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Circuit Explanation

  1. Jun 20, 2012 #1
    Hey all,
    This newb needs help understanding the posted circuit. Real quick, I'm going to post what I understand from left to right.
    R53 and R62 = Voltage Divider
    D2 = Clamping Diode (Clamping to 3.3V)
    R45 = R1 in basic (R2/R1)Vin = Vout op amp formula

    This is where I get stuck.
    Let's call the node b4 R45 Vin. We would have [(Vin-Vref)/R45] = [(Vref - Vo)/R2] - however no R2. Instead, there's a diode that is in the off (depending on how you look at it) preventing any output from the first op-amp. Where then I guess the second OP Amp equation looks like Vo = [(Vin-Vref)/R45]. I am pretty sure I am missing some inverse logic in all of this, but regardless, I hope I got what I was trying to say across. Click Here for the Op-Amp datasheet. Any help would be greatly appreciated. Namely, what is that diode doing? What's the purpose of this circuit? Thank you.
     

    Attached Files:

  2. jcsd
  3. Jun 20, 2012 #2
    And when I say newb, I am referring to myself.
     
  4. Jun 20, 2012 #3
    U10B work as clamping circuit. The output voltage can changer from 0 to 3V.
    Higher voltage will be clamp.
    Vout = Vin if Vin<3V, D4 is off and U10B is in positive saturation because inverting input voltage is smaller then non-inverting input.
    And if Vin > 3V then Vout = 3V. Now D4 is ON and and closes the negative feedback loop. And non-inverting input is equal to inverting input voltage.
    https://www.physicsforums.com/showpost.php?p=3866238&postcount=3
     
    Last edited: Jun 20, 2012
  5. Jun 20, 2012 #4
    What is this circuit used in? Do you know what Vfilter comes from?

    I miss Altium by the way.
     
  6. Jun 20, 2012 #5

    AlephZero

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    Homework Helper

    It's a circuit to monitor the voltage at Vfilter. Presumably the output "Vbat detect" either tells you if flying mammals or superheroes are in the area, or (more likely) that some sort of storage battery (maybe a backup powier supply) that is connected to Vfilter somehow, is outside its correct voltage range. The 3.3V supply for this circuit would be independent of what was being monitored.

    To get 3V output from the potential divider, Vfilter would be at 3 x (84.5+5.1)/5.1 = 53V, so presumably that is the critical voltage level for whatever the circuit is monitoring.
     
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