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Help to review a schematic with MSP430

C
Expert 2065 points

Hello

I have finished a schematic and I need a second pair of eyes to review the schematic done.

What we are trying to achieve here is to have a MSP430 to pilot a relay and two leds after a sensor piloted via UART send us a validation signale. 

We also have 4 keys, 3 on the pcb and one on wire. 

We have created a major connector to connect the sensor, one key and the two leds.

The device is running using 9V commercial squared battery  (non rechargeable). 

The sensor is powered in 3.3V, the relay on 3V, and the MCU on 3.3V. We have used BC817 NPN transistor to pilot most of the output.

Does anybody see any problem ont this schematic ?

Thanks for your second eye.

SCHEMATIC.pdf

  • 0
    Guru 74080 points
    I just had a quick look at it - Q203 will short your supply.
  • 0
    Guru 283480 points
    Using NPNs for Q200 and Q201 is unusual, because in the common-collector configuration, the collector/emitter voltage drop cannot become smaller than that over base/emitter.

    Q203 tries to short power and ground. This does not make sense. If you want to switch power to some external device, you must drop the ground connection, and use a PNP (or consider a P-channel MOSFET).
  • 0 in reply to Clemens Ladisch
    Expert 2065 points

    Hello Clemens, 

    Thank you for bringing your contribution. How would you replace Q 200, Q 201 and Q203 ? Do you have any circuit to propose for replacement even on a piece of paper. It would be well appreciated so I can correct our circuit.

    Have a good day.

    Regards

  • 0 in reply to Dennis Eichmann
    Expert 2065 points
    HI Denis,
    Thank you for your contribution. How would you modify this circuit to avoid this ? Would you add a resistor in between and if yes what value ?
    Thank you in advance for your feedback.
  • 0 in reply to C
    Guru 283480 points
    Well, replace them with PNPs (and note that the signal to the base must be inverted).
  • 0 in reply to Clemens Ladisch
    Expert 2065 points
    Do you have any reference or part number in mind which can use with SMD ?
  • 0 in reply to C
    Guru 283480 points
    The PNP complement of the BC817 is the BC807.
  • 0 in reply to Clemens Ladisch
    Expert 2065 points
    Hello,
    There is something I don't understand; For the LED circuit which is commanded by the NPN transistor in my schematic, the source of the circuit comes from here: electronics.stackexchange.com/.../how-to-drive-a-20ma-led-from-a-4ma-max-gpio-pin

    So I don't understand why Q200 and Q201 are wrong !? Can you clarify ?
  • 0 in reply to C
    Guru 283480 points

    I said these PNPs are unusual, not that they are necessarily wrong.

    But the base resistor indicates that you do not actually understand the circuit.

    You can use NPNs for the LEDs, but then you should drop the base resistors (R200, R203). And you have to consider the additional voltage drop over the transistors when calculating the current limiting resistors (R202, R205).

    The sensor surely does not like a power supply with only 2.6 V. Here, you must use a PNP (or better, a P-channel MOSFET).

  • 0 in reply to Clemens Ladisch
    Expert 2065 points

    What about if I change the circuit this way: 

  • 0 in reply to C
    Guru 74080 points

    What do you want to do? I googled the number under J201 and found out it is a connector. Is there something connected that you want to power under control of Q203? You normally control the supply rail, not the common GND. Furthermore having the transistor in the GND path will cause a higher GND potential for the connected electronics because you have a voltage drop across C and E of the transistor. Better use a PNP transistor (or P-FET) and control +3V3.

  • 0 in reply to Dennis Eichmann
    Expert 2065 points
    I want to control (using a GPIO of MSP430) a sensor being powered with 3.3V.
    I want to do the same with 2 led
  • 0 in reply to C
    Guru 74080 points

    As Clemens said, for the LEDs your configuration is not wrong, but it is more common to have a NPN transistor at the cathode of the LED or a PNP at the anode. You can use the NPN like you have it now, but then you do not need a base resistor at all. And the voltage after the transistors emitter is roughly the voltage at the base minus 0.7V, so if you control the transistor with 3.3V you only have about 2.6V at the emitter resistor. Keep that in mind when calculating the resistor.

  • 0 in reply to Dennis Eichmann
    Expert 2065 points
    Ok, what about the transistor which control the power management of the sensor ?
  • 0 in reply to C
    Guru 74080 points

    See my earlier post here:

    Better use something like this:

    (Here a reference voltage is powered that then powers a sensor)

  • 0 in reply to C
    Guru 46710 points
    >So I don't understand why Q200 and Q201 are wrong !? Can you clarify ?
    It is unusual to drive LED with emitter follower. Is it wrong? - perhaps not. I have no time nor wish to analyze particular circuit. Anyway typical LED driver still is NPN or N-FET transistor as switch.
  • 0 in reply to Dennis Eichmann
    Expert 2065 points
    This use a mosfet, correct?
  • 0 in reply to C
    Guru 74080 points
    The given schematic uses a P-FET, right.
  • 0 in reply to Dennis Eichmann
    Expert 2065 points
    I read here and there that a FET is more suitable for controlling power device. Our sensor is only 100mA@3.3V . do we really need to go on a fet or should we consider a BJT ?
  • 0 in reply to C
    Guru 74080 points
    0.1A is quite much - I would not use the shown BSS84 for this current. The voltage drop would be too high. But you can use a BJT, of course. No one said that a FET is always the better choice. It depends on your application.
  • 0 in reply to C
    Guru 283480 points

    With a collector current of 100 mA, the BC807 has a voltage drop of about 50 mV. This corresponds to a FET with an RDS(on) of 500 mΩ. There are FETs that are much better. (If you care; 3.25 V might be OK.)

  • 0 in reply to Dennis Eichmann
    Expert 2065 points

    SCHEMATIC_V2.pdf5432.SCHEMATIC.pdfHello

    I have read about BJT transistor and i understand better why NPN and PNP now.

    So I have made some replacement and I have also kept some value too. I am going to explain here what I have changed and how.

    1./ First I have modified the led by red one from another manufacturer where the forward voltage seems to be 2V and the forward current 20 mA. 

    the transistor BC807  have VCEsat at -0.7V, I am powering the cicruit with 3.3V and I have decided to use current at 25mA to have some margin security on the LED forward current.

    my calculation goes then as follow when the transistory is acting like a wire in closed switch

    3.3 = - 0.7 +Vres + 2

    so Vres = 2

    I decided 25mA for current so R should be 80Ohm as per Vres= R * I

    Let me know if I am wrong

    2./ For the relay, which the datasheet is also below, i need to give him 3V to click. at 3V the related curren tis 50mA

    I power in 3V3 this time I have chosen another transistor PMBT2369 with a 0.25V VCEsat. so my equation goes as follow: 

    3.3 = 2 +0.25 + Vres

    Vres = 0.05V

    so Vres= R x I where I=50ma

    so R= 1ohm

    Let me know if I am wrong

    3./ I have a problem with my sensor. its voltage is supposed to be 3.3V and I power in 3.3V. I don't know any transistor who got VCEsat = 0 so how to do in this case ? I have use also a PNP here, but I am not sure this is the right things to do

    4../ For the limiting resistor in the GPIO I have chosen 1K everywhere considerng that our transistor need 0.7V on VBE to switch and considered the current to be between 2mA and 10mA on the GPIO pins

    www.nxp.com/.../PMBT2369.pdf

    )

    LED KENTO 3MM.pdf

    EDIT: I forgot to add the schematic

  • 0 in reply to C
    Guru 283480 points

    LED forward voltages have large production variations, and 20 mA is probably too bright with any modern LED. Be prepared to adjust that resistor.

    VCEsat is smaller than 0.7 V when the collector current is smaller then 500 mV. And while these voltages are specified as negative values for PNP transistors (because the current flows in the other direction), the LED calculation still must use the absolute value for the voltage drop.

    1 ohm is nothing; you could just as well leave it out. What is the current at which the relay is specified to blow up, i.e., the absolute maximum rating?

    How much current does the sensor need? You should be able to use a P-channel MOSFET with low RDS(on), such as, for example, the DMP2066, IRLML6401, or FDN338P.

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