Do 0.1 uf and 100 k ohm act as low-pass filter in here?
Answer
Do 0.1 uf and 100 k ohm act as low-pass filter in here?
No, since there is no current through the parallel RC network, there is zero volts across the network and so
$$V_{out} = V_- = V_+ = V_{in}$$
thus, the circuit implements a unity gain buffer:
$$V_{out} = V_{in} $$
Another way to see this is to consider the canonical non-inverting configuration in the frequency domain:
The transfer function (frequency response) is:
$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{Z_2}{Z_1} $$
In the schematic you provide, \$Z_1\$ is an open circuit, i.e. \$Z_1 = \infty \$. Thus, the transfer function for your circuit is:
$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{Z_2}{\infty} = 1$$
which is, as before, just a unity gain buffer. Now, if you place a resistor \$R_1\$ to ground from the inverting input, the transfer function is:
$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{100k\Omega || \frac{1}{j \omega (0.1\mu F)}}{R_1}$$
which will provide some filtering - the details of which I will leave as an exercise for the reader.
:)
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