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SN74LVC1G17-Q1: Schmitt trigger Device Current

ArTzy
Mastermind 18670 points
Part Number: SN74LVC1G17-Q1


Hi Experts,

Seeking your assistance on this query:

We are using SN74LVC1G17-Q1 Schmitt trigger and We tie the VCC to 3V3 and output 'Y' is going to IMXRT 117xx GPIO Pin.

We are limiting the 'Y' Output current through 3K3 Resistor as IMXRT 117xx GPIO Pin can sink upto 25mA and SN74LVC1G17-Q1 also Source upto 24mA for 3V3 Supply.

So want to know the device current in above case ? As per our understanding IO = 1mA (Y Current - sinking into IMXRT GPIO Pin) + 10uA(Icc) + 20uA(A input Current) = 1.03 mA ; is this understanding is clear.

As there is internal circuitry given in the datasheet, if that is available then please share.

Thank you.

Regards
Archie A.

  • 0
    Guru 318050 points

    A GPIO pin configured as input has a very high impedance. (The RT1170 datasheet specifies the maximum leakage current as 400 nA.) A resistor is not necessary.

    The SN74LVC1G17-Q1's supply current (which flows into VCC and out of GND) is at most 20 µA. The load current is the RT117x's leakage current, it flows either into the RT117x's VCCIO, through the GPIO and the Y pin, and out of the SN74LVC1G17-Q1's GND, or into the SN74LVC1G17-Q1's VCC, through the Y pin and the GPIO, and out of the RT117x's GND.

  • 0
    TI__Guru* 97066 points

    Hi Archie,

    This is what I understand the question to be:

    How much current will flow through the 3.3kohm resistor if the LVC device is in the high state and the GPIO is in the low state?

    -

    Assuming I am correct, the current will be approximately I = V / R = 3.3V / 3300ohm = 1mA

    There is some internal resistance for each device, but it will be very small compared to the 3.3kohm resistor. LVC devices operating at 3.3V will have an output resistance around 15 ohms, and since your GPIO has a similar current rating, I'll estimate it also at 15 ohms.

    This estimates the output current for the LVC device as about 991 uA, which is equal to the input current to the GPIO.

    ArTzy said:
    So want to know the device current in above case ? As per our understanding IO = 1mA (Y Current - sinking into IMXRT GPIO Pin) + 10uA(Icc) + 20uA(A input Current) = 1.03 mA ; is this understanding is clear.

    If by "device current" you mean the maximum current into the VCC pin, ie the supply current, then it will be up to 10uA (which the maximum across all conditions) plus the output current of 991uA, or 1.001mA.

    This calculation assumes that the input of the LVC device is at either VCC or GND. if it is not, then the current will increase from the input circuitry.

    ArTzy said:
    As there is internal circuitry given in the datasheet, if that is available then please share.

    No, we cannot share the complete internal design of our device - however we do have a detailed video on how a Schmitt-trigger input works, and you can see the paths for current there. https://training.ti.com/schmitt-trigger-cmos-inputs