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Understanding the Gain and Bode Plot of a Low Pass Filter Circuit with C1 > C2
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[QUOTE="Henryk, post: 5394515, member: 568354"] Hi, I'm afraid your answer is not correct. The simplest way to analyze the circuit is to note that ## C_1## and ##C_2 ## form a capacitive voltage divider connected to ## V_{IN} ##. The divider is equivalent of ## C_1## and ##C_2 ## in parallel connected to the voltage source ## V_{eq} = V_{IN} \frac {C_1} {C_1 + C_2} ##. Then connect the resistor between the output of the two capacitors and the ground. What you have is a high pass filter with the corner frequency given by ## \omega _c = R *(C_1+C_2)## If ##C_1 \gg C_2 ##, then you can simply ignore the second capacitor. The equivalent circuits is showing you why you can do that. I'm attaching a drawing of the equivalent circuit. [ATTACH=full]96561[/ATTACH] [/QUOTE]
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Understanding the Gain and Bode Plot of a Low Pass Filter Circuit with C1 > C2
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