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operational amplifier ---comparator

Discussion in 'General Electronics Chat' started by boro3, Mar 2, 2018.

  1. boro3

    boro3 Member

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    connect it with the oscilloscope or µC..
    dknguyen says its not floating and analogkid says the opposite,
    i'm a bit confused :D
     
  2. dknguyen

    dknguyen Well-Known Member

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    Oh, if that's what you are doing then yeah, it should work because although the differential inputs of the op-amp are not floating, the secondary of the transformer is floating so the secondary of the transformer will float to whatever the op-amp is.

    But in general, differential does not mean the signal does not need ground or that the the inputs of the component function independent of ground. It ONLY means the measurement is independent of ground. Truly independent of ground = floating.
     
    Last edited: Mar 5, 2018
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  3. dknguyen

    dknguyen Well-Known Member

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    I do not recommend a 1:1 transformer if you're connecting to a computer (even if you use protective measures such as resistive dividers or diode rail clamps) because if you do anything wrong, or those components fail, you lose the computer and everything connected to it. Galvanic isolation (like that in a stepdown transformer or even opto-couplers) are a lot more reliable and safer than something like diodes and resistors. Not to mention you might get shocked.
     
  4. dave miyares

    Dave New Member

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  5. boro3

    boro3 Member

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    dknguyen,
    your explanation with the build was really helpfull, thak you :)

    But if i use it without connecting secondary side with ground, (because its floating,) how does the current flow ?
    where does it come back to the transformer?



    dif.amp.png
     
    Last edited: May 25, 2018
  6. dknguyen

    dknguyen Well-Known Member

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    The signal current from the transformer flows in one input pin of the diff amp and flows out of the other input pin in the diff-amp so the current in the secondary circulates in a loop with the two input pins on the diff-amp.

    Since the secondary is floating it's common mode voltage will float around and automatically adjusts itself to be within whatever range the diff amp wants it to be (since the diff-amp is connected to ground and this somehow relates back to the inputs inside the op-amp circuitry). This is ONLY able to happen because the secondary is floating.

    If the secondary was something that was not floating (i.e. if it was another type of device) then it would mean it had it's own ground voltage, separate from the diff-amp. In that case, the source has it's own ground and the diff-amp has it's own ground and if they are not connected then the two grounds voltages (measured with respect to the surroundings...probably the earth itself) become different from each other and never adjust to each other. The grounds could become different enough that the common mode voltage on the two source outputs got too high for the diff-amp to handle.

    Do you understand? Each circuit has it's own "ground" which is 0V for everything in the circuit. But all voltages are relative so two circuits isolated from each other can have their own "0V ground" are in their own little universe. But that can cause problems if you try to connect two circuits together without connecting the grounds so that the grounds are the same. The grounds are measured with respect the surroundings (i.e. the earth itself). which is considered to be the "true 0V".

    I guess each circuit is like a different planet and each one has it's own buildings (voltages and circuits) sitting on the surface of each planet (ground). But if you try to build a bridge between two buildings on two different planets, it's a problem unless you somehow ensure that the surface of the planets do not move with respect to each other or else your bridge will break apart (i guess connecting the grounds in two circuits together is a bit like...bolting the surface of two planets together?).
     
    Last edited: Mar 5, 2018
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  7. boro3

    boro3 Member

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    Thank you,
    i think your explanations is worth its weight in gold for many newbies like me :D
    and how is ist about the current through the feedback resistor?
    how does it flow back?
     
  8. dave miyares

    Dave New Member

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  9. dknguyen

    dknguyen Well-Known Member

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    It would flow how it normally does. There's nothing special there. The purpose of tying grounds together between two different circuits is to make sure they both have the same reference so they can properly communicate with each other to make sure one circuit doesn't. Everything else is the same.

    In really bad cases, the ground on circuit A might go up to 300V and 1000V relative to the ground on circuit B so that when circuit A sends a 0V signal, circuit B actually sees the signal as 300V or 1000V and gets damaged. But long before the ground voltages get that high, circuit B won't be able to properly process the signals anymore.
     
  10. boro3

    boro3 Member

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    could you please draw the "ways"of the current through the feedback resistor of the opamp ?
     
    Last edited: Mar 5, 2018
  11. dknguyen

    dknguyen Well-Known Member

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    Well, current from the secondary will mostly only flow around the secondary through the resistive divider. A very small amount will leave the resistisve divider and flow into one of the inputs (depending on the polarity of the secondary) and back out the other input to return to the secondary.

    The currents from the op-amp supply would flow through both feedback resistors and the center resistor in the divider.
     
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  12. boro3

    boro3 Member

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    so there is no current flowing from secondary through the fedback resistor like in my drawing in blue?
    i thought, becaue of the high input impedance of the opamp,
    the current schould flow in the direction which i have drawn...
     
  13. dknguyen

    dknguyen Well-Known Member

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    When I say currents flow into and out of the inputs of the op-amp, I am talking about bias currents of a non-ideal op-amp. If we are talking about an idea op-amp then that does not happen.

    For the current you have drawn, think about how the current makes a loop to return to the sources. It doesn't have a way to do that.

    I think that current in the feedback resistors around the center resistor in the divider, not into the secondary itself. I know it's a little weird. It's because half your circuit is floating. Your diff-amp doesn't really care about anything except for the voltage across the center resistor. If that voltage is zero, your diff-amp is happy.

    The only place where current from both the secondary and diff-amp flow is the center resistor in the divider. It's the only way the currents can form a loop. The secondary drives a current through the entire resistive divider (and in a non-ideal op-amp a tiny little bit of bias current in and out of the inputs of the diff-amp so that the diff-amp can actually have signal voltages to function), while the diff uses it's +/-5V to drive a counter-current through the center resistor so that the voltage across the center-resistor is zero. So basically the diff-amp tries to make the three-resistor divider look like a two-resistor divider to the secondary.

    Replace the secondary with a voltage source. That might makes things clearer for you.
     
    Last edited: Mar 5, 2018
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