Beginner Problems / Reading a Schematic / Female Jacks

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Hey everybody!

Wassup? Here are a couple of beginner questions, which I hope some of you experts might help with.

I'm making a Marshall Guv'nor in accordance to a schematic found on Experimentalists Anonymous : https://www.experimentalistsanonymo...rtion Boost and Overdrive/Marshall Guvnor.gif

And have the following problems : First, the input appears, to my beginner mind, to be a stereo jack entry... Is that true? I'm assuming, unless using a Guv'nor requires a specific jack chord [which sounds a little unsettling], that my mono jack chord won't interact correctly with the stereo jack entry? Correct? I've googled something along the lines of "jack entry schematic symbols" to try to find out what the differences are between different symbols for jacks, and have found nothing but confusing information. It appears to me that on the Guv'nor Schematic, there are two electric paths beside the one that goes to ground. Can anyone help with that? Does that mean I have to buy a stereo entry? A stereo jack chord? Does that mean something else? Thank you.

Second of all, very relatedly, a jack entry [female jack] on the schematic for the "loop" entry has a little arrow drawn into it! Now what does that mean? I'm assuming contact is meant to be kept while no jack is plugged into this entry [the loop circuit is optional]. Is that correct? Does that require a specific socket?

Furthermore, I've been doing unsuccessful tests with this guitar pedal circuit, and am wondering why they have been unsuccessful. Now, just to make things a little harder for everyone, my very limited budget means I do not detain much troubleshooting equipment (multimeter, etc.). So, I can get very lost as soon as things don't go well : troubleshooting has always been very difficult with pretty much everything I've done. I'm assuming that if there's something I got wrong with those jack entries mentioned above, that may answer the problem. However, a part of me does believe it to be wise to ask the following questions : is there any reason that kicking out the EQ circuit [bottom right of the schematic] might make the entire circuit not work? Or kicking out the variable resistors [to replace them with nothing at all]? My simple representation of electronics tells me that this should not alter the functioning of the diagram, but, making this kind of deduction whilst knowing nothing of electronics is sometimes a starting point to f***in' everything up. Obviously, I've also kicked out the loop entry, supposing I should have, with the mentioned configuration, a Guv'nor with max gain, max volume, max Highs, Lows and Mids, and no additional effect, before moving on to adding these.

If anyone sees this and has an altruistic minute to spare, I'll take it no problem!
May the force be with you!
Jonah. (aka "Jonague")
 

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So...

The input jack is arranged so that it grounds the battery negative when a plug is inserted. That will power the unit on. You could use an ordinary mono input socket and have a separate power switch if you wish.

It is however drawn as a stereo type and so the mono plug (your lead) that you fit would short out the inner connection to ground and turn the thing on.

Output is just a mono socket.

DC adaptor (you've no need to fit this unless you intend using a separate adaptor) is just a mono socket that has the ability to link from one side of the socket to the other. When you insert the plug it opens the connection and the adaptor takes over. Such sockets are common and typically would have four connection tags, two on each side and with an obvious ability to switch as the plug is fitted.

I would say 'loop' is a similar type of socket but this time stereo (so 6 terminals, 3 on each side). You link L and R on the switched side and so there is a 'short' between those two points. Insert and plug and the short is broken and the plug connections take over.

The electronics I can tell you anything you want to know regarding voltages and what to expect. One quick check is to confirm that half the supply voltage is present on all opamp pins excluding the two supply pins.
 

PRR

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> my mono jack chord won't interact correctly with the stereo jack entry?

Don't think. Many guitarists are technically clueless, especially when the gig is about to start, so the gear has to be designed to work without thinking.

This is a cheap "improper" trick used on MANY guitar pedals. When you stick a mono plug in a stereo jack, the Ring finger is shorted to Shell (ground). When you wire the battery to this, the power is on when a plug is in, off when the plug is removed. This is cheaper and more memorable than a switch.

FWIW, these days a DC wart is more common than actual batteries. It has become common to have a dozen pedals on the pedalboard, often semi-firmly attached, so battery upkeep becomes overwhelming.

> reason that kicking out the EQ circuit ..might make the entire circuit not work?

Sorry, you have to draw a picture what you "kick out". Yes, it is possible to remove part of a plan but, like my surgery, or a mystery a friend is writing, you have to stitch the loose-ends properly.

Mooly has said the same theory-details his own way. And I don't usually tout one forum on another. But *this* type question (a specific popular pedal and only a smatter of tools/theory) is well-handled at DIYstompboxes.com - Index There's even a Debugging thread specifically adapted to pedal-work.
 

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Hey there!

I'm back! Just a quick question, I've already published it on the long thread started by Mooly on LTSpice, but realized only afterwards there were about a billion posts on it.

I'm running an old computer with OS 10.6.8. And, obviously, with a little help from Murphy's law, minimum requirements for LTSpice are OS 10.7. Damn. Does anyone know of an alternative? I downloaded MacSpice, but gather it's more complicated to use. Or does anyone know where I can find a decent MacSpice tutorial? Their website is cool, but I'm guessing this isn't the most intuitive software there is in this universe. So...

Thank you!
Jonah.
 
Rereading this found:
DC adaptor (you've no need to fit this unless you intend using a separate adaptor) is just a mono socket that has the ability to link from one side of the socket to the other. When you insert the plug it opens the connection and the adaptor takes over. Such sockets are common and typically would have four connection tags, two on each side and with an obvious ability to switch as the plug is fitted.
Sorry but no ... you are more of a Hi Fi guy I guess :)

In the Guitar Pedal World a DC adaptor jack is almost universally a "a Supply Jack for a hollow plug connector", size 5.5mm x 2.1mm

s-l1600.jpg


as you can see, they have 3 legs

they expect this kind of plug, not regular guitar ones:

ppbar.jpg


2.1mm Straight Barrel Cable
The most widely used cable with 2.1mm straight barrel connector on each end and standard (center negative) polarity. Used with a majority of popular pedal brands including Boss, MXR/Dunlop, TC Electronic Nova, Line 6 ToneCore, Ibanez and more.

This picture shows all 3 jack types you will fin d on seame pedal ... different functions of course:

* 6.5mm for regular "guitar plugs"
* 3.5mm for headphones
* 5.5 x 2.1mm 9v supply

preview1-c90ed2da59e3c268b215210604093151-1024-1024.jpg
 
'Sup everyone?
Thanks to those of you who've helped! I'm afraid I still don't have a circuit simulator for software purposes, but I appear, after a little bit of tryin', to have got my circuit to work on a breadboard. That is a big relief. However, a cool thing about mistakes is learning a little from them. Mooly said something about all the pins but supply pins on the diagram being at 4.5 V. Very strangely, it seems that as soon as I connect the tension divider with pin 5 of the TL072 (the non inverting input of the second opamp of the IC) the tension of the tension divider is greater than the expected 4.5 Volts. I'm a bit startled.
Anyone wanna show their knowledge off?
Thanks!
Jonague.
 
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The fact you have a working circuit shows you probably have it correct :)

You should connect your meter negative to the junction of the 47k divider as that is our 'virtual ground'. All voltage are measured from here.

From this point pin 8 should read plus 4.5 volts and pin 4 should read negative 4.5 volts given a total supply of 9 volts.

All other opamp pins should read zero volts.
 
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