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schematic

simulate this circuit – Schematic created using CircuitLab

I have faced a problem in understanding, the following circuit. But after deep search, I expect to be Differential Low Pass Filter. If so, why we have make the first Capacitor common with Vin+ and Vin-, and the next two capacitors, we have put a 'ground' between them ?

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The filter is both a differential filter and a common mode filter. That is why the caps are configured the way they are. Think about both inputs being joined together - the two caps to ground service the common mode filter function.

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  • \$\begingroup\$ what is the benifits that i get , to make a filter work on both mode ? \$\endgroup\$
    – hbak
    Jun 22, 2014 at 15:10
  • \$\begingroup\$ You don't have to worry about out-of-band common mode noise making it to your differential filter. \$\endgroup\$ Jun 22, 2014 at 15:17
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If you had a differential filter then you'd have differential outputs. You don't. In fact this isn't even a low pass filter in the signal domain, it is a Common mode filter in that it extracts the common mode of the differential inputs and cancels everything else (\$R_3 , R_4)\$.

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    \$\begingroup\$ I suppose, in reality both outputs are not connected. It seems the diagram is a mixture between circuit diagram and block diagram because the last block is called "diff. amplifier". \$\endgroup\$
    – LvW
    Jun 22, 2014 at 15:15
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    \$\begingroup\$ @LvW the Vin+ and Vin- are being combined by \$ R_3 & R_4\$ and exactly cancel. the only remaining signal is the common mode. \$\endgroup\$ Jun 22, 2014 at 15:18
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    \$\begingroup\$ Since the downstream component is labeled "differential amplifier", I suspect OP didn't draw the schematic accurately. \$\endgroup\$
    – The Photon
    Jun 22, 2014 at 15:35
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    \$\begingroup\$ @ThePhoton The circuit however, is simply extracting the Common mode. In some circuits this would then feed into the differential amplifiers CMFB pin. So it's also possible that the OP also misunderstood it's purpose and the drawing is fine. That just isn't the differential input. In either case, the circuit is not what he thinks it is. \$\endgroup\$ Jun 22, 2014 at 16:38
  • \$\begingroup\$ @placeholder: I am aware that the circuit (as shown in the figure) extracts the common mode signal only. However, the question is if that is really the purpose of the circuit (intention of the questioner)? In any case, it is not clear how the the circuit is connected to the next stage (diff. amplifier). \$\endgroup\$
    – LvW
    Jun 23, 2014 at 8:13

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