Need some advice on a LTP input stage.

[Shoog]

Hi all,
I have attached a plan for two different LTP phase splitter input stages. The idea is to hang the whole of the input stage off of a -300V negative rail. I am having some real difficulty conceptualising all of the issues involved so I have a few questions. the LTP is expected to draw about 10mA (20mA for two channels) so it should be possible to create a very stable neg supply with a simple CLC filter. The intention is to go with the pentode version, the triode version is just there to illustrate the basic concept.

1- In the first triode based schematic is the bottom resistor functioning to extend the tail of the LTP, or is it just burning up voltage.
2- In the second schematic will the voltage across R10 stay stable and so not effect the screen voltage.
3- If I substitute a CCS for R10 in the second scheme, how will it find its own anode voltage, and again will that voltage stay stable.

Am I barking up the wrong tree. Are there any issues I have failed to spot.

Shoog

[Jeb-D.]

Your using an input transformer to split the phase. There is no need for the negative supply, or a long tail. Just have the two tubes still share a cathode bias resistor. Ground makes a better voltage reference than a negative supply. So it would be better to bump up the B+ and drop the - supply, if your circumstance permits. Personally, I'd replace V4 with a capacitor, it's the lesser of two evils. Also, you may get better results hooking it to the -250 rather than -300 if keeping the bipolar power supply.

[Shoog]

Unfortunately the need for such a large negative rail is because the anode loads on the triodes/pentodes are taken off the anodes of the output stage (ala plate to plate feedback). Pushing the +B up means achieving the same result, but with much more heat generated in the cathodes of the output stage. This scheme is less wasteful - if it works.

Also, you may get better results hooking it to the -250 rather than -300 if keeping the bipolar power supply.

Unfortunately this will result in the tail been shortened to accommodate the extra current through V4, also there is the possability that the voltage over R10 will become less stable - though this shouldn't matter that much as the reference point should become more stable. I had hoped to use one VR150 for both channels.

Shoog

[Jeb-D.]

What is it going to be driving? Pentodes can put out a lot of swing with just that 350V supply.

If you must have 450V total you may want to consider using a separate low current, high voltage transformer for that stage. You can usually find some small open bracket ones that are small enough to fit in the chassis. If what you are doing is getting your + and - off the same transformer. Make sure your using a CT transformer and using the two diode Full wave config for your + and another two diode Full wave for the -. That's the only way I know to pull off that asymmetrical load with getting minimal mechanical hum.

You don't need the tail anyways, And if you use a cap insted of V4 you won't have the extra DC current to sink. The - rail can be filtered down to -250 instead of -300.

[Wavebourn]

Originally posted by Shoog
Unfortunately the need for such a large negative rail is because the anode loads on the triodes/pentodes are taken off the anodes of the output stage (ala plate to plate feedback). Pushing the +B up means achieving the same result, but with much more heat generated in the cathodes of the output stage. This scheme is less wasteful - if it works.

Too high and unstable output resistance for a parallel feedback. Get more swing from the tranny and use cathode followers instead.

[Shoog]

If you must have 450V total you may want to consider using a separate low current, high voltage transformer for that stage. You can usually find some small open bracket ones that are small enough to fit in the chassis. If what you are doing is getting your + and - off the same transformer. Make sure your using a CT transformer and using the two diode Full wave config for your + and another two diode Full wave for the -. That's the only way I know to pull off that asymmetrical load with getting minimal mechanical hum.

Thats exactly how I intended to derive that neg rail.

The whole question of swing is massively complicated by the plate to plate feedback arrangement. Since the DC for the pentode's needs to come through the load/feedback resistor you need a lot of voltage in order to be able to keep the feedback resistor reasonably large. Swing is really not the issue, current to the drivers is. In this case the 400V across the resistor with 5mA of current gives a feedback/load resistor of 80K, which is on the lower limit of what is acceptable without loading down the driver.

Shoog

[Shoog]

Too high and unstable output resistance for a parallel feedback. Get more swing from the tranny and use cathode followers instead.

I have built a SE configuration of this concept (the RH807) and though I didn't much like the result, output resistance didn't strike me as been the issue.
Essentially this is a proven concept (Gary Pimms Tabor amp) I'm trying to simplify his excellent design. I may not succeed, but I don't think it will be for the reason you suggest.

Shoog

[rdf]

Is the intention direct coupling to the output stage? I've often wondered about using equivalant topologies that way.

[Shoog]

Is the intention direct coupling to the output stage? I've often wondered about using equivalant topologies that way.

The Tabor amp is direct coupled. I had given up on the idea of direct coupling for the sake of simplicity, but the voltages are so near that I think I may just try to get it to work that way.

Shoog

[Wavebourn]

Originally posted by Shoog


I have built a SE configuration of this concept (the RH807) and though I didn't much like the result, output resistance didn't strike me as been the issue.
Essentially this is a proven concept (Gary Pimms Tabor amp) I'm trying to simplify his excellent design. I may not succeed, but I don't think it will be for the reason you suggest.

Shoog;

output resistance of the driver does not need to satisfy you; it needs to satisfy a low input resistance of the output stage with parallel feedback by voltage. If your driver had low output resistance amplification factor of the output stage will depend on plain linear resistors; but in your case it is determined by one feedback resistor and output resistance of your driver that is not linear.

MS Windows is a proven concept (Bill Gates et all), you may try to simplify his excellent design, but what really helped them, was design of DEC team that developed NT concept to replace what Microsoft did before...