The circuit below shows my setup. The resistor and capacitor act as passive low pass filter. In my setup the C1 is grounded to GND (0 V) but the op-amp operated with dual supply (+-5V). My question is this, do I need to ground the C1 to -5V for negative voltage inputs to make the filter work well with negative voltages? basic circuit


The correct place to ground it is to 0 V. It is the reference for all voltages in your circuit.


simulate this circuit – Schematic created using CircuitLab

Figure 1. For perfectly regulated power rails C2 and C3 would perform in this application exactly the same as C1.

Theoretically you could also connect the bottom of the capacitor to -5 V or even +5 V provided they are very stable. The capacitor in any of the three cases (+5, 0, -5 V) will provide an identical impedance to AC signals but will just have a difference in the DC voltage across it.

  • The reason for using 0 V normally is that the power-rails usually have some noise on them due to variations in current demand and the voltage drop that this causes along the wiring / connections.
  • Another reason would be to prevent switch-on transients. If the C2 option was used then OA1 non-inverting input would be pulled to V+ on switch on. Typically we want to avoid this kind of action. In an audio circuit, for example, it would result in a large DC "thump" from the loud speakers.
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  • \$\begingroup\$ +1 for thump. I would also mention though, if for some reason you are using a polarized capacitor for C1, e.g. electrolytic or tantalum, you are pretty much forced into tying it to the lowest rail. I can't imagine why you would use one in a filter though unless you live somewhere that makes parts procurement difficult. \$\endgroup\$ – Trevor_G Aug 11 '17 at 13:29
  • \$\begingroup\$ Hi, @Trevor. Do I detect an audio interest in this and other recent comments of yours? \$\endgroup\$ – Transistor Aug 11 '17 at 14:45
  • \$\begingroup\$ LOL.. yes audio stuff always fascinates me. It's one of those EE areas that, on the face of it, seems simple, but in reality it is quite an art. \$\endgroup\$ – Trevor_G Aug 11 '17 at 14:48

The capacitor just needs to have one of its ends at a fixed voltage, so you could use GND or even VDD.

If you however are going to output negative voltages then you need to feed the op-amp with a negative voltage, otherwise it won't be able to make that voltage.

If however you are only going to use AC voltage, then you can put another capacitor straight after \$V_{out}\$. That way the DC voltage will disappear and you will be able to make AC signals (positive and negative) without needing to use a negative voltage source for the op-amp.

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  • \$\begingroup\$ For clarity: "*... straight after and in series with \$V_{out}.*\$ \$\endgroup\$ – Transistor Aug 11 '17 at 13:00
  • \$\begingroup\$ @Transistor Ah yes, you are correct, thank you for clarifying that. \$\endgroup\$ – Harry Svensson Aug 11 '17 at 13:04

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