# Filter of DC does not work?

I have the following simple circuit: I thought that the capacitor under R1 can be used to filter off the offset of the DC, but instead I am still measuring (steady) voltage on the output.

Is this the correct way to filter the DC? What am I doing wrong?

P.S. Capsule=Microphone

• What is connected to your output? If nothing is connected then the capacitor will do nothing to the signal. Add a pull down resistor to ground to the output and the DC offset should go away. – Tom Carpenter Oct 31 '15 at 17:03

You would need a resistor from the output to Ground to remove the offset. Without a resistor to ground, leakage in the capacitor will hold the output voltage near the voltage at the bottom of R1.

Depends on the value of the capacitor and resistor, and the load applied to the other side of the capacitor. The "charge" and "discharge" of a capacitor promotes a current into and out of the capacitor plates. If there is enough current to load the capacitor plates, at a time this current will reduce to virtually zero (never zero), and you will measure almost zero volts.

Your voltmeter has a resistance, normally in MegOhms range. When measuring voltage, the current that flows through the voltmeter is the responsible to develops a voltage that the voltmeter will shows.

If the capacitor is fully loaded, no more current will flow through it, so you will not measure any voltage at its output. But if the capacitor is not fully charged yet, what must be the case in your example, you will measure voltage.

If you multiply the value of the resistor by the capacitor (ohms and farads), you will have a value in seconds that means how long the capacitor will take to charge 66% of its total charge.

If you don't have a load at the right side of the capacitor, the circuit is incomplete, so the capacitor never charge.

However, if you add a small resistive load from that capacitor lid to ground, lets say, 10k to 100k ohms, the capacitor will charge fast and you will measure no more voltage.

Nearly all capacitors will have some small amount of DC current leakage. So with a high impedance voltmeter at the output the small leakage current will register as a voltage. Large value polarized capacitors can have fairly large amounts of leakage, especially electrolytic types. The manufacturer's specifications will give the value of DC current leakage at various operating conditions.

If you were to use a high quality low value ceramic, disc, or mica capacitor you would most likely measure 0 VDC with most standard voltmeters.

As already mentioned, adding a pull down resistor after the capacitor can reduce the measured voltage to near zero (as the voltage divides passing through the high DC resistance of the capacitor).