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CSD25484F4: MOSFET-P not switching Kindly advise

Naveen K1
Intellectual 870 points
Part Number: CSD25484F4
Other Parts Discussed in Thread: CSD17382F4

Hi,

We using P-MOSFET, Source connected to 24VDC source, Drain connected to Relay coil one end, Relay other connected to Ground. MOSFET Relay will be always ON(Drain=24V) for any of gate voltage VGS=5V or VGS=0V.

Please find attached the Schematic, MOSFET not switching for given Gate Voltage, it's always Source-Drain closed state only. Not able to turn off the relay/Load.



Kindly advise to solve this switching issue.

Thanks and regards,
Naveen K

  • 0
    TI__Mastermind 29645 points

    Hello Naveen,

    Thank you for your interest in TI FETs. Based on your schematic and circuit description, it looks like you're exceeding the absolute maximum VDS (-20V) and VGS (-12V) ratings for the CSD25484F4 P-channel MOSFET. When the photo-transistor is off, the gate is pulled up to 5V thru R28 while the source is connected to 24V and VGS = 5V - 24V = -19V. While this exceeds the abs max VGS rating, it also means the FET will never be turned off. The gate needs to be pulled up to the source in order to turn off the device. When the photo-transistor is on, the gate voltage is (5V - Vce(sat)) x (1k/11k) = (5V - 0.25V) x (1/11) = 0.43V and VGS = -23.57V. Again, this exceeds the abs max VGS rating of -12V. I believe the FET will be damaged under these conditions. Because the input voltage is 24V, this application requires a higher voltage P-channel FET. Also, in order to turn off the FET, the gate needs to be pulled up to the source. VGS needs to be limited (a resistor divider or zener diode might work) to prevent it from exceeding the abs max rating of the device. Unfortunately, TI P-channel devices only go to -20V VDS and -12V VGS abs max. TI cannot guarantee reliable operation of the FET if the absolute maximum ratings specified in the datasheet are exceeded. Let me know if I can be of any further assistance.

    Best Regards,

    John Wallace

    TI FET Applications

  • 0 in reply to John Wallace1
    Intellectual 870 points

    Hi John Wallace,

    Thank you for the information and Appreciate your quick response.

    Sorry for my mistake on the Schematic(24V named instead 19V), Source Voltage was 19V only from power Adapter, not 24V.

    Kindly advise,

    1. Given Schematic is correct?

    2. MOSFET is not switching  to OFF. To Switch off the MOSFET, 5V on the R28 will not works? Need higher Voltage? 

    3. If our Schematic is correct, what will be the voltage range require on R28 to switch the MOSFET? If we provide between  +11V to +19V through R28, MOSFET will Switch? 

    4. If we change to other MOSFET with same schematic, What will be the MOFET VGS we need to select. Kindly advise, what are the changes we have to do to work this?

    Thanks and regards,

    Naveen K

  • +1 in reply to Naveen K1
    TI__Mastermind 29645 points

    Hi Naveen,

    The VDS (-20V) rating of the CSD25484F4 should be OK with a 19V max input although there is not much margin for transients or any other variations.

    To answer your questions:

    1. No, the schematic is not correct. To turn off the FET, |VGS| < 0.7V. The gate must be pulled up to the source voltage thru R28.To turn on the FET the gate must be pulled lower than the source by at least 1.8V but not more than 12V.
    2. The 5V pull up is not enough as VGS = 5V - 19V = -14V and the FET will remain on. This also exceeds the abs max VGS rating of -12V specified in the datasheet and will damage the FET.
    3. R28 should be connected to the source of the FET which will turn the FET off when the photo-transistor is off. I recommend changing R29 to 10k which will form a resistor divider and limit VGS when the photo-transistor is on. VG = Vin - Vce(sat) x (10k/20k) = (19V - 0.25V) x 0.5 = 9.375V and VGS = 9.375V - 19V = -9.625V which should be OK.
    4. Making the changes detailed above should allow the CSD25484F4 to work in your application: connect R28 to the input voltage at the source and change R29 to 10k.

    Best Regards,

    John

  • 0 in reply to John Wallace1
    Intellectual 870 points

    Hi John,

    Thank you very much for your advice.

    We will connect R28 to 19V i.e Source.

    1. But R29 need to be change to 10K? Mean, Gate pin should not be 0V, right? If gate pin is 0V, then VGS=0V-19V=-19V, which exceeds abs Max rating of VGS=-12V, am correct?

    Thanks and regards,

    Naveen K 

  • 0 in reply to Naveen K1
    TI__Mastermind 29645 points

    Hi Naveen,

    When the photo-transistor is off, the voltage at the gate is 19V as is the voltage at the source and VGS = 0V. When the photo-transistor is on, the voltage at the gate = (Vin - Vce(sat)) x (R29/(R28+R29)) = (19V - 0.25V) x 10k/20k = 9.375V and VGS = 9.375V - 19V = -9.625V which is within the abs max rating for VGS of the FET. Does this make sense? Let me know if you have additional questions.

    Thanks,

    John

  • 0 in reply to John Wallace1
    Intellectual 870 points

    Hi John,

    Thank you for the information.

    We need your advise to understand more on following options can be used or not.

    1. I hope, for existing schematic can not replace MOSFET with Transistor, right?

    2. We have another application with 12V source(Coil Voltage), Can we drive 12V by 5V on R28 as in below schematic will woks?

    3. We have another application with 12V source(Coil Voltage), Can we drive 12V by 12V on R28 as in below schematic will woks?

    4. Can we replace P-Channel MOSFET with N-Cannel as in High side ?

    Thanks and regards,

    Naveen K

  • 0 in reply to Naveen K1
    TI__Mastermind 29645 points

    Hi Naveen,

    1. A PNP bipolar transistor could work but similar to a PFET, the base (gate) must be pulled up to the same voltage as the emitter (source) to turn off the device. R28 must be pulled up to the same voltage as the emitter (source).
    2. No, R28 must be pulled up to the same voltage as the source.
    3. Yes, this will work. I would increase the value of R29 (2k to 5k) to reduce VGS when the FET is on to provide some margin on VGS abs max rating.
    4. No, the gate of the NFET must be pulled higher than the input voltage to turn on. The CSD17382F4 requires minimum VGS = 1.8V above Vin to turn on in a high side switch configuration. If you can configure this as a low side switch, then a N-channel FET would work. Connect the relay and diode from Vin to the drain and the source to GND.

    Let me know if you have any additional questions.

    Thanks,

    John

  • 0 in reply to John Wallace1
    Intellectual 870 points

    Hi John,

    Thank you for your advise.

    Last Query, Kindly advise.

    2. You told to connect Gate(R28) to source(then, VGS=12-12=0V) instead Source=12V and Gate(R28)=5V. Please correct me if am wrong. In the datasheet table, VGS range mentioned in RDSON is -1.8V to -8V. So, if "Source=12V & Gate=5V", VGS=5-12V=-7V, -7V is within the mentioned VGS range and MOSFET will be Turn OFF, right? Kindly advise why you are suggesting to connect Source voltage to Gate Voltage?

    2.b. What will be the VGS ranges can be used to Turn OFF the MOSFET (As per the datasheet/Table)?

    3. When Phototransistor is ON, If (R28)=12V and Source=12V. R29=5K, Gate=3.92V & VGS=3.92V-12V=-8.08V, and it cause, MOSFET OFF(Always) only right? So, R29 should be <4K right?

    3.b. Mean, We should not apply 0V(Ground) to Gate pin to OFF the MOSFET, right? it causes, VGS=0V-12V=-12V, it will exceeds VGS abs max Voltage of ±12V if source=>12V, right?

    * We are connecting Opto-Isolator/Phototransistor Anod/Input to CC3220MODSF12MODR Wi-Fi Module through 1K resistor. We are driving MOSFET source of 19V. If any damages on MOSFET/Phototransistor causing issues/failure to Wi-Fi module also?  

    Thanks and regards,

    Naveen K

  • 0 in reply to Naveen K1
    TI__Mastermind 29645 points

    Hi Naveen,

    For the CSD25484F4, P-channel FET, it is OFF (not conducting) when VGS ≅ 0V and it is ON (conducting) when -12V ≤ VGS ≤ -1.8V which is the minimum value where on resistance is specified in the datasheet.

    2. VGS = -7V, FET is ON

    2b. VGS ≅ 0V, FET is OFF

    3. VGS = -8.08V, FET is ON

    3b. Correct. The table from the datasheet shows typical on resistance at different values of VGS where the FET is ON

    The opto-isolator isolates the WiFi module from 19V. Damage to the FET will not affect the module.

    Best Regards,

    John

  • 0 in reply to John Wallace1
    Intellectual 870 points

    Hi John,

    Thank you for your time and advise. Well appreciated your support and it's very helpful.

    Thanks and regards,

    Naveen K