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Speaker question ?

Started by SpareRibs, December 24, 2013, 04:17:10 PM

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Roly

Keep the switchpot.  Goes in your junk stash of potentially useful parts.

Before you go drilling any holes or adding a Gain pot, first just connect the -ve end of the cathode bypass to ground and do some initial tests, we'll add complications in small steps if you don't mind.

Limiter bulb (wattage?, supply voltage?)

VHT1,2,3 =?

We need table of Va and Vk for all valves.

6V6a   Va, Vs, Vk
6V6b   Va, Vs, Vk
6AU6   Va, Vs, Vk
12AX7a Va,     Vk
12AX7b Va,     Vk
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

Hello,
      I used a light bulb limiter with 60 watt bulb. Light came on bright then went to nothing. Second try light came on dim also went to nothing. I unhooked the speaker changed to 25 watt bulb now I have maybe half the
output of the bulb constant glow.
      I checked the voltage on the 6V6's it was 95v with light bulb. Is it ready to plug directly into the wall ? Or should I take more readings first ?

J M Fahey

Tube amps have intrinsic high power consumption, even at idle, so the 25W bulb is definitely too small, and probably the 60W one too.
Use at least a 100W one and if everything looks normal, plug it straight into the wall.

Roly

It should go quite dim after a bright start, 60W or 100W.

Then post...
Quote
VHT1, 2, 3 = ?

We also need table of Va and Vk for all valves.

6V6a   Va, Vs, Vk
6V6b   Va, Vs, Vk
6AU6   Va, Vs, Vk
12AX7a Va,     Vk
12AX7b Va,     Vk
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

Hello,
   With limiter still on the voltages are as follows. I only have voltages as I am not sure what Va,Vs,or Vk,are.
1.6AU6 pin 1= (-2)                                    1. 12AX7 = Anode pin 1 (286) / Cathode pin 3 (259)
2.6AU6 pin 5=(293)                                   2. 6AU6 = Anode pin 5 (282) / Cathode pin 7 (0)
3.C3 = (293)                                            3. U1-6V6 = Anode pin 3 (265) / Cathode pin 8 (15)
4.R15 = (293)                                          4. U2-6V6 = Anode pin 3 (265) / Cathode pin 8 (16)
5.12AX7 pin 6 = (61)                                 5. 6X4 = Anode pin 6-1 (fluctuates 1-4) / Cathode (295)
6.12AX7 pin 7 = (0)
7.6V6 pin 5 = (0) both tubes
8.6V6 pin 4 = (284) both tubes

         Removed resistor 560K right of pin 5 6AU6 going to pin 6 of second half of 12AX7 on original schematic
as it was missing from appended cct.
          Is it now ready to plug directly into the wall ?
         


Roly

Quote from: SpareRibsIs it now ready to plug directly into the wall ?

NO!

We have a way to go yet, and the supply voltage is only down 5% on spec; from an initial glance at your voltages you worked it out, Va = voltage anode, Vk = voltage kathode (don't ask), Vs = voltage screen, which is what you've done. 


I'm going to do some LTSpice stuff with those numbers and see what that shows up.  Stand-by.

Meanwhile...
What do the voltages around the first triode, 12AX7a, tell you, (259) volts on the cathode for example?

You've got 15V on one 6V6 cathode and 16V on the other, what do you make of this?

What is the actual marked and measured value of the 6V6 cathode resistor?
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

Hello,
      Glad to know things are somewhat normal. I will do all of the readings over again using alligator clips. The
way I took the readings I posted was using an alligator clip for ground and a probe to get the readings, that may have led to the one volt difference. I will check them ALL again just to be sure.
       I will also be sure to do the 6V6 cathode resistor. 

SpareRibs

#37
Hello,
      OK, here are the resistor readings using alligator clip. Some are slightly different but should be in spec.
Master wiper = (234)                           12AX7 = PIN 1 = (294)        6AU6 = PIN 5 = ANODE (291)           
C3 = (295)                                                     PIN 2 = (20)                    PIN 7 = CATHODE (0)
R15 = (295)                                                    PIN 3 = (270)
6AU6 = PIN 5 = (291)                                       PIN 4 = (0)           U1 6V6 = PIN 3 ANODE (261-269)*
6AU6 = PIN 1 = (-2)                                         PIN 5 = (0)           U1 6V6 = PIN 8 CATHODE (16)
12AX7 = PIN 6 = (295)                                      PIN 6 = (295)        U2 6V6 = PIN 3 ANODE (271-275)*
12AX7 = PIN 7 = (0)                                         PIN 7 = (0)           U2 6V6 = PIN 8 CATHODE (16)
6V6 = PIN 5 (0) Both tubes                                PIN 8 = (271)
6V6 = PIN 4 (285) Both tubes                             PIN 9 = (0)           6V6 CATHODE RESISTOR 16V ACROSS
                                                                                               VALUE = 300-OHMS
                                                                                            COLOR -ORANGE-BLACK-BROWN-GOLD
                                                                                             
                                                                                       *Indicates fluctuation between two values
   Hope this helps. Alligator clip was a tight squeeze in some places. One more question. I am working on this
out of the cabinet. Should the speaker be hooked up ? It is of great concern to me as it is said not to power it up with the speaker disconnected. However I am not putting under a load, but I am thinking it should be there to complete the circuit.



       

Roly

{I actually asked you those questions to get you to think about them, and what they might imply.}


With your latest readings a couple pop out at me.

Quote from: SpareRibsMaster wiper = (234)

Yawot?  You've got 234V on the wiper of the Master pot?  Really?  Take a look at the circuit - how can this be?
What do you read on the other two lugs, particularly the nominally "grounded" lug?
Are there any voltages on the tonestack pots?

Secondly, the 12AX7, both sections
Quote from: SpareRibs
PIN 1 = (294)
PIN 2 = (20)
PIN 3 = (270)

The cathode is at 270 volts?  Really?  Look at the circuit - there's (supposed to be) a 1k5 cathode resistor to ground.
Apply Ohm's Law, how much current is that?
Apply the Power Law, how many watts is that in the cathode resistor?

If it isn't cooking then my best guess is that it's open circuit somewhere.

(pins 4, 5, and 9 are part of the heater circuit and normally not quoted unless there is a specific heater problem, then in AC volts which they typically are.)


Now the other section...
Quote from: SpareRibs
PIN 6 = (295)
PIN 7 = (0)
PIN 8 = (271)

Again, we are supposed to have a 1k5 cathode bias resistor to ground, but we clearly don't.


What we want is for the anode voltage to be about half the supply, and to get that we need to pass a bit over 1mA though the 100k anode resistor, and therefore also through the 1k5 cathode resistor, which should then be creating a drop of around 1.5 -> 2 volts on the cathode - ideally.  Both sections.


So you need to find out why the 12AX7 stage (and maybe the Master pot) doesn't seem to have a ground return to the power supply.


An open output shouldn't be a problem but if you are worried about it just put a clip lead or tack a link across the output to short it, or a 8 ohm 10W w.w. resistor if you happen to have one to hand.   ;)
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

#39
Hello Roly,
       Well I found the grounding problem.  A is strip riveted and soldered to the chassis. It has 3connections.
I thought they were all grounds, I was wrong. Only the middle is ground. so I soldered the grounds to that. I now have some new numbers and they are radically different. So here goes.
C1 = 296
C2 = 289
C3 = 226
R15 = (55)
12AX7 = ANODE (67) CATHODE (0)
6AU6 = ANODE (67) CATHODE (0) PIN 1 (.718)
6V6 = U1-ANODE (289) CATHODE (16)
6V6 = U2-"               " "                    "
Master Volume (0)
       I also had to change a couple of wires on the tone controls. As I was working on it I was looking at it backwards. As I said before to most of you guys on this forum, this is all second nature to you. I am not as well versed in electronics. I am sorry if you get impatient with me.

Roly

Whoops. That'll do it every time.  The new voltages look more like it.

Quote from: SpareRibsAs I was working on it I was looking at it backwards. As I said before to most of you guys on this forum, this is all second nature to you. I am not as well versed in electronics. I am sorry if you get impatient with me.

Not so.  I've been doing this stuff most of my life and I frequently have to double check things like pot wiring to make sure I've got it right.

Quote from: SpareRibs12AX7 = ANODE (67) CATHODE (0)

I assume that this is the Phase Inverter section.  I find the cathode = 0 a bit curious.

On the supply end of the anode load R10/270k we have around 230 volts, and on the anode end we have around 70 volts, meaning that there is 230 - 70 = 160 volts across the resistor.  The current through this resistor is then given by Ohms Law as;

I = E / R

160 / 270 = 0.5925925mA

If this current is going through the valve section, as we would hope, then it must also be going through the cathode resistor, R11/1k5.  Again Ohm's Law gives the expected cathode voltage;

E = I * R

0.59 * 1.5 = 0.885 volts.  (On high voltage ranges your DMM will show shuch a small voltage as "0", so you'll need to use a lower voltage range for measuring cathode voltages).

BTW, because the resistance of the grid circuit is typically many times the resistance of your multimeter there isn't much point in measuing grid voltage (except in the output stage) because the meter will so seriously load the grid circuit as to make the reading meaningless.

I'll leave it to you to use the same method to work out what the cathode voltage of the 6AU6 should be, but it will also be (very) small.

You don't give the voltage of the preamp section of the 12AX7, but I'll make a couple of guess-timates;

I guess the valve current at 1mA

Therefore the anode voltage will be (1mA * 100k = 100V drop),  230 - 100 = 130 volts.

1mA through 1k5 gives 1.5 volts cathode bias.


Getting there.   :tu:
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

#41
Hello,
      Thanks, it certainly feels good to be making progress. I guess that grounding strip was causing most of my
problems all along. No matter what I tried I was fighting an open circuit. Without knowing where or what to check I was just beating my head against a wall. Oh well onward and upward.
       I will check the readings once again. There is a location on the meter for lesser readings. I will utilize it for the cathode reading, and post it back.

        I set the meter on the more sensitive setting and here is the result, (12AX7) Cathode (0.532). It may not be as accurate as it should be as the meter is not high dollar,







90.532

SpareRibs

Hello,
      I am not sure I did this right, but here goes.

6AU6 Supply side R7 68 - Anode side 67
68-67= 1
1/68=0.014
Actual meter reading - 0.010
Preamp voltage - Anode 68 - Cathode .526

Roly

With audio electronics 10% accuracy is fine.  For all the crud that goes on in Hyper-Fi forums about parts-in-a-billion effects, audio ain't rocket surgery, and guitar audio even less so.

Quote from: SpareRibs6AU6 Supply side R7 68

Really?  But isn't this supposed to be HT3 @ 230-odd volts?

Now the anode load is 680k and the screen feed resistor is 1.9Meg, and I'm going to introduce you to something very important.

Pull the 6AU6 out of its socket, power up, and confirm the supply voltage (~230V), then measure the voltage on the valve end of the 680k anode resistor, then the valve end of the 1.9Meg resistor.

With the valve out there should be no current flowing in these resistors and therefore no voltage drop across them, right?

Is that what your measurements say?

Why do you think you are getting the result you are getting?

(yes, there will be a test next period.  ;) )

Quote from: SpareRibsPreamp voltage - Anode 68 - Cathode .526

Okay, some harmless sums;

0.526V on the cathode, and 1k5 cathode resistor, therefore the current is;

I = E/R
0.526/1.5 = 0.35..mA

The same current must be flowing in the anode resistor, R15/100k, off the 230V supply, so does this add up?

Voltage drop;

E = I * R
0.526 * 100 = 52.6V  (mA * K-ohm = Volts)

230 - 52.6 = 177.4V on the anode.  But you measure only 68V, so that obviously doesn't add up.  In Ye Olde Days of analogue meters I'd say you mis-read your meter, but in these days of digital accuracy I'll say that you weren't measuring what you thought you were measuring (wrong side of the twin triode perhaps, confusing the preamp with the phase inverter maybe?)

On the preamp section I expect around 1.5V on the cathode and around 130V on the anode (for a 230V supply).
If you say theory and practice don't agree you haven't applied enough theory.

SpareRibs

Quote from: Roly on January 01, 2014, 10:19:34 AM
The attached circuit is unfinished, potentially full of value errors because the original was so lo-def; hasn't run yet in LTSpice because the OPT is still unknown
http://www.r-type.org/pdfs/6au6-1.pdf

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.