SN74LVC2T45: SN74LVC2T45

singam vamshi krishna

Part Number: SN74LVC2T45

Hi Team,

I am using SN74LVC2T45 in my application. I am translating from 3.3V to 5V. I have connected VccA to 3.3 V along with DIR pin and VccB to 5V. I have fed A1,A2 with 3.3V , But I don't see any output at B1,B2. Could you please help me with what went wrong. please find the circuit below and kindly help me out with what went wrong.

Thank you in Advance

over 3 years ago

0 Sebastian Muriel over 3 years ago

TI__Mastermind 18010 points

Hi Singam,

Once the switch is opened the capacitance at the input discharges until the diode is no longer forward biased, after that there is only a path for leakage current to discharge, so it will take a very long time for the input to reach 0 V. You need to place a resistor in parallel with the diode to allow discharging to ground. Try a 10kΩ.

Regards,

Sebastian

0 singam vamshi krishna over 3 years ago

Intellectual 775 points

I agree. But in my case the diode is conducting and the voltage at the input to the IC (A1, A2) is 3.3V. In that scenario I am supposed to get 5V at output right, I mean at B1, B2. Assuming that I am bypassing the diode.

0 Sebastian Muriel over 3 years ago in reply to singam vamshi krishna

TI__Mastermind 18010 points

Hi Singam,

Are you able to swap a new device into your board and place the old device onto a known working board? I want to see if it's the board or the device causing the issue.

Regards,

Sebastian

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SN74LVC2T45: SN74LVC2T45