I thought of filtering out my DC adapter power supply for better radio reception, and I think it worked somewhat better.

Here is how I made my filter:

I took two fat diodes (diode #1N5 something), and connected the anode of one diode to +ve of the battery, and the cathode to the +ve connection point on my circuit board. Then I connected the anode of the other diode to the -ve connection point on my circuit board, and the cathode to the -ve of the battery. I connected the opposite ends of the two diodes together through 1000uF capacitors. It's like almost making a bridge.

I wonder if I should make a bridge.

maybe someone has a better filtering circuit I can make.

 

That is a filter circuit? I'm not sure how diodes are helping things, I'm sure it is attributed to the capacitors alone.

If you want better filtration, then use a PI filter. This is made using a large capacitor across the supply, then an inductor in series with your circuit, then a capacitor across the circuit.

http://en.wikipedia.org/wiki/Pi_filter

Steve

 

Get rid of the diodes. All they are doing is dropping your supply by 1.4V.

If your current draw is low enough, use a linear regulator.

 

if I use the PI filter, what are the best values to use for C1, C2, and L1?

I think C1 and C2 can remain at 1000uF, but I'm wondering about L1. I think that if L1 is too high, then it may end up being a high in-series resistor.

 

Choose L such that it represents a large impedance at or above frequencies you wish to attenuate. Inductive reactance (Xl) is equal to 2*pi*f*L where f is the frequency and L is the inductance.