Mosfet tho, so IDK.In the sim, removing NAN doesn't make much difference to the L4 current, (L4 is the main switching transformer, I just left out the 12V secondary side, and with PFC active it would have 340V, not 170V).Putting a 2N7002, with Vgs on of 1.6V, drastically lowers the V3 PWM current from 0.735A to 92mA, without much difference to the L4 current. But the NPN base Vbe and current go from ~1.7V and 35mA to 3V and 91mA.It would be easy for me to find a sot23 2N7000 on my junk PCB collection. The only JFETs I found, are J111 TO92, closer to -8Vgsoff in 1 I checked. I have a J310 somewhere, and some Pch J176.I'll try a 2N7002, I found 1 on a mobo, but IDK.With middle pin of the NAN pointing at youthe left most pin is on GNDthe middle pin is on D1 and L1 pin4and the right pin is on the top of the 10k

author: jwet
author_rank: Frequent Contributor
timestamp: June 16, 2023, 03:27:25 pm
content: That makes more sense- thanks- I'll look at the DS for the CM6802 controller. I wonder if that circuit came from a reference design or app note- that's kind of a strange circuit. I may have a look at the patents. I come on EEVblog when I get up in the am and have coffee for a diversion- this is a worthy adversary.Those PFC/PWM outputs from the CM6802 are FET driver outputs like a big CMOS inverter. The datasheet give Routs for high and low drivers at 30/15 ohms respectively for PMOS/NMOS. These values will soften the edges of the drives a bit and are as low as required given the speed.I think your first test with the NAN was upside down and the 11 mA is exactly what I would expect having the body diode conducting and the SD reversed. There are other configurations but the fact that you a 1V drop across the device is significant. A JFET source and drain are symmetrical first order so I wouldn't expect a difference with polarity. This test leans towards a MOSFET to me.The second test of the NAN is testing the JFET hypothesis I guess. 1.8v is in the ball park for a small mosfet or a low threshold JFET unfortunately. Kind of tough when you don't know pinouts and polarities! Too may permutations. JFETs really hate having their gate forward biased. Bare MOSFET are more ESD sensitive but this seems less likely as the failure. This test leans towards smoking a JFET to me.Those two tests are kind of contradictory and now I think you have a dead device.In looking again at things, it looks like the function might be something like the following. When the drive is high, the emitter follower is sourcing the inductor and the FET is on (assuming NMOS FET) pulling the far side to ground- charging up the inductor and coupling to the load. When drive is low, the PNP is still active sinking from the inductor pulling current up through the diode with NAN fet off. Its kind of assymetrical drive, strong source, weak/slower sink. I'll think about this some.You're got me now- Keep us informed on what you learn.

author: jwet
author_rank: Frequent Contributor
timestamp: June 16, 2023, 03:49:11 pm
content: In looking in more detail at the DS- I'm more confused. I'll study a bit more and maybe return tomorrow. That emitter follower connection is a common poor man's FET driver. Its show in the data sheet but only as a FET driver. Is it directly drving an inductance in your circuit or was this just something for doing sims? In the data sheet circuit- I see a similar circuit with an L and diode in series. Note that the inductors in that circuit are coupled like a tranformer- I think this is some gate drive symmetry stuff. I'm pretty confused. Curious as to outcome.

author: MathWizard
author_rank: Super Contributor
timestamp: June 17, 2023, 01:53:33 am
content: Thanks, yeah that NPN/PNP driver it's just like that on the PCB. That totem pole has been on most of these computer PSU's I've mapped the circuits of, but those PSU's had a capacitor, coupling a pulse into the inductors, with the primary low side on GND.This one, all the sot23-3 diodes are marked D, and all the transistors are marked Q, including the NAN. Now it could be a mistake, or a bodge. But the driver circuit is off by itself, so not too hard to map, and I've gone over it a few times, and I'm 99% sure it's right. At the same time I checked the sot23 BAV70/99.The sim says a 2N7002 would have a PWM Id of ~0.9A or an average of 0.45A. The datasheet says it's only rated for 0.21A, and the traces are small. So I won't be trying that. And that's about the only sot23 MOSFET I could find on my junk circuits. So at some point I might still order something.D1 has ~0.75Aavg with or without any FET, and with 20ohms added in V3. To me, that doesn't sound very safe either.Yeah I forgot about the body diode, that sounds about right.When it was on the PCB, I did diode checks, and it only seemed to have 1 diode junction, the rest were open. I don't remember the polarity.Right now on a breakout board, middle pin pointed at mea 0.600V PN junction is from Left(+) to middle(-) pin. And in reverse, that reads, 1.8-2.30V so far. So I was calling the middle the drain, and the left one the gateThats with 4V on diode mode. Nothing else seems to do anything, and unlike a J111 JFET, I can't open any possible DS path from gate charge. On the J111, I could read a few tens of mV sometimes, so I guess a few 10's or hundreds of ohms.The Nch, low mOhm mosfet I have near me, I can only read the 0.54V body diode, and I can charge the gate and open the DS to nearly a dead short.As for the PSU, a few years ago it had a resetting problem, and I started to take it apart, but life got in the way, and it sat in a box for ages after. I was kind of saving it for fun, and for when I learned a bit more. But I still have a pile of magnetic and control/feedback theory equations to learn to really understand these.So last week I mapped 99% of the circuits, but never powered it up yet. I have an auto-transformer, and isolation transformer. So pretty soon I'll have to try it, and see which controller chip is doing the resetting. It looks super clean, if the output caps are worn out, I'd expect to see some dust in there too.

author: jwet
author_rank: Frequent Contributor
timestamp: June 17, 2023, 07:57:13 am
content: With an Variac and an Iso transformer you're likely safe and you can bring it slowly and watch things. Still be careful with an earth grounded scope. I have an OWON HDS242s battery scope that I use for servicing modern stuff- $150 bucks. I'd hate to take out a nice Tek scope trying to repair some unrepairable POS. Good luck.

author: Konkedout
author_rank: Frequent Contributor
timestamp: October 25, 2023, 04:21:11 am
content: There are many sources of doubt in what you say and what I see. All that I see is a simulation circuit showing what looks like gate drive. But anyway; in my many years of designing power supplies I do not remember encountering JFETs as power switches. Only MOSFETs and what came before that was bipolar transistors.One thing is that silicon JFETs have no gate isolation oxide, so the gate must be driven negative and they are depletion mode FETs. These are much less convenient to design with because you normally need a negative voltage. In theory a MOSFET can be either enhancement or depletion mode, but the enhancement mode MOSFETs are by far the most common.