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UCC27201: Issue with turning on high side MOSFET

Anmol Jaiswal
Prodigy 210 points
Part Number: UCC27201
Other Parts Discussed in Thread: DRV8300

Issue with turning on high side MOSFET using UCC27201 Before going to the original operation, I was trying to verify that am I able to turn on any particular MOSFET one by one, I was able to turn on low side mosfets when ever turned it ON but when I am trying to turn on any high side mosfet I was unable to turn it ON I am getting 6V on the gate of high side mosfet without switching it on, I thought MOSFET should be ON so I checked resistance between drain and source and found very high resistance(in 30s of Kohm), so I found it to be off then I checked voltage on the source of the MOSFET, it was also around 6V, Now VGS was almost 0V so the mosfet should be off as it was.

1. How am I getting 6V on gate and source of the high side mosfet.

2. Why am I not able to turn on only any particular high side MOSFET.

  • 0
    TI__Intellectual 2100 points

    Hi Anmol,

    Thank you for reaching out! For the high side MOSFET to turn on, you must go through a full switching cycle between the low side and high side MOSFETs so that the boostrap circuitry can properly charge as well as discharge. This means you will not be able to verify operation of the high side separately from the low side.

    I've included a link to an article discussing Bootstrap Circuitry Selection:  Link

    Section 2 of this article discusses operation of the bootstrap circuit and includes images to help illustrate the bootstrap charging and discharging path.

    I hope this helps answer your questions! Please feel free to reach out if you are still experiencing trouble verifying operation of the high side MOSFET.

    Best,

    Alex Weaver

  • 0 in reply to Alex Weaver
    Prodigy 210 points

    Hi Alex,

    How am i getting 6V on the on the phase with nothing connected on Phase.

  • 0 in reply to Anmol Jaiswal
    TI__Intellectual 2100 points

    Hi Anmol,

    Can you provide a schematic of the gate driver + MOSFET circuitry? If this isn't available, can you provide a drawing of the system? This will help the team to better analyze the root cause here.

  • 0 in reply to Alex Weaver
    Prodigy 210 points

    Hi Alex,

    It is just a half bridge circuit using MOSFET directly connected to Half bridge driver UCC27201

    Input to the driver is 12V, HI and LI pins are connected to a controller and HO, LO pins are connected to gate of the MOSFETs through a series resistor of 10Ohm. A bootstrap capacitor is connected in between HB and HS pins and HS pin is connected to the phase.

    I checked the voltages on HO and LO pins, They are at 0V.

    I designed a different circuit i.e. 3 Phases circuit using DRV8300, Because that circuit was working fine so for the reference I checked the phase voltages, there also 2.5 to 3V I could see.

    Everything is working fine in the circuit but the question is how I am getting voltage on the Phases and almost the same voltage on the gate of the MOSFET 

  • 0 in reply to Anmol Jaiswal
    TI__Genius 11770 points

    Hi Anmol,

    Alex Weaver is out today, so I will reply in the mean-time.

    Is there anything on VBUS? In this condition, the HS/PHASE node is floating, so small leakage currents can move the voltage. For example, the two FETs forming the half-bridge will have a certain leakage current, and therefore form a voltage divider (IE if VBUS is 12V, I would expect about 6V on HS). There is also some small leakage from the driver itself, and the bootstrap path can also cause the node to float.

    Similarly, there is a path through the body diode of the high-side's pulldown FET that will bring HO to roughly match HS if HS is higher. This is probably why you see the same voltage on HO. In any case, the fact that HO and HS are floating to the same voltage, means the Vgs on the FET is close to zero, and therefore will not turn on. 

    Because of the fact that the Vgs balances to zero, the floating voltage is usually harmless. If you would like to get rid of it, a pulldown (10-100K) from HS to GND can get rid of it. 

    Please let me know if you have any further questions.

    Thanks,

    Alex M.

  • 0 in reply to Alexander Mazany
    Prodigy 210 points

    Hi Alexander,

    Yeah, There is voltage on VBUS but it is 50 Volts so according to voltage divider it should be 25V on HS which is not the case.

    i didn't understand second paragraph that how low side FET will bring HO to roughly match HS if HS is higher.

    Bootstrap capacitor charging path is through 12V as 12V is applied to power up the driver(UCC27201) so ideally bootstrap capacitor should charge through 25V HS pin to gate of high side MOSFET.

    If i consider capacitor divider that is bootstrap capacitor(100nF) and Cgs(9.1nF) so Voltage at gate should be atleast 23V.

    Please clarify if i am wrong

    Thanks.

    Anmol Jaiswal

  • 0 in reply to Anmol Jaiswal
    TI__Genius 11770 points

    Hello Anmol,

    Here is what I mean, sorry for the rough drawing:

    Inside the driver, there is a diode path from the pulldown FET's body diode. If HS is higher than HO, it should conduct until the difference is too small for the diode. 

    In the proper operation, whenever the driver turns on the low-side FET, the HS node will be pulled to ground. When that happens, the HB-HS capacitor charges from VDD (12V) through the diode. The bootstrap capacitor will eventually charge to roughly VDD (12V), when the low-side turns off and the high-side turns on, HS will rise to 50V in this case, but HB-HS will still be 12V and thus the HB-GND voltage would be 62V. The HS voltage would be 50V, and the VGS of the high side FET would therefore be 12V.

    This circuit requires that the low-side FET be switched occasionally to recharge the bootstrap capacitor. This is what is causing you to see strange behavior in the DC test. The exact voltage you see on HS and HO is probably coming from various leakage current in the driver. For example, there is a current draw from the HB pin to power various circuitry on the high-side like the level shifter. I don't think you can use the reasoning about the voltage divider, as the capacitors are not really in series (when HO goes high they are in parallel). 

    There are ways to enable 100% duty cycle or DC operation, but you cannot do it with the bootstrap circuit as is. 

    Thanks,

    Alex M.