Hi, I'm working on a low power PP amp, and am obsessing over the phase splitter. I need about 10V to drive the 6Y6 pentode outputs.
A couple of years ago, SY posted a thread called "fun little PP input" with a neat circuit I hadn't seen before. If anyone - SY?- has tweaked this further, I'd be interested in any late breaking developments / thoughts.
A couple of years ago, SY posted a thread called "fun little PP input" with a neat circuit I hadn't seen before. If anyone - SY?- has tweaked this further, I'd be interested in any late breaking developments / thoughts.
Time Marches On
Nothing new, it's the same circuit I've been using for decades. It works quite well and is the input stage of my "big" amp.
If I were to redo it, I'd probably combine it with an input transformer, use higher current tubes (e.g., ECC88), and replace the biasing resistors with LEDs.
Nothing new, it's the same circuit I've been using for decades. It works quite well and is the input stage of my "big" amp.
If I were to redo it, I'd probably combine it with an input transformer, use higher current tubes (e.g., ECC88), and replace the biasing resistors with LEDs.
Re: Time Marches On
SY, .. you did have a Jensen input transformer in the original circuit http://www.diyaudio.com/forums/showthread.php?s=&threadid=20889
Could I use a CD player on the primary of the transformer, with a volume pot, or would I need a line stage.
I'm confused about how much current a jensen or cinemag input transformer can handle. . I don't need much swing on the secondarys and am pondering using an input transformer to drive the grids of a push pull EL84 pair
Thanks for any input.
SY said:
SY, .. you did have a Jensen input transformer in the original circuit http://www.diyaudio.com/forums/showthread.php?s=&threadid=20889
Could I use a CD player on the primary of the transformer, with a volume pot, or would I need a line stage.
I'm confused about how much current a jensen or cinemag input transformer can handle. . I don't need much swing on the secondarys and am pondering using an input transformer to drive the grids of a push pull EL84 pair
Thanks for any input.
Hey-Hey!!!,
I just put together a PP 6V6 amp. B+ of 260. It is two stage, a 6BK7B LTP/diff amp riding a DN3545N3 CCS. The negative rail is RC decoupled and reduced from the -30V bias supply to -15V.
The amp is two-stage, and runs a Chicago B0-series OPTx rigged for CFB with its tertiary winding. The front end runs the amp very well, with adequate gain to run a buffer linestage with the CD machine. There is no other NFB, and the front end is probably delivering 50-60V at full power.
For most listening I can put any valve I have with that base/pinout, all the way down the mu ladder to the 6GU7.
cheers,
Douglas
I just put together a PP 6V6 amp. B+ of 260. It is two stage, a 6BK7B LTP/diff amp riding a DN3545N3 CCS. The negative rail is RC decoupled and reduced from the -30V bias supply to -15V.
The amp is two-stage, and runs a Chicago B0-series OPTx rigged for CFB with its tertiary winding. The front end runs the amp very well, with adequate gain to run a buffer linestage with the CD machine. There is no other NFB, and the front end is probably delivering 50-60V at full power.
For most listening I can put any valve I have with that base/pinout, all the way down the mu ladder to the 6GU7.
cheers,
Douglas
Fascinating. I'd love to see your schematic. How to bias the output tubes is a decision I haven't made yet.
Sy, when you bypassed the cathode resistors in your input stage / splitter: did you tie one end of the cap to cathode, and the other to signal ground? Or did the cap just go across the resistor? Forgive me for asking for so much hand holding.
Sy, when you bypassed the cathode resistors in your input stage / splitter: did you tie one end of the cap to cathode, and the other to signal ground? Or did the cap just go across the resistor? Forgive me for asking for so much hand holding.
Yet another question!
Prehaps you will tolerate one more question, then. I confess that I'm lost when it comes to biasing transistors. Does the tube act as an active load for the JFET, and set the current flow all by itself? Can I choose a value for the cathode resistor - to set the bias for the tube - just like I would if it was connected directly to ground, or is there more to it than that ? (If the source follower wasn't there, I'd expect about 1.85ma of current in your 400V B+ 100k plate resistor 6SL7 1K cathode resistor circuit). Does the voltage drop of the follower come into play? I can't quite figure out how to look at this circuit to come up with the current flow.
Prehaps you will tolerate one more question, then. I confess that I'm lost when it comes to biasing transistors. Does the tube act as an active load for the JFET, and set the current flow all by itself? Can I choose a value for the cathode resistor - to set the bias for the tube - just like I would if it was connected directly to ground, or is there more to it than that ? (If the source follower wasn't there, I'd expect about 1.85ma of current in your 400V B+ 100k plate resistor 6SL7 1K cathode resistor circuit). Does the voltage drop of the follower come into play? I can't quite figure out how to look at this circuit to come up with the current flow.
It's actually a little more complicated. The tubes and FETs are both depletion devices. That means that for the tube, the grid must be negative wrt the cathode. Equivalently, the cathode must be positive wrt the grid.
Now, the FET is a p-channel device, so depletion requires that the gate be positive wrt the source. Equivalently, the source needs to be negative wrt the gate.
If the gates are at DC ground, then the sources will be negative wrt ground for a given standing current. That negative voltage is applied to the tube grids. For the sake of simplicity and proper circuit operation, I assume that the FETs have been matched, so the negative voltages on each tube's grid will be equal.
OK, at the chosen standing current, the tube cathodes will have to be positive wrt the grid voltage. But that's equal to the source voltage, so you have to drop a constant voltage to bias things up. Thus the resistor.
With a set of curves and the usual load-line procedures, the resistor's value is straightforward to calculate for a chosen tube/FET/load/operating point. The fact that FETs vary enormously means that not only will you need to match them carefully, you also will see big spreads between pairs. The characteristics in the databook will almost certainly fit neither. So guessing, trimming, selecting, and measuring will get you there just about as fast and certainly more accurately.
Now, the FET is a p-channel device, so depletion requires that the gate be positive wrt the source. Equivalently, the source needs to be negative wrt the gate.
If the gates are at DC ground, then the sources will be negative wrt ground for a given standing current. That negative voltage is applied to the tube grids. For the sake of simplicity and proper circuit operation, I assume that the FETs have been matched, so the negative voltages on each tube's grid will be equal.
OK, at the chosen standing current, the tube cathodes will have to be positive wrt the grid voltage. But that's equal to the source voltage, so you have to drop a constant voltage to bias things up. Thus the resistor.
With a set of curves and the usual load-line procedures, the resistor's value is straightforward to calculate for a chosen tube/FET/load/operating point. The fact that FETs vary enormously means that not only will you need to match them carefully, you also will see big spreads between pairs. The characteristics in the databook will almost certainly fit neither. So guessing, trimming, selecting, and measuring will get you there just about as fast and certainly more accurately.
forced bias?
Splitter fans might look up the 6Y6 amp circuit on page 12 of the old Acrosound catalogue. The cross-coupled splitter is connected to a driver via a technique that's new to me: the splitter's plate is connected to the grid resistor of the driver tube via a 1.2 meg resistor AC shorted by a 0.05U cap. Looks like the plate resistor of the splitter, in series with this one, creates - in DC terms - a voltage divider with the grid resistor of the subsequent tube. In this circuit with a B+ of 200V the 220 meg plate resistor plus 1.2 meg coupling resistor would make a divide of about 25V with the 220K grid resistor, to provide bias to the driver. Anybody used this technique? Forgive me if it's old hat to you - if so, does it work well?
Splitter fans might look up the 6Y6 amp circuit on page 12 of the old Acrosound catalogue. The cross-coupled splitter is connected to a driver via a technique that's new to me: the splitter's plate is connected to the grid resistor of the driver tube via a 1.2 meg resistor AC shorted by a 0.05U cap. Looks like the plate resistor of the splitter, in series with this one, creates - in DC terms - a voltage divider with the grid resistor of the subsequent tube. In this circuit with a B+ of 200V the 220 meg plate resistor plus 1.2 meg coupling resistor would make a divide of about 25V with the 220K grid resistor, to provide bias to the driver. Anybody used this technique? Forgive me if it's old hat to you - if so, does it work well?
Yes, I use it - it's sometimes called a step network. In my case, the coupling is between a 6SL7 differential splitter and a 6SN7 differential driver. It falls between AC and DC coupling and is designed to have the advantages of both and the disadvantages of neither.
This arrangement avoids having too high a voltage on the grid of the driver, which DC coupling would cause. At the same time, it avoids the LF instability that could be caused, with AC coupling, by having too many coupling caps within the NFB loop.
An additional advantage is that it reduces the effect of any DC imbalance between the first pair of triodes, which would be passed on to the second pair of triodes, with pure DC coupling, and amplified, possibly causing a serious DC offset between the plates of the second pair.
I also use feedback resistors between 6SN7 and 6SL7 plates, to improve the balance even further. It works very well indeed.
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