Direct Coupling with Zener Diodes

[J M Fahey]

Being primarily a bass amp, sticking a large (or small) value cap in the coupling stages of overdriven valves would seem to be a big no-no with the low B going down to something like 31Hz.

Bass amps (specially tubed ones) have happily lived for ages with coupling caps designed for the 42 Hz lowest frequency generated by 4 string ones.
If you need downward extended response to 31Hz, simply doubling their value will more than compensate for that.
Since I infer that the power amp will be tubed also, I would rather worry about its frequency response, how to direct couple the phase inverter to the power tubes and even more about the output transformer.
Not to mention the speakers.
On the contrary, taking he SS path would easily allow a DC coupled preamp and a DC coupled power amp.
The speaker problem still remains.

[redelephant]

I haven't had time to read all posts, so apologize if I'm doubling up here...(I'm at work and not supposed to spend time doing this )

I was perhaps a little negative, only thinking 'cons' about your idea. All unusual ideas are good so hope you still try it out. Your challenge is keeping the current thru the zener constant, and if you can add a negative rail this is easily acchieved with a constant current source. The zener drift will go positive, menaing zener voltage increases with temp, so that is not too damaging to the tube's quiescent state.

You mention school break, so I gather your studying electronics, and nothing better than to study zeners, CCS, tubes, and all the pros and cons when you have the time.

Heres a simplified circuit showing how I'd start off. The CCS keeps the zener steady, and the high impedance of the CCS means you lose no signal and the signal can swing more than enough to drive the following stage.

To figure out the quiescent conditions for the input tube and proper load resistor, look at where the current is common for zener and tube, and where they separate. The plate voltage depends on the plate current and bias voltage you get on Rk. Use a datasheet to figure the rest.

[Steve Conner]

The problem with the DC coupled approach is that the DC gain is the same as the AC gain. The circuit becomes hopelessly sensitive to the bias point of the first stage. In solid-state land this is dealt with by putting a global NFB loop around the whole amp to haul it into alignment. When this is only active at low frequencies it is called a DC servo.

Also, if you want to overdrive the preamp, the shifting bias voltages on the coupling capacitors are a big part of the tube sound.

I've also seen the neon lamps used, I suspect they might work better than zeners, unless they turn out to be noisy. (you can't put capacitors across them to filter the noise, they will just flash)

Finally, look up the Loftin-White circuit, the original DC coupled tube amplifier.

[redelephant]

Yeah. I completely agree, I already said my view on zeners-dont like em at all, but as a learning prosject I figure why not and good luck...

[kg]

Because it's much closer to the cathode than any other part of the tube.

agreed!

When the grid goes positive it pulls in electron current(in conventional flow it means "the grid conducts"),

agreed!

it in fact becomes a diode to ground from zero volts on up.

a diode with a non-zero impedance.

But the grid isn't designed for electron flow, it's just a voltage barrier, in fact it's a very delicate structure.

not as delicate as you'd think... besides what kills internal elements (other than an oxide cathode itself, which is easily damaged by overcurrent) is power dissipation. keep the Vgk low and quite a bit of current may flow with no damage whatsoever.

I don't know if I undertsood your question correctly, or if I'm just being reduntant/too basic here, I hope this is what you asked....

that is what i asked.. however i disagree with your assertion that Vgk can never be positive, even in small signal tubes, without severely damaging the tube. what generally happens is that the ubiquitous RC coupling, along with the typically very high grid circuit DCR, clamps positive g1 excursions to Vgk=0, but that is a function of the DRIVER and CIRCUIT, not the local tube. direct coupling removes both of these impediments to pulling grid current.

in fact, i would argue that the very common direct coupled "marshall style" cathode follower preamp tone stack driver pulls g1 current on every large signal/clipped positive excursion. consider what happens with the first plate loaded stage cuts off entirely... the plate immediately goes straight to b+. so the directly coupled CF right after it now has its plate at b+, its grid at b+, and its cathode is being yanked as hard as possible in the positive direction. it's NOT going to reach b+! so there you go--small signal/preamp positive grid current. how much depends on how large the plate loaded stage's load DCR is.

[exclamationmark]

Steve - I have those bias concerns too, especially with the drifting plate voltages and zeners. Though by using a 12au7, (I hope) a bias drift of a couple volts shouldn't be totally catastrophic for the following stage compared to a 12ax7 or the like which would turn the tube into a half wave rectifier. Part of the fun is to see what it would sound like without capacitors too

Redelephant - That circuit is pretty much what I was going to attempt to do. Was gonna choose a bias point for the stage before the zener and measure the plate voltage and choose a zener that essentially nullifies the DC offset (any remaining offset I can account for with a lower/large bias resistor in the following stage). Also what's nice is that I can use two terminal off the shelf CCS's designed for LED's that cost like 50c instead of cooking up my own.

My experience has been that large coupling capacitors don't particularly like overdriven valves and low frequencies. I don't intend to directly couple ALL the stages, only those that are heavily overdriven, with the rest having generous coupling caps.

[pdf64]

I'm totally comfortable with silicon in the signal path (in fact LED cathode bias is my favourite mod to my existing amps

Is an LED used for cathode bias in the signal path? Pete.

[exclamationmark]

Is an LED used for cathode bias in the signal path? Pete.

Any noise generated by the LED (or non linearities) will basically be multiplied by the mu of the valve. Pretend the LED is replaced by a signal generator with the grid at 0 volts, and it's easy to see by drawing load lines that the changing bias (the cathode to grid voltage) caused by the signal generator causes the plate voltage to fluctuate at whatever the signal generator generating times the mu of the valve (provided it doesn't clip, etc). So in a roundabout sort of way, I would consider the LED to be in the signal path.

At least that's my rationalisation of it. Anyone should feel free to prove me wrong though.

[jmaf]

kg, if what you described ever happened that'd be an entirely different scenario, I understood your example for theoretical purposes, but the system works precisely to never let that scenario happen and consequently let the grid go positive.

It's a very interesting circuit and entirely worth this discussion. I remember the day I understood the Fender dc-coupled pair, it was one of those eureka moments because I always thought the next grid went positive during swings, when it does not. The driving stage is always more positive than the next.

The plate resistor for the previous stage and the cathode resistor for the dc-coupled stage must be matched. This is something I often overlooked - it is a trick Fender discovered way back, because of what I'll explain next. On most their schematics they note the voltages at the previous plate and the following cathode to be equal, when that is not true, not a single Fender amp has that cathode at the same voltage as the preceding plate. You can go and test all the Fenders you find, the 2nd cathode is always positive with respect to the driving plate, which makes its grid negative.

Here's the catch. You can see the driving triode has a cathode resistor, and that the cathode follower does not have such a corresponding resistor at the plate. The consequence is that the previous stage plate is always around 2.xxV more negative than the driven cathode follower cathode - remember the 100k resistors are matched perfectly. The cathode follower grid is always negative with respect to its cathode, that is the genius of the Fender style dc-coupled pair, no matter how far up it swings, the next grid is always negative, because the two 100k resistors are matched, and the driving stage has a cathode resistor which the cathode follower does not. If that cheap little cathode resistor wasn't there, the grid would go positive, that is something that took me a long while to understand and that most folks I know get wrong.

Positive grids may damage the tube, but it's not the damage that keeps this from working or that should concern us...it's the fact that a positive grid simply kills all signal. Tubes do not work as amplifiers with a positive grid, the only reason preamp tubes modulate a tiny bit is because random electrons keep the grid slightly negative when you apply a zero volts to the grid. Push it up 0.5 v positive even and the tube is a diode, and there is no signal modulation at all.

[kg]

it's the fact that a positive grid simply kills all signal. Tubes do not work as amplifiers with a positive grid,

what the hell are you talking about?

[jmaf]

what the hell are you talking about?

??

[redelephant]

Man, I dont get it either...
What's the catch?
Do you think Fender figured out that cathodes have potentials above the grid all by himself...or is it something simple inherent in tubes, basic tube101?
What do u mean the resistors must be matched? Nonsense.
Grids can go positive if provided with enough drive.

[Steve Conner]

I have no idea what jmaf is talking about either. :O

Are we talking about the cathode follower that drives the tone stack in a Marshall?

[jmaf]

Grids cannot be biased positive on any tube. If you mess up those two resistors the DC operating point will be set positive. kg simply dismissed everything I wrote, perhaps my English is that bad or my electronic ignorance is that big, anyhow for those who understood what I took the time to write and didn't simply dismiss it, I hope the info was useful, whatever part of it made any sense.

[jmaf]

I have no idea what jmaf is talking about either. :O

Are we talking about the cathode follower that drives the tone stack in a Marshall?

Yes, I was talking about Fender, but it's the same subject (that at least I am referring to). I don't know if I'm messing the topic up, but I've never heard of biasing a 12AX7 positive, I have never heard of "enough drive" for a 12AX7 positive grid and as far as I know a positive grid makes any tube a diode.