Question about direct coupling a anode follower into a cathode follower.

[SY]

For a preamp output with the Z you're targeting, understanding that you don't need scads of current or voltage, the transconductance is the most important spec. A 5687 will barely get you the transconductance you need, but you'll have to run it pretty hard (12 ma or more). I don't have the 6BL7 data sheet in front of me, but I recall the transconductance as also being pretty marginal. These are probably not optimum tubes for this application.

Save the 6BL7 for a tiny triode amp output stage- or as a cathode follower driver for an AB2 amp, where you can make use of its major virtue, the ability to drive in lots of current.

 

[G]

For a preamp output with the Z you're targeting, understanding that you don't need scads of current or voltage, the transconductance is the most important spec. A 5687 will barely get you the transconductance you need, but you'll have to run it pretty hard (12 ma or more). I don't have the 6BL7 data sheet in front of me, but I recall the transconductance as also being pretty marginal. These are probably not optimum tubes for this application.

Save the 6BL7 for a tiny triode amp output stage- or as a cathode follower driver for an AB2 amp, where you can make use of its major virtue, the ability to drive in lots of current.

Well it appears that the ECC88 is the one to go with. I've searched my little preamp tube database and I can't find a better candidate. High transconductance and low voltage requirements. Although a 5842 would work also but they are much harder to come by.

 

[SY]

Among commonly-available tubes, it's a great choice. If you want more exotic, there are frame-grid RF triodes with high transconductance and mu (like 6HA5-types), and even Nuvistors. But the 6DJ8/ECC88/7308 family will get you there efficiently and with no fuss.

 

[G]

Among commonly-available tubes, it's a great choice. If you want more exotic, there are frame-grid RF triodes with high transconductance and mu (like 6HA5-types), and even Nuvistors. But the 6DJ8/ECC88/7308 family will get you there efficiently and with no fuss.

Well I guess that's settled. I would like to keep the plate voltages below 200v as I have some dual section Elna Cerafines that I would like to use. They are rated at 350 volts. Would you like me to start the calculations and go from there or do you have something specific in mind? I'm by no means a direct coupling expert and there are still a few things that I'm not sure about as far as the CF stage and biasing it go but I can certainly design the gain stage. I'm not ebtirely sure that I want to use tube rectification either. I have a couple of IXYS integrated hyperfred bridges that I would like to try. Even though they are grossly overrated for the task.

 

[fdegrove]

ONLY 50V HIGH...

Hi,

Funny you'd decide on the 6DJ8/ECC88 for that:

DC COUPLED LINE STAGE USING GUESS WHAT...

I'll leave the PS to you but I think you know what I used ...

Cheers,

 

[SY]

Would you like me to start the calculations and go from there or do you have something specific in mind?

I had a few options in mind, but by all means, throw out some proposals. They'll either be good as-is or at least be useful as starting points or as pedagogical exercises.

 

[SY]

Frank, something along those lines was very much one of the options I had in mind. The gain would need to be knocked down a bit, but hey, more degeneration never hurt anyone.

 

[G]

Frank, something along those lines was very much one of the options I had in mind. The gain would need to be knocked down a bit, but hey, more degeneration never hurt anyone.

The first one that Frank posted is about what I had in mind so let's start there. No need for the input cap though.

 

[SY]

Well, I'd probably keep the input cap just because I don't trust the offset of my sources. Heck, my soundcard has a nice 60 mV of DC riding on one channel.

But... I'd have the input cap feeding the top of a potentiometer (you DO want to control volume, don't you?), something in the 100K range, then take the wiper tap to the tube grid through a 1K resistor (for RF suppression). If it were commercial product, I'd also tie a safety resistor between the first stage grid and ground (maybe a meg or so), just in case the wiper of the pot goes out. You'll also want to tie a large (>1M) resistor from the input side of the cap to ground just to prevent bangs and pops when things are (accidently, of course!) hot-plugged.

Finally, the output should be shorted to ground during warmup and particularly on shut down. You can use an NC relay or (horrors!) a bipolar transistor coupled to a timer circuit to do this task. The bipolar has the advantage of speed, and when it's cut off, the resistance from collector to emitter is really, really high and totally swamped by the shunt impedances across it.

Again, all this is optional and probably not going to make an audible difference, but it cleans things up a bit. You can indeed go DC in, there's a nice simplicity to that, but you'll have to cross your fingers that nothing feeding it is going to have an output offset ever.

The advantage of the input cap where Frank has drawn it is that you can play a few degeneration tricks with that first stage to knock down the gain without worrying about what the DC level on the grid is.

 

[G]

Well, I'd probably keep the input cap just because I don't trust the offset of my sources. Heck, my soundcard has a nice 60 mV of DC riding on one channel.

No offense but I would prefer to live dangerously. One cap in the circuit is enough for my taste.

But... I'd have the input cap feeding the top of a potentiometer (you DO want to control volume, don't you?), something in the 100K range, then take the wiper tap to the tube grid through a 1K resistor (for RF suppression). If it were commercial product, I'd also tie a safety resistor between the first stage grid and ground (maybe a meg or so), just in case the wiper of the pot goes out. You'll also want to tie a large (>1M) resistor from the input side of the cap to ground just to prevent bangs and pops when things are (accidently, of course!) hot-plugged.

I plan on using a 100K 24 step ladder attenuator at the input. I will definately use a suppression resistor to the grid.

Finally, the output should be shorted to ground during warmup and particularly on shut down. You can use an NC relay or (horrors!) a bipolar transistor coupled to a timer circuit to do this task. The bipolar has the advantage of speed, and when it's cut off, the resistance from collector to emitter is really, really high and totally swamped by the shunt impedances across it.

I'm not following why this is needed. I understand doing this on SS outputs but I've never seen it done on a tube preamp. At least not on a DIY one. I guess if I use a SS rectifier this is a necessary precaution?

Again, all this is optional and probably not going to make an audible difference, but it cleans things up a bit. You can indeed go DC in, there's a nice simplicity to that, but you'll have to cross your fingers that nothing feeding it is going to have an output offset ever.

All of my sources (all two of them) use caps on the outputs so I think I will chance it.

The advantage of the input cap where Frank has drawn it is that you can play a few degeneration tricks with that first stage to knock down the gain without worrying about what the DC level on the grid is.

I don't have a problem with local feedback. I just don't know how to implement it properly yet. I had planned on not using a bypass cap on the gain stage triode's cathode to hold the gain down but I'm not sure how effective that will be nor how it will sound. I just want to have a good sounding tube linestage that is capable of driving a 20K - 50K load with grace. I was considering buying a Foreplay and customizing it a little during the build but after researching it I decided that I don't like the operating points they use so now I'm back to square one and more frustrated than ever.

 

[SY]

Well, in Frank's circuit, the cathode resistor is indeed unbypassed. That gives some local degeneration. And you've got plenty more, in fact 100%, in the cathode follower. You can alter the degeneration in the first stage if you use an input cap, but if you don't, you'll just have to live with more gain. I haven't run the calculations, but this circuit as it is drawn looks like the gain will be about x10 (20 dB).

As far as the power-on/power-off mute, it's a really good idea even if you're running into a tube power amp. For one thing, it is just a matter of time until the turn-on or turn-off sequence is done wrong. Especially after a nice evening of wine. There are no doubt quite a few examples of commercial tube preamps that do this, but one I know offhand (because I've got one) is the classic Berning TF10. And Berning has the schematics posted on his web site, so you can swipe it easily.

 

[G]

Well, in Frank's circuit, the cathode resistor is indeed unbypassed. That gives some local degeneration. And you've got plenty more, in fact 100%, in the cathode follower. You can alter the degeneration in the first stage if you use an input cap, but if you don't, you'll just have to live with more gain. I haven't run the calculations, but this circuit as it is drawn looks like the gain will be about x10 (20 dB).

As far as the power-on/power-off mute, it's a really good idea even if you're running into a tube power amp. For one thing, it is just a matter of time until the turn-on or turn-off sequence is done wrong. Especially after a nice evening of wine. There are no doubt quite a few examples of commercial tube preamps that do this, but one I know offhand (because I've got one) is the classic Berning TF10. And Berning has the schematics posted on his web site, so you can swipe it easily.

I will have to explore my options on the muting. I may just use a time delay off relay as the signal will not be passing through the contacts. Or maybe a vacuum tube delay relay.

I noticed the symmetry in Franks design. In that I mean he has dropped the same voltage across the anode resistor on the gain stage as he has the cathode resistor on the output stage. Doesn't that mean that the grid of the second triode is at the same potential (in relation to ground) as the cathode? Does it not matter because the tubes are direct coupled? What controls the current through the second triode? Since it seems to be at 0 volts potential.

 

[SY]

The cathode of the CF will be slightly more positive than the grid, just enough so to set the current to the (approximate) ratio between the plate voltage of the preceding stage and the cathode resistor of the CF. That's feedback in action for you!

Look at the feedback this way: if the cathode voltage with respect to the grid of the CF is more positive than required, the tube will conduct less. That reduces the drop across the cathode resistor, thus bringing the cathode less positive wrt to the grid. Feedback.

 

[G]

The cathode of the CF will be slightly more positive than the grid, just enough so to set the current to the (approximate) ratio between the plate voltage of the preceding stage and the cathode resistor of the CF. That's feedback in action for you!

Look at the feedback this way: if the cathode voltage with respect to the grid of the CF is more positive than required, the tube will conduct less. That reduces the drop across the cathode resistor, thus bringing the cathode less positive wrt to the grid. Feedback.

Let me run this by you. Using Kirchoff's law there are three elements in the first stage that have a voltage drop across them. The anode resistor, the triode and the cathode resistor. The cathode resistor of the first stage would seem to be the key to me. Let's say there are two volts dropped across it. That would in turn mean that there are two less volts at the grid of the second triode than is being dropped across the cathode resistor of the second triode. That would mean that the cathode resistor of the first triode sets the current through the second tube as well so long as both tubes are fed from the same value of B+. Is that correct? I'm not vey good at AC analysis but I will never understand how this circuit works if I can't get a handle on the DC analysis.

 

[fdegrove]

Hi,

Is that correct?

It has to...
Admittedly I look at it in a different way than you do but the net result is still the same...

Looking at the circuit, the cathode resistor of the first triode is causing a voltage drop X across itself, this will inevitably reappear in the next stage as the cathode resistor of that triode has the exact same value as the plate resistor of the first triode causing a voltage difference equal to X.

So if we measure, say a 100V, on the plate of the first triode and direct couple that plate to the grid of the next triode we'll measure 100V + X on the cathode of the second triode, ergo the grid of the seconde triode will be negative by the amount equal to the difference between the two, namely X.

If we'd want a different bias for the second triode we'd have to change the value of it's cathode resistor but by doing so we'd also lose much of the inherent AC stability of the original circuit.
Why?
To borrow John Broskie's analogy of the see-saw, the use of equal plate and cathode resistors for first and second stage respectively forces equilibrium on our see-saw: whichever side is pulling or pushing, the other is going to try to regain equilibrium using the same amount of force.
Changing either side will break that equilibrium and our see-saw will struggle.

It's a little oversimplified but I hope it helps nonetheless.

Cheers,

 

[G]

Hi,



It has to...
Admittedly I look at it in a different way than you do but the net result is still the same...

Looking at the circuit, the cathode resistor of the first triode is causing a voltage drop X across itself, this will inevitably reappear in the next stage as the cathode resistor of that triode has the exact same value as the plate resistor of the first triode causing a voltage difference equal to X.

So if we measure, say a 100V, on the plate of the first triode and direct couple that plate to the grid of the next triode we'll measure 100V + X on the cathode of the second triode, ergo the grid of the seconde triode will be negative by the amount equal to the difference between the two, namely X.

If we'd want a different bias for the second triode we'd have to change the value of it's cathode resistor but by doing so we'd also lose much of the inherent AC stability of the original circuit.
Why?
To borrow John Broskie's analogy of the see-saw, the use of equal plate and cathode resistors for first and second stage respectively forces equilibrium on our see-saw: whichever side is pulling or pushing, the other is going to try to regain equilibrium using the same amount of force.
Changing either side will break that equilibrium and our see-saw will struggle.

It's a little oversimplified but I hope it helps nonetheless.

Cheers,

Hi Frank. I guess the design below won't work very well then. I was trying to keep the voltage down so I can use my stash of Elna Cerafine caps. Nice big cans with low ESR. I think this would work alright into a 20K or so load. 6SN7s are supposed to sound nice. I could raise the B+ to 250 I think and run both stages at 5mA instead. My caps are rated for 350v. What is your opinion. Granted it is a simple circuit but I think it has potential.

 

[SY]

Let me run this by you. Using Kirchoff's law there are three elements in the first stage that have a voltage drop across them. The anode resistor, the triode and the cathode resistor. The cathode resistor of the first stage would seem to be the key to me.

Keep it simple. The CF draws essentially no current from the first stage. So for the purpose of DC analysis of the first stage, you can ignore the CF completely. Using load lines or CAD or whatever, you can determine what the plate voltage of that stage is going to be. It will be determined by the cathode resistor, the plate resistor, and the B+ rail. They're all part of the equation. But the point to take away is that you can calculate what you need about this stage in isolation. Let's call the voltage at the plate of this stage Vp. Doesn't matter what it is, we can go calculate it.

Now, let's move to the CF. Clearly, the voltage at the grid will be Vp because it's direct coupled from the previous stage. So with that as a given, we can look at this stage in isolation. The voltage at the cathode will be Vp plus a couple of volts, which are needed for biasing. But here's the trick: the required bias voltage in this circuit is very low compared to the drop across the CF's cathode resistor! So we can, to a very good approximation, say that the cathode voltage is also Vp. Sure, sure, it's actually going to be Vp plus something like two volts. But Vp is going to be something like 60-70 volts, so saying that 62 = 60 is not too bad an approximation.

Now... with this in hand, we can either calculate the value of the CF's cathode resistor from the desired current in that stage OR we can figure out the current in that stage from a given cathode resistor. Let's say, for argument, that Vp is 60 V. So the voltage at the cathode of the CF will be slightly higher than 60, but not significantly so. Then for the 20K value of CF cathode resistor, the current will be about 60/20 or 3 ma. You don't have to worry about whether that CF is going to need to sit 2 volts above Vp or 3 volts above Vp or whatever- the current will take care of itself. The variation in required biasing voltage will cause the current to be not exactly 3 ma, but who cares if it's 2.94 ma instead? Not me.

Is that a little clearer?

 

[fdegrove]

Hi,

Granted it is a simple circuit but I think it has potential.

Sure, but you know me, right?

BTW, if the figures on the diagram are correct (I didn't cross check the math) then the CF is biased at 0V...
Not good but it's most likely the drawing that's not quite right.

As for the 6SN7s, yes most of them sound fine but I would only use them if I had plenty of them...
There are still sooo many sleepers out there that are just as good and dieing to be revived.

Cheers,

 

[SY]

BTW, if the figures on the diagram are correct (I didn't cross check the math) then the CF is biased at 0V...

I didn't do the math either, but didn't need to. On inspection, one can see that the voltage values are not correct. See my above post on biasing. The CF will be running at approximately 5 ma current, so the cathode will have to sit a few volts (4 or 5 maybe) above the grid.

 

[G]

As for the 6SN7s, yes most of them sound fine but I would only use them if I had plenty of them...
There are still sooo many sleepers out there that are just as good and dieing to be revived.

Cheers,

You mean like a 6C8G?

I can change the cathode resistor on the CF to 22K. But that is where I'm getting lost. Does the resistor set the current through the tube in the CF? If so then a 22K resistor would make the cathode 10v more positive than the grid.