# Low pass filter issue - is the 10k resistor part of the circuit?

This may be a really simple question but Op Amps still confuse me a little. I have a low pass filter circuit as shown below.

My question is whether the 10k resistor is part of the filter circuit or is it just for input current biasing reasons. If it is part of the filter circuit how do you derive the transfer function?

From looking at it i initially thought that the filter was the 47 Ohm resistor and capacitor then using 1/2*piRC that would give the cutoff frequency but i am not sure this is correct.

Thanks for your help - its really appreciated!

• Really, it i just for input current biasing reasons. – Circuit fantasist Feb 2 '15 at 14:46
• @Circuitfantasist ...yes, and it's useful only if the (not shown) internal resistance of the signal source is around 10k, too. Unless there is a subtler reason (coupled with parasitic input capacitance for example to shape the HF feedback response). – Rmano Feb 2 '15 at 18:57

The filter is effectively the R-C output network with a 33.8kHz roll off (0.707 x Vin). It would be better to put a 470 pF cap across the 10K feedback resistor and omit the shunt 0.1uf. In that way, the opamp output current requirement (current output slew rate) is eased (output impedance is also fixed at 47 ohms) and the cost of the capacitor is greatly reduced. Buying a 0.1 temp stable (COG or NPO) cap is a  problem and if you use a ceramic 0.1, you also need to verify that it is voltage stable. Original configuration is just a voltage follower followed by RC lowpass.