1
\$\begingroup\$

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!

\$\endgroup\$
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ So, if I don't pull it high, then no translation takes place, and its just high impedance on the B pins? What's the point of this switch then? Because its not actually allowing a switch if I have to have a resistor and track to it on a PCB....? I won't ever be driving the chip in both directions, it will always be A > B \$\endgroup\$ – RenegadeAndy Jan 7 at 22:32
  • \$\begingroup\$ The Enable is a control signal to this translator so if you have it as low, then both A and B are high impedance which means this IC doesn't set their values but any external circuit does. the PULL up resistor is required in case you want it to be always on. However if you have a control signal on your circuit which can be high or low to turn on or off the translation, you can use that signal directly to the EN pin but it should be in Vref_b domain ( 5V in your case), I would still put a pull up to be safe. \$\endgroup\$ – Ash Jan 7 at 22:37
  • \$\begingroup\$ Ah. SO if i wanted to turn A1 & A2 to B1 & B2 OFF , then I would send LOW to EN, if I wanted it on, then I would send whatever B1 & B2 vRef (in this case 5v). You only use the pull up resistor in the case that you always want the chip to always be enabled? \$\endgroup\$ – RenegadeAndy Jan 7 at 22:40
-1
\$\begingroup\$

The datasheet indicates EN enables 200k pullup on the Vb side. If you need a lower R pullup due to pF load and rise time, you must decide what RC =T value you need. This subtracts from available pulldown current shown in Table 4.

Ron spec is 10 Ω at 5V so 64.5mA results in a rise of 645mV (TYP) tolerance is not given for Ω nor have you given a tolerance for Vol.

(MORE DETAIL)

This shifter is a bi-directional switch , therefore NON-INVERTING enter image description here

Your table is Correct here shown re-arranged to prevent our misunderstanding.

 Pin 1 - GND
 Pin 2 - VRefA = 3.3V

 Pin 3 - A1 -- input channel 1 (in this case could be 3.3V)
 Pin 6 - B1 -- output channel for input 1 (in this case would be 5V)


 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 7 - VRefB == 5V
 Pin 8 - EN = Switch enable input; connect to Vref_B and pull-up through a high resistor (200 kΩ) 
  • A1=B1 either 0 or 1
  • A2=B2 either 0 or 1

B side open drain when= "1" , 10 Ohms when "0"

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Hi @sunnysky I find your answer incomplete and hard to comprehend. Please explain the role of the EN pin, and how the pullup resistor operates as an explanation. Additionally, confirm that my other pins are understood correctly. \$\endgroup\$ – RenegadeAndy Jan 7 at 22:31
  • \$\begingroup\$ A1=B1 either 0 or 1? I am not exactly clear on your syntax here. I am also not clear what you are referring to with the mention of 10 ohms, or the 200kOhm pullup. \$\endgroup\$ – RenegadeAndy Jan 7 at 22:59
  • \$\begingroup\$ 10 Ohms is the switch Ron resistance as stated in the datasheet \$\endgroup\$ – Sunnyskyguy EE75 Jan 7 at 23:03
  • \$\begingroup\$ Perhaps what I am asking is, what is 'Switch Ron Resistance'? What is 'Ron' \$\endgroup\$ – RenegadeAndy Jan 7 at 23:11
  • \$\begingroup\$ I wonder who thinks my answer is worth -1. Ron exist for EVERY switch. Sometimes called RdsOn in FETs. It is what causes drivers to to have voltage drop from current flow. or switching to 0V from a pullup load like Vol/Iol="Ron" approx. In a level shift this can shift the levels lightly. but if you demand 61.5mA load and it is rated for 60mA. Then it will shift more than rated and run a bit hotter. \$\endgroup\$ – Sunnyskyguy EE75 Jan 11 at 20:55

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.