A little help getting started

[instantaphex]

I'm just getting started building a blue collar clone and I've realized that I'm having a bit more trouble following the schematic than I had anticipated.  Here is where I am starting:



I'm breadboarding this first because I know it's going to be a bumpy road.  I'm trying to look at this and figure out exactly how to lay this part of it out but I'm missing some fundamentals.  After quite a few google searches here I am.  I assume the +9V is the red lead from the battery.  So I have a 10K resistor plugged into the positive row on the left hand side of my breadboard.  The other side goes into the "a" row of the breadboard followed by the anode of the 22uF cap.  On the same row I have another 10K resistor with the other end connected to the cathode, which is connected to ground.  As far as I can tell, the hole between the anode of the cap and the second 10K resistor is where my VREF junction is.  Here is a picture of what I've got so far:



First of all, have I laid these components out correctly?  When I connect things to V+, should they be getting a direct 9v or should all V+ components hook into the positive row after the 10K resistor on row 55?  I've got an image of a veroboard layout that I can use as reference when i actually go to perfboard it, I just want to make sure that I actually get some experience building the circuit from the schematic.  Should the 4.7K resistor be getting 9V as well or should it be above row 55 after the 10K resistor as well?  I just want to make sure that my assumptions about what the schematic is saying are correct.

[deadastronaut]

looks fine...

the 4.7k :   9v  >> 4.7k >>led>>>switch.

vref  :    the 2 x 10k's act as a voltage divider :  should measure 4.5v at the vref point. 

have fun.

[GibsonGM]

You're doing fine!  As astro said, you will get 4.5V at "VRef".    You place your black meter probe on the battery "-" terminal (or a wire connected to the ground bus on your breadboard - the slot the minus is connected to, where the orange jumper wire is), and with the meter set to read "DC Volts", you should get APPROX.  4.5V by touching the red probe to the VRef spot in the pic.   Sorry if you know how to use a meter, just put that out there in case you don't.

After a while, it becomes second nature...but getting started, using the board can be confusing.  Give it a week, you'll be a pro...

[instantaphex]

Thanks guys.  I sat there looking at the schematic for like an hour before I put anything in and then second guessed myself.  I've since set up most of the rest of it.  My main point of confusion is still the power busses on my breadboard.  Do I need to plug in my LED and switch before the power section so that it's not limited by the resistors?  I forgot my multimeter at work so I wasn't able to test anything last night.

[duck_arse]

the 2 resistors are there to provide a reference voltage to another part of the circuit. they are not meant as a "supply" for anything. any part of your diagram showing "9V" or "V+" connects to the 9V rail, ie straight to the battery (more or less).

[instantaphex]

the 2 resistors are there to provide a reference voltage to another part of the circuit. they are not meant as a "supply" for anything. any part of your diagram showing "9V" or "V+" connects to the 9V rail, ie straight to the battery (more or less).

Ok that's what I was wondering.  I didn't know if giving an op-amp 9V was going to fry it or if that part was there to smooth out the current flow.

[smallbearelec]

Do I need to plug in my LED and switch before the power section so that it's not limited by the resistors?

You don't have to connect the LED and the switch at this point. Actually, that switch is one pole of a stomp switch, which you'll add after all is working without bypass. At the beginning of this article:

https://www.smallbearelec.com/Projects/UrsaMinor/UrsaMinor.htm

you'll find a description of one standard way to wire a stomp switch, you'll see the same use of the third pole to control the in-use LED.

[instantaphex]

Do I need to plug in my LED and switch before the power section so that it's not limited by the resistors?

You don't have to connect the LED and the switch at this point. Actually, that switch is one pole of a stomp switch, which you'll add after all is working without bypass. At the beginning of this article:

https://www.smallbearelec.com/Projects/UrsaMinor/UrsaMinor.htm

you'll find a description of one standard way to wire a stomp switch, you'll see the same use of the third pole to control the in-use LED.

Yeah, I guess there is really no reason to hook that stuff up yet.  I think I need to just start testing to make sure I actually know what I've got going where. 

On a side note.  I totally forgot to order some op-amps on my order from you, which is fine because you had recommended I try out some others besides the JRC4558.  Before I put in another order, do you have any suggestions of some to try?  Also what is the difference between your 8-pin sockets?  Thanks!

[smallbearelec]

you had recommended I try out some others besides the JRC4558...do you have any suggestions of some to try?  Also what is the difference between your 8-pin sockets? 

Any of the ones that will work in the TS will work here: TLC2262, TLC2272, LM833, TL072, RC4558P, NE5532, OP2134PA, OP2604AP.
Though coupling caps and clipping diodes are waaay more important to the tone. The choice of machine-pin vs. leaf socket is designer's preference.

[GibsonGM]

I prefer "leaf socket", I guess they're called. A lot easier (for me anyway) to get IC's in and out of the socket.  The pin type seems to work best if you're just going to insert the IC one time and done.    Just my 2 cents!

Read up on opamps (search engine, type in "opamp tutorial" or something); just a quick run-thru will do you wonders on how this stuff operates!

The 4.5V reference voltage "biases" the opamp to rest in the middle of its possible amplification range when in a quiescent (no signal input) state.    So, when a guitar note comes in which goes both plus AND minus, it will do so around that point.  If you just left it at zero, you would cut off the negative-going parts of any input signal.

This eliminates the need for a dual supply that offers plus AND minus polarities - which you will see if you read about opamps.   It's a sensible work-around for what we do with them.  You'll see this setup all over the place.