DC adapter noise cancelling board

[ghiekorg]

Hallo everyone,
I would like to build a simple board to filter the noise of cheap DC adapters. I found this online:



I have absolutely no knowledge of circuits, do you think i can do it like this? I have substitute the 4007 with a 4001 (i don't have a 4007) and instead of a 220uf i placed 2x 100uf and 1x 20uf in parallel. Does it make sense?



Thanks a lot

[Rob Strand]

Don't bother with the 20uF.   The values are not critical.   One guy might use 20uF alone, another guy might use 100uF.   Usually the larger the better.

You can use 1N4007.    With the 100 ohm present you would be better off moving the diode to the output side since then the diode will protect the circuit and the 100 ohm will protect the diode.

You should draw the circuit with +V on the top line and 0V (gnd) on the bottom.   That's a more conventional way to draw it.

The 100ohm does limit the amount of current draw.   Perhaps only 10mA.   You can reduce the resistor value to get more current but then the filter isn't as good.

For high currents you can replace the resistor with an inductor,

https://www.diystompboxes.com/smfforum/index.php?topic=124690.0

(With this circuit it also has electrolytic caps on the input side.    You could consider this optional.
With the electrolytics on the input side you really need to move the diode back to the input side.)


I'm sure the issue of filtering switch-mode wall warts will come-up more often as the old transformer based warts are becoming harder to get.

[ghiekorg]

thank you for the quick answer

Don't bother with the 20uF.   The values are not critical.   One guy might use 20uF alone, another guy might use 100uF.   Usually the larger the better.

Ok thanks, that's better so i can make it smaller

You can use 1N4007.

You mean 1N4001? I only have 4001 atm...

With the 100 ohm present you would be better off moving the diode to the output side since then the diode will protect the circuit and the 100 ohm will protect the diode.
You should draw the circuit with +V on the top line and 0V (gnd) on the bottom.   That's a more conventional way to draw it.

Is it correct now this way?

For high currents you can replace the resistor with an inductor,

I am not sure i understood this point. I mean, i am just using this to test stompboxes at home and even if i would put it in a pedalboard (i think in case i would just buy a new DC adapter with an already made filter) i would use it to give power to not more then 3 pedals at the same time. I don't know if this has anything to do with what you are talking about (i am really ignorant about that)

[11-90-an]

>You mean 1N4001? I only have 4001 atm...

Yes, that can be used also

Maybe move the 470uF cap after the diode... (might explode on you... )

[ghiekorg]

So like that?



Thanks a lot

[Rob Strand]

Ok thanks, that's better so i can make it smaller

2x100uF would be good.

You mean 1N4001? I only have 4001 atm...

Sorry, yes.  I got it around the wrong way.

Is it correct now this way?

Yep, looks good.

I am not sure i understood this point. I mean, i am just using this to test stompboxes at home and even if i would put it in a pedalboard (i think in case i would just buy a new DC adapter with an already made filter) i would use it to give power to not more then 3 pedals at the same time. I don't know if this has anything to do with what you are talking about (i am really ignorant about that)

If you budget 10mA per pedal then 3 pedals is 30mA and that's 3V drop across the 100 ohm resistor.  So your circuit only gets 6V.
The drop is calculated as Vdrop = 100 ohm * Current in amps;  mA  = milli-amps which is amps / 1000  so 10mA is 10/1000 = 0.01A

I think 10mA per pedal is a good budget unless you are doing digital pedals, which is going to be a lot more current, or old-school transistor pedals, which are going to be a lot less.

https://www.harmonycentral.com/forums/topic/1510243-pedal-power-consumption-list-in-milliamps-ma/

I think 100 ohms per pedal is good.  So three pedal means  R = 100 / 3 = 33.   So a 33 ohm  would be good

For higher currents it's hard to get good filtering because the resistor needs to be too small.   An inductor can give good filtering but doesn't  have the voltage drop.

Maybe move the 470uF cap after the diode... (might explode on you... )

So like that?

No need.  It's 470nF (non-polar) not 470uF.

[Ben N]

Rob: Any suggestion on inductor value range?

[11-90-an]

>No need.  It's 470nF (non-polar) not 470uF.

Sorry... didn't see it properly.. 

[Rob Strand]

Rob: Any suggestion on inductor value range?

Something like 330uH to 470uH is a good place to start.

Cheap ferrite bobbin types, just make sure the current rating is enough.


Just to be absolutely clear:  There's two types of wall-warts.   
- The old-school transformer types, which are a little bit heavy (and are usually unregulated).   
- The new type switch-mode power supplies, which are quite light, are regulated and usually have somewhat higher current ratings.   The phone chargers are the latter type.

The noise from the first type is hum.     The simple filter with the resistor will remove some of the hum.   However it will also remove noise from switch-modes.

The inductor based filter probably won't do much for the old-school warts with the hum.   The filter will only remove noise from the modern switch-mode types.

[ghiekorg]

Sorry last question, is the draw and the direction of the capacitors correct? I went a bit random on them
Thank you

[antonis]

Draw correct..
Directions semi-correct..
(flip 180^o left 100μF cap..)

For polarized caps, (+) plate should always face to more positive point..

[ghiekorg]

Thank you 😊

[mozz]

Tack a .01 in there somewhere. Or .001

[ghiekorg]

Tack a .01 in there somewhere. Or .001

Thank you. Doesn't matter where, just aligned to the other? and can i ask why?

[mozz]

Tiny or smaller value cap will get rid of higher frequency noise.

[ghiekorg]

Thank you. I just built one, As soon as i get the DC plugs i ordered I will tell you how it works 😊

[Sooner Boomer]

Just one small thing to add: make sure the resistor is large enough to handle the maximum current that will EVER go through it.
current x votage = wattage
Resistors (typically) come in 1/8, 1/4, 1/2, 1, 2 watts (and larger). If you have a doubt, go the next wattage larger (or double the resistance and use two in parallel).

[ghiekorg]

i built one to try. doesn't work what is wrong? i can't understand. It doesn't power on the pedal at all...


Thanks for your help

[antonis]

I think it powers the pedal with reverse polarity.. 

Output plug is wired as center pin positive where should be negative..

[ghiekorg]

omg i am so stupid... I was just thinking "center is hot, outside is ground", like an audio cable. I totally forgot about the - center
Thank you