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pauldune 14th April 2012 10:54 AM

I need help with ccs based LTP
 
3 Attachment(s)
First schematic attached is a standard LTP; it is a great differential gain stage, and if you use only one output, also a great differential-to-SE converter.

Second schematic is a LTP with CCS's in the tail and the anode circuits. Normally the problem with this schematic is that it it nearly impossible to set the currents right. The current set for the top ccs should be exact 50% of that for the cathode ccs. I tried to solve it with setting the top ccs'es for a little bit more then 50% and a bleeder resistor to bleed the extra current to ground.

But my real question is: does it stil function as a differential amplifier?
I fear not; the current in the tail looks to me as always 50% triode A and 50% triode B. So I fear that it behaves more or less as two separate stages.
Am I right?

Third schematic is a possible solution; but has it advantages over schematic 1, or am I making it unnecessary difficult for myself?

What I would try to accomplish is a better PSRR; I need only a few mV swing.

godfrey 14th April 2012 11:11 AM

Quote:

Originally Posted by pauldune (Post 2984794)
Second schematic .... does it stil function as a differential amplifier?

Yes, with high value resistors from anodes to ground it should work well and give good PSSR. The third circuit should work OK too, but I think I prefer the second one. YMMV
:2c:

jan.didden 14th April 2012 11:22 AM

Quote:

Originally Posted by pauldune (Post 2984794)
First schematic attached is a standard LTP; it is a great differential gain stage, and if you use only one output, also a great differential-to-SE converter.

Second schematic is a LTP with CCS's in the tail and the anode circuits. Normally the problem with this schematic is that it it nearly impossible to set the currents right. The current set for the top ccs should be exact 50% of that for the cathode ccs. I tried to solve it with setting the top ccs'es for a little bit more then 50% and a bleeder resistor to bleed the extra current to ground.

But my real question is: does it stil function as a differential amplifier?
I fear not; the current in the tail looks to me as always 50% triode A and 50% triode B. So I fear that it behaves more or less as two separate stages.
Am I right?

Third schematic is a possible solution; but has it advantages over schematic 1, or am I making it unnecessary difficult for myself?

What I would try to accomplish is a better PSRR; I need only a few mV swing.

It might help to realize that there is no conceptual difference between #1 and #2 - only a difference in quantities. If in #1 you increase the supply voltage and the anode load inpedance you get closer and closer to #2. #2 allows you to have a very high anode load impedance but with a reasonable anode voltage. Not fundamentally different from #1.

jan didden

SY 14th April 2012 11:25 AM

Use a very stiff cascode CCS in the cathode, then less stiff (e.g., one transistor) CCS on the plates.

bigwill 14th April 2012 01:29 PM

As has been suggested, you want a high but matching impedance in the anodes of #2, I think it's best to use a stiff CCS so this is easily definable with 1% resistors in parallel (220k sounds like a good value, I think much higher than that then you're at the whim of various leakage capacitance to set things off balance)

bigwill 14th April 2012 01:33 PM

In fact, I think gyrators would be better suited for the upper loads (in parallel with high value resistors), I think this would make the DC operating point a bit more predictable. You'd even be able to set the plate voltages independently, or even adjust for the same current (via plate voltage) through each half.

pauldune 14th April 2012 02:13 PM

I have a lot of 10M45S in my partsbin; I would like to use them for this.
Since most of you seem to prefer v2, i'm going to use that. Could I run into problems because they are all equally "stiff"?

@bigwill: I think I can use the bleeder-resistors to set the plate voltage. If I send say 2mA through them; I can choose the value of them so that I get the wanted plate voltage.

Thanks for all your replies so far!



Paul

pauldune 14th April 2012 02:17 PM

Haha! I see now I made a mistake in all 3 schematics; the current setting resistor in the cathode ccs is connected wrong...


FYI:
Cathode ccs current=40mA
Anode Currents 22mA.
Tube's D3a triode connected

Yvesm 14th April 2012 03:14 PM

Quote:

Originally Posted by pauldune (Post 2984794)
First schematic attached is a standard LTP; it is a great differential gain stage, and if you use only one output, also a great differential-to-SE converter.

Second schematic is a LTP with CCS's in the tail and the anode circuits. Normally the problem with this schematic is that it it nearly impossible to set the currents right. The current set for the top ccs should be exact 50% of that for the cathode ccs. I tried to solve it with setting the top ccs'es for a little bit more then 50% and a bleeder resistor to bleed the extra current to ground.

But my real question is: does it stil function as a differential amplifier?
I fear not; the current in the tail looks to me as always 50% triode A and 50% triode B. So I fear that it behaves more or less as two separate stages.
Am I right?

No !
Even with different currents in each tube, the DIFFERENCE will always be the same.
The sum of the currents being imposed by the CCS in the cathode, individual currents in each tube will always be equal but of different polarity.
So, the stage remains purely differential as long as the load resistors remains equal regardless of the tube's caracteristics, even if the current repartition if far from 50/50 % !

But if you really bother about that, why not put a CCS allowing exactly 50% in only one anode ? The other tube will then be forced to draw the 50% remaining ;) . . . more or less the variations imposed by the signal applied between the grids . . . I guess :2c:

Yves.
Mmmmh, Perhaps should I patent that ? :D

pauldune 14th April 2012 04:10 PM

@yves: in case of the single ccs in anode: would both differential input signals be equally represented in single ended output?
After all I am using it as differential to se converter....

Paul


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