Anode loaded 6080 giving less than unity gain, Why??

[Shoog]

I am using a 6080 as a push pull output stage with a transformer anode load of 600ohm. The cathode is 280R resistor per triode, unbypassed. grid is at -30V and the current is about 90mA per triode.
I am putting in about 110Vpp but only getting out about 80Vpp. Any ideas why??

Shoog

[coresta]

Shoog, because of degenerative feedback of unbypassed cathode resistors

[Shoog]

I tried 1000uf of cathode bypass and it increased gain a bit, but still slightly less than unity. The 6080 should have 2x gain.

Shoog

[Shoog]

I think I have the answer.
The optimum load for the operating point I have would be 1000ohmplate to plate. I have 1300ohm, that would account for about a 1/3 drop in output.

Shoog

[jeapel]

hi
The 6080 should have 2x gain

no!

6080 ~= 6as7

like any triode 2x (or 20x for 6sn7) (or 4x for 300B)is the maximum
gain with very high R load but the output power will be very low

the standard for triode is
R load =3 x plate resistance to get low distortion with tube harmonics with some power

i don t remenber exact math formula but for exact
gain you need

plate resistance=280ohm
R load=
u=2
gm=
rk=

[coresta]

Originally posted by Shoog
I tried 1000uf of cathode bypass and it increased gain a bit, but still slightly less than unity. The 6080 should have 2x gain.

Shoog

or your 6080 are tired

[SY]

Look at that load (the load on each tube is 1/4 the plate to plate load). Look at the expected plate resistance at your operating point (from the datasheet). Understand that the gain is the mu multiplied by the plate load resistance divided by the sum of the load resistance and the plate resistance.

What would you predict for the gain?

[jeapel]

with bypass cap on rk:

gain=mu(RL/(RL+rp))


with no bypass cap on rk:

gain=mu RL / ( rp + RL + Rk ( mu + 1 ) )


mu=2

rp=~=280ohm

it s gain for single end triode no push pull but
the principle is the same maybe a factor 2
but not sure

[jeapel]

gain=mu RL / ( rp + RL + Rk ( mu + 1 ) )

i just try :

80/110=.72

2x600/(280+600+280(2+1))=.6977

.72 ~= .6977

seem ok to me

[jlsem]

Look at that load (the load on each tube is 1/4 the plate to plate load). Look at the expected plate resistance at your operating point (from the datasheet). Understand that the gain is the mu multiplied by the plate load resistance divided by the sum of the load resistance and the plate resistance.

For a push-pull output transformer, each tube sees half the plate to plate impedence. To predict the gain of two tubes in a PP output stage accurately, you need to work it out graphically. 1500 ohms P-P is really the correct load. With a transformer load, the output stage gain should be damned close to 4. I didn't check it graphically myself, but you may have it biased such that the 6080's cut off with that steep load before the full voltage swing is realized.

John