# Opamp DC offset Question

Just need some help understanding what's happening in removing the DC offset from two band-pass filters.

VG has a value of 4.5V. When I added in the virtual ground the circuit had an offset of 4.5V at the Out and Out2 nodes. I added in capacitors C7 and C8 at 1$\mu$F understanding that it will block the DC signal. This worked for the node at Out but not Out2. Later I saw a similar problem where the solution was to put a resistor between the output and opamp which worked but I do not understand why the resistor is needed and why Out didn't need one but Out2 did.

The resistor that was added was R7.

• Explain specifically where the resistor was placed and also explain how you measured the voltages. Jul 9, 2018 at 11:12
• Oh yep sorry, I'm talking about R7 was the resistor that was placed and measured the voltages using the Transient analysis on LTSpice. So Vin is a 1kHz sine wave with an amplitude of 1 and I'm measuring the signal at Out and Out2. Jul 9, 2018 at 11:14
• You shouldn't need C7 or C3. The output of C8 will probably float up to 4.5VDC without R7.
– JRE
Jul 9, 2018 at 11:21
• Capacitors have a large (but not infinite) resistance at DC.
– JRE
Jul 9, 2018 at 11:21

## 1 Answer

When you run a simulation, in most circumstances, the default starting voltage across a capacitor is 0 volts. This means, that at the instant of beginning the simulation, if 4.5 volts is forced onto one plate of that discharged capacitor by the op-amp output, 4.5 volts is also seen on the other plate. Without a discharge resistor (such as R7), the OUT2 node will remain with an average level of 4.5 volts.

• Oh ok I understand. So if I were to hypothetically build this circuit with some sort of output load at Out2, the resistor wouldn't be required correct? Jul 9, 2018 at 11:23
• Any reasonable DC path to 0 volts will achieve your aim. Jul 9, 2018 at 11:26