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  1. #1
    Quote Originally Posted by cropwell View Post
    It is. but they quoted an impedance which is generally used for AC signal inputs. 100K impedance means that the current from the pot wiper is very low.
    OK - (probably got this wrong) are you saying the input impedence is basically circuit to restrict the ampage coming in???? and are you saying this circuit is already built in?
    Every time I am wrong - the World makes a little less sense.

  2. #2
    It seems so. If the input were the full 10v, it would give a current dependant on the output capability of the 10v source. The input terminal has to have some protection against too much current. It is also filtered to get rid of any induced voltage spikes to give a more stable control of the speed.

  3. #3
    Quote Originally Posted by cropwell View Post
    It seems so. If the input were the full 10v, it would give a current dependant on the output capability of the 10v source. The input terminal has to have some protection against too much current. It is also filtered to get rid of any induced voltage spikes to give a more stable control of the speed.
    You see that's where it is confusing... looking at the diagram
    Click image for larger version. 

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    The circuits (to me) at the bottom reopresent the circuit a user has to add. This is because the switch from 5 - 7 is not on/in the board, the user adds that.

    Does that make sense?
    Every time I am wrong - the World makes a little less sense.

  4. #4
    I'm in agreement with Cropwell's interpretation. There's any engineering "wet thumb" rule that you design the input impedance at least 10 times the source impedance in order to not substantially influence the input. So, 10k input pot, 100k input impedance - rule is met. (ignoring balanced input design etc).

    Don't get hung up on impedance over resistance.

    And Nemo - Because I'm in a good mood I'm going to agree with you too - at least re. the circuit outside of the box being that that the user has to add.

  5. #5
    Quote Originally Posted by Doddy View Post
    I'm in agreement with Cropwell's interpretation. There's any engineering "wet thumb" rule that you design the input impedance at least 10 times the source impedance in order to not substantially influence the input. So, 10k input pot, 100k input impedance - rule is met. (ignoring balanced input design etc).
    I'm still lost... "Set speed Term 3: Potentiometer minimum or 0 to +10V, impedence 100K ohm filtered"


    Do I need the 10k pot if using 0 - 10v from a spible card?
    Every time I am wrong - the World makes a little less sense.

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