LED Into an MXR Envelope Filter ??

Started by akg414, September 11, 2010, 11:19:34 AM

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akg414

I'm trying to install an LED into my 70's envelope filter. I have the 3PDT switch and a diagram of a MXR E.F. (clone).

Everything is lining up, EXCEPT where to solder the long-leg of the LED to?

The CLONE diagram has a place for it, but the actual circuit board doesn't.

Anyone know where I should tap into on the MXR's circuit board?

I've posted some pictures below to help illustrate my point.  Any help would be greatly appreciated - as I'm kinda new at this  :icon_wink:






Govmnt_Lacky

The switching of the LED on and off is controlled by the ground (short leg of the LED) which you should have attached to one of the 3PDT lugs.

The long leg of the LED (power) should be connected to your power input (Battery+ or DC jack+) via a pull-down resistor. If you notice in the layout, the LED is connected... in some form... to power but it goes through a 4K7 resistor. This resistor is NECESSARY to adjust the voltage from the power supply/battery to be used by the LED.

Just solder the long LED leg to one side of a resistor (4.7K for green, 1K for red, 10K for blue) and the other side of the resistor to your power in.
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

Govmnt_Lacky

#2
Good luck and let us know how it goes!  :icon_biggrin:
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

akg414

Quote from: Govmnt_Lacky on September 11, 2010, 11:49:02 AM
The switching of the LED on and off is controlled by the ground (short leg of the LED) which you should have attached to one of the 3PDT lugs.

The long leg of the LED (power) should be connected to your power input (Battery+ or DC jack+) via a pull-down resistor. If you notice in the layout, the LED is connected... in some form... to power but it goes through a 4K7 resistor. This resistor is NECESSARY to adjust the voltage from the power supply/battery to be used by the LED.

Just solder the long LED leg to one side of a resistor (4.7K for green, 1K for red, 10K for blue) and the other side of the resistor to your power in.

The red (center) wire is where the battery goes in on the actual board. So I need to solder the resistor to (THAT) point...  and the connect the LED according?

Govmnt_Lacky

Unless you are going to add a DC Jack I would leave the LED out. If you really insist on it then your V+ from the battery would need to be split into two. One would go to the RED wire that you are talking about. The other would go, via the resistor, to the LED.
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

akg414

Quote from: Govmnt_Lacky on September 12, 2010, 02:16:44 PM
Unless you are going to add a DC Jack I would leave the LED out. If you really insist on it then your V+ from the battery would need to be split into two. One would go to the RED wire that you are talking about. The other would go, via the resistor, to the LED.

Absolutely, I'm definitely going to include the power-supply jack as well.

akg414

Quote from: Govmnt_Lacky on September 12, 2010, 02:16:44 PM
Unless you are going to add a DC Jack I would leave the LED out. If you really insist on it then your V+ from the battery would need to be split into two. One would go to the RED wire that you are talking about. The other would go, via the resistor, to the LED.

And thanks man for the help, much appreciated  :icon_wink: