Nurse Quacky Effects Pedal Questions

Here are the relevant references:
1. Original schematic: http://www.home-wrecker.com/nurse-quacky.html
2. Wiring diagram: https://www.electro-tech-online.com/custompdfs/2011/06/ggg_nurse_quacky.pdf

Preface:
I'm not installing the bypass switch, just going straight from the original schematic.

My first question:
The original schematic shows a 9V DC source being split between two 47k resistors. The first node seems to be the +9V and the second seems to be obviously +4.5V and these voltages are shown to go to their respective places in the rest of the circuit. What doesn't make sense to me is why then, in the wiring diagram, does it show a 9V battery being connected to a DC jack? I'm assuming that just has to do with their use of the DPDT bypass switch.

My second question:
Would I be able to use Mono jacks for input and output? It seems like I need a stereo jack and it seems to be for the input, so if this is so, where do I connect the sleeve, ring, and tip of the jack?

Thanks for any help.

-Palbert

Sorry I can't view your PDF schematic, if you post a image (GIF PNG etc) of the schematic more people can help.

The 9v battery in guitar effects pedals almost always goes to the input jack, specifically to the power switch that is part of the input jack, so the battery is disconnected when the input jack is unplugged. Generally this type of jack socket is called a "switched jack socket" or similar.

Apparently that's just a power input jack (what they have labeled as "DC jack"), for an external power source. It's hard to tell from the wiring diagram, but it looks as if they may be using the input jack as a power switch, so when you unplug from it it disconnects the battery. You could copy that, or you could add a power switch (which of course, according to Murphy's Law, will never be turned off, so the battery will drain ...).

Apart from that you can use mono jacks, since it's just a single-channel device.

Just curious: what kind of effect box is this?

The 4.5 volts is just to bias the op amp at 1/2 the supply voltage (9 volts) so the signal can swing above and below this voltage. This lets the amplifier run from a single 9 volt battery.
This is a single channel circuit so a mono plug should be fine.. The sleeve in that case would go to ground and the tip to "in".

Thanks for the replies, very helpful. Mr RB, that is exactly what I want my pedal to do. How would I connect a stereo input jack to accomplish that? I'm guessing the tip and ring would go to +9V and GND respectively and then the sleeve would go to the input of the pedal?

Carbonzit, this is a simplified auto-wah pedal. This is mainly just a fun project for me so I'm not expecting stellar quality, I just want some wah.

Palbert said:

How would I connect a stereo input jack to accomplish that? I'm guessing the tip and ring would go to +9V and GND respectively and then the sleeve would go to the input of the pedal?

Click to expand...

You don't need a stereo jack, just a switched mono one. In this case, you'll be switching the negative side of the battery to ground to disconnect it. Otherwise it'll be wired as usual: tip to the + of the input, sleeve to ground. (No ring.)

Carbonzit, this is a simplified auto-wah pedal. This is mainly just a fun project for me so I'm not expecting stellar quality, I just want some wah.

Click to expand...

Aha, I suspected as much.

carbonzit said:

You don't need a stereo jack, just a switched mono one. In this case, you'll be switching the negative side of the battery to ground to disconnect it.

Click to expand...

I'm not sure what you mean by negative side of the battery to ground to disconnect it. Where are the ring and tip connected in the Nurse Quacky circuit?

Here's how the jack would be wired:

**broken link removed**

No ring. The tip goes to input +, the sleeve to input ground and power ground (V-). The switch goes to battery -. You need to get a jack with the switch (N.O., normally open) on the sleeve.

When you pull out the plug, it opens the switch and disconnects the battery.

Actually, now that I think about it, you probably could use a stereo jack which would function exactly as described above. Connect the battery - to the ring contact, which will short to the sleeve when a mono plug is plugged in, connecting the battery - to ground. In this case, the jack doesn't need a switch.

Ok so just to be sure, tip goes to +9V of battery, sleeve to ground, and ring to -V of battery?

Nevermind, I re-read your last comment and got it I think. Tip to input, sleeve to ground, ring to - of battery?