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Discussion in 'Amp Central Station' started by ce24, Jun 3, 2020.
Thanks! Are you talking about the trem pots?
Ahhhh! Its a mounting plate for ???
Is there anything mounted to it? What is the black square thing?
It looks to me like a shielding partition between two circuit sections.
You got me.....I didn't pay much attention to it. Junction in some medium. I'll take the top off tomorrow and find it on the scheme then share it.
Just curious is all.
Answering my own question:
The square is most likely an optocoupler, inside a dashed square by V5A here marked "national semiconductor." This is shielded with the little partition because the circuitry involved tends to couple "ticking" from the trem oscillator into the audio circuit nearby.
It a semiconductor.....the square outline in the middle. Next to the 12ax7.
Holy crap......this seems to be very hot bias! JJE34L. Using eurotubes simple bias probe.....i did purchase and install a bias pot but i didnt hook it up inside yet. Will this works from robrob since its a bassman hes referencing. Only difference is mine doesnt have the IN4007.
I have a 39k Instead of the 27k from gnd to lug 2 of Bias pot.
And a 47k insted of the 15 k across the top. BUT br/grn/blk is not 15K its just 15???
Is the blue "bullet" shape your bias diode? (Not familiar with diodes that look like that but maybe?)
If so, then I have a question.
The two blue wires from the power trans are the bias secondary and I see one soldered to the chassis - the ground/positive side of the bias circuit. I see the other going to the eyelet directly below the diode.
I highlighted what looks to me like a connection between that eyelet and the one just to the left. There should NOT be anything connected at that eyelet other than the blue wire and the banded end of the diode.
The pointed end of the bullet-shaped diode is pointing in the correct direction but it looks to me like that diode is shorted out by a jumper from one eyelet to the other. Here's some bullet-diode info:
Nice..good eyes.. yes the blue bullet is the diode...that possible short wasn't or isn't now at least with the recap.
Thanks for the help... I'll work on it and see how it goes.... What your take on my bias number.... From my research it's a little high but in range and no redplating.... Not even close.
If bias is too negative, (too big a number) that's cold, not hot. The closer bias gets to zero, the more current pulled through the tube.
I'll check.the expected numbers and get back to you.
Oh wow...I didn't know that inverse..... So it's a little cold then? Makes sense because my research on this amp said they biased them a little cold for tube life.
Triode. 3 elements: plate, cathode, grid. (Save pentodes for later.)
Also we need a heater (some tubes have heater & cathode integrated as the same thing, but not many since the 1930's.)
Heater in the middle of the cathode. Heats it up. This excitation knocks electrons loose.
They just hang about, UNLESS, a positively charged object attracts them. So let's apply a positive voltage to the plate. Result: current flows through the tube. Electrons are resupplied at the cathode because ground (literally, the earth) is a limitless supply of more electrons, and they fly across and into the power supply via the plate connection.
If we stop there, we have a diode. The current flow is governed only by how positive the plate is compared to the cathode.
But if we add a third connection - the grid - we can now have a powerful effect on the electron cloud. The grid is between the cathode and the plate, closer to the cathode so it has a strong effect on the charge that the cathode is exposed to. If we charge the grid negatively compared to the cathode, the electrons are repelled back towards the cathode. If the grid is very negative, almost no current flows, we call that "cutoff."
We can go for an inbetween value where a controllable current flow occurs. More negative, less current. Less negative, more current. A small change of grid voltage can have a large effect on current flow.
Those basics make it clear that the grid must be held negative compared to the cathode, and the more extremely negative, the less current will flow.
Now get this trick: If we pull the currrent through a resistor at the plate and keep one end of the resistor steady at the power supply's voltage, the variations in current flow translate to variations in voltage at the other end (where the resistor is connected to the plate.) A small fluctuating voltage at the grid results in a larger fluctuating voltage at the plate: we have amplification.
At that moment, we have just moved from the world of electricity into the domain of electronics, which depends on amplification for all its usefulness.
What values should these resistors be changed to. I will be using a 10k bias pot.