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I am constructing an operational amplifier as shown in the following figure. I use a batter as supplier for the OP Amp and set it up as a non-inverting amp circuit. I saw that the output was clipped while I feed it with a sine at about 1kHz. I wonder if there is any thing to do with the power supplier, what supplier should I use if so. enter image description here

p.s. the intpu signal is of sine form, the peak-to-peak voltage is about 650mV, the frequency is about 1KHz or less

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  • \$\begingroup\$ What's the actual input voltage of the sine wave? Also what's the output load? \$\endgroup\$
    – AndrejaKo
    Commented Aug 27, 2013 at 21:51
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    \$\begingroup\$ You need to put a Vcc/2 bias on the noninverting pin, and eliminate the amplifier's DC gain. Also, the TL08x is a horrible choice for a 9V single supply. Look for a CMOS rail to rail op amp such as the TLC2272, LM7322, or LMC6462. \$\endgroup\$
    – Matt Young
    Commented Aug 27, 2013 at 21:52
  • \$\begingroup\$ I think the TL081 designation comes from the default component labels in CircuitLab -- I would verify what opamp the OP is really using. For some component types, CircuitLab allows you to change the part number; for others, it doesn't, which can be quite confusing! \$\endgroup\$
    – Jon Watte
    Commented Aug 28, 2013 at 5:11

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The gain of the circuit is +11 (Gain = 1 + \$\frac{R_F}{R_1}\$). This means when the input sinewave is positive 0.1V you'd expect an output that is +1.1V and when the input is negative 0.1V you'd expect an output that is -1.1V. These numbers are all relative to ground. The first problem is that your op-amp has supplies that are +9V and 0V so immediately it's clear that the op-amp can't produce the -1.1V or indeed any negative voltages.

What to do? Both the input signal and the ground connection of \$R_1\$ need to be set to halfway between 9V and 0V - at the moment you're throwing darts at a board that is 2 foot higher than you thought - you are just hitting "three" when you should be aiming for bullseye (sorry about the darts analogy but it seems apt!).

OK, so the problem is that your input AC signal is 0V (or ground) referenced - if you just wish to amplify AC signals then you can use a capacitor (that blocks the steady 0V connection) and then have a resistor divider that sets midrail on the input at 4.5V. You have to do the same for \$R_1\$ - I would make R1 2 x 200ohm resistors; one going to +9V and one going to 0V and joined together at the -Vin input.

That's a start; then get a better op-amp that can deal with rail-to-rail outputs - you might achieve 8.8Vp-p from this circuit with a 0.8Vp-p input.

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  • \$\begingroup\$ Thanks a lot. I got some idea now. I try to connect +9V to Vss and -9V to -Vss pin instead of using 9V and 0V. But the output are all going now. So in this case, where should I have the R1 connected to? (it is connecting to IN- and ground) not the ground replaced with -9V, so where should it go? \$\endgroup\$ Commented Aug 28, 2013 at 17:00
  • \$\begingroup\$ @user1285419 if you now have a split power supply of +9V, 0V and -9V then R1 and the input remain at 0V. \$\endgroup\$
    – Andy aka
    Commented Aug 28, 2013 at 17:04
  • \$\begingroup\$ thanks.So should I have all my function generator, scope common ground to the split power supplier? And I am pretty wonder now we have +12V and -12V connected to Vss and -Vss of OP Ampm so what about ground? no ground needed for the OP amp? \$\endgroup\$ Commented Aug 28, 2013 at 17:49
  • \$\begingroup\$ @user1285419 The op-amp doesn't need a specific ground connection - all it needs are reference points for amplification and signals and R1 connected to 0V ensures the device works fine. Should you have had offset power rails like +7V and -15V, all the op-amp knows is that there is a refrnce point somewhere near the middle of its range. \$\endgroup\$
    – Andy aka
    Commented Aug 28, 2013 at 19:25

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