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I'm working on a project with lots of op amps and so far had no problem but this one is driving me nuts. I need a 1:2 non-inverting amplifier which I built on breadboard but doesn't work as expected. The circuit is trivial, I tried different op amps (LM741,TL072,LM358), use a rail-splitter or just a voltage divider for the bias DC, with our without the DC decoupling caps and yet it doesn't work. It seems that negative half of the signal is gone (yellow being the output signal while the incoming signal is from a signal generator). Thank you for your help in advance!

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What power supply voltages are you using and what is the DC offset at the opamp output? –  jippie Jan 19 at 7:08
What is the label "virtual ground" connected to? Is it left open? –  Tom L. Jan 19 at 7:28
The power supply is 9V DC and the virtual ground comes from a TLE2425 rail splitter and it's 4.5V. –  alkopop79 Jan 19 at 7:35
The signal comes from the scope's signal generator. It's 3.3V and I tried 0-3.3V DC and did not make much difference. –  alkopop79 Jan 19 at 7:36
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2 Answers

up vote 7 down vote accepted

You have nothing setting the operating point of your positive input to the op-amp. As such the input bias current of the op-amp is causing C1 to charge up to one of the rails, and it's therefore cutting off part of your signal.

This issue will likely be exaggerated in simulations as well, since few simulations properly account for the input bias currents and input networks of their virtual op-amps. If you are using a simulation, try measuring the voltage on the + input of the op-amp. I suspect you'll find it exceeds the supply rails when the output does so as well.

You need a large-value resistor from the + input of the op-amp to the virtual ground. That resistor will set the operating point for the + input pin, and fix the issue.

Issue as described
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It seems to work fine with audio range signal but unfortunately the signal I need the amp for is stepped voltages (control voltage) and the frequency is very low. I'm concerned that the cap on the input might block the DC (which is here unwanted!). Any ideas? –  alkopop79 Jan 19 at 9:00
The cap on the input will block DC, as will the cap on the output. Isn't that why you put them there? If you need your circuit to work at DC, you can't have any caps in the signal path at all. –  Connor Wolf Jan 19 at 11:11
Thank you!!! I needed to add a resistor from the positive input to the ground (V-) to 'shift' the signal down. You were right! –  alkopop79 Jan 20 at 7:27
If you are trying to measure a signal that exceeds your supply voltages, you will indeed need some way to apply a DC offset correction to it. Your resistor to V- is likely doing this, but you should be aware if you don't have a series resistor inline with the input, you are relying on the signal-source's impedance to form part of a resistor divider. This can be a bad idea, if the source impedance can vary, or you have multiple source-devices with different internal resistances. –  Connor Wolf Jan 20 at 7:34
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It appears that you are referencing ground for V1 (which implies that V1 goes negative at some point), and you are using a virtual ground for a reference and a rail-splitter to set the reference at 4.5V. I am assuming that this means you are using a single-supply setup (op amp Vcc is connected to 9V and Vee is connected to ground). If this is the case, then the op amp is clipping because it can't drive the signal below ground when the sine wave goes below 4.5V. When Vin is below 4.5V (or the virtual ground reference value), the input is negative relative to it, so the op amp will try to drive a negative voltage so that V+ and V- (voltages at the non-inverting and inverting inputs, respectively) are equal.

You have the input source referenced to ground right now (ex. the sine wave input has the minus side connected to ground). Try connecting it to 4.5V. That should allow you to get to 0V on the input source before clipping occurs. Ideally, you would offset it by half of the peak-to-peak voltage, which gives you equal room on both sides. Keep in mind that with a gain of 2 (though as you have drawn the circuit, you have a gain of 4), you should ideally offset the input source by 4.5V, and the peak-to-peak voltage should be no larger than 4.5V, or an amplitude of 2.25V.

Hopefully that helps.

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