P-channel across schottky, body diode still conducts. - Page 1
Forum section: electronics / beginners

author: step_s
author_rank: Regular Contributor
timestamp: September 18, 2016, 08:39:57 pm
content: Hi EEV'ers.Just wanted to test out something simple: I'm using a premade boost converter board from ebay, using the SX1308. I then added a P-channel across the output schottky diode (the source to the output, and drain to the SW note, so the body diode is same direction as the schottky).My thought was that when the gate is kept at the source pin (output of boost converter), the MOSFET is off, and the schottky diode would keep the body diode from starting to conduct, but this doesn't seem to be the case.The body diode seems to cause problems, and the MOSFET gets hot when the converter is running.Am I missing something here?Hope you can help

author: danadak
author_rank: Super Contributor
timestamp: September 19, 2016, 01:20:23 am
content: Consider posting a schematic, use the snipping tool to capture itin windows. So we can see the drive interface, etc..Regards, Dana.

author: Marco
author_rank: Super Contributor
timestamp: September 19, 2016, 02:08:56 am
content: Add a shunt and measure the reverse current through the MOSFET when the SX1308 switch turns on, I assume it's the reverse recovery losses heating the MOSFET up rather than any conduction losses when the SX1308 switch is off.

author: step_s
author_rank: Regular Contributor
timestamp: September 19, 2016, 09:10:20 am
content: @DanaThought it was a pretty simple setup.http://www.datasheetspdf.com/PDF/SX1308/921054/1the SX1308 shows the normal configuration. All that i added was a P-channel across the D1 schottky diode.@MarcoYe, I figured as much, since this is without load. But that would mean that the body diode still conducts, and the losses are from the diodes reverse recovery. Is it possible to prevent the body diode from conducting in the first place?Thanks for the answers

author: David Hess
author_rank: Super Contributor
timestamp: September 20, 2016, 12:24:53 pm
content: Quote from: step_s on September 19, 2016, 09:10:20 amIn applications where it is imperative that the body diode not conduct because of reverse recovery, another diode can be placed in series with the MOSFET. With a schottky diode as a shunt though, this is not usually a problem.So you have the gate tied to the source at the output?

author: step_s
author_rank: Regular Contributor
timestamp: September 20, 2016, 03:22:16 pm
content: So the diode in series would be pointing the same direction? Before or after the MOSFET?Ye, the gate is clamped to the source pin of the MOSFET, which is connected to the output of the boost converter.I just read somewhere that the schottky diode shunt would prevent the body diode from conducting, due to it's lower voltage drop, and faster reaction time. This is why I'm wondering if it's not possible to prevent the body diode from conducting in the first place.

author: David Hess
author_rank: Super Contributor
timestamp: September 20, 2016, 05:38:57 pm
content: Quote from: step_s on September 20, 2016, 03:22:16 pmIt could be in series with either the source or drain but usually the drain. Then the original external schottky or other diode is placed in parallel with the series combination. This configuration is used where the off voltage is high enough that a schottky diode cannot be used.QuoteThat is certainly what is suppose to happen; the lower forward voltage drop of the schottky diode prevents the body diode from conducting. I do not know what is going wrong in your case; maybe the MOSFET was damaged by ESD? It should not be conducting with the gate shorted to the source.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2016, 06:56:40 pm
content: A PMOS has three terminals. I don't get how you "just add one across it". Schematic please.And, how have you determined that the body diode conducts, if it has a diode across it?Tim

author: step_s
author_rank: Regular Contributor
timestamp: September 20, 2016, 07:46:35 pm
content: Okay, made a simplified schematic. Some things are left out, like caps etc.The gate of the MOSFET is in test circuit, clamped to the +5V output, which is also the source of the MOSFET.@DavidIt might have taken some kind of damage in the process. I will try to change it and see if it has any effect.@TimThe MOSFETs active component should be completely off when gate is as high as source, but it gets hot, even when the converter is under no load. This should only mean one thing, that the body diode is conducting, but too slow to turn off in reverse bias, making current flow backwards in the system. Might be something else?

author: Marco
author_rank: Super Contributor
timestamp: September 20, 2016, 08:00:39 pm
content: You are tying the gate to the source with a direct solder bridge right? Not by just putting it at the same voltage with some long wire to the power supply?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2016, 08:36:34 pm
content: What's "gate control" from? A driver?Tim

author: step_s
author_rank: Regular Contributor
timestamp: September 21, 2016, 12:01:40 pm
content: @MarcoYes, it's connected directly with a alligator clip wire. Nothing external@TimThe gate control is meant to be controlled by a PIC that its connected directly to the output 5V. It would be able to pull the gate control to +5V.In this setup though, I have connected it directly as I wrote.Just tried with a new P-channel, and no successGonna try and put an N-mos on and see what happens if the gate is kept at 0V.

author: Delta
author_rank: Super Contributor
timestamp: September 21, 2016, 12:19:37 pm
content: What MOSFET is it? Is 5v enough to turn it fully on?

author: step_s
author_rank: Regular Contributor
timestamp: September 21, 2016, 06:29:34 pm
content: @DeltaIt's an SI2301, so there should be no problem. Also, it doesn't matter, since it's not about turning it on, but keeping it off

author: Marco
author_rank: Super Contributor
timestamp: September 21, 2016, 07:03:51 pm
content: Quote from: step_s on September 21, 2016, 12:01:40 pmThat's still a reasonably amount of inductance and it has a really low threshold voltage. The drain transient might be turning it on for a moment.

author: step_s
author_rank: Regular Contributor
timestamp: September 21, 2016, 08:23:32 pm
content: @MarcoIt seems you are rightThe only way it will be off, is by a direct connection soldered on the gate to the source. Even a short path will have too much inductance it seems.This kind of negates the idea of having it be controlled by a microcontroller, since the output and the path will without a doubt have some inductance, and make it impossible to keep completely closed

author: David Hess
author_rank: Super Contributor
timestamp: September 21, 2016, 10:19:19 pm
content: Quote from: step_s on September 21, 2016, 08:23:32 pmWhat exactly is the purpose of the MOSFET? If it is to bypass the forward voltage drop of the diode when the regulator is shut down, then I think your circuit will just require a slight modification.Tie the gate to the source close to the MOSFET using a resistor *and* a parallel bypass capacitor. Now the gate will be shorted to the source at high frequencies. When you want to turn the MOSFET on when the regulator is not operating, pull the gate to ground. The resistor can be sized for the maximum acceptable current when the MOSFET is on.

author: step_s
author_rank: Regular Contributor
timestamp: September 26, 2016, 12:00:21 am
content: Hi guys, sorry for the late answer.@DavidThe point of this setup was actually to try something a bit silly.Wanted to test out if I could make a 5V battery booster, with the ability to charge from the input. It's a longer storyThe Pmos is switcing in a buck-converter setup, and charges the battery, while the SX1308 takes care of the boosting.The thing is, that it actually works, but after testing the different setups, I found that the impendance of the PIC is too high to pull up the gate. A bypass cap works wonders, but makes it impossible to switch the gate at any useable frequency. The same with a low value pull-up resistor.The only solution would be to use a mosfet-driver or a mosfet with a low impedance, to pull up the gate, when the buck setup was off, but since all I'm saving by doing all this, is a second inductor for the buck-setup, it would be stupid to add more active components.I think the simple answer is to find a small package inductor

author: nuno
author_rank: Frequent Contributor
timestamp: September 26, 2016, 12:22:12 am
content: I've seen a power system or 2 where the FETs are "far" (lets say 10cm (4inch) or so) from their drivers, and the solution to reduce the high path inductance is to use twisted wire (2 wires, soldered on gate and source, go twisted to connect at the driver's output and GND/Vcc).

author: Towz
author_rank: Contributor
timestamp: September 26, 2016, 02:31:22 am
content: In this configuration the body diode will always conduct. From what I read, you seem to want to charge the battery from the 5V rail when powered externally. In that case, you should add a diode from the mosfet's drain (anode) towards the current node (cathode)(the schottkey diode anode+inductor)You may also want to add a 10k-ish resistor between the uC output and the mosfet's gate.

author: David Hess
author_rank: Super Contributor
timestamp: September 26, 2016, 03:05:52 am
content: It is not difficult to implement a bidirectional switching regulator which can transfer power in either direction. I think Linear Technology has some designed to do exactly this.

author: step_s
author_rank: Regular Contributor
timestamp: September 26, 2016, 11:23:14 am
content: @nunoEven when the PIC is connected directly, very close to the gate to prevent inductance, it helps, but still not good enough.Also, wouldn't it be hard to twists traces on a PCB?Could ofcourse use a choke.@TowzI left out the buck-setup diode from my drawing, but it's there. No worriesAlready put a 2K resistor between output and gate, and it keeps it in check, but doesn't due much when the SX1308 starts switching.@David HessIt would be easier if it was a synchronous setup, but the 8bit PIC can't drive it fast enough, to have a decent resolution at a higher frenquency. PIC12F1571 max 32MHz. Would require a higher value inductor if used to boost, and that defeats the purpose.I did some testing, and it seems that an N channel is better at this job, since the gate can be kept BELOW source, all the way down to 0V, preventing noise from opening it in the same way. Problem is, that this would require some kind of high side driver, and again, defeats the purpose of the simple design :/

author: nuno
author_rank: Frequent Contributor
timestamp: September 29, 2016, 02:10:02 pm
content: Quote from: step_s on September 26, 2016, 11:23:14 amRun the 2 tracks very close on the same pcb side.A choque? Isn't that, an inductor?