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Why is there a DC offset in an amplifier?
– That resistor from amplifier side to ground is to ensure that any input bias currents from the amplifier’s input terminals (if it’s an op-amp) will not charge the capacitor with DC. When this happens, the output disappears because the input is pegged to a constant DC value. Usually, something like 100 K for a typical op-amp will due just fine.
How to remove DC offset from a signal?
If you wanna get technical, you can look up the worst-case bias current in the datasheet and do the maths necessary to derive the “nominal” resistor value…or just stick 100 K in there Thanks! I added a diagram above which from what has been said, I think may work.
What kind of gain control does a microphone have?
Some microphones even have automatic gain control (AGC) ICs. When the mbed is in debug mode the internal analog noise is a bit higher, an interface powerdown can force it out of debug mode (but you need to hold down reset to download new code, so that can get a bit messy).
What is the output voltage of a microphone?
No additional op amp is included, the output peak-to-peak voltage has a 0.67V DC bias and about 100mVpp (peak-to-peak) when talking near the microphone, which is good for attaching to something that expects ‘line level’ input without clipping. The peak-to-peak can be as high as 1Vpp when there is a very loud sound.
What happens when you remove the DC offset?
So without a DC path the offset current will charge the cap so that input pin will wander way off ground, and the amp will amplify that DC. So eventually the output will tend to one power rail or the other. The solution is to add a resistor from the input going to the cap to ground. The value of the resistor depends on the value of the cap.
How does DC come through an op-amp?
It’s not really floating, it’s attached to an op-amp. The op-amp has a high impedance, but it’s not infinite. A change in charge on one plate causes a movement of charge on the opposing plate. So AC comes through, but not DC.
How does a capacitor remove DC from the source?
Since the second leg of the capacitor is floating, the voltage of the source will just pass right by the cap. The capacitor will remove any DC from the source. But you shouldn’t let the input float like that.