Understanding Passive Probes

แชร์
ฝัง
  • เผยแพร่เมื่อ 21 พ.ย. 2024

ความคิดเห็น • 9

  • @erikchavarin6085
    @erikchavarin6085 5 หลายเดือนก่อน +4

    If I'm watching these videos, generally it's because I'm studying and stressed out. The birds chirping in the background really inspire serenity.

  • @mariofilippi3539
    @mariofilippi3539 9 หลายเดือนก่อน +1

    Thanks for taking the mystery out how probes work.

  • @yuvarajv758
    @yuvarajv758 4 ปีที่แล้ว +3

    good work... well presented....

  • @ahmedalshalchi
    @ahmedalshalchi 4 ปีที่แล้ว +2

    Very good basic knowledge presentation for the topic... I appreciate the presenter , fruitful one !!...

  • @Momibaily
    @Momibaily 2 ปีที่แล้ว

    how does the probe tip cancel out scope input capacitance ? C1=C2 therefore net cap is C/2?

    • @pauldenisowski
      @pauldenisowski 2 ปีที่แล้ว +2

      (See diagram at 2:51) At higher frequencies the capacitance the scope input will decrease signal level (essentially shorting the signal to ground). At higher frequencies, the parallel capacitance in the probe tip will "bypass' some of the resistance in the probe tip, increasing overall signal level. If we choose / adjust these capacitances properly, then we can balance out these two effects. Hope that helps!

    • @Jnglfvr
      @Jnglfvr ปีที่แล้ว +2

      Basically what you want to do is to maintain the 10:1 ratio of probe signal to input signal because that is what the scope is "expecting." The inevitable input capacitance of the scope acts like a low pass filter at higher frequencies attenuating them. Higher frequencies will reduce the net parallel input impedance of the scope and, being frequency dependent, you have no control over this. So the signal "seen" by the scope is lower than expected. So the probe impedance needs to be reduced to maintain that 10:1 ratio. To reduce the probe tip impedance one needs to increase probe tip capacitance (which will lower net parallel probe impedance). In this situation we say that the probe is undercompensated. Conversely if we reduce the probe impedance too much a higher voltage will appear at the input terminals and the probe is overcompensated.