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I am working on the non-inverting amplifier using a single supply, which can amplify bipolar input signal. Non-inverting amplifier is working fine without any issues, only the negative portion of the input is clipped. The schematic is shown below.

Fig. 1

To amplify both positive and negative portion of the circuit, I gave DC bias voltage at non-inverting terminal, but in the output I was not getting any signal. Schematic is shown below. I even simulated the circuit and I got proper waveform amplifying both positive and negative signal, but while making, it is not working. I am getting flat line in the oscilloscope.

Fig. 2

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  • \$\begingroup\$ Which op-amp are you using? Link to the datasheet? What's your source impedance? Why 1K resistors for your non-inverting input bias, you could use much higher values for higher input impedance. Try 100K or more instead of 1K. \$\endgroup\$ – John D Jun 30 '19 at 19:18
  • \$\begingroup\$ The correct circuit already explained in detail by Olin Lathrop here: DC offset of amplifiers for 12V single supply operation. If you also care about the DC offset to be amplified or adjustable, or even also gain less than one, then see my answer here: differential and inverting op amp problems \$\endgroup\$ – Unknown123 Jul 1 '19 at 22:50
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Good start, just one more thing to add... R1 cannot be DC-grounded. It must be AC-grounded with a capacitor. You would choose the capacitor value so that its reactance is equal to (or less than) R1 at the lowest frequency that's important to you. For example, if R1 is 1000 ohms, and you are amplifying audio where 20 Hz is the lowest audio frequency, C1 (below) is about 10uF.

With such a low DC supply voltage, a rail-to-rail opamp is a good choice...many common opamps cannot provide enough output signal swing with a +5V supply.

schematic

simulate this circuit – Schematic created using CircuitLab

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Your circuit still has DC gain. You need to add a cap between R1 and GND. The output will have a DC offset. If that needs to be removed, add a series cap between the output and whatever is downstream. Since you have not provided any frequency information, you're on your own to determine the values.

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Those two 1k input resistors will probably be too low in value. They give an input resistance of 500R (1k//1k) which, unless the signal source resistance is very low, will reduce the amplitude of the input signal quite significantly.

I would recommend swapping them for 100k resistors (more typical value) giving an input resistance of 50k. The amount of signal then lost, by the potential divider formed by the source resistance and input resistance, will be much less and the gain of the amp will be closer to 1+(R2/R1).

The input resistance and source resistance in combination with the input capacitance form a high pass filter the cut-off frequency of which is given by:-

fc = 1/(2*piC(Rs + Rin))

So if the source resistance is known (or assumed to be 0), the input resistance is known (say 50k) and the cut-off frequency is required to be say 20Hz then the required value of the input capacitor can be calculated.

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