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I was watching this video by w2aew on the function of ceramic bypass capacitors, and at 6:00 he states that high-frequency currents take the path of least-impedance and make their way to the power ...
#2: Post edited
Hello, everyone!- I was watching this [video](https://youtu.be/9EaTdc2mr34?t=361) by w2aew on the function of ceramic bypass capacitors, and at 6:00 he states that high-frequency currents take the path of least-impedance and make their way to the power rail through the ceramic capacitor next to the IC.
Okay, so I get the fact that the high frequency content wants to avoid the high-Z path created by the return loop, but I also thought that current has to return back to its source through the negative terminal, to the power supply, and then flow out of the positive terminal. So why is the capacitor shunting it back directly onto the supply rail? Could someone please explain to me where my understanding of current flow, or whatever else, is wrong?Thanks.
- I was watching this [video](https://youtu.be/9EaTdc2mr34?t=361) by w2aew on the function of ceramic bypass capacitors, and at 6:00 he states that high-frequency currents take the path of least-impedance and make their way to the power rail through the ceramic capacitor next to the IC.
- Okay, so I get the fact that the high frequency content wants to avoid the high-Z path created by the return loop, but I also thought that current has to return back to its source through the negative terminal, to the power supply, and then flow out of the positive terminal. So why is the capacitor shunting it back directly onto the supply rail? Could someone please explain to me where my understanding of current flow, or whatever else, is wrong?
#1: Initial revision
Bypass Caps and High Frequency Current Return Path
Hello, everyone! I was watching this [video](https://youtu.be/9EaTdc2mr34?t=361) by w2aew on the function of ceramic bypass capacitors, and at 6:00 he states that high-frequency currents take the path of least-impedance and make their way to the power rail through the ceramic capacitor next to the IC. Okay, so I get the fact that the high frequency content wants to avoid the high-Z path created by the return loop, but I also thought that current has to return back to its source through the negative terminal, to the power supply, and then flow out of the positive terminal. So why is the capacitor shunting it back directly onto the supply rail? Could someone please explain to me where my understanding of current flow, or whatever else, is wrong? Thanks.