Using high pass filter in front of an opamp

[preethi19]

Using high pass filter infront of an opamp

Hi i have attached an image of an op-amp circuit amplifying the signal from the microphone.

The link is https://www.youtube.com/watch?v=TQB1VlLBgJE
My question is in the video it is told that the microphone inputs 20mV signal and it is powered by the 9V supply. So what happens is der occurs a DC offset with the 20mV input signal shifting from 0 to 9V. So in order to remove this DC component they have used a high pass filter.. My question is


why use a high pass filter to block the DC component. A capacitor also blocks DC. So why not just use a capacitor in the path of the input signal to the opamp input to remove the 9V DC.

 

[BradtheRad]

Re: Using high pass filter infront of an opamp

I suspect that the resistor has the purpose to tie the input terminal to ground. It should not be allowed to 'float'. The capacitor alone is insufficient for that purpose.

The 100k resistance still allows the mic signal to get through to the op amp input. The mic signal has a much lower impedance than 100k.

 

[crutschow]

Re: Using high pass filter infront of an opamp

An op amp input must have a ground path to carry the small input bias current and establish a ground reference.
With only a capacitor in series with the input, it will drift to to one of the rails and the output will saturate.

 

[preethi19]

Re: Using high pass filter infront of an opamp

thank you for clearing my doubt!!! I understand why the resistor is used. inorder to tie the input to a ground path. But is der any reason why 100k value is chosen. (apart from cutoff frequency f=1/2*pi*R*C)..
In the video it was mentioned that "having a high resistance value will ensure that the microphone's un-amplified output is not overloaded".
what does this mean??? The microphone's unamplified output is 20mV. So using a high valued resistor helps in wat way???

 

[zorro]

Re: Using high pass filter infront of an opamp

The lower that resistor, the lower the voltage at the "+" input of the OA.

Look:
The microphone can be modeled with its Norton equvalent, i.e. a current source Is in parallel with a resistance Rm.
Let's call Ra the microphone bias resistor (5k) and Rb the OA bias resistor (100k).
At frequencies well above the cutoff frequency, the voltage on the capacitor can be neglected and the voltage at the "+" pin is Is*(Rm||Ra||Rb).
As usually Rm>>Ra, V=Is*(Ra||Rb).

But if Rb is too high, the bias current if the OA input (LM324 has a bipolar transistor as input stage) will cause a significant negative value at the input. That voltage, amplified 101 times would appear at the output and could saturate the OA.

 

[BradtheRad]

Re: Using high pass filter infront of an opamp

But is der any reason why 100k value is chosen. (apart from cutoff frequency f=1/2*pi*R*C)..

No particular reason to choose 100k. It should be much larger than 5k (resistor at far left). It should be much less than the op amp's input resistance for which a typical value is 1M.

In the video it was mentioned that "having a high resistance value will ensure that the microphone's un-amplified output is not overloaded".
what does this mean??? The microphone's unamplified output is 20mV. So using a high valued resistor helps in wat way???

The 100k loads the mic signal. But the load is slight. However if it were a 5 or 10 or 20k, then it would load the mic signal to such an extent to severely reduce its amplitude.