equation for working out tube circuit impedance

Hi


i am try to expand my knowledge of tube circuitry , and i would like to know how work out the impedance of a output stage to get the best transfer between the tube and a transformer.

What are the variables needed to work this out .



cheers


skal1

Tube manual.

Seriously.

which one PRR, i have RDH3 and 4, should it be in their?


skal1

hi


is this the formula for output impedance ?

Output impedance:

    Rout = (Ri + Rf) * [Ro / (Ri + Rf + Ro - Ri*A)]


can you explain this formula


skal1

What sort of output stage? Push-pull? Single-ended? Plate-follower? Cathode-follower?....... Too many variables.

push pull i think , i just want to understand how the altec engineers calculated the impedance of the 6CG7 stage of the 436 to get the impedance for the transformer they used.


skal1

I have learned a lot about Valve Impedances from Merlin:
http://www.freewebs.com/valvewizard/

His book gives more in depth calculations than the website.  I highly recommend his book.  It's a great read to understand various calculations of tube circuits. 

You will need to know the Plate resistor value, the tube's internal resistance, Cathode Resistor (depending on weather it's bypassed or not) and the Grid leak resistor of the following stage.

i have been on that site may a time  before , did not see the stuff about impedance , i will take another look.


skal1

> i have RDH3 and 4, should it be in their?

Tube MANUAL: ; from the guys who wanted to sell more tubes.

However using small triodes transformer-loaded was out of fashion when 6CG7 was introduced, so they don't give you a suggestion.

> how the altec engineers calculated the impedance of the 6CG7 stage

Transformer coupled triode power stage: select load 2*Rp or higher, play with the curves for biggest swing with lowest distortion. Then try to find ANYthing close.... Altec surely used whatever transformer was handy in stock.

> i have RDH3 and 4, should it be in their?

Yes, both have the complete work-up of push-pull triode power amplifier. If you can wade through it (it aint easy), re-do it with 6CG7 curves.

However the fact they used 6CG7 means it may not be a POWER stage, cuz the 'CG is a small bottle. Actually it can probably do a whole Watt push-pull, whereas a studio line doesn't need even a half-Watt. So exact analysis is not needed.

And here's the thing. 6CG7 is 7K(?). Load with 14K per tube. In push-pull, the output winding will be specified as 56K. But there are NO 56K power OTs. 10K, 20K, maybe 30K.... but the cost rises and the treble performance droops. So it isn't so much about the tube as about the limits of practical transformers.

The local feedback section:
http://www.freewebs.com/valvewizard/localfeedback.html

Thanks every one .


skal1

For lots of tube manuals look here:
http://www.pmillett.com/tube_data.htm

what does the sign highlighed in red mean in this impedace formula Zoa = Ra || ra


skal1

Ra and ra are in parallel, the usual formula for parallel resistors applies.

Thanks volker

This stuff is really greattttttttt  just like frosteys  sugar coated


skal1

ok have done some reading and i am a bit confused about what circuit topology is used for the vari stage and the the output stage of the  altec 436.

can some one help


is  the output stage pushpull?

is the vari stage long tail ?


skl1

Hi.

If I remember well, both push-pull.

Laurent.

pyjaman said:

Hi.

If I remember well, both push-pull.

Laurent.

Click to expand...

Sweet

ok  i have been trying to do some maths  on this 436 circuit but i think my maths is results are way of.


so i am trying to find the current  at the junction of r3 and r4 {210/10k=0.021} correct me if i am wrong.

so now i want to find the voltage drop through  r3{0.021*47000=987v} i know this is wrong because the schematic shows 65v , where am i going wrong?


skal1

I = voltage at one end minus voltage at other end divided by resistance.
Current in R3 & R4 is (210 - 65)/47k = 3.08mA; this is the small-signal current (Vg = -0.5V - high bias - no compression)
Under high-signal (Vg = -30V - low bias - mucho compression) (290 - 285)/47k = 0.1mA
When resistance is expressed in ohms, the result is in Amps; when expressed in kilo-ohms, the result is in milliAmps.

abbey road d enfer said:

I = voltage at one end minus voltage at other end divided by resistance.
Current in R3 & R4 is (210 - 65)/47k = 3.08mA; this is the small-signal current (Vg = -0.5V - high bias - no compression)
Under high-signal (Vg = -30V - low bias - mucho compression) (290 - 285)/47k = 0.1mA
When resistance is expressed in ohms, the result is in Amps; when expressed in kilo-ohms, the result is in milliAmps.

Click to expand...

so if i did not have the idle voltage which 65v {no compression) on the schem how would i work out the voltage drop across R3 AND R4 , would the formula be

voltage drop =current *ohms ?


skal1