Calculating and Metering Total Impedance (Guitar related)

I am going to be a bit coy on this one....I am not trying to be offessive.....I do this mainly to get you thinking about what your learning in school....

What is in parallel with the volume pot in the circuit?

Now what is the formula for parallel impedances?

Draw the schematic, and think about it.....

I think your forgeting something else in the circuit....nudge nudge

Plus you need to draw out the circuit.

Link is giving you very good clues...

Question 1, what do you think the DC resistance of the pick up is?
Question 2, what is the DC resistance of any resistance in series with a capacitor?

Guitarfreak you're getting there, Draw the circuit... MrEase is making it easy for you ;)

What you have here is a parallel circuitry of the pickup, 1 MOhm, 500 kOhm, 333 kOhm and what ever the resistance of the tone pot to the lower lead ( of which you did not say if it is ground or anything else...).
Not to forget that the 1000 pF and the tone pot build a parallel curcuit to the same electric potential of the lower wire, as well.
Now, you must also consider that Ohmic load is not the same then impedance..(inductive & capacitive reactance). On high frequencies a condenser has low resistance (with DC it has infinit resistance) and a choke or coil (with DC almost close to a short circuit ) have high resistance. Pure Ohmic resistors stay pretty close to their value widely independent from the frequencies. This is different in the MHz and GHz area, though..Not interesting, here.

You find all necessary formulas in the internet... usually it takes a few years of apprenticeship to understand and handle this math. Just start with Ohm's law... that one is easy... More complicated it gets with the effects of alternating current and those passive components.

M2c..

Big K

your circuit is not complete, its a good start but you need to include everything in the current loop in order to see why your meter reads what it does. Although with all the hints given I am sure you can guess the main culprit that is lowering the impedance already.

Your missing the most important part of the circuit in your questions.....Not trying to be a pain, but in my opinion the only way you learn is by working through it yourself..

Now that said.....Mr Ease asked what is the impedance of the Pickups......hint hint

Oh and on your circuit draw it from pickup to guitar jack, map out every part, switches, pick ups,caps resistors....etc

The circuit's fine. If you now try to answer the two questions I posted ages ago you should have all your answers!

Guitarfreak, post: 352249 wrote: You all seem to be implying something else in the circuit downstream is causing this

Not me... I'd suggest going as far "upstream" as you can!

Guitarfreak, post: 352249 wrote: Are you saying that since there is parallel resistance later in the circuit that the earlier stuff doesn't matter as much as the later stuff?

It doesn't matter where it is (remember Kirchoff's law from your other thread?), what is important is the individual DC resistance of every element in the circuit. A big clue will be to vary the volume pot and see if that affects your resistance reading. How can that reading of 5.11k possibly occur - there is nothing wrong with that reading BTW.

I'm trying to work with Link555 here as if we just tell you (which we are getting ever closer to) then you will not be thinking it through for yourself and your learning will be limited.

Close but not quite! The reading of 5.11k is ALL the resistances in parallel so the pick up itself is somewhat over 5.11k as there are other resistors in circuit. This is evidenced by the reading rising to 14k (although that sounds a tad high for many pickups). What is clear is that the lowest resistance around is the pickup as this is clearly dominating the reading you are getting. The only way to measure the total load on the pickup is to disconnect the pickup and then measure what is left. This is hardly worthwhile though as you already know the circuit and calculation should be quite adequate. Knowing all the other values actually enables you to calculate the pickup resistance directly. Now there's another exercise for you! :<)

BTW with the circuit as you have drawn it, the minimum resistance should occur with the pot at minimum and the maximum when the pot is at maximum. There should be no mid range peak in resistance. I notice that your 1 meg resistors have disappeared too!

I suspect that either your circuit is wrong or your measurements are based on the information you have given.

From your original description the circuit you did with the 1 meg in parallel with the pot was correct. Either way you should not get an impedance peak at the mid volume position.

Guitarfreak, post: 352306 wrote: A few things on my mind:

While your analysis of the reading being a combination of all resistances in parallel, the tone knob still reads ~500k.

I very much doubt that! Either your circuit is wrong or you are measuring in the wrong place. EDIT: Woops, see the next post!

Guitarfreak, post: 352306 wrote: The 1M notation I posted first seems like it is wrong because that would imply that the 1M is there to show more parallel resistance to the circuit and get a brighter signal, while in actuality its function is to darken the signal by working against the pot resistance to lower the pot value.

Resistors in parallel have a resistance LOWER than the lowest valued resistor, therefore there is MORE load on the pickup with the 1 meg in place, ignoring the 333k resistor for now, the pickup load would be 500k without the 1 meg and 333k with it. This should indeed "darken" the sound.

While we are doing our best to help you here, there is no doubt that you are struggling with the basics of electronic networks. You really need to find some good tutorial books or find another course with a tutor who is actually going to teach you something worthwhile.... I think there is only so much support we (or at least I) can give on this forum - we (I) have to earn a crust as well...

Sorry, in my first quote I mistook the tone pot for the volume pot! The thing there is that the tone pot has a capacitor in series with it and that will look like an open circuit at DC hence the pot value has nothing effective at DC in parallel with it and therefore you will measure just the pot value.

Hey, Guitarfreak,

I was just browsing thinking of your "basics" and came across this site. I think you should find it very useful.
[[url=http://[/URL]="http://www.electron…"]Basic Electronics Tutorials and Revision[/]="http://www.electron…"]Basic Electronics Tutorials and Revision[/]

I had a quick read through and it seems mostly OK although I did have some serious issues with their section on transistors. The author seems to think that transistors are current operated devices which is quite wrong! I have contacted them to point out the error but have no idea if they will take any action. It could be a good pointer to the integrity of the site if they react to such a blooper!