What is this MOSFET good for? - Page 1
Forum section: electronics / projects

author: compet17
author_rank: Regular Contributor
timestamp: February 16, 2018, 10:57:52 am
content: Hi thereI'm in the progress of re-engineering a PCB that I want to build myself. Part of the circuit is an A/D converter that monitors two voltages of the supercap. This A/D has a 3.3V power supply that is connected via transistor Q3 to the chip.And there is my problem: It seems Q3 is installed backwards, current flows from source to drain...I tought that current in a N-MOSFET always flows in at drain and out at source. Is it possible to run it other way round?Anyway how should this circuit work? If the transistor is off, current will still flow because of the reverse-diode.It seems not to make any sense...And yes, I'm sure I drawn that right. I measured it multiple times, checked the transistor and compared it with identical parts on the same board. I would say this is a design error. What would you say...?PS: Check my Youtube channel "Play with Junk" - an interesting video about this project will follow. More vids already there :-)

author: bson
author_rank: Unknown
timestamp: February 16, 2018, 12:05:55 pm
content: The symbol is backwards... pin 2 is S and 3 is D, so the body diode is actually the other direction.http://www.onsemi.com/pub/Collateral/NTR4501N-D.PDFThe MOSFET permits power cycling the ADS1015 if it wedges, or to shut it off to save power (which may not work as intended).

author: tszaboo
author_rank: Super Contributor
timestamp: February 16, 2018, 12:17:00 pm
content: N channel FETs are bi-directional. If they are conducting, they conduct both directions.If they are off, one direction it is an open circuit, the other direction it is a diode.And yes, your symbol is backwards.

author: chris_leyson
author_rank: Super Contributor
timestamp: February 16, 2018, 12:23:26 pm
content: The symbol isn't backwards it's shown correctly, the N channel mosfet is being used like a synchronous rectifier. Initially the body drain diode conducts and after power up SYSTEM-2 is pulled high to short out the body drain diode. Check the rest of the board and trace SYSTEM-2 signal, it's probably pulled up to 5V or higher. If it was used to switch 3V3 to the ADS1015 it would make more sense to use a P channel fet. Measure the gate voltage, SYSTEM-2 signal. Using a mosfet in this way is also a common technique used for reverse polarity protection.

author: ArdWar
author_rank: Frequent Contributor
timestamp: February 16, 2018, 01:11:32 pm
content: The schematic is still confusing either way, cross-check with the actual circuit implementation and correct it immediately before it causes more confusionIf it's implemented like what the pictorial symbol depicts (current flow from S -> D) then most likely it's a reverse polarity protection.If it's implemented like what the pinout number depicts (surrent flow from 3 -> 2) then, uhh.. most likely it's a power switch, or less likely ramp-up control, or even (unlikely) surge protection.

author: David Hess
author_rank: Super Contributor
timestamp: February 16, 2018, 06:35:17 pm
content: Assuming that the MOSFET gate signal comes directly from the controller, they used the wrong MOSFET and it should have been a p-channel.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 16, 2018, 06:52:44 pm
content: Observations- not enough information- MOSFETS are bidirectional (as noted by another post). Drain/Source designations are determined by terminal conditions for a four-terminal MOSFET, but with body tied to one leg (as in this case), it is customary to refer to that terminal as the source.- Body diode will supply current to VDD with an offset 3.3V - 0.7V more or less- If System-2 is 3.3V, the voltage at VDD will be approximately 3.3V - VTH (a source follower)- If System-2 is 5V, the voltage at VDD will be about 3.3V (as another post implied)- I2C pullups to 3.3V which seems to indicate this is a 3.3V system, so the I can only guess that System-2 signal is also 3.3V but it is a guess.- Youtube is a big place. Would love to see your video but it is generally helpful to provide a link to said video

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2018, 06:53:33 pm
content: I've seen that from time to time, some idiot draws the schematic AND the symbol wrong -- but manages to get the pinout right.Tim

author: bson
author_rank: Unknown
timestamp: February 17, 2018, 05:27:01 pm
content: Quote from: chris_leyson on February 16, 2018, 12:23:26 pmNo, it's drawn incorrectly. Compare the schematic and the datasheet.

author: bson
author_rank: Unknown
timestamp: February 17, 2018, 05:39:09 pm
content: Quote from: David Hess on February 16, 2018, 06:35:17 pmWell, observe that pin 2 is actually S... which means when the gate is held high the source can't drop (much) below a logic high without becoming forward biased. (The body diode actually points away from S.) It's kinda screwy and sorta works (Vdd will be unreliable, not ideal for an ADC), but a P channel would have worked better.

author: chris_leyson
author_rank: Super Contributor
timestamp: February 17, 2018, 05:57:44 pm
content: Well spotted, I should haven checked.So it's not in there for reverse polarity protection but to switch VDD. It's still a little strange I would of expected a P-channel fet if everything is run from 3V3. Makes sense now if there are multiple I2C devices on the bus, switch VDD to enable each device in turn.

author: compet17
author_rank: Regular Contributor
timestamp: February 18, 2018, 08:26:16 pm
content: The video is not made yet. But I have many other videos that might be interesting :-)My interpretation of this circuit is that when the MOSFET is on 3.3V is connected to the A/D and when it is off only 2.7V will be supplied. But is that good for anything?

author: compet17
author_rank: Regular Contributor
timestamp: February 18, 2018, 08:30:15 pm
content: I hope you did not mean ME as an idiot :-)No, I'm sure that everything is drawn correctly. I measured the circuit and the MOSFET multiple times and checked the data sheet twice and triple. That's the reason why I'm asking here.But it is possible that the manufacturer of this board has made an error. But they did not notice because it works anyway.

author: compet17
author_rank: Regular Contributor
timestamp: February 22, 2018, 09:14:40 am
content: I did some checks on that circuit.... 3.3V is always on (drain and source). Gate is at 6.8V.There is also an EEPROM on the same I2C bus that has no switch attached. So switching individual bus components off makes no sense. The EEPROM is always on.I think they simply use the transistor as a diode.

author: David Hess
author_rank: Super Contributor
timestamp: February 22, 2018, 04:02:31 pm
content: Quote from: compet17 on February 22, 2018, 09:14:40 amIf the gate voltage is that high, then they got the source and drain swapped. An n-channel MOSFET will work in this application if the gate voltage swing is higher than the drain voltage. If the gate voltage swing is equal to the drain voltage, then a p-channel MOSFET is needed.

author: compet17
author_rank: Regular Contributor
timestamp: February 22, 2018, 06:56:49 pm
content: Yeah I think they simply designed that thing wrong. My next try will be to leave that transistor away and put a jumper in to supply 3.3V directly. I bet there is no difference in function...By the way, it's an Equallogic (Dell) circuit

author: compet17
author_rank: Regular Contributor
timestamp: February 27, 2018, 05:11:15 pm
content: OK.... thanks for all the answers and tipps but it is now the moment to answer the question myself.THIS MOSFET IS GOOD FOR NOTHING!!!How did I get to this conclusion? Well, I removed the transistor from a working PCB and replaced it with a jumper wire, connecting 3.3V directly to the A/D converter. Guess what... no difference. Works exactly as before. I can read and write the EEPROM which is on the same I2C bus and I also get the voltage readings from the A/D. The unit works and sees a healthy supercap board.

author: David Hess
author_rank: Super Contributor
timestamp: February 27, 2018, 06:57:39 pm
content: If they had used the MOSFET correctly, then it would allow externally powering down the ADC although this is questionable because the ADS1051 can be configured to power down between conversions anyway. Were they trying to save 5 microamps?