Driving a P-channel MOSFET - Page 1
Forum section: electronics / projects

author: mattko
author_rank: Contributor
timestamp: December 09, 2017, 06:58:18 pm
content: Hello!In the circuit I have a three phase MOSFET bridge to power a BLDC motor. (Not included in the picture). N-channel MOSFETs are not a problem to drive directly from the 3.3 V processor, as it delivers at least 20 mA of output GPIO current, which turns on and off the MOSFET fast enough and it opens the MOSFET enough as it is a logic-level MOSFET.I have a problem at closing the P-channel MOSFETs. I have made 3 circuits to discuss advantages and disadvantages of each one.Adding a MOSFET driver to the circuit would be a solution, but a more expensive oneInput comes from a processor pin.Output is one of the three phases of the BLDC motor.Option 1Bipolar Push-pull option. Here the P-channel MOSFET doesn't completely turn off because of Vce voltage of the NPN transistor (around 0.8 V). This option would probably be OK, if MOSFETS were used instead of bipolar transistors (T1 and T2) as MOSFETS would have a much smaller Vds voltage.Option 2The NPN transistor T4 can close the P-MOSFET completly. But! In my understanding it can't provide more than 5V / 1kOhm = 5 mA of gate current, which closes the P-MOSFET too slowly for my application. The MOSFETS would get too hot.If we decrease resistance of R3, then we get higher gate current, but losses on the resistor R3 get quite big ( P = U^2 / R ), which again is undesirable in this application.Option 3It is similar than option 2, just another resistor is added (R5), to limit the gate current when P-MOSFET it opening.______My doubts are:1) At option 3, would current go through R4 to close the P-MOSFET and through R5 to open the P-MOSFET?2) Is there another good solution for this problem?I hope I was clear enoughYour input is highly valued!PN MOSFET datasheet:https://www.diodes.com/assets/Datasheets/DMC1030UFDBQ.pdfSome MOSFET theory:http://www.st.com/content/ccc/resource/technical/document/application_note/68/cd/c6/ab/ef/17/41/06/CD00003900.pdf/files/CD00003900.pdf/jcr:content/translations/en.CD00003900.pdf

author: slugrustle
author_rank: Frequent Contributor
timestamp: December 09, 2017, 08:50:28 pm
content: Hi mattko,In option 2, since the gate of the P mosfet forms a capacitor relative to the source, T4 will sink extra current to lower the voltage on the Gate-Source capacitor when you initially turn on T4. This is a good thing, and Option 3 gets rid of it. The important part about sizing R3 in Option 2 is the turn-off time for the P mosfet, which depends on its gate capacitance.Option 1 is a little confusing. It seems that T2, the PNP transistor, is never properly biased. The emitter should be at a higher voltage than the base in order to turn it on.I've attached a short drawing to give you even more options! I'm not sure if it's the greatest idea though, and it requires an extra I/O pin.

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 10, 2017, 04:39:35 am
content: option2 = pfet will quickly turned on but slowly turned off due to 1Kohm resistor. this reduce efficiency and hence hotter fet in hi-amp application. not a problem in low-amp application, more preferable due to circuit simplicity.option3 = pfet will both equally slowly turned on and off. you will have problem turning on a pfet whose Vgth threshold is greater than 2.5V. even hotter fet in hi-amp application not a problem in low-amp application, more preferable due to circuit simplicity.option1 = is push pull emitter follower and is much preferable as pfet (or nfet switch) in hi-amp application since it will turned on and off fet quickly, but your circuit T2 is upside down. T2's emitter should meet T1's emitter as in the following diagram...R1 is to bias fet gate voltage so it will not exceed maximum Vgs. in your circuit is using 5V Vds so R1 may not be needed as most fet can deal with 5V Vgs.

author: Omar Mekkawy
author_rank: Contributor
timestamp: December 10, 2017, 10:05:22 am
content: You can also use a diode in parallel with the gate resistor in order to make the discharging faster. I have attached a picture for an example.

author: RES
author_rank: Regular Contributor
timestamp: December 10, 2017, 01:38:33 pm
content: If you need very fast rise and fall times of high speed PWM signals, put a high side current mirror in front of push-pull driver (+Rg, depends on Ciss mosfet), and in front of the current mirror one NPN driver transistor (100pF over base resistor)

author: Wimberleytech
author_rank: Super Contributor
timestamp: December 11, 2017, 03:48:19 pm
content: How about this? The new resistor needs to be large enough so that T2 can overpower the new mosfet/resistor path.

author: mattko
author_rank: Contributor
timestamp: December 16, 2017, 12:17:42 pm
content: Thank you for your inputs! Many good ideas, also for future design improvements.At the end, I used option 2 with R3 being only 100 Ohms. Which makes for fast switching, but high losess over R3 (but currently losses are not a priority).As I resoved this issue, a new one emerged, that I will post it to the forum as well in a few minutes (it doesn't have much with this topic IMO).Thank you!