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Old 27th January 2006, 06:37 PM   #21
EC8010 is offline EC8010  United Kingdom
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Oh, you don't need to worry about the replacement transistor being a 300V device, a 30V one would be fine in this application.

There are seven types of bipolar transistors: Small NPN, medium NPN, big NPN, small PNP, medium PNP, big PNP, and blown-up ones. With FETs, you get lots of different types; trouble is, they all have the same device number stamped on them (FET tolerances are legendary).
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Old 27th January 2006, 07:38 PM   #22
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In reality there is a conversion process where any of the first 6 types can be converted to the seventh type. It involves allowing that little bit of smoke that is built into the device to escape. This process works on mosfets also. I have performed this conversion many times. It seems to be irreversible.
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Old 27th January 2006, 08:34 PM   #23
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Quote:
Could this be more likely?
Yes..that one is perfect..oops...I did not see that you had a CT.
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Old 27th January 2006, 10:41 PM   #24
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Quote:
Originally posted by Giaime
I've simulated and resimulated the PSU with PSU Designer, and I seem that I can't get under 350VDC with my 300VCT PT

Use a bigger choke. I think you need at least 3H for critical inductance. 5H would probably be better.
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Old 28th January 2006, 09:50 AM   #25
Giaime is offline Giaime  Italy
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Thank you all!!!

I think I've settled on this http://www.diyaudio.com/forums/showt...043#post829043

I have to determine what wattage I need in that resistor... I guess that 2 100ohm 10W in series would do.
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Old 28th January 2006, 09:58 AM   #26
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Ian,

Brilliant design - I'd expect nothing less. Particularly enjoyed the EL84 bias networks.....

Cheers,

Hugh
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Old 29th January 2006, 09:25 AM   #27
AJT is offline AJT  Philippines
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Quote:
Originally posted by Bas Horneman
I have taken the liberty to put them on my site with the changes of the heater and pot. Makes it easier to post elsewhere as well.

http://basaudio.net/schematics/pp/ppel84_baby_huey.gif
Click the image to open in full size.

and

http://basaudio.net/schematics/pp/pp...y_huey_psu.gif
Click the image to open in full size.


hi,

can anybody give more explanation as to how this circuit topology works? i understand ccs, but the plate load resistors of the 1st input differential pair is in series with what seems to be feedback resistors from the plate of the EL84's, then is bridged by a single resistor.

thanks
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Old 29th January 2006, 11:13 AM   #28
ilimzn is offline ilimzn  Croatia
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It is actually very simple, and elegant - and once you have figured it out, like most elegant designs, obvious

Consider this:
What would the voltage be on the joining point of the 47k and 22k resistors between the plates of the output and input LTP, IF the 16k resistor was a short curcuit? Answer: it would be B+ reduced by the voltage drop of the OPT winding resistance, so close to B+. Why? Simple: because the amp is symetrical PP, the OPT ends always swing symetrically around +B, and making the 16k resistor a short, would just create a divider establishing a mean of the voltages on the output plates - which is exactly B+ lowered by the voltage drop in the OPT winding resistance. This establishes a virtual B+ for the LTP, and the impedance of the virtual B+ is 47k / 2.
Now, since the LTP has a CCS in the tail, the sum of both plate currents in it is always equal to the current through the CCS, i.e. a constant - this means there is a constant drop in the 47k/2 impedance of the virtual B+ - in essence, if the 16k resistor was a short, it becomes a 'common mode' node.

Finally, what happens now if the 16k resistor is NOT a shor any more? Well, immagine it was made out of two equal resistors - 2x8k say, connected in series. Their center node remains the common mode opint and is always at a voltage equal to the virtual B+ reduced by the drop in the 47k resistors. The big difference is now that the node between the 220k and 47k resistors is not the common mode node any more, but instead there is the AC component from the output tube plates, divided down by the 47k and half of the 16k resistor - i.e. there is feedback. The beauty of this is that changeing the feedback can easily be made by the choice of the 16k resistor, without signifficantly changeing the DC operating point of the LTP, because 47k in parallel with 1/2 of 16k is a rather small value compared to the 220k triode plate resistor, so DC currents branch just a little different and the virtual B+ remains almost the same.

One important point already made by the author: the input triodes are high Rp - and this is because the actual feedback is again scaled down by the divider formed from the 220k and the triode Rp. With a low Rp, feedback would always be low - and could only be increased by reducing the feedback network resistors, which means a higher current through the LTP and more losses for the output as this is where the current for the feedback network comes from. Alternatively, if a pentode LTP was used, then the high Rp of the pentodes would make the 220k resistor virtually insignifficant in comparison, and feedback would nto be scaled down.

Finally, SY asked about running the amp in pentode mode - it appears to me that no changes to the rest of the circuit would be needed, as long as G2 DC component remained the same, i.e. ~~ B+. Also, since ultralinear is a form of feedback, open loop gain would increase, but cinsidering there are two more feedback loops in addition to the ultralinear, I don't see the total gain changing dramatically, nor the gain without global feedback. It would be an interesting exercise to try plate to plate feedback signifficant enough that global NFB would not be needed at all
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Old 29th January 2006, 12:01 PM   #29
SY is offline SY  United States
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Actually, my question was about using pentodes for the input tubes, since the desiradata there are high plate resistance and high mu. Seemingly, they would work better than triodes for this application.
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Old 29th January 2006, 09:31 PM   #30
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Well I'm back after 5 days of no internet access and am now about to dissappear for another 3 days.
Thanks to EC8010, BAS, SY and ILIMNZ for fielding the equiries and to BAS for fixing my drawing error(s).

There was aquestion about setting the EL84 bias. The Bias current is set by the 16R resistor in the bias blocks. From memory this gave 38mA idle current. There is about 10 to 12 volts DC across the bias network. The MJE340 was used, as EC8010 suggested "because I had a bag full of them in the parts bin". a lower voltage device can certainly be substituted - use something with good hfe for best results.

Cheers,
Ian
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