LT6000 Saturated Characteristics

Hello

                 Please refer to below circuit. If 3.32V of IN-  and 3.27V of IN+, output of LT6000 is 0V. If 3.21V of IN-  and 3.27V of IN+, output of LT6000 is 3.3V. The results looks normal. 

                  But for below red circle. If 0V of IN-  and 3.27V of IN+, why is output of LT6000 0V ? Is it correct? Thank you. 

BR

Patrick



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[edited by: patrickchen@morrihan.com at 9:06 AM (GMT -4) on 28 Aug 2020]
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  • +1
    •  Analog Employees 
    on Aug 27, 2020 3:57 AM

    Hi,

    If I understand it correctly, you have used the LT6000 as comparator with IN+ as your reference pin.
    Your reference voltage is fixed at 3.27V coming from the Voltage Divider Network.

    When your IN- voltage is higher than IN+(3.32 in your 1st case), the Vout is 0V.
    When your IN- voltage is lower than IN+ (3.21 in your 2nd case), the Vout is 3.3V.
    This results are normal and expected.

    For your 3rd case, you have 0V at your IN- pin and that is lower than your IN+ pin, the output should be 3.3V.
    I would like to know how did you implemented 0V in your IN- pin here?

    Thanks and regards.

Reply
  • +1
    •  Analog Employees 
    on Aug 27, 2020 3:57 AM

    Hi,

    If I understand it correctly, you have used the LT6000 as comparator with IN+ as your reference pin.
    Your reference voltage is fixed at 3.27V coming from the Voltage Divider Network.

    When your IN- voltage is higher than IN+(3.32 in your 1st case), the Vout is 0V.
    When your IN- voltage is lower than IN+ (3.21 in your 2nd case), the Vout is 3.3V.
    This results are normal and expected.

    For your 3rd case, you have 0V at your IN- pin and that is lower than your IN+ pin, the output should be 3.3V.
    I would like to know how did you implemented 0V in your IN- pin here?

    Thanks and regards.

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