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
I am putting in about 110Vpp but only getting out about 80Vpp. Any ideas why??
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
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
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=
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=
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?
What would you predict for the gain?
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
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
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
i just try :
80/110=.72
2x600/(280+600+280(2+1))=.6977
.72 ~= .6977
seem ok to me
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
Impedance looking into the primary of the OPT is the square of the turns ratio. 1/2 the turns (one leg) means 1/4 the impedance... all other things being equal.
Not for a push-pull scheme. It is a more complicated matter than what appears to be two single-ended OPT's in series. As a matter of fact, the load each tube sees varies along the load-line (each tube sees approximately one half the plate to plate impedence at the quiescent point. As the grid for one tube swings to its most positive value [while the other tube's grid swings the other way] the impedence it sees approaches the full a-a value. Conversely as it swings toward its most negative value [to 0 volts], the impedence it sees approaches [a-a]/4), but to keep it simple, we'll just call it 1/2 the plate to plate impedence.
John
Sorry... just being a goof.
The transformer impedance is what it is. Voltages and currents are transformed by the turns ratio and impedances by the square. The Zin is simply a function of the load resistance and the turns ratio (perfect transformer).
You are talking about a method for PP analysis though, and I would like to learn more. I'm having trouble following why the impedance changes. You are talking about the effects of primaries driven simultaneously and a non superposition way of looking at the circuit? Are you talking about transistions in operating class?
Can you elaborate?
😕
The transformer impedance is what it is. Voltages and currents are transformed by the turns ratio and impedances by the square. The Zin is simply a function of the load resistance and the turns ratio (perfect transformer).
You are talking about a method for PP analysis though, and I would like to learn more. I'm having trouble following why the impedance changes. You are talking about the effects of primaries driven simultaneously and a non superposition way of looking at the circuit? Are you talking about transistions in operating class?
Can you elaborate?
😕
The analysis is rather complicated, but with two tubes "running in opposite directions" on the same primary of a transformer, let's just say there is an autoformer effect going on.
This is a class A situation. Other modes of operation get even weirder.
I suppose I could go down to my garage and dig out an old textbook and quote verbatim a longwinded explanation complete with mathematical equations, but if you don't already accept the rule of thumb of each tube seeing 1/2 the plate to plate impedence of an OPT, maybe you ought to study the problem yourself.
John
I'm OK with that... and I will dig into RDH4 as well. I was, however, under the impression that the rule of thumb was (a-a)/4.
The autoformer effect seems clear to me; but that would be based on whether currents are flowing through the "other" primary, and, the Zin, or effective Zin, would be a function of this. Hence my question re operating class.
Curiouser and curiouser...
The autoformer effect seems clear to me; but that would be based on whether currents are flowing through the "other" primary, and, the Zin, or effective Zin, would be a function of this. Hence my question re operating class.
Curiouser and curiouser...
"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"
That looks to be about exactly the situation I have, so I have nothing to worry about.
Thanks everyone, I understand a little bit better now.
Shoog
i just try :
80/110=.72
2x600/(280+600+280(2+1))=.6977
.72 ~= .6977
seem ok to me"
That looks to be about exactly the situation I have, so I have nothing to worry about.
Thanks everyone, I understand a little bit better now.
Shoog
poobah, it's (a-a)/4 when you're in Class B mode. Either actually operating that way, or when you hit a peak in Class AB.
When one of the tubes goes into cutoff, the other one sees only half the winding, the other half is open-circuited by the cutoff tube. Half the winding open circuit halves the rated winding ratio so the other tube sees 1/4 the rated impedance.
I *think* this is the right way to look at class A operation. If you look at each tube individually, you can see it. Basically each tube sees the full primary TWICE, once directly, and again through the coupling of the two halves of the primary together. So it sees the full primary (the a-a) value, in parallel with the full primary again. I'm having trouble visualizing the voltage analysis that shows it directly but I'm pretty sure that's the mechanism, rather than each tube somehow seeing 1.414 of its winding. 🙂
When one of the tubes goes into cutoff, the other one sees only half the winding, the other half is open-circuited by the cutoff tube. Half the winding open circuit halves the rated winding ratio so the other tube sees 1/4 the rated impedance.
I *think* this is the right way to look at class A operation. If you look at each tube individually, you can see it. Basically each tube sees the full primary TWICE, once directly, and again through the coupling of the two halves of the primary together. So it sees the full primary (the a-a) value, in parallel with the full primary again. I'm having trouble visualizing the voltage analysis that shows it directly but I'm pretty sure that's the mechanism, rather than each tube somehow seeing 1.414 of its winding. 🙂
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