Single Supply Op Amp Questions

Discussion in 'The Projects Forum' started by GatorCpE, Aug 25, 2011.

  1. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    I'm working on a project in which I have to use an op amp in a single supply configuration. I understand that in order to get this to work, I have to ground the negative supply rail AND connect the + input terminal to 1/2 the supply voltage. I know that I have to do this in order to get the circuit to work. My question is why?

    I understand that when I ground the negative supply, the output can only swing between (theoretically) 0v and the voltage of the positive supply.

    I'm having trouble understanding what applying half the supply voltage to the non-inverting terminal does.

    If someone could provide a thorough explanation I will be most grateful.

    Thank You
     
  2. JingleJoe

    Member

    Jul 23, 2011
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    Applying half Vs to the non-inverting input biases the output to half Vs, creating a virtual ground at that point which allows the amplifier to work in class A operation (if one inputs a waveform to the inverting input).
     
  3. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    Ok let me see if I understand this correctly. Considering an inverting amplifier:
    When 1/2 the supply voltage is applied to the + input terminal, that same 1/2 supply voltage appears at the - input terminal because the voltages at both terminals are equal if an ideal op amp is assumed.

    With a sinusoidal input applied to the -input terminal, the voltage that gets amplified is the sinusoidal voltage and the 1/2 DC supply voltage. Because the 1/2 DC voltage is present at the input, the output voltage is "shifted up" by 1/2 of the supply voltage. Now, the output signal will be centered around 1/2 the supply voltage and the bottom part of the sinusoidal output will not be clipped at 0 volts.

    If this correct?

    If my previous analysis was correct, I have another question. Since the output waveform is able to swing freely without being clipped, how do I re-center the waveform around 0 volts? Can I insert a coupling capacitor on the output to remove the DC bias?

    Thanks for the help
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    So which of the two attached circuits would work?
     
  5. GatorCpE

    Thread Starter New Member

    Jul 14, 2010
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    I'm going to say that the one on the right looks like it will work. To me, it looks like the +input in the left circuit is still tied to ground.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Actually the one on the right won't work as it stands. The issue is the effective signal common point and the effective DC offset of the input signal relative to that common point. This is something one needs to keep in mind with unipolar supplies to op-amp circuits. If the signal source in the right hand circuit was DC offset to the mid-point potential (7.5V) then the circuit would work. In that case you could couple the output via a suitable capacitor to remove the DC offset from the output.

    Similarly if the input signal source in the left hand circuit had any DC offset it too may not work, as the output might (depending on the DC offset polarity and magnitude) then be driven to either of the supply rails. Of course that will be true of any op-amp (bipolar powered or otherwise) driven by a DC shifted AC input.
     
  7. Hi-Z

    Member

    Jul 31, 2011
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    We'll assume the input is connected via a capacitor, because then your statements are mostly pretty correct. If you want your ac output to be referred to ground, then you can block the dc component with a capacitor (and then you can "anchor" it to any voltage you like - including ground - using a resistor).
     
  8. JingleJoe

    Member

    Jul 23, 2011
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    Almost correct, the half Vs on the non-inverting input does not appear at the other input, but is applied to the output because op amps sort of mix the two signals together, one is subtracted from the other or added depending which way you look at it.
    Additionally your wave may clip 0V, it all depends on the gain, for example in a circuit with Vs of 10V and input wave of 1V peak to peak, you should have a gain of no more than 10 to avoid clipping (9 to be safe).

    Regarding your two diagrams; bump those 1k resistors in the divider up to 10k to draw less quiescent current.
    The one on the right has worked for me in the past, using real components (not simulation).
     
  9. bertus

    Administrator

    Apr 5, 2008
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