• State Resolved
  • Locked Locked
  • Replies 5 replies
  • Subscribers 20 subscribers
  • Views 2928 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Questions of LSF0102

Eric Tian61
Intellectual 890 points
Other Parts Discussed in Thread: LSF0102

Hi,

We are tring to use LSF0102 to translate both 2V and 5V pulse wave to 3.3V pulse wave, but it does not work as expected.

The schematic we use is similar to fig.2 of slva675b,Vref_A=1.8V and Vref_B=VBPU=3.3V, A1 and A2 are inputs, B1 and B2 are outputs, no pull-up resistors on A side and 360ohm pull-up resistor on B side.

The results are shown below

(1)   Input (blue): Pulse wave Vpp=2V, pulse width = 40ns(application requires such narrow width),rising edge=5ns

Output (yellow): low-volatage=0.4V and high-voltage=2.6V, The high-voltage can go up to 3.3V only if pulse width > 120ns

(2)   Input (blue): Pulse wave Vpp=5V, pulse width = 40ns(application requires such narrow width),rising edge=5ns

Output (yellow): low-volatage=0.4V and high-voltage=2.9V, The high-voltage can go up to 3.3V only if pulse width > 100ns

(3)   Input (blue): 0V DC

Output (yellow): 0.45V DC

My questions:

(1) Why is the low output is about 0.4V, is it the right result or something wrong with our schematic?

(2) Can we get a 3.3V high output with a 40ns width pulse wave input and how?

(3) Is pull-up resistor on A side needed? I am confused because there is no pull-up resistor in fig.2 of slva675b, however in this post

pull-up resistor is needed.

Regards,

Eric

  • 0
    TI__Guru 62240 points
    Hi Eric ,

    Sometimes the schematic connection from Vref and En could cause issues when the LSF wouldn't be properly biased but I see the translation is happening but not upto the desired voltage level .
    1) regarding the low output , LSF is a passive translator which passes the low as it is to the output and the high level is through the pullup resistor .When you have the 360ohm pull up resistor at B side , the A side driver must be able to handle the IOL current flowing into it during low state . With the present conditions , it would be 3.3/370 =~9mA . If the input driver is not rated for IOL at 9mA , then the Vol rises . Even if it is rated for 9mA , it will still not be at 0V exactly .
    2) When you say that the 3.3V can be reached with wider pulse width , it presents the speed requirements. The speed is dependent on the output cap load and higher the cap load ,the slower the speed . Ofcourse , it also depends on R . Lower R results in higher speed but also means higher current flowing into A side driver . To achieve the desired speed , lowering output parasitic cap would be desirable if output load cap is constant.
    3)Pull up on A side is not required if your input is push pull . If it is open drain , then yes , you will need a pullup resistor for desired voltage . I do not think with the scope shots provided ,you will need a pull up . However, you can still put one if you are worried about signal integrity .Prepare for the current flowing when output is low .
  • 0 in reply to ShreyasRao
    Intellectual 890 points

    Hi Shreyas,

    Thank you for your reply.

    I got higher speed on B side by lower R.

    My input on A side is an TTL device of output impedence=50 ohm, I am not sure it is push pull or open drain. However, If I add a pull up R on A side, the low output on B side goes higher. If I add a pull down R on A side, the low output on B side goes lower. So here is what I thought:

    When input is low, current flows from 3,3V to P point, assume A side driver cannot handle IOL current flowing,

    (1) If a pull up R is presented, I1 will make the voltage of P point higher than 1.2V

    (2) If a pull down R is presented, I2 will make the voltage of P point lower than 1.2V

    So I think a pull down R is needed instead of pull up R, please correct me if I am wrong.

    Regards,

    Eric

  • 0 in reply to Eric Tian61
    TI__Guru 62240 points
    Hi Eric ,

    Essentially , the pulls ups or pull downs are not required if the A side driver is push-pull. In case of open drain , pull ups are required for appropriate level setting .
    The pull ups or pull down will result in additional current into the driver side through it which it needs to handle . If pull ups are used , it would be about the IOL , and with pull downs it would be about IOH .
    If you are satisfied with your high level with the pulldown resistors, you can use this . It should not affect your output low voltage now .
  • 0 in reply to ShreyasRao
    TI__Guru 62240 points
    I am hoping that you have not connected your circuit as shown in your post resulting in short between 3.3 and 1.2V supplies .
  • 0 in reply to ShreyasRao
    Intellectual 890 points
    Hi Shreyas,
    Thank you for your reply. I didn't short 3.3 and 1.2, I just borrowed the image from another post.
    Regards,
    Eric