need your help with Op-Amp...

Discussion in 'General Electronics Chat' started by Piscu, Dec 25, 2009.

  1. Piscu

    Thread Starter New Member

    Dec 25, 2009
    Hi guys, I am new member, I hope we all benefit from each other

    I really dont understand why does the amplifier require both positive and negative supply voltages? Why provide balanced supplies, i.e. ±15 or ±10 VDC?!

    and what is DC-offset in the output voltage waveform of the integrator or defferentiator?!

    I appreciate your response, thanks
  2. kdillinger

    Active Member

    Jul 26, 2009
    It doesn't. In fact, no op-amp ever requires a balanced supply to operate, but it may require a balanced supply for it to operate correctly in ones circuit depending on what that circuit does.

    If you go back 10 years or more, others older guys here can correct this, op-amps were not optimized for single supply. Any op-amp could still be used, but the typical problem is the input and output rail performance. If you had an amplifier specification of ±15V stating an input common mode range of ±12V and and output swing of ±12V then that amplifier would be very limited in operation with a single +5V supply or even a +10V supply.

    To restate the above, and op-amp's input common mode range is ±12V when the power supplies are ±15V. This means that input voltages higher than +12V or lower than -12V will not be 'seen' by the op-amp; the output voltage will not be a linear function of the input voltages because the input voltage is outside of the input range. This example op-amp can only use voltages that are 3V from either supply. If the supply was a single +10V, then the input range of the op-amp is now between 7V and 3V. This means your signal could not be higher than +7V or lower than +3V. Not very good when a lot of people need an input range to include 0V!

    This can be compounded by the limitation of the output stage. Even if you are within the input common mode range of the op-amp, the output voltage could limit against the output swing capability, called clipping.

    Say we still use our example op-amp on a ±15V supply and our input voltage is a 1kHz sinewave with an amplitude of ±2 volts, and the op-amp is configured with a gain of 7. Clearly, ±2V is within the ±12V input range. One would expect that the output voltage swing would be ±14V (±2V * 7). The problem is the output swing is ±12V so the output can only swing between +12V and -12V; any higher or lower voltage and the output will clip the sinewave causing distortion.

    Using a single supply of +10V then again, the output clips even more-so on the negative cycle because the output swing is again limited to voltages between +7V and +3V.

    This is just an example. You need to look a the op-amp datasheet to understand what the input common mode voltage range is and the output swing limitation. They will not always be symmetrical like my example; i.e., ±12V for input range and output range on a ±15V supply.

    Over the last 10 years op-amps have been optimized for single supply operation. Even on a +5V supply the input common mode range can be 0V to 4.5V and the output swing can be 0V to 4.99V in some cases. Most of these are CMOS op-amps and there is no free lunch. Many of these rail to rail op-amps are limited in the amount of current they can provide to a given load. The more current the load demands of the op-amp then the poorer the output swing will be.

  3. SgtWookie


    Jul 17, 2007
    If you want to use an opamp with a single rail, you'll need to use a capacitor on the input to block the DC level, and resistors to divide the supply voltage by 2 to get the input within the range of the opamp.

    You'll also need to use a cap on the output to isolate the DC from the AC signal.

    This can be problematic if your goal is to monitor DC levels.

    Some opamps can sense inputs down to their negative rails. Some are rail-to-rail input and output. One has to take the particular application in mind as to what is most cost-effective and appropriate. In today's world, if it is not cost effective, it is not viable.

    A differentiator with a square wave input will output a series of positive-going and negative-going spikes that correspond with the leading and trailing edges of the square waves.

    An integrator with a square wave input will output a triangle wave.