Resistance through a circuit

Discussion in 'The Projects Forum' started by jerseyguy1996, Apr 24, 2011.

  1. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
    206
    9
    I am building the circuit on page 5 of this appnote:

    http://focus.ti.com/lit/ug/sluu023b/sluu023b.pdf

    I have not included resistors R16, R15, or R17 in my circuit because they are not necessary for my application.

    Looking at the schematic I can only see one path for current to take when all of the transistors and are off and that would be from:

    +VDC-->D6B-->R21-->R14-->R12-->R20-->Ground

    I add those up and I get a resistance 331K but when I check the resistance of my circuit across VDC+ and VDC- I get a resistance of only 500 ohms. I'm scared to plug this thing in until I know for sure what the resistance should be. Would the PNP transistors and P-Channel FET be conducting when measuring resistance with an ohm meter?
     
  2. designnut

    Member

    Apr 21, 2011
    33
    1
    You are measuring the input diodes that are inherent to the circuit. are you sure you don't need the resistors, sounds like you are a real beginner.
     
  3. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
    206
    9
    Look closely at the schematic. Those resistors aren't connected to anything if you don't have a jumper connecting them across JP5.
     
  4. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
    206
    9
    Not sure what you mean by measuring the input diodes. Shouldn't current flow from VDC+ through D6B (I used D6B instead of D6A) through R21 through R14 through R12 through R20 and back to ground?
     
  5. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
    206
    9
    Well found the problem. I had the circuit already hooked into the power supply and I was reading the resistance cross the V+ and COM of the power supply. Actually I guess I was reading the parallel combination of the resistance of the power supply and the resistance of the circuit.
     
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