Wednesday 8 June 2016

operational amplifier - Is this correct for non-inverting op-amp?


op-amp sheme



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:



enter image description here


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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