Just done and it turns out to be infinitely high. So this is broken. I'm going to replace it. Do I have to worry about C59 ?!
Yes it worked. That was the problem. Now a neat signal comes out and volume is also at the right level again. Hum has also disappeared.
Do I still have to think about how this resistor died? (Didn't look black)
Or was this just a bad luck?
Do I still have to think about how this resistor died? (Didn't look black)
Or was this just a bad luck?
With a 12AX7 in there the dissipation in the anode resistors is less then 100mW but with the 12AT7 it goes up to 400mW.If those resistors are ment for the AX they overheat with the AT.
Take 2W resistors for the two anode resistors.
Mona
And if I replace the AT for an AX, would that also be a good solution? And then I don't have to change any other components?
Or what would be a good reason to place a 12AT7 in this position? That is not entirely clear to me.
Or what would be a good reason to place a 12AT7 in this position? That is not entirely clear to me.
Doesn't make much difference for this amp.
I prefere the AT in this position, it has better drive power.But to take full advantage of that tube you have to change more then two resistances.
Mona
I prefere the AT in this position, it has better drive power.But to take full advantage of that tube you have to change more then two resistances.
Mona
Can you tell me how you calculated these values? These now indeed correspond with reality. I also try to calculate it but I can't. I may overlook something, but hopefully you can help me.
To find those voltages you have to draw a loadline.The load is what comes between the supply (444V) and ground for the tube.
Starting from ground there is R99, R98, R93, R94//R95 and R96.
4k7+10k+0k47+45k+4k7=~65k (I am lazy, like to round-off 😉 )
The 45k is 100k//82k.
Since this load is used by the two triodes each sees 130k.
Now we go back to post#14 where you see the loadline 440V-130k and a line for where you get with a cathode resistance of 1k (also double R93).
The intersection of those two lines gives the working point, about Va=150V,Ia=2mA and Vg=-2V.
The current for most resistors is 4mA exept the anode resistors, one (100k) a bit less then 2mA the other (82k) a bit more.
Nowing the current it's easy to find the voltages now (thanks mr.Ohm).
There is a little error because R96 is also used by another circuit giving extra voltage drop, but who cares 😀
Mona