Speaker question ?

[joecool85]

Hello,
      Here is where this began some time ago, I have made no progress, I am just as confused now as when I first started. The constant mathematical equations are overwhelming. To completely redesign the circuit and change all of the values and voltages seems a bit of overkill when the amp was functional at one point. All I wanted to do was install a tone stack between the two existing tubes, and it has spiraled out of control to the point that I cannot plug it directly into the wall to see if any progress is being made or not.
      I thank you deeply for all your time and effort, but I have to bow out of this project. I am going to restore the circuit to its original specifications and make small changes one at a time, the first being the tone stack in the LTSpice drawing drafted into the original circuit. I am going to replace the 180K resistor with a potentiometer. Change the value of the 560K resistor from power supply to #1 pin on preamp side of the 12AX7.
       I think those changes will help. Once again thank you very much for your time and effort but I feel as if
I am in over my head.

I commend you for your ability to realize your situation and know when you need to stop.  Sometimes just taking a break (a day or even a few months) can help.  Good luck with it in whichever direction you end up going  :tu:

[Roly]

it would require a fair bit of modification

The constant mathematical equations are overwhelming.

This isn't "mathematics" - it's basic shopkeeping dollars-per-pound arithmatic; you aren't even doing algebraic transformations, much less using the Operator j or Calculus, just value substitution and calculator number-crunching.  This is the application of Ohm's Law, multiplication and division, and as basic as it gets in electronics.

A couple of significant reasons to modify the original circuit are;

- there is nowhere to insert a tonestack (unless the mic preamp triode is modified to suit a guitar signal)

- the input sensitivity and impedance are seriously wrong for guitar (whichever input is used - you will either get serious front-end overload, and/or excessive loading on the guitar pickup making it dull and thin sounding).

You have been provided with a circuit with ball park values.  All you have to do is build it.

[SpareRibs]

Hello,
      Given a circuit with ballpark values to build, is a far cry from suggesting a circuit to duplicate. Previously I had explored other options. The Fender Princeton, or Tweed Deluxe tone stacks. There is a circuit with ballpark values that already works and may need some modification.
      The value substitution and calculator number crunching is the mathematics I am talking about. It seems to be endless. I don't know where to start what to crunch and to what end or for what purpose. I told you I am not as well versed as you are in electronics.
       So as opposed to endless riddles, questions, and dead ends , I have decided to go in another direction. If the solution was as simple as modifying the preamp triode to suit the guitar why was that not the target from the start as opposed to starting from the opposite end of the amp increasing all of the voltage causing
the need to change all of the resistance values and make all sorts of calculations. It all just seems pointless
as the amp was working after a fashion before all of this started.
       I have put it back to stock and will start from there adjusting values to get the input similar to one of the Fender schematics.
       Once again Thank you for all your time and effort, but I am simply not capable of the disassembling and completely redesigning one component into another

[Roly]

So as opposed to endless riddles, questions, and dead ends , I have decided to go in another direction. If the solution was as simple as modifying the preamp triode to suit the guitar why was that not the target from the start as opposed to starting from the opposite end of the amp increasing all of the voltage causing
the need to change all of the resistance values and make all sorts of calculations. It all just seems pointless
as the amp was working after a fashion before all of this started.

That has been the target since post #8, but 50 years of doing this stuff has taught me that you have to get the basic starting point right before you start changing things and bamboozle yourself.

The voltage that you posted in post #37 showed there was something basically wrong, see my post #38, and indeed there was a missing ground resulting in crazy voltages.

But by post #43 you still have a problem...

But you measure only 68V, so that obviously doesn't add up.

At this point we are still trying to get a clear idea of what you have with the amp as it originally stands, before any modification, and trying to answer the question "does the amp as original have any faults/problems?".  If we don't first make sure that a) the supply voltages are good, and b) the amp doesn't have any pre-existing faults, then we will get lost chasing our tails.  Persisting in trying to start a car that has no fuel in the tank will only get you a flat battery as well.

In fact we had a missing ground that would have stuffed you up big time if it hadn't been identified and fixed before proceeding.

We still seem to have a voltage, 68V on the "preamp" anode, which can't be right (most likely due to you mistaking which dual triode anode you were probing; this is almost certainly the anode voltage on the Phase Inverter section not the preamp section).

You must remember that you have the amp right under you nose, we don't, and are totally depending on the accuracy of what you post, in this case not the value of the voltage, but what voltage you were actually measuring.  Posting a voltage from the wrong point is a "fault" as serious as the open ground connection, and must also be sorted out before we can proceed.  Do we have a firm foundation on which to build the next step?

Given a circuit with ballpark values to build, is a far cry from suggesting a circuit to duplicate.

Perhaps my "ballpark" is a lot smaller than yours.  One variable here is, for example, that the supply voltage in your amp is not identical to a 5e3, so some minor tweeking of the preamp cathode resistor value might be required for best conditions, but just using the values on the circuit I gave will get some sort of operation near to what you want.

So let's take a look at the 5e3 circuit you nominated;



Check out the values around the second stage, the 12AX7 (left), 100k anode load, 1500 ohm cathode resistor, 25uF cathode bypass.  Now look at the values around the preamp section of the 12AX7 I drew for you, 12AX7 check, 100k anode load check, 1500 ohm cathode resistor check, 25uF cathode bypass check.  In fact the values I nominated as your starting point are identical to the values Leo fender used in his 12AX7 stage.  Yes the grid resistor is different but that is of no significance (other than it may give you a richer tone).  That's not just in the ballpark, that's a home run.

Even if we look at the actual preamp using a different dual triode we see that the values are almost the same (the 820 ohm cathode resistor is shared by two triodes so the effective value for a single section will be double, and 1640 ohms is pretty damn close to 1500 ohms).


There are two ways of measuring the current through a valve and its anode and cathode resistors;

a) direct - cut the circuit and insert a milliammeter, or

b) indirect - measure the voltage drop across either resistor and apply Ohm's Law.

We do b) because it is much easier than cutting and rejoining wires (and because inserting a meter in series can produce other problems that confuse the picture).

There is an awful lot of long-haired maths in electronics, but almost all of it can be avoided, however once you pick up a soldering iron Ohm's Law is part of the turf that cannot be avoided - pounding some numbers through your calculator is elemental to what you are doing; trying to avoid Ohm's Law in anything electrical is like trying to drive a car blindfolded.

In fact you have already made considerable progress by finding and fixing a missing ground, and despite the fact that the amplifier in question is somewhere on the other side of the planet I was able to point you to the problem simply by using Ohm's Law - that's how powerful just one multiplication or division is.

At this point I can only suggest that you Google "Ohms Law" and review pages until you find one that resonates with you, e.g. http://www.allaboutcircuits.com/vol_1/chpt_2/1.html and http://www.allaboutcircuits.com/vol_1/chpt_2/2.html

Then you need to do the same with "voltage divider", e.g. http://www.allaboutcircuits.com/vol_1/chpt_6/1.html

If learning Ohm's law was like being introduced to the ABC's, learning about voltage dividers would be like learning how to spell cat.

Why voltage dividers?  A triode (or transistor) amplifier is basically the same as three resistors in series, the anode resistor, the valve/transistor itself, and the cathode resistor.  To get the best out of the stage, indeed to get anything out of the stage, we need to get the voltages on the anode and cathode right, or at least close, and that requires the repeated application of Ohm's Law.

You have a very simple choice here, get comfortable with Ohm's Law, turn on the light, or grope around in the dark and get nowhere fast.  You can be your own boss, or just copy what others have done and never understand why things don't work out for you.


Otherwise just copy what I (and Leo) have laid out for you.

[SpareRibs]

Hello,
    I am going over this whole thread again to try to come to grips with it. I went back to the very beginning to try to get it all in context. When I gave you the reading of 68 it was off of pin five of the 6AU6 as that is the next in the circuit from the plate of the 12AX7 to the grid of the 6AU6.
    I will take a bit of time to reread all of the thread and ponder it in depth. I will get back to this thread after I go through a download of (A Glossary of Common Amplifier Terms), Volume 2 of Basic Electricity by Neville, Nooger & Van Valkenburgh. It covers Ohm's & Kirchhoff's Laws. It is illustrated so it is not just blah, blah, blah, reading. That should help a bit
    I guess I let all of this get to me by being impatient, sorry about that. I will try to understand things in
relation to Ohm's law.

[Enzo]

I have been in electronics now almost 60 years.  I can tell you that without a doubt Ohm's Law is THE most central idea in electronics.   And as foreign as it might start out, it really isn;t all that complicated.    it simply states that voltage, current, and resistance are related, and how so.   I use it every day, in fact I have a cheap little pocket calculator sitting next to me so I can do the arithmetic to apply the Law.   Seriously, not a day goes by I don;t calculate some current or voltage or power dissipation in some resistor.

And close on its heels is the idea of voltage divider.  Also pretty simple, but they are everywhere.  A basic volume control is a voltage divider.   The gist of that is that given a couple resistors (or resistances) with a voltage across them, the voltage across each will be proportional to the percentage of each of the total.

In other words, If I have a 40k ohm resistor and a 60k ohm resistor in series, or 100k ohms total, and I put 100v across them, then i will wind up with 60v across the 60k and 40v across the 40k.    If I started with only 50v instead of 100v, then the ratio is the same but the voltages are lower, like 30v across the 60k and 20v across the 40k.

[SpareRibs]

Hello,
      Yes Enzo, I agree with the fact that Ohm's Law is the basis of any project, and am sorting through all of the changes with that in mind. It takes me longer at times because of some of the abbreviations and terms you guys use. It just becomes frustrating trying to do measurements and doing it in the wrong area. Or worse not knowing where to measure what, to find out why or where the problem is.
       I will do the measurements and post them on a drawing from now on to make it clear what I am talking about, that should help everyone involved.
       Thank you for your input, it is encouraging to know I am not just plain stupid, I am just not going about it in the proper order. 

[SpareRibs]

Hello,
      Can someone tell me, after calculating these numbers using Ohm's Law what are they telling me and of what use are they. I just do not know what to be looking for or where to look.
R2-V/284-divided by-.0022=.129090,apparently amperage or current.
R3-V/275-divided by-.056=.001339
Going out to the circuit-V/208-divided by-.027=.007703
       Any help or explanation would be greatly appreciated.

[g1]

Please provide more info.  What schematic are those calculations referring to?
What do those numbers represent?  As straight math they do not work out.

[SpareRibs]

Hello,
       It is the Newcomb Pathfinder schematic, being converted using the LTSpice diagram contained in this thread.
       I used a calculator to derive those numbers, but am at a total loss as to what they mean. I divided the voltage into the resistance at each junction in the power supply C2 and C3 then out to the PI side of the 12AX7. of what use they may be I have no idea. I can read the voltage drop across the resistors with my VOM meter. I was trying to find the current, using Ohm's Law. It is all very confusing.
       Thanks for your response.
       

[Enzo]

I wondered what we were still working on also, why not just put up another link so we don;t have to go sort back through four pages to find the schematic.

If you have 250v at one end of a resistor and 230v at the other, then 20v was dropped across it.    or just measure the 20v end to end as it sits.  Ohm's Law says I = V/R.  SO if we have for example a 4700 ohm resistor (4.7k), then 20/4700 = .004.   The units of Ohm's Law are amperes, volts and ohms.  SO that 0.004 is in amperes.  0.004A is the same thing as 4milliamps, 4ma.   So in my made up example, if I have 20v across a 4700 ohm resistor then 4ma current is glowing through it.

I don;t know what all the numbers you posted are.

Ohm's Law works for any combination of two of the three components.  SO if I have a 6800 ohm resistor and I know it has 14ma flowing through it, I can figure  V = IxR = 0.014 x 6800 = 95v.  I'd have 95v across that resistor.

A word about numbers.  Your calculator might come up with a number like 0.007703, but your figures cannot be that precise.  We have what we call significant digits in a number.  If I have 45v, that has two significant digits.  If I have occasion to divide 45 by 7, my calculator will tell me it is 6.4285714.  But the most I have is two significant digits, so I cannot use more than 6.4 of my answer.    If my original measurements were 45.137v, then I would have 5 significant digits, and I can make more precise calculations.



But really, this is a guitar amp, approximate is more than enough.  If you have 0.001339 amperes, that is 1ma for any discussion we might have.  The extra 339 microamps will not make any difference.

[SpareRibs]

Hello,
      Thank you that really helps a lot explaining how to derive the calculations. I have made all of the modifications and all of the readings are pretty close as the transformer does not put out the same amount of volts as the LTSice drawing. I am going by the Newcomb schematic for the incoming voltage and resistor readings between C2 and C3.
       I will open a new thread if I don't get results soon. I am going to take the readings one more time and if it
all checks out I am plugging it in directly to the wall.
       Thanks again sorry to be such a PITA.

[g1]

Ok, here is a repost of the schematic.
As Enzo mentioned, the voltage you use should be what is measured across the resistor.  And somehow you are going the wrong way with your decimals.  A 270K resistor is 270,000 not .027 or anything, for calculations use values in ohms.
  So, according to the values on the schematic:
R2 is 2200 ohms.  There is 312V on one end, 303V on the other.  That means 9V across R2.  9V divided by 2200 = .004  That is the current in amps, or can be called 4mA (milliamps).  Do you know the metric system?  It uses a lot of the same abbreviations and makes electronics easier to understand.
  Anyway, we have 4mA of current flowing through R2.
R3 has 71V across it.  It is a 57K resistor.  So 71 divided by 57000 = .001A or 1mA
So what do these numbers do for me?  Well, there is a difference of 3mA.  So some of the current that was flowing through R2 is not flowing through R3.  Where did that 3mA go?  Look at the original drawing and you will see that there is a line from R2/R3 junction to the power tubes.  This is the screen grid.  So that is where the extra 3mA went.  The spice drawing is missing that line so it is a typo.
  So in this case, those numbers helped me find a typo, but generally we use them to see if tubes are drawing proper current etc.
  If you like, figure out the current through R10.  Then, if I tell you that plate current of a tube is pretty much the same as cathode current, you should be able to figure out what the voltage should be across R11.

[SpareRibs]

Hello,
      THANK YOU. You have explained more now than I have been able to glean from any of the 56 previous posts. That brings it all together. I could not understand how to find the amperage carried forward from one resistor to the next. All of the calculations I made on the calculator didn't come up with any useful answers.
       Now I can go back over the numbers and it will make more sense. Thanks again.

[phatt]

Hey SpareRib,
                   You may find something here will help to get your head around how it all works.
http://www.valvewizard.co.uk/

Comes with pics and not overloaded with maths.
Phil.