Turn on a microcontroller using a high-side Mosfet switch











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I am using a P-Channel Mosfet to control the power of an STM32F103VETE ARM microcontroller (MCU1). The source of the Mosfet is connected to 3.3 V and the drain goes to the MCU1 Vdd pins. The gate of the Mosfet is controlled by another microcontroller (STM32F030RET (MCU2)) which is directly connected to 3.3 V net.



Now the problem is that I can't turn off MCU1:




  • When I put a Logic 1 on the gate of the Mosfet, that makes its Vgs = 0 but I still get 2.4 V at the drain of the Mosfet which is enough to turn on MCU1.

  • When I put a Logic 0 on the gate of the Mosfet, its Vgs = -3.3 V, it turns on properly and I can read 3.3 V at the drain.


Can anyone please help me to solve this? What can cause such a problem?



Here is the schematic of the circuit I have used:



The "power_on" is the signal from MCU2. The Mosfet is STS3DPF20V



"POWER_ON" is the signal from MCU2. The Mosfet is STS3DPF20V.










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    I am using a P-Channel Mosfet to control the power of an STM32F103VETE ARM microcontroller (MCU1). The source of the Mosfet is connected to 3.3 V and the drain goes to the MCU1 Vdd pins. The gate of the Mosfet is controlled by another microcontroller (STM32F030RET (MCU2)) which is directly connected to 3.3 V net.



    Now the problem is that I can't turn off MCU1:




    • When I put a Logic 1 on the gate of the Mosfet, that makes its Vgs = 0 but I still get 2.4 V at the drain of the Mosfet which is enough to turn on MCU1.

    • When I put a Logic 0 on the gate of the Mosfet, its Vgs = -3.3 V, it turns on properly and I can read 3.3 V at the drain.


    Can anyone please help me to solve this? What can cause such a problem?



    Here is the schematic of the circuit I have used:



    The "power_on" is the signal from MCU2. The Mosfet is STS3DPF20V



    "POWER_ON" is the signal from MCU2. The Mosfet is STS3DPF20V.










    share|improve this question









    New contributor




    MoHaMaD InSoMnIaC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I am using a P-Channel Mosfet to control the power of an STM32F103VETE ARM microcontroller (MCU1). The source of the Mosfet is connected to 3.3 V and the drain goes to the MCU1 Vdd pins. The gate of the Mosfet is controlled by another microcontroller (STM32F030RET (MCU2)) which is directly connected to 3.3 V net.



      Now the problem is that I can't turn off MCU1:




      • When I put a Logic 1 on the gate of the Mosfet, that makes its Vgs = 0 but I still get 2.4 V at the drain of the Mosfet which is enough to turn on MCU1.

      • When I put a Logic 0 on the gate of the Mosfet, its Vgs = -3.3 V, it turns on properly and I can read 3.3 V at the drain.


      Can anyone please help me to solve this? What can cause such a problem?



      Here is the schematic of the circuit I have used:



      The "power_on" is the signal from MCU2. The Mosfet is STS3DPF20V



      "POWER_ON" is the signal from MCU2. The Mosfet is STS3DPF20V.










      share|improve this question









      New contributor




      MoHaMaD InSoMnIaC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I am using a P-Channel Mosfet to control the power of an STM32F103VETE ARM microcontroller (MCU1). The source of the Mosfet is connected to 3.3 V and the drain goes to the MCU1 Vdd pins. The gate of the Mosfet is controlled by another microcontroller (STM32F030RET (MCU2)) which is directly connected to 3.3 V net.



      Now the problem is that I can't turn off MCU1:




      • When I put a Logic 1 on the gate of the Mosfet, that makes its Vgs = 0 but I still get 2.4 V at the drain of the Mosfet which is enough to turn on MCU1.

      • When I put a Logic 0 on the gate of the Mosfet, its Vgs = -3.3 V, it turns on properly and I can read 3.3 V at the drain.


      Can anyone please help me to solve this? What can cause such a problem?



      Here is the schematic of the circuit I have used:



      The "power_on" is the signal from MCU2. The Mosfet is STS3DPF20V



      "POWER_ON" is the signal from MCU2. The Mosfet is STS3DPF20V.







      microcontroller high-side stm32f1






      share|improve this question









      New contributor




      MoHaMaD InSoMnIaC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




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      share|improve this question




      share|improve this question








      edited 1 hour ago









      SamGibson

      10.8k41537




      10.8k41537






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      asked 5 hours ago









      MoHaMaD InSoMnIaC

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      111




      New contributor




      MoHaMaD InSoMnIaC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      MoHaMaD InSoMnIaC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          1 Answer
          1






          active

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          up vote
          4
          down vote













          You have the MOSFET connected correctly.



          Most likely something else is driving a pin on the MCU high, which is partially powering it through the protection diodes (hence the 0.7V difference). This is not a good situation and can damage the chip.



          You have to make sure that all inputs are low before removing power from the MCU, and similarly wait for the Vdd to rise before driving any one of them high.



          This can be bit messsy, and often it’s better to just put the MCU in the lowest power sleep mode and keep power on it.



          Note that your switch only opens the supply, and it may take some time for Vdd to fall if there is a lot of capacitance on the switched Vdd. A brief interruption my not reset the MCU, for example.






          share|improve this answer



















          • 1




            Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
            – MoHaMaD InSoMnIaC
            4 hours ago






          • 1




            Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
            – Spehro Pefhany
            4 hours ago








          • 1




            @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
            – Andy aka
            2 hours ago








          • 1




            @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
            – MoHaMaD InSoMnIaC
            2 hours ago








          • 1




            The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
            – Spehro Pefhany
            2 hours ago











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          up vote
          4
          down vote













          You have the MOSFET connected correctly.



          Most likely something else is driving a pin on the MCU high, which is partially powering it through the protection diodes (hence the 0.7V difference). This is not a good situation and can damage the chip.



          You have to make sure that all inputs are low before removing power from the MCU, and similarly wait for the Vdd to rise before driving any one of them high.



          This can be bit messsy, and often it’s better to just put the MCU in the lowest power sleep mode and keep power on it.



          Note that your switch only opens the supply, and it may take some time for Vdd to fall if there is a lot of capacitance on the switched Vdd. A brief interruption my not reset the MCU, for example.






          share|improve this answer



















          • 1




            Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
            – MoHaMaD InSoMnIaC
            4 hours ago






          • 1




            Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
            – Spehro Pefhany
            4 hours ago








          • 1




            @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
            – Andy aka
            2 hours ago








          • 1




            @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
            – MoHaMaD InSoMnIaC
            2 hours ago








          • 1




            The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
            – Spehro Pefhany
            2 hours ago















          up vote
          4
          down vote













          You have the MOSFET connected correctly.



          Most likely something else is driving a pin on the MCU high, which is partially powering it through the protection diodes (hence the 0.7V difference). This is not a good situation and can damage the chip.



          You have to make sure that all inputs are low before removing power from the MCU, and similarly wait for the Vdd to rise before driving any one of them high.



          This can be bit messsy, and often it’s better to just put the MCU in the lowest power sleep mode and keep power on it.



          Note that your switch only opens the supply, and it may take some time for Vdd to fall if there is a lot of capacitance on the switched Vdd. A brief interruption my not reset the MCU, for example.






          share|improve this answer



















          • 1




            Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
            – MoHaMaD InSoMnIaC
            4 hours ago






          • 1




            Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
            – Spehro Pefhany
            4 hours ago








          • 1




            @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
            – Andy aka
            2 hours ago








          • 1




            @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
            – MoHaMaD InSoMnIaC
            2 hours ago








          • 1




            The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
            – Spehro Pefhany
            2 hours ago













          up vote
          4
          down vote










          up vote
          4
          down vote









          You have the MOSFET connected correctly.



          Most likely something else is driving a pin on the MCU high, which is partially powering it through the protection diodes (hence the 0.7V difference). This is not a good situation and can damage the chip.



          You have to make sure that all inputs are low before removing power from the MCU, and similarly wait for the Vdd to rise before driving any one of them high.



          This can be bit messsy, and often it’s better to just put the MCU in the lowest power sleep mode and keep power on it.



          Note that your switch only opens the supply, and it may take some time for Vdd to fall if there is a lot of capacitance on the switched Vdd. A brief interruption my not reset the MCU, for example.






          share|improve this answer














          You have the MOSFET connected correctly.



          Most likely something else is driving a pin on the MCU high, which is partially powering it through the protection diodes (hence the 0.7V difference). This is not a good situation and can damage the chip.



          You have to make sure that all inputs are low before removing power from the MCU, and similarly wait for the Vdd to rise before driving any one of them high.



          This can be bit messsy, and often it’s better to just put the MCU in the lowest power sleep mode and keep power on it.



          Note that your switch only opens the supply, and it may take some time for Vdd to fall if there is a lot of capacitance on the switched Vdd. A brief interruption my not reset the MCU, for example.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Spehro Pefhany

          201k4146401




          201k4146401








          • 1




            Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
            – MoHaMaD InSoMnIaC
            4 hours ago






          • 1




            Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
            – Spehro Pefhany
            4 hours ago








          • 1




            @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
            – Andy aka
            2 hours ago








          • 1




            @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
            – MoHaMaD InSoMnIaC
            2 hours ago








          • 1




            The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
            – Spehro Pefhany
            2 hours ago














          • 1




            Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
            – MoHaMaD InSoMnIaC
            4 hours ago






          • 1




            Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
            – Spehro Pefhany
            4 hours ago








          • 1




            @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
            – Andy aka
            2 hours ago








          • 1




            @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
            – MoHaMaD InSoMnIaC
            2 hours ago








          • 1




            The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
            – Spehro Pefhany
            2 hours ago








          1




          1




          Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
          – MoHaMaD InSoMnIaC
          4 hours ago




          Thank you very much for your response. I have several pins of the mcu1 pulled up to 3.3v (not to the mcu1 vdd itself) with a 10k resistor. Can it be the problem?
          – MoHaMaD InSoMnIaC
          4 hours ago




          1




          1




          Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
          – Spehro Pefhany
          4 hours ago






          Yes, that could do it. It's unlikely to damage the MCU with 10K, but it will draw power and (perhaps) prevent it from resetting, depending on the brownout reset (BOR) circuit and its tolerances/settings.
          – Spehro Pefhany
          4 hours ago






          1




          1




          @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
          – Andy aka
          2 hours ago






          @MoHaMaDInSoMnIaC Why don't you have those 10k pull-ups on the switched side? Also the MOSFET leakage current could be contributing so having a push-pull load switch would clear that up (or maybe just a pull-down resistor AND if that pull-down resistor is too low in value for normal operation and gives too much current draw then use the switched MCU to switch it out of circuit when it begins to operate).
          – Andy aka
          2 hours ago






          1




          1




          @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
          – MoHaMaD InSoMnIaC
          2 hours ago






          @Andyaka well I didn't assume that it would make a problem since the absolute maximum rating for the stm32f103 gpio is vdd-0.3 and vdd+4.0 volts. About the leakage current I checked the datasheet of the Mosfet and it's value is very low and needs a resistor in order of giga ohms to produce such a voltage. The pull down resistor however seems to work in a way that it brings down the off time voltage to about 1.5 volts and stm32 will turn off but is it OK to have 1.5v on vdd when the MCU is off?
          – MoHaMaD InSoMnIaC
          2 hours ago






          1




          1




          The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
          – Spehro Pefhany
          2 hours ago




          The exact circuits used in protection networks are seldom well described in the datasheets, unfortunately. We something similar in the analog world where maximum differential voltage of an op-amp is something reasonable such as 30V however the inputs draw a lot current for differential voltage of more than a diode drop.
          – Spehro Pefhany
          2 hours ago










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          MoHaMaD InSoMnIaC is a new contributor. Be nice, and check out our Code of Conduct.
















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