I'm building up a no-name kit amplifier. 6SL7 driving P-P 6V6s, two identical channels for stereo. I haven't started wiring it up yet, was just going over the schematic before doing so and see what looks like an inaccuracy:



The way its drawn, looks like the sections of the 6SL7 are directly coupled to one another...the grid of the phase inverter would have the same B+ as the plate of the driver.

Did they leave out a cap, or should C2 (39pF) be moved above the tie point, or am I just not understanding something about that tube?

No, there's not likely a mistake. I have a Mason guitar amp, similar circuit, 6SL7 (iirc), preamp direct coupled to the inverter (iirc). It plays great! I'll try to find the schematic for you.

cheers

John

Read this, lots of good info. http://www.angelfire.com/electronic/fun ... Heart.html

Thats a very common and cheep way to make a phase inverter

That's a normal concertina arrangement and intended to not need a cap. V1a current and load resistor are chosen so that plate volts will properly bias the concertina, otherwise you'd need an extra cap and another resistor (grid leak) under V1b to set bias. The second approach is also done but it's desirable to eliminate the cap since they make it more difficult to stabilize the amp after negative feedback.

The C and R from grid to ground are for just that, rolling off high frequency (the R forms a shelf limiting the cut) so the amp is more stable with the negative feedback.

Glad I asked! Now I'm that much smarter. Thanks for the input everyone.

Could you post the values of the resistors: They are not too clear for me. The grid connected
to the plate is interesting but of more interest would be the cathode voltage of the
phase inverter section of the 6SL7.

Sure...I annotated the schematic and cropped to just the signal path, leaving the power supply off.



I'm also somewhat unclear about why the decision to use two 680 Ohm resistors in parallel with each other for the cathode resistors on the 6V6s. Two 680R / 5W in parallel is 340R 10W...maybe they just didn't want to use a 10W resistor there.

I ran a quick simulation and, assuming B+ comes out around 330V, V1B cathode is about 80V.

B+ comes off a transformer with a 255V secondary (non-CT), into a bridge rectifier...B+ to the output transformer comes right off there (with a 470uF cap); B1+ to the front section through a 3.3K resistor in the power supply also filtered with a 470uF cap. There's a 220K bleeder between B+ after the first filter (and before the 3K3) and B-.

For what that's worth.

That matches what I was able to gleen from the original posted pic.
At the end of the day actual B+ will depend on winding resistance, loading, etc, so I just used a 1.3 'rule of thumb' factor to get the 330V ballpark figure.

Cathode volts will also vary from tube to tube so that's just a ballpark number too.

I'm surprised at what they used in the cathode circuit of the PP 6V6s. Likely they could have gotten away with just ONE 330R instead of the pair of 680s. Plus the bypass cap shouldn't be necessary as the current will be pretty steady as one tube starts to shut down and the other conducts more heavily.

Scott Todd

That only applies Class A.

Here's the schematic and parts list that I drew up before I recapped my Mason V amp:






Mason Model 5 Guitar Amplifier

The Galt Musical Instruments Company

(made in Canada circa 1950)

Resistors :

1) 100k
2) 100K
3) 560
4) 470k - 1W
5) 100k
6) 500k - Volume
7) 200
8 ) ??? - NFB
9) 1.2 meg
10) 2.2 meg
11) 220k - 1W
12) 220k - 1W
13) 470k - 1W
14) 470k - 1W
15) 500 - 10W
16) 50k - 1W
17) 4.7k- 1W

Capacitors :

a) 10 / 40v
b) .05 / 600v
c) .005 / 600v
d) .05 / 600v
e) .05 / 6300v
f) 50 / 40v
g) 20 / 450v
h) 20 / 450v
i) 8 / 200v
j) .01 / 600v