DC Impeadance Problem?

Discussion in 'The Projects Forum' started by Shultz1e, Jun 9, 2014.

  1. Shultz1e

    Thread Starter New Member

    Feb 6, 2013
    I believe this is my first post on the forum so excuse my lack of terminology.
    I work in the mobile hydraulics field, and a particular product we sell is a proportional electro-hydraulic valve. DC voltage only, 12 or 24vdc
    The electronic solenoid is a ratio-metric voltage controlled circuit. It has a 25%-50%-75% actuation signal. This means @ 25% supply voltage, I will get full spool travel in one direction, 50% puts my spool in neutral, 75% provides full spool travel in the opposite direction. Simple so far right?
    Well, the solenoid has fault monitoring so that if the voltage drops below the 25% or above the 75% (there is a +/- margin) it will shut the coil down for safety.

    To make an 'at home' controller for this coil, the manufacturer recommends a 500ohm resistor on +, 1K pot, 500 ohm resistor on ground. This all makes sense to me at this point, after some considerable reading. This works fantastic on my 12vdc power supply. However, I only have a 10K pot, so I substitute this with a 5k resistor, 10k port, 5k resistor. On the power supply, still works fine. However, when I plug it into my 'PVE' (coil) it drops 2+ volts, immediately when my pot is all the way up. This drop in voltage prevents full spool travel. Then I read (in the part book)the coil has 12K impeadance. Then I read (online in various places)and it says that your output current should be 10% of the input current, which leads me to believe that is why the manufacturer specifically said 1k pot vs. any ratio available.

    Does this make sense to anyone who can break it down? I prefer stupid terms, however the more technical you can get on voltage dividing and impeadance, the better. I found some great sites online
    And then use the www.falstad.com/circuit/ to check things out, and it sounds like I need to drop my pot down to a 1K, but wanted to get some education on the subject.

    I am fascinated with electronics and I look forward to your kind replies.
    Thank you!
  2. MrChips


    Oct 2, 2009
    Voltage dividers are great for doing just that - dividing voltages.

    The voltage divider needs current flowing through the resistors for it to function.

    When you connect a load to the divider, it too will take current which will upset the operation of the divider. So a general rule of thumb is: don't take more that one-tenth of the divider's current.

    When you translate this to resistance, it means that the resistance of the load must be ten times greater than the divider resistance.

    Hence if your load has a resistance of 12kΩ, use a voltage divider that is 1.2kΩ or lower. That is why your 10kΩ with 5kΩ resistors on the arms don't work.

    BTW: You may notice that I use the word resistance instead of impedance. Impedance is used when we are discussing AC circuits. For DC, we use resistance.
  3. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    You can get this to work using a 10k pot, provided you change the other resistor values and add a transistor, like this:
  4. Shultz1e

    Thread Starter New Member

    Feb 6, 2013
    Mr. Chips, thank you for your reply. Your comments confirmed my 'readings' I found online. I only used the term impedance as this is what the technical literature called it. Thanks for the tip though! I went down to radio shack and found a 1k trim pot just for testing purposes with a 500 and 1K resistor and it definitely improved the circuit. Enough to satisfy our requirement anyway..

    Alec_T, thank you for your suggestion. I personally am interested in checking other theories and becoming more familiar with electronics so I very well may take the time to A) try it B) learn to understand it.. however time is critical so I must stick with the generic route.
  5. inwo

    Well-Known Member

    Nov 7, 2013
    If you want to improve (lengthen) the deadband (neutral), Try a 5K pot with a 1K resistor across it.

    I just built a similar test circuit for a -3>0<+3 volt control.

    A resistor across the pot changes the linearity.
    Good for testing.

  6. Shultz1e

    Thread Starter New Member

    Feb 6, 2013
    Thank you inwo

    My project appears to work the way I want it to with the 1k pot.
    1k resistor on one side and 2.2k on the other side. I have a DPDT switch feeding the +/- of the resistors.
    So in one position I have roughly a 3-6vdc range and in the other I have a 6-9vdc range. Luckily for me the accuracy can be fairly crude due to the application.
  7. MaxHeadRoom


    Jul 18, 2013
    I have always used the term in a general sense.
    Even 'The Art of Electronics' condones it.;)

    "Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied.

  8. Shultz1e

    Thread Starter New Member

    Feb 6, 2013
    Thanks MaxHeadRoom,
    As I am a newb in the electronics field.. I know using the wrong terminology really bothers some folks, and for good reason. But not always knowing the proper terminology I just have to use what I have in front of me. In fluid power we constantly have customers call their pumps, motors, and vice versa.. so your conversation starts at the wrong end. So I understand the suggestion of using the write terminology. I am happy to know that I used industry accepted terms however!

  9. MrChips


    Oct 2, 2009
    Note: I did not say that the use of the word impedance was incorrect, right or wrong.

    There are times when resistance and impedance are used interchangeably.

    Impedance tends to be used when there is reactance in the circuit.