I am starting to learn to use operational amplifier recently. I construct a simple circuit with an amplifier (http://www.ti.com/product/op07c) with is driven by 9V (on VCC and GND). I have the V+ and V- connected to a function generator providing a sinusoidal signal of 500mV and 100kHz (also tried 100Hz). I have the output (Vout) feedback to the V-. T Vout connect to the V- via an 200ohm resistor (as shown in the fig). But if I observe the output with scope, it is not really a sinusoidal wave (it more like the positive section of the sine is flat while the negative section of the sine is like sinusoidal. But if I connect the Vout to V+ with an 200ohm resistor, the output is pretty well sinusoidal but the amplitude is attenuated instead of magnified. enter image description here

  • \$\begingroup\$ Pics and schematics, or it didn't happen! :) \$\endgroup\$
    – user3624
    Commented Aug 26, 2013 at 1:06
  • \$\begingroup\$ A schematic and a picture, as David said, would be very helpful. \$\endgroup\$
    – arthur.b
    Commented Aug 26, 2013 at 1:08
  • \$\begingroup\$ Your text says, "The Vin+ and Vin- are connected by a 200ohm resistor", but your schematic doesn't show this. Could you fix whichever one is not telling us how you built your circuit? \$\endgroup\$
    – The Photon
    Commented Aug 26, 2013 at 3:57
  • \$\begingroup\$ Don't use "MSPAINT.EXE" for schematic capture. :) \$\endgroup\$
    – Kaz
    Commented Aug 26, 2013 at 6:05

2 Answers 2


Just looking at the behavior you described and the datasheet there are a couple of points I want to present.

  • Since this doesn't look like a rail-to-rail OpAmp you will have some trouble with the available range in the output. This means in your case that it will probably only be able to swing to something like 1.5v to 7.5v (assuming a 1.5v drop by the OpAmp). If you try to generate an output higher than those values it will simply clip the value, something like your "flat top".
  • The other problem that you may encounter is that you are operating near the Unity-gain bandwidth of this IC (which is typically 400kHz, by the datasheet). This means that your OpAmp will operate different than a ideal opamp (which has inf. gain).

Try to reduce the gain of the circuit, or the amplitude of the signal. Also you can try to ad an offset to your signal (shift it up or down).

Also for a great reading about electronics take a look at the book The Art of Electronics. You can go directly to the first chapter about OpAmps, and if you got interested devour it from begin to end.

  • \$\begingroup\$ Thanks. I think I need to learn something new, I really have no idea what's rail to rail mean but just like what you said, the "flat top" was observed. I tried to decrease the freq to 100Hz but still the same. \$\endgroup\$ Commented Aug 26, 2013 at 1:21
  • \$\begingroup\$ Take a look at the last paragraph I just added \$\endgroup\$
    – arthur.b
    Commented Aug 26, 2013 at 1:25
  • \$\begingroup\$ I am testing the circuit again based on your suggestion. The peak-to-peak voltage of the signal is just 500mV and since I chose the feedback resistor to be 200ohm, I think the gain is just two (1+R1/R2 = 1+200/200=2). When you mentioned the output is probably within [1.5v, 7.5v], do you mean the model of the OP limited the output or there is something like the driver voltage (here I use 9V) will affect that also? Will it work if I drop the driven power (Vcc) to 5V instead? \$\endgroup\$ Commented Aug 26, 2013 at 1:49
  • \$\begingroup\$ The opamp can only swing it's output up to near the voltage rails (in your case 0v to 9v), I assumed in that calculation that it could go near up to 1.5v to the rail so [1.5v,7.5v]. Increasing the supply voltage would wield a greater excursion, dropping it will only make it worse. \$\endgroup\$
    – arthur.b
    Commented Aug 26, 2013 at 3:14
  • \$\begingroup\$ Another thing that is happening is that your Opamp doesn't have a infinite gain open loop gain, so the equation (1+R1/R2 = 1+200/200=2) isn't valid. You have to consider the open loop gain that in your case should be about 4 ( using the GBWP of 400khz, divided by your signal freq of 100 = 4), taking that in consideration your gain will be something like 1.33 (users.ece.gatech.edu/mleach/ece3050/notes/OpAmps/opampbw.pdf) \$\endgroup\$
    – arthur.b
    Commented Aug 26, 2013 at 3:17

Aside from the issues brought up in the other answers, your circuit has some more basic problems. By connecting the V- input directly to the inverting pin of the op-amp, you are effectively shorting out the resistor between that node and ground. This will cause the op-amp to act as a comparator, rather than produce a linear gain.

(Note: putting part designators in your schematic would make it much easier to discuss it)

If your function generator has a differential output, the classic op-amp circuit to amplify a differential signal looks like this:

enter image description here

(picture from the excellent Op-Amps for Everyone from TI)

Here, V1 and V2 would be the two output pins of your function generator. Notice that each input has a resistor between it and the op-amp input pins.

Edit in response to edits on the question.

OK, your circuit now shows explicitly that the op-amp's inverting input is tied to ground. The 200 Ohm resistor, now connected between ground and ground, can do nothing in this circuit. It will not work as you wish.

The classic non-inverting amplifier configuration looks like this:

enter image description here

Notice that the negative terminal (BNC outer conductor, in your case) of the input source is only tied to this circuit through the ground symbol. Having R1 between the inverting input of the op-amp and ground allows feedback to drive the inverting input equal to the non-inverting input. This is usually what we want to happen in an op-amp circuit with negative feedback. In your circuit it can't happen because the inverting input is tied directly to ground.

  • \$\begingroup\$ Thanks. Here when you say "V1 and V2 would be two output of your function generator" do you mean positive and negative terminal of the output? My function generator's output is a single BNC terminal, so I assume V1 is the positive of BNC and V2 is the ground. And in my real circuit, I put all grounds (function generator, scope and OP amp) to common. Is that correct setup? \$\endgroup\$ Commented Aug 26, 2013 at 4:21
  • \$\begingroup\$ If your function generator output is a single BNC terminal, that is probably not a differential output. The outer conductor of the BNC should probably be tied to your circuit's ground. Please update the schematic in your question to reflect your real situation. \$\endgroup\$
    – The Photon
    Commented Aug 26, 2013 at 4:23

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