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**Enzo** · 10-24-2007, 04:05 AM

With no power tubes, the speaker load becomes irrelevant. Unfortunately without the power tubes the plate voltage readings are also irrelevant. No power tubes means the majority of hte current draw on the high voltage supply is gone, so the voltage will drift very high. it will tell you nothing about t=what the tubes face in operation.

if this amp ever worked - and we will assume that this brand and model were not sold in such a bad design condition that they don't work - then teh tubes were not subjected to undue voltages.

DO the old 6BQ5s work? From your description of improving sound, I'd have to guess they do. Then leave them in the amp and measure the voltages you find. They will not change appreciably with a different pair of tubes.

NEVER operate a tube amp with no speaker load.

**milothicus** · 10-24-2007, 04:02 PM

the old ones do work. i just got excited last night after drawing out most of the schematic, and wanted to find out what plate voltages are being supplied.

I'll resolder the speaker and put the old tubes back in to check it.

A couple of other questions about making bias checking easier...

would it be ok to install normally-closed 1/4" (or 1/8") jacks between the cathode and cathode resistor? this way when no plug is inserted, the socket is shorted, but inserting a plug could run the current through my multimeter.

seems like it would work to me, but i'm not sure if that would run too much power through the jack when no plug is inserted...

**Bruce / Mission Amps** · 10-24-2007, 05:05 PM

Originally posted by milothicus View Post

the old ones do work. i just got excited last night after drawing out most of the schematic, and wanted to find out what plate voltages are being supplied.

I'll resolder the speaker and put the old tubes back in to check it.

A couple of other questions about making bias checking easier...

would it be ok to install normally-closed 1/4" (or 1/8") jacks between the cathode and cathode resistor? this way when no plug is inserted, the socket is shorted, but inserting a plug could run the current through my multimeter.

seems like it would work to me, but i'm not sure if that would run too much power through the jack when no plug is inserted...

If it is cathode biased just measure the voltage drop across the biasing resistor and use simple Ohm's Law to find the current.
EX:
12v across a 150 ohm resistor = 80ma
80ma / 2 tubes = 40ma per tube.
40ma times the measured plate to cathode voltage of say, 310vdc = 12.4 watts per tube.
ETC.

**milothicus** · 10-24-2007, 05:58 PM

I understand the ohm's law method, but i figured since i've got the thing apart, i might as well install a method to check the bias more easily later.

I'll be reconnecting the speaker and tubes and checking voltages tonight, hopefully.

**Enzo** · 10-25-2007, 01:24 AM

But Bruce is right. What is easier than one voltage reading and a few strokes on your calculator? Adding such a cutout jack is just one more potential trouble spot when the contacts oxidize. If you want as test point, run a wire from the cathode to a test point jack.

If you didn't want to calculate anything, then add a 1 ohm resistor in series with the cathode resistor, and run wires from either end of it to test point jacks. Them read milliamps as millivolts on a meter across the 1 ohmer. Adding 1 ohm to whatever the existing resistance is will not affect the amp.

But that seems like way more effort than is needed jist to avoid a little calculator arithmetic.

**milothicus** · 10-25-2007, 01:30 PM

i have no problem with arithmetic (i'm a mechanical engineer), but i know i've read that calculating the current from the voltage drop across a resistor has some inherent flaws. mainly due to resistance fluctuations with heat, i believe.

i figured the most accurate way to measure current is to measure current, so i might as well put in a method to do that. I've put the amp back together now with no test points, so I'll be using the resistor method, anyway.

Thanks for the help. i'm sure i'll have more questions...

**Enzo** · 10-26-2007, 02:42 AM

I can appreciate attention to detail, but I don't think the variation in resistance over temperature is large in these amps. In any case, measure it cold, then warm the amp up and take a new reading immediately after power down when the resistor is still fully warm. I'd be surprised at more than a couple percent change. And keep in mind this is just a little guitar amp, no sensivie circuit. Variation in mains voltage will far outweigh any thermal changes. I think any inherent flaws are likely due to bad assumptions. Someone puts a "470 ohm" resistor in there, calculates the volatge across it for a current reading, then measures current directly and finds them not the same. WHat they didn't do was measure the 470 ohm resistor in the first place. A 10% resistor could be anything from 423 to 517 ohms and be within spec. 5%? 446 to 493 ohms. Never assume it really is what the nominal resistance might be.

**milothicus** · 10-26-2007, 01:02 PM

ok, i have another question about what to expect when i put the new tubes in...

if i measure the cathode current (using the resistor method

) and multiply by the measured plate voltage, i'll get the power dissipation for the current tube (pun intended).

which values are likely to change when i put a new pair of tubes in?

the old tubes are marked 6BQ5 and look pretty old, and the new ones are groove tubes EL84S, rated at 3. are the old tubes likely to be drawing more current than the new ones will need? i'm worried about burning out the new ones before i find out their bias is way off...

**Enzo** · 10-27-2007, 04:13 AM

Who knows? The new tubes won't be burning up in a few minutes anyway. You have plenty of time to see if the amp works. if it works with the old tubes, it will work with the new ones. Remember, the tube makes its own bias with the cathode current. SO if one tube set draws more current naturally, that means a higher cathode voltage, whuch in turn means more bias, which limits the tube more. In other words to a certain extent it is self correcting. It just is not all that critical.

**milothicus** · 10-29-2007, 03:35 AM

Ok, i took some measurements, and this is what i got...

old tubes (Hit Ray 6 BQ5 Japan):
V5 (plate to ground) = 368v
V6 (plate to ground) = 370v

common cathode resistor: (labelled 150 ohms) 157 ohms
voltage across it = 16v

result: 51 mA per tube, approximately 19w/tube? seems high, but it's been working well enough for the last year or so...

new tubes (Groove Tubes GTEL84S -3)
V5 (plate to ground) = 377
V6 (plate to ground) = 383
Cathode resistor voltage = 13v

result: 16 w per tube...

so i let it sit for a few minutes and went back to it.

V5 = 362v
Cathode resistor voltage = 16v

and the metal inside the tube was turning red... 18w.

so i'm guessing i should increase the size of that resistor... any recommendations for a ballpark of how much?

**Enzo** · 10-29-2007, 08:06 AM

Before you do that, some thoughts. First, remember to subtract the cathode voltage from the plate voltage for the voltage across the tube. Multiply THAT times the current. I get about a watt less than you do, and I think that is why.

You checked the plate volts and saw it drop as the tubes conducted more current. (That makes the cathode voltage rise.) What you didn't do was watch the grid voltage while this happened. When it warms up and strats to red plate, check the grids. If they are not staying at ZERO volts DC or even a tiny bit negative, then I'd replace the coupling caps from the PI. Any positive voltage there is leakage from somewhere.

The old tubes didn't red plate did they? Does the current through them rise after a warm up? Whether they red plate or not, does it?

Just because the new tubes are new, doesn't mean one of them isn't bad. Was just one tube red plating? if so, swap places with the other one. Now does the same tube red plate in its new home? Or does the other tube now red plate in that same socket? This determines if the tube was bad or if there is somehting about that socket and its circuit.

You don't want to be modifying an amp to cure a leaky coupling cap.

**milothicus** · 10-29-2007, 04:12 PM

It was just one tube redplating...

and an old redplated in the same position, but then neither did when i reversed them.

i'm not sure which caps would be from the PI. that's a phase inverter, right? unfortunately i still don't have a full schematic of the amp (working on it), or an understanding of the phase inverter...

i'll take some more measurements tonight.

**milothicus** · 10-31-2007, 01:37 PM

If interested, i've attached a pdf of the schematic i drew up. i know the HV label terminology is wrong, but i haven't corrected it yet.

i'm going to change the coupling caps tonight or tomorrow. should i be replacing all the coupling caps? i see that as C1, C8, C10, and C11.

are the caps in the tone stack at risk too?

Attached Files

20071031-LifcoModel630.pdf (37.2 KB, 681 views)

**milothicus** · 11-02-2007, 03:41 AM

ok, old and new tubes redplating in the same location. i'll change the coupling caps, but i have to wait until i get get some. .01 microfarad capacitors rated at 600v seem to be hard to find in person.

i have another few questions, though.

My power transformer's output is 370-0-370. this seems way to high for EL84's from what i've read. Am I wrong about this? is there a simple way to bring this down about 100v to what most other amp designs are showing?

The tremolo on my amp doesn't work. is there a simple explanation of how this tremolo circuit works, to help debug it?

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