Single tube linestage schematic wanted: 8CG7

[DF96]

Yes, an LED at the bottom is equivalent to a bypassed resistor.

[Bigun]

Originally Posted by DualTriode

I have the hard copy and did not download. Put this in google.

Mit Radiation Lab V18 G Valley H Wallman Vacuum Tube Amplifiers


I got a lot of hits, perhaps there is a good link.

DT

I had no luck with this, the link was so slow it just clogged up my computer. Maybe you can post an image of the pages that you were quoting - the thing is so old it's clearly out of Copyright.

[DualTriode]

Originally Posted by Bigun

I had no luck with this, the link was so slow it just clogged up my computer. Maybe you can post an image of the pages that you were quoting - the thing is so old it's clearly out of Copyright.

Bigun,

See post #56 that link seems to work!

DT

[Bigun]

Nope, it's just toooo slooow my machine just stops working, the web browser goes blank, the hard disc ticks away, the high speed modem flashes away, my beard grows longer.........

[DualTriode]

Hey team,

I do not have a scanner. Will someone scan and post a copy of page 464 from the Valley and Wallman text. Fig. 11.46 is on that page.

Thank You

DT

[DavesNotHere]

how about a common cathode on the first section with a directly coupled cathode follower afterwards.

[rongon]

here it is

[DualTriode]

Originally Posted by rongon

here it is

rongon,

Thank You

DT

[rongon]

I just ran a few tubes through those equations and got consistent answers.

rp + (mu + 1)R

where R = Rk

When you do those equations, is it better to use the published specs, or the characteristics of the tube in question?

For instance, a 12AX7 has datasheet spec for mu of 100, but we all know it's always about 70 in circuit.

If I use mu = 100, I get

62500 + 101 * 1500 = 214,000

or

62500 + 71 * 1500 = 169,000

That's a big difference.

Zp = the results of the equation above divided by 2. So either about 107k or 85k, depending on the figure for mu that you use.

What's funny is that you could just as easily put a 220k resistor as the plate load and get the same or better results, right?

--

Same thing for 6DJ8:

3500 + 26 * 330 = 12,080

so Zp = 6k

That's not so great... Or what did I miss here?

--

It comes out to Rload = approx rp*3 in all cases.

--

[DualTriode]

Rongon,

I am not sure that I followed you, I could be way behind.
I like those musty old books. You also look at the data sheets. You plug numbers into a calculator and get 3.5 places of accuracy. You get a feeling of precision. Do not get real comfortable and feel precise. The two triodes inside the envelop may not match vary closely. Different brands of the same tube number are constructed differently. So sort of close is the best we can assume. Using the data mu, rp and gm is a place to start. If you want to know what numbers to use for the circuit you are building plug it in. measure the voltage drop across a known resistor, say the cathode resistor. You know the grid voltage and you can calculate a circuit idle voltage. Now you can plot the operation point on the plate curves in tube data sheet document. Now you can draw a line tangent to the operating point on the plate cure. You now have enough to calculate the rp and gm. The set of parameters you measure and calculate may be close to the assumed operating point in the data sheet or not. Use the numbers that you come up with for your circuit to use in the musty old formulas. They may be a close estimation. Remember that each tube you pull out of your plastic shoebox will be different.
The average man is 5’ – 10” and wears size 9 shoes.

DT