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Simple EQ Circuits, where's the power come from?

Started by icecycle66, November 13, 2012, 09:24:05 PM

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icecycle66

I was looking though a bunch of EQ schems and came across these two pretty simple and easily adjustable circuits.

I don't have any formal schem reading education, but it doesn't look like the power source and power sections of the circuits are there. 

Where should they be? i feel like the input and output  power should be coming in around R7 and R21, but that could just be my ass talking.

Shouldn't the EQing opamps all have power coming into them too?


http://experimentalistsanonymous.com/diy/Schematics/Tone%20Control%20and%20EQs/10%20Band%20Graphic%20EQ.gif



http://experimentalistsanonymous.com/diy/Schematics/Tone%20Control%20and%20EQs/Versatile%20Multi-Band%20EQ.pdf


jkokura

The second schem shows you what's going on. The ICB in that diagram has the V+ and V- symbols, which are typically going to be +9V and 0V in most pedals.

I would be tempted to do a circuit like that with a +9V to -9V power arrangement, but that would require adjusting the circuit a fair amount...

Jacob
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icecycle66

So, for circuit 2, there would be little copper lines going straight from my 9v source to the "+" of the op amps without any sort of resistor conditioning or anything?

Or would there be a little power circuit like the "9VDC Adaptor Jack" on this Rat schem?  Then from that little power circuit, little copper lines to the op amps.

http://experimentalistsanonymous.com/diy/Schematics/Distortion%20Boost%20and%20Overdrive/Proco%20Rat.pdf

Scruffie

Quote from: jkokura on November 13, 2012, 09:29:22 PM
The second schem shows you what's going on. The ICB in that diagram has the V+ and V- symbols, which are typically going to be +9V and 0V in most pedals.

I would be tempted to do a circuit like that with a +9V to -9V power arrangement, but that would require adjusting the circuit a fair amount...

Jacob
Look at where the non inverting input of the output opamp is connected in that schematic, that is in a +9v/-9V arrangement.

For stomp box i.e. 9V & 0V use you're going to have to introduce a V.Ref for those opamps to work.
Works at Lectric-FX

icecycle66

I think I am learning something important to pedal building.

Looking at the above schematics, I'm trying to figure out how you get "+" and "-" voltage from one source.

I read this:
http://www.eng.yale.edu/ee-labs/morse/compo/sloa058.pdf
Sections 1 and 2

To me this means:

Red battery line goes to a copper trace
the copper trace splits into two copper traces

trace 1 goes to the the corner of the triangle (opamp) marked "-".

-and-

trace 2 goes to something called a virtual ground.  This virtual ground is made up of a resistor and capacitor arrangement as indicated in the text and wraps around to the corner of the triangle (opamp) labeled "+".

There is also something called AC coupling that has to do with grounding. I reads like it's just the common grounding done with a jack sleeve.


The opamp section in the above schems looks incredibly similar to the inverting gain stage at the top of page six in my link in this post.

All this to say, in the first schematic up top; the negative pole will be receiving hot and unregulated 9v action.  The bottom corner "+" will be connected to the virtual ground.  True?



JakeFuzz

The + and - symbols on the opamp in the schematics you have shown are not for powering the amplifier. I agree it is confusing to omit but what you are looking for is typically not shown on the symbol (or in that first schematic you posted in the schematic at all!). That second schematic you have linked shows the power rails which are actually two separate pins that power is applied to directly. Take a look at the pinout for a standard dual opamp package.



The + and - are where the signal inputs are applied. Depending on what configuration you are using your guitar signal is usually applied to one terminal and a reference voltage is applied to the other. If you are using +9v supply only, the input is usually biased or dc offset to half the supply voltage. This gives the opamp the maximum room to amplify the signal without clipping the supply rails. This is why you sometimes see Vb or Vcc/2 connected to one of the + or - terminals.

For bipolar supplies (+9 and -9) the center of the opamp swing is already zero ((-9+9)/2=0) so you don't need to bias the input.

That resistor conditioning you see in line with the suppy is a current limiting resistor. This prevents the power supply from burning itself out if the circuit shorts somehow. It is usually very small so it doesn't significantly drop the supply voltage going into the pedal.

icecycle66

#6
Quote from: JakeFuzz on November 15, 2012, 09:18:11 PM

For bipolar supplies (+9 and -9) the center of the opamp swing is already zero ((-9+9)/2=0) so you don't need to bias the input.

So each of the power legs, pins 8 and 4 on a TL072, both get red wire power so that each of their internal ciruits are energized?

-or-

Pin 8 gets red wire power and pin 4 gets plain ol' ground? (this is the o ne I think is right, otherwise there wouldn't be a ground body for the IC internal circuits)

-or-

You split the source voltage 9v into two 4.5 volt channels and send them into the IC.  4.5 volts to pin 8 and 4.5 volts to pin 4.  Somehow pin 4 voltage equals negative voltage and lkfmka;skdmfasdgb......

---

I'm trying not to let the magic smoke out of my brain.  This doesn't seem like it should be so dificult to understand.  How do I know where to send electricity if one location says "+" (which should mean it wants more elcetricity) and another part wants "-"(which means it should wants less than none electricity).



I'm getting the feeling that "+" and "-" don't mean what they say they mean?


I'm also reading the TI manual "Op Amps For Everyone".  If the want it to be for everyone one, they need to put and equation free application section.

JakeFuzz

Quote from: icecycle66 on November 15, 2012, 09:33:47 PM
Pin 8 gets red wire power and pin 4 gets plain ol' ground? (this is the o ne I think is right, otherwise there wouldn't be a ground body for the IC internal circuits)

Yes! That is how we typically connect it.

I wouldn't worry about bipolar supply as not many pedals use them. If you want to look more at how it is done with a +9v supply, take a look at the kingslayer schematic. You have to add an additional charge pump to generate -9v. Then you connect +9v to pin 8 and -9v to pin 4. Notice in that schematic how also there is no Vb voltage divider because it is not needed.

Quote from: icecycle66 on November 15, 2012, 09:33:47 PM

I'm trying not to let the magic smoke out of my brain.  This doesn't seem like it should be so dificult to understand.  How do I know where to send electricity if one location says "+" (which should mean it wants more elcetricity) and another part wants "-"(which means it should wants less than none electricity).

I'm getting the feeling that "+" and "-" don't mean what they say they mean?

Yep, trust me it confused the hell out of me when I first saw the symbol too. The + and - are more indicators of how the opamp amplifies the signal than the potential it wants to see. The + terminal is called the non-inverting input and the - terminal is the inverting input. When we use the amplifier configuration for the inverting mode (guitar signal goes into the - terminal) the phase of the guitar signal is shifted by 180 degrees (or inverted using amplifier terminology). This makes sense if we go back to the most basic definition of what an opamp does; all an opamp wants to do is make the potential (voltage) at the + and - terminals equal. If we put the same guitar signal on both inputs the potential difference is going to be zero. To show zero on the output the opamp must invert the phase on the inverting (-) terminal and internally add it back to the non-inverting (+) terminal. The two waves will cancel each other and we get no amplified signal out of the amp.

icecycle66

Quote from: JakeFuzz on November 15, 2012, 09:50:07 PM
Quote from: icecycle66 on November 15, 2012, 09:33:47 PM
Pin 8 gets red wire power and pin 4 gets plain ol' ground? (this is the one I think is right, otherwise there wouldn't be a ground body for the IC internal circuits)

Yes! That is how we typically connect it.

Then consider this case closed. (Or is it? Mwa-ha-ha-ha-ha-ha-haaaaa)

Also, should this thread be moved to the "Beginner's Paradise" due to the new conversation outcome?