LED and Switch for circuit

[SparkyCal]

As some of you know, I have finally suceeded in building a guitar effects box, thanks to all the help I received here

This is the schematic I used. https://www.electronicspoint.com/forums/attachments/screen-shot-2020-06-22-at-10-21-12-pm-jpg.48602/

I am now in the process of putting it into a box. let me explain though.

At this point, I just want to build a few of these circuits, experimenting with different op amps, sound etc. So, I am not going to put a regular stomp box switch in these boxes. The reason for this is two fold:

1. I found some perfectly sized wooded boxes at the dollar store, that I will repurpose for use as the stomp box housing. As such, I don't want a switch that you have to stomp on, because the box won't be able to withstand it.

2. I'm just using a cheap on/off switch for now.

Here is my question:

Using the above circuit, can I add a red LED that will light to signify that the box is active (actually producing distortion), and is not lit when it is off?

I plan on using a simply switch that I will mount on the front or side of the box. I'd like it to work so that it kills the connection to the 9 Volt battery when not in use, and lights up the LED and sends power to the necessary parts of the circuit when in use. In that way, it will also serve to save the battery when not in use.

Can anyone suggest an addition to the schematic that will do that?

Thank-you

 

[Martaine2005]

If you are not ‘stomping’, then any latching switch will work.
Simply put the switch in series with the battery positive lead. OFF disconnects the battery. Put the LED and current limiting resistor after the switch and to GND (negative).

 

[SparkyCal]

Thank-you. How do I figure out the value of the limiting resistor? I know OHMs law. I know the voltage 9V...but I am not sure how to figure out the current

 

[bertus]

Hello,

To calculate the current limiting resistor use the following formula.
Rlimit = (Supply_Voltage - Led_Voltage) / Led_current.
In your case the Supply_Voltage is 9 Volts.
The red led will have a Led_Voltage of about 2 Volts.
The Led_current can be 20 mA.

So we get ( 9 Volts - 2 Volts ) / 20 mA = 7 Volts / 20 mA = 350 Ohms.
To be on the safe side, use the next higher resistor value of 390 Ohms.

Bertus

 

[Martaine2005]

It will depend on the LED. Typical red, 2v, 20mA.
So forward voltage of LED is ‘say’ 2v. 9-2=7.
Lol
@berus beat me

 

[SparkyCal]

Thanks guys. Good to know.

 

[dave9]

If it's a generic ~100mW LED, I would shoot for lower than 20mA for long life, but then again I don't suppose it's going to get many hours if running off a 9V battery. Just suggesting that it would be bright enough to see at 10mA for just a power indicator.

 

[Audioguru]

The original old schematic still has the polarity of C3 backwards.
The circuit produces distortion only if the output signal level is high enough which depends on the input signal level and the gain setting. The LED will light and slowly kill the little battery all the time the power switch is turned on.

Don't use the tone control. It might kill the opamp and/or battery when it connects the 22nF C29 through C3 to the output of the opamp.

 

[Harald Kapp]

Got a question about driving LEDs?

 

[SparkyCal]

I built the circuit with Bertus's schematic which placed the capacitors with the correct polarities. I think i will still to an old school switch that says On/Off on it anbd avoid the LED thing altogether, although good to know how to calculate resistor value.