Home Tube DIY Asylum

Do It Yourself (DIY) paradise for tube and SET project builders.

RE: Broskie is correct on this one.

"The difference in gain between the two outputs is given by: A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))]"

This is consistent with the answer Crowhurst would have got if he hadn't made an algebraic error. He has transcribed his formula from Chapter 12 for the input impedance of a grounded grid amplifier incorrectly. The correct formula from Chapter 12 would say Rin = (RL2 + Rp2)/(1 + mu2), but first of all this has ended up in the garbled form appearing in the line above eqn (96), and then he has evidently interpreted this, incorrectly, as (RL2 + Rp2)/mu2. This gives him a wrong formula (96) for Rx, and then in turn this gives him a wrong formula (97) for the required ratio RL1/RL2 of anode load resistors for getting balanced outputs.

His formula (97) clearly cannot be right because if you take the limit when Rk goes to infinity (the ideal CCS limit), it gives RL1/RL2 = 1+ 1/mu, which would say that the anode resistors should be unequal even in the ideal CCS limit. As has been discussed extensively already, this is simply impossible.

Correcting his calculation, one instead gets

RL1/RL2 = (1 + mu) Rk/ [RL2 + Rp2 + (1 + mu) Rk],

and this does indeed imply RL1/RL2 becomes 1 in the large Rk CCS limit. It is also consistent with Ralph's formula (his (A1/A2)^{-1} is the corrected expression for RL1/RL2 above).

Out of curiousity I tried the repeating the (corrected) Crowhurst calculation in the case where one does not make any assumptions about the two tubes being the same. So I allowed them to have different mu values mu1 and mu2, and different internal plate resistances Rp1 and Rp2. If I did my sums correctly the required ratio RL1/RL2 for getting balanced outputs is now

RL1/RL2 = (1 + mu2) Rk/ [RL2 + Rp2 + (1 + mu2) Rk]

In other words, the characteristics of tube 1 do not enter at all into the formula for the ratio of load resistors one needs in order to get balanced outputs! This seems a bit remarkable, but I think it is correct; at least, if Crowhurst's other formulae are correct. It happens because Crowhurst's eqns (92) and (93) imply Ep1 = RL1 Ek/Rx, and so when one equates Ep1 to Ep2 and solves for RL1, the Rp1 and mu1 parameters characterising tube 1 do not enter.




This post is made possible by the generous support of people like you and our sponsors:
  Signature Sound   [ Signature Sound Lounge ]


Follow Ups Full Thread
Follow Ups
  • RE: Broskie is correct on this one. - cpotl 13:32:00 06/15/21 (0)

FAQ

Post a Message!

Forgot Password?
Moniker (Username):
Password (Optional):
  Remember my Moniker & Password  (What's this?)    Eat Me
E-Mail (Optional):
Subject:
Message:   (Posts are subject to Content Rules)
Optional Link URL:
Optional Link Title:
Optional Image URL:
Upload Image:
E-mail Replies:  Automagically notify you when someone responds.