Need Help understanding a triode fed LTP.

[SGregory]

I am having a very difficult time understanding how a triode fed (for that matter a pentode fed) LTP circuit works with respect to input signal.

In this case the triode is current source loaded and the LTP has a CCS in the tail. The LTP is a 12at7. To me, it would seem that the LTP can take a direct signal from the source without the need for amplification as the Vg will be around +1V.

In a mullard 5-20 configuration, (as in the recent thread(s)) the LTP is driven by a pentode or triode with a pretty high gain. (it also has 30db feedback so I understand why some gain is required to overcome that.)

In my pea brain, I think the answer lies in the behaivor of the ltp with the cathode potential moving with current. I can't get my thoughts around how the cathode potential moves to accomodate such a large signal from the driver.

Is there an easy explanation that a material engineer can understand?

Thanks

[TheGimp]

The current source in the triode drive circuit acts as a near infinite load impedance allowing the full Mu of the triode to be utilized for voltage gain. So you have a lot of voltage gain available.

I think of a LTP (particularly with a current source in the cathode) as a sea-saw. Since the cathode current source provides a fixed current to both tubes (fulcrum), if one draws more, there is less left for the other. This results in the LTP not seeing it's cathode swing as much as one would expect.

With a grid drive requirement of about 30V, the EL34 does not put an extreme requirement on the LTP. If the LTP had a gain equal to mu, it would only need 30/19.7 = 1.523V swing at the grid to drive the output to clipping. The cathode of the LTP does swing since one grid is tied to ground, but it only swings through this 1.523V requirement.

If the gain of the LTP (12AU7) is 19.5, and the gain of the input triode (12AX7) is 100, then you have one boatload of gain to drive the EL34s. Open loop the input requirement is only 30/100/19.5 = 15.4mV. Obviously GNFB reduces the input sensitivity and it will take a considerably larger input signal.

In addition, the gains will not be the full X100 and X19.5 so this also will result in more input dive.

[jrenkin]

Wow, thanks Gimp! Very useful post. I am working on a similar idea in an EL34 PP rebuild.
I am planning to use 1/2 6sl7 without cathode bypass into a 6sn7 LPT with a CCS, but am concerned that it just will have too much gain, unless I want to swat it down with loads of feedback.
My source is a 2Vrms DAC (Red Wine)
Could I just skip the 6sl7 and input directly into the LPT?

I will post more regarding this at a later date when I have a schematic and can ask the forum helpfuls like yourself for reassurance and advice (I will have a lot more q's), but i am not quite ready yet.

[Arnulf]

Originally Posted by jrenkin

My source is a 2Vrms DAC (Red Wine)
Could I just skip the 6sl7 and input directly into the LPT?

LTP has ~1/2 the voltage amplification of straight common cathode stage which in case of tube with mu = 20 is 20x at best, and likely to be lower. This might just be enough with 2V RMS input (~25V P-t-P swing at the grid of EL34), depending on how you intend to operate your EL34s.

[jrenkin]

So maybe I should use a lower mu triode input tube...

[TheGimp]

That would work also. You might need to adjust your bias current for the different tube.

[jrenkin]

Well, since I abm rebuilding the entire front end of this amp (it was using a 6u8 in pentode input, triode concerta) /i can work out the bias as I wish.
I see options based on what tubes I have.
1. use the 6sn7 as the input mu 20 and the 6sl7 as the LPT (problem, I only have one 6sl7)
2.6688 (E180F) input mu 50
3. 6cg7 input mu 20
4. 6u8 input mu 40

Any thoughts on what might be worthwhile out of this gamish? I am kind of saving the e180f's for a 300b design....

[SGregory]

In the attached schematic (I am sure it is missing a few grid stoppers) lets assume the input triode has a mu of 50. Without feed back a +/- 1V [I] input will result in +/- 50V output at the anode [A]. (Theoretical ). How does the LTP eat this voltage without clipping? Or do I have to have enough NFB to drop the gain so that the triode output [A] is within the limits of the input grid on the LTP.

For this example lets assume the LTP is a 12at7 running at 460V and 3mA per side. Anode resistor is 50k. The gain in the LTP would be somewhere around 27dB assuming it is lightly loaded.

Please ignore that I wrote mu=100 on the drawing. Not close enough to the real world.

[m6tt]

Since you are DC coupling, I believe the grid of the non-inverting half of the LTP needs to be at the same DC potential as the input tube's anode, approximately, but with none of the AC signal. Connect 1-2 Meg from first LTP grid to second LTP grid. Alternately, a fixed voltage source could be used for better fidelity but with more adjustment requirements. There always needs to be a DC path to ground from the grids. Obviously the cathode voltage of the LTP will need to be higher than the anode voltage of the preceding stage, by exactly the intended Vgk bias for the LTP.

In terms of clipping, any voltage higher than the difference between cathode of LTP and anode of first stage will force grid current. No coupling cap will allow this to occur, however it will distort the up-going cycle of a sine wave by "rounding" it out. Ideally, the peak to peak voltage input of the LTP should be no more than 2*vgk, with LTP center biased.

If you have a PC, or linux machine go and get LTSpice, it's free and easy, and lets you test the circuit (approximately) before you build it. The learning curve isn't too terrible if you can already draw a schematic & google error messages.

[m6tt]

Another idea is to use a local feedback resistor around the first voltage gain stage to drop its output down to the desired level, as well as increasing bandwidth and decreasing distortion before clipping. It will respond "quicker" than a big gNFB loop.