Back to the Breadboard

[Audioguru]

The original schematic posted today uses a Jfet-input TL082 dual opamp which has an extremely low input current. I do not know why it is a dual instead of a TL081 single one.
The older NJM4558 dual opamp uses ordinary input transistors with 2500 times more input bias current which reduces the input voltage from the 1M resistor -0.5V maximum which reduces the output voltage a maximum of -0.5V.

The 470nF input capacitor driving the 1M resistor produces a cutoff frequency of 0.34Hz which is way too low, it will pass earthquake or percussion vibrations. I recommend 10nF which has a 16Hz cutoff frequency that passes all audio frequencies and cuts subsonic vibrations. 6.8nF into 1M produces a cutoff frequency of 24Hz and 15nF produces 11Hz which are fine.
Do not use your 1uF capacitor as an input capacitor because its value is 100 times too high (double the 470nF one).

The 1uF capacitor C3 shown is a backwards polarity polarized electrolytic. I would use a smaller 330nF (0.33uF) film capacitor to pass all audio frequencies. Film capacitors are not polarized.

C28 is shown as 220nF which is too low and produces a cutoff frequency of 727Hz. I recommend 3.3uF to pass frequencies above 49Hz.

 

[SparkyCal]

@AG- I have used your recommendations along the way. These are the changes I am using, based on your advice:

1. I am using your voltage divider circuit, rather than the original

2. I am using a 10nF capacitor in place of the 470nF

3. I used a 3.3uF capacitor instead of the 330nF originally called for.

4. I inserted a 270R resistor just before the 1uF capacitor

All of these changes have worked fine, and I like the sound of the pedal.

I am confused about your comment re: the 1uF (C3) capacitor. I have been using the 1uF (C3) as shown in the schematic, except I have the polarity correct (he shows it in a backwards state). I think you are suggesting I use a 330nF (.33uF) film capacitor instead? I can do that going forward, but on this build, I already installed the 1uF prior to seeing your post.

My current status is as follows:

I have everything in place, including the Stomp Box switch. I am getting proper readings and am using a new battery.

My next step, which I have never done- is to install the LED light and the DC Barral jack (I ordered the ones that VB suggested and they got delivered today).

I will probably attempt this next part later tonight.

 

[SparkyCal]

Attached is my plan to wire the LED and the DC barrel to my circuit. Do you think it will work?

The idea is to have the LED light up when the effect is active and go off when it is not.

The idea is also to have the DC barrel plug power the circuit when it is plugged in, and, at the same time, disengage the battery when it is being wall-powered.

So, to summarize:

When battery powered, the LED lights up when effect is engaged and is off when it is not.

When wall powered, the LED lights up when effect is engaged and is off when it is not. When wall powered, the battery is disengaged.

Thank-you. Just want to make sure this will work before I try it.

BTW- it is supposed to say 9V battery clip (I know it looks like TV battery clip)

 

[Audioguru]

1uF is 1000nF and the output capacitor does not need to be that much. A 1uF film capacitor is fairly large and is expensive and a 1uF electrolytic capacitor is even larger but a 330nf film capacitor is small and inexpensive. Use a 1uF electrolytic if you want, it will sound the same.

I forgot that the original schematic is missing the important 270 ohms resistor I added between C3 and the tone control.

 

[VenomBallistics]

Ummm nope.
You will apply voltage to the battery rather than switch to the barrel

 

[SparkyCal]

Hmm. How do I get the led powered irrespective of whether the barrel or the battery is being used?

 

[Audioguru]

A 4.7k resistor in series with a 3.2V blue LED and powered from 9V draws (9V - 3.2V)/4700 ohms= 1.2mA for a fairly dim LED. Use 470 ohms or 270 ohms to be much brighter.

A good LED is bright over a wide angle. A cheap old LED from the other side of the world is dim but has a case that focuses its light into a narrow brighter beam that can barely be seen unless it points at your eyes.

 

[VenomBallistics]

The switch is normally closed when nothing is plugged into the barrel port. So .. the switch lug goes to battery, giving it a path to the board.
As soon as you plug into it, the switch opens, PSU takes over while the battery is disconnected.

 

[VenomBallistics]

Basically, delete the lead from the switch lug to 4.7 and you should be alright

 

[SparkyCal]

Thanks guys!

 

[SparkyCal]

@AG- I changed the resistor to a 270R and you're right, it's better.

@VB- I am not sure if I understood correctly. You wrote: basically, delete the lead from the switch lug to 4.7 and you should be alright

On my drawing, the negative end of the LED, goes to the upper middle lug on the stomp switch.

Are you saying to remove that connection and instead, route the negative lead of the LED to the negative battery terminal? I breadboarded that quickly, and it didn't work the way I expected. Maybe I misunderstood you.

OR

Did you mean for me to remove the wire going from the middle lug of the DC barrel, to the resistor?

 

[VenomBallistics]

You have 3 lugs on your power connector. One of them is a switch.
What you have drawn at the right lug ( assuming this is the switched lug ) will not allow the battery to disconnect when the plug is used.
This will still work, until the battery explodes. Reliability may suffer thereafter.

 

[SparkyCal]

Ok….when you referred to the switch, I thought you meant the stomp box switch,

 

[SparkyCal]

I'm confused. VB, or someone- can you guys look at my drawing and draw on it, as to what I have to correct. I breadboarded this, but obviously, I am not getting it.

Thank-you

 

[VenomBallistics]

I'm confused. VB, or someone- can you guys look at my drawing and draw on it, as to what I have to correct. I breadboarded this, but obviously, I am not getting it.

Thank-you

here is a schemmy diagram of your power jack. If, you connect your battery to post 3 and your board to post 2, the normally closed switch gives the battery a path to power the board in condition 1 (no external power connected)
In condition 2, your power plug opens the switch so that the battery at 3 is disconnected while the plug takes over supply duty. Assuming all else is correct ( prehaps I shouldn't but will anyhow ) you have a lead running from 3 to your 4.7 ohm that also has a connection from 2 to the same 4.7 ohm, shorting across this switch, rendering it useless.
Removal of the 3 to 4.7 lead restores it's function as a battery disconnect.

 

[SparkyCal]

Thanks VB

 

[SparkyCal]

Call me an idiot, but I still can't get this. I have attached a drawing of my latest attempt, based on my understanding of VB's instructions.

When I wire it this way, the LED is lit immediately when I plug the battery in. Clicking the stomp switch on or off has no effect on the LED.

When I plug an adapter into the DC barrel, the battery goes out.

Obviously, this is hooked up wrong. I likely misunderstood VB's message.

Have a look at the attached drawing please.

 

[VenomBallistics]

MMMkay .... when you plug in with an adapter ... is the adapter powered or are you plugging in non live?
If not live, you have the jack switch in proper use .. if powered, "2" and "3" are backwards.
Now your stomp switch is just a show piece in your current drawing. but that's okay for now, because you need to get that jack dead right first

 

[Audioguru]

The datasheet for the 3-terminals jack should show a photo explaining which terminal is the switching one.

 

[SparkyCal]

Despite the explanations, I was getting confused.

So......I forced myself to think this through. Then I breadboarded my thoughts and came up with the attached diagram which I believe works.

I verified that the barrel and the battery clip were sending power, when one of them was disengaged and that the plug-in in of the DC barrel, resulted in the battery clip being disconnected.

Once I knew it was working as such, I wired it to the stomp box switch "on the way back"- by that, I mean that I inserted the switch between the ground wire on the way back to the battery clip. So, the only way for the circuit to work, is for the stomp switch to be in the On mode.

Thanks for all of your advice. I think that thinking it through, forced me to understand how the components worked,


Thoughts?