I am considering using this TI LSF0102 level shifter in my circuit – but would like to understand, based upon its datasheet, if I have my understanding correct.
My aim is to be able to supply 3.3V and shift it up to 5V, in the case that my input is LOW, then my output should be LOW. It needs to handle at least 61.5mA of current, and I need to be able to control 4 channels (in this case I would buy 2 of these chips for this purpose).
I believe I can do the following in the 8 pin LSF0102 DCT or DCU Package.
Pin 1 - GND
Pin 2 - vRefA == 3.3V
Pin 3 - A1 -- input channel 1 (in this case could be 3.3V)
Pin 4 - A2 -- input channel 2 (in this case could be LOW)
Pin 5 - B2 -- output channel for input 2 (in this case would be LOW)
Pin 6 - B1 -- output channel for input 1 (in this case would be 5V)
Pin 7 - vRefB == 5V
Pin 8 - EN = Switch enable input; connect to Vref_B and pull-up through a high resistor (200 kΩ)
I am not sure of the purpose of the EN pin, how is it acting as a switch? How does it work? Additionally, are my other understandings correct?
Is there a better alternative chip for my purpose?
Many thanks!
Best Answer
Yes, if you are planning to translate logic signals from 3.3V to 5V domain. This level shifter should do the job. EN signals controls the internal switch, so if you pull it high to ref B the translation will take place, so if A1 is high(3.3V) or low(0V), B1 will be correspondingly high(5V) or low(0V) same goes for A2 and B2.
If EN is low, all I/O pins will be high impedance.
The translation works in both direction but make sure Vref_b > Vref_a +0.8 V. So you can't put 5 V on Vref_a and 3.3 V on vref_b.