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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.

enter image description here

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!

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    \$\begingroup\$ Really, it i just for input current biasing reasons. \$\endgroup\$ – Circuit fantasist Feb 2 '15 at 14:46
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    \$\begingroup\$ @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). \$\endgroup\$ – Rmano Feb 2 '15 at 18:57
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If you have an ideal OpAmp the 10k resistor would carry no current and could be replace by a resistor with an arbitrary value, including a piece of wire.

Therefore the answer is no, the 10k resistor plays no role wrt. the low pass filter.

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The opamp acts only as a simple buffer with unity gain (100% feedback). Because there is no voltage divider, the value of the feedback resistor has no influence on the buffer operation and - hence - not on the filter.

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The 10k resistor has no affect on the passive filter pole, but the resistor, together in feedback with the op-amp do have an effect on the filter by increasing the input impedance.

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    \$\begingroup\$ The input impedance is enhanced due to negative feedback. However, the amount of feedback does not depend on this resistor - the feedback factor is always F=1. Hence, this resistor does not influence the input resistance. \$\endgroup\$ – LvW Feb 2 '15 at 17:39
  • \$\begingroup\$ @LvW you are absolutely right - effectively an emitter follower. \$\endgroup\$ – docscience Feb 3 '15 at 3:48
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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.

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