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2nd order Butterworth High Pass filter

Alex_Bam

Sep 28, 2020
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Hello,
Could anyone review my 2nd order Butterworth HPF design? I have mentioned design details along with schematics and initial results.

Design specifications for 2nd HPF: Cutoff frequency= 100Mhz, Design method: by considering equal component, Input signal= 10V,120MHz, sinusoidal.

Design specification for OPAMP: Slew rate= 5V/usec , CL=10pF, ICMR+= 4.5V, ICMR-=1.5V , Vdd=5V, GBW>=5MHz, W/L ratio is mentioned with each transistor.

According to the calculations, it should work but I do not know where I am making mistakes...

Thanks
 

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Harald Kapp

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I do not know where I am making mistakes...
Please explain your problem.

Design specification for OPAMP: Slew rate= 5V/usec
This is absolutely inadequate for a 120 MHz, 10 V peak signal.
The max. slew rate of a sine wave occurs during zero crossing with Slew rate = 2 × π × f × V (link) = 7540 V/µs.
Your slow opamp with 5 V/µs is not able to follow the signal.
This is also obvious from the GBW >= 5 MHz, a far cry from the 120 MHz signal you want to process.
 
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