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    Connecting the load to VCM or to ground in single supply operation

    Dear friends,

    I usually work with single supply operational amlifier when I design the op-amp, The voltage I am using is 3.3 V, thus with VDD =3.3 V and VSS= 0. In this case I connect the load (resistor and capacitor) from the output to the reference voltage VDD/2 = 1.65 which I consider it as the analog ground. However I see in some texts that op-amp is single ended and the authors are connected the load with reference to ground.

    Can you please explain me the difference ?

    Thank you in advance

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    Re: Connecting the load to VCM or to ground in single supply operation

    Hi,

    If you want bipolar output you need to use VDD/2.
    If you just need unipolar output you may use GND.

    ... it depends on your application's requirement.

    Klaus
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    Re: Connecting the load to VCM or to ground in single supply operation

    Quote Originally Posted by KlausST View Post
    Hi,

    If you want bipolar output you need to use VDD/2.
    If you just need unipolar output you may use GND.

    ... it depends on your application's requirement.

    Klaus
    Dear Klaus

    Thank you for your reply

    I need to use the op-amp as a general purpose amplifier, say for example I want to connect it as an inverting amplifier.

    Do you mean by unipolar that I will not be able to see the negative part of the output signal ? for example if my input is Sin signal then output will be only the positive part ? which kind of application accept this ?



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    Re: Connecting the load to VCM or to ground in single supply operation

    Hi,

    There are numerous unipolar applications.
    * speed control
    * heating cintrol
    * light brightness control

    Every measurement, amplification, regulation where there is no negative result.
    Example: there is no negative brightness.
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    Re: Connecting the load to VCM or to ground in single supply operation

    Quote Originally Posted by Junus2012 View Post
    Do you mean by unipolar that I will not be able to see the negative part of the output signal ?
    It's helpful to think in terms of current direction. Negative output voltage implies the device sinks current (that is, current flowing into the output terminal). To make this possible we need to arrange for the power supply to provide such a current path, as well as polarity.


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