Direct Coupling with Zener Diodes

[exclamationmark]

With the uni holidays approaching I'm looking to fiddle with a pseudo bass (*cough* 8 string guitar *cough*) pre-amp using direct coupling. I know I have two options (unless there are more ) in regards to level shifting, which is to effectively use a voltage divider or zener diode to shift the quiescent DC grid voltage.

I can get low current zener diodes (50uA) for 5 cents a piece, but I'm kind of wondering, how do I actually calculate the current flowing through the grid resistor, and hence, the zener. Is it simply just a case of the Rg + Ra resistor in series, or does something funky happen with the tube in the mix. If some weirdness does occur, how would the extremes of the signal through the valve affect the current through the zener.

Another odd question is what would happen if the quiescent voltage of the previous stage was something like 150v and I use a string of zeners to drop 160v. Would it be the same as having the grid voltage at 0 (like an ac coupled stage) and sticking a 10v diode (or equivalent resistor) at the cathode? I think this is referred to as grid bias, but I've never heard or seen it in a guitar amplifier.

I'm planning to stick with a stash of 12au7's I have so I can make the plate/grid resistors fairly small. The data sheet doesn't really state the impedance of the zener diodes but I can assume it's probably around 1k maximum at 50uA, so it shouldn't affect things too badly. The voltage drop at 50uA is... varied... to say the least but at 5 cents a pop I can afford to measure around a little.

[jmaf]

The voltage at the resistor under the zener will be Vp - Vzener. About the effect of the tube, when the plate swings Va VAC, output will swing Va - Vzener, possibly distorted and clipped due to silicon's infinite maze of details(I have no idea, but gut tells me it'd sound bad).

[kg]

With the uni holidays approaching I'm looking to fiddle with a pseudo bass (*cough* 8 string guitar *cough*) pre-amp using direct coupling. I know I have two options (unless there are more ) in regards to level shifting, which is to effectively use a voltage divider or zener diode to shift the quiescent DC grid voltage.

I can get low current zener diodes (50uA) for 5 cents a piece, but I'm kind of wondering, how do I actually calculate the current flowing through the grid resistor, and hence, the zener. Is it simply just a case of the Rg + Ra resistor in series, or does something funky happen with the tube in the mix. If some weirdness does occur, how would the extremes of the signal through the valve affect the current through the zener.

honestly i would stay away from 50 microamp zeners. breathe on them the wrong way and they'll pop.

the top of the "grid resistor" will be whatever the plate voltage is, minus the zener voltages.

Another odd question is what would happen if the quiescent voltage of the previous stage was something like 150v and I use a string of zeners to drop 160v. Would it be the same as having the grid voltage at 0 (like an ac coupled stage) and sticking a 10v diode (or equivalent resistor) at the cathode? I think this is referred to as grid bias, but I've never heard or seen it in a guitar amplifier.

won't work quite the way you're expecting. the "bottom" of the zener will never go below whatever the grid resistor is connected to--in your schemo, ground, or zero volts.

grid leak bias uses very very high grid return resistance so that electrons intercepted by g1 build up and form a negative charge, thus effectively applying a bias voltage.

I'm planning to stick with a stash of 12au7's I have so I can make the plate/grid resistors fairly small. The data sheet doesn't really state the impedance of the zener diodes but I can assume it's probably around 1k maximum at 50uA, so it shouldn't affect things too badly. The voltage drop at 50uA is... varied... to say the least but at 5 cents a pop I can afford to measure around a little.

if you want to increase your signal swing excursion below 0v then you'll need to return the zener string + grid resistor to some negative voltage rail.

[redelephant]

Not that I've tried myself, but this is what I know about zeners in the signal path:
1. zeners are nonlinear, much worse than capacitors, not in my signal chain thanks (as the signal current swings thru them, the zener voltage will vary)
2. zeners are noisy, espescially at low currents
3. zeners drift, the larger voltage the worse they get
4. shit, can't come up with anything else...

[exclamationmark]

if you want to increase your signal swing excursion below 0v then you'll need to return the zener string + grid resistor to some negative voltage rail.

So you're saying that if I drop the quiescient grid voltage to 0 through the use of a zener, I've effectively made a half wave rectifier (unless I connect a negative source so it can swing negative?). Of course it seems obvious to me now after reading about how grid bias actually works, and well, using my brain... I've seen some pretty crazy designs with a string of LED's at the grid providing the bias fueled by a constant current source + negative voltage rail, with the cathode straight to ground.

Another thing I forgot to mention is that the HT is only ~50v in my circuit, so exploding zeners would be rather entertaining, rather than dangerous or annoying. The good news is since my HT is so low I should be able to 'modify' the negative rail of an op-amp supply from the effects loop pretty easily.

I'm totally comfortable with silicon in the signal path (in fact LED cathode bias is my favourite mod to my existing amps) but I know zeners can be slightly troublesome. By keeping the current above 50uA at all times it should keep the voltage drop relatively linear... I can see why people say direct coupling isn't worth the effort, but what else am I gonna do with my 2 week holiday

[jmaf]

So you're saying that if I drop the quiescient grid voltage to 0 through the use of a zener, I've effectively made a half wave rectifier (unless I connect a negative source so it can swing negative?). Of course it seems obvious to me now after reading about how grid bias actually works, and well, using my brain... I've seen some pretty crazy designs with a string of LED's at the grid providing the bias fueled by a constant current source + negative voltage rail, with the cathode straight to ground.

Another thing I forgot to mention is that the HT is only ~50v in my circuit, so exploding zeners would be rather entertaining, rather than dangerous or annoying. The good news is since my HT is so low I should be able to 'modify' the negative rail of an op-amp supply from the effects loop pretty easily.

I'm totally comfortable with silicon in the signal path (in fact LED cathode bias is my favourite mod to my existing amps) but I know zeners can be slightly troublesome. By keeping the current above 50uA at all times it should keep the voltage drop relatively linear... I can see why people say direct coupling isn't worth the effort, but what else am I gonna do with my 2 week holiday

Heh, now there's a trick question... But contrary to what it may seem, it'll never swing negative... The swing will have a midpoint, at some voltage between Vplate and +B, but never below Vplate for obvious reasons.

Tubes are always depletion mode, "N channel" devices(only electrons flow, no holes), attaching a negative rail underneath is the same as increasing +B above...there is no enhancement mode tube. The maximum theoretical swing is zero to +B, though that'll never happen for obvious reasons(the tube and the load pose resistance).

The negative voltage you apply to grids have zero relation with a symmetrical complementary pair as with transistors. Negative voltages in tubes only serve to control the electron flow. Tubes are depletion mode devices, they will conduct maximum current when the grid is at zero volts with respect to the cathode of the tube. So, in a circuit, if the cathode is at zero volts(power tubes, for example), you then need a voltage below zero to keep that tube from conducting its maximum current and burning up.

IF, on the other hand, your cathode is above zero, and you tie the grid to ground, then the grid is negative with respect to the cathode. Which is how most class A preamp tubes work. Tubes will never swing below zero, unless their cathode is at a reference below zero, which is possible but uncommon.

About having diodes at the cathode, etc... Take a good signal generator and run a perfect sin wave through a regular red LED. The result is a noisy aberration. Run the same sin wave through a 1n4007 diode. Noisy aberration. Run the same sin wave through a zener in forward bias mode just for kicks, noisy aberration. That same noise becomes a part of your signal whenever you put such sillicon on cathodes and such. It ruins your tube sound, IMO.

[mooreamps]

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

thank you....

-g

[exclamationmark]

The negative voltage you apply to grids have zero relation with a symmetrical complementary pair as with transistors. Negative voltages in tubes only serve to control the electron flow. Tubes are depletion mode devices, they will conduct maximum current when the grid is at zero volts with respect to the cathode of the tube. So, in a circuit, if the cathode is at zero volts(power tubes, for example), you then need a voltage below zero to keep that tube from conducting its maximum current and burning up.
.

This is precisely the idea behind the negative voltage rail, in that "Negative voltages in tubes only serve to control the electron flow". I'm not actually making the tube swing negative per se, but the voltage at the grid to appear as an AC sine wave with a DC offset of 0 volts (because of the zener dropping the quiescent grid voltage to 0). The output at the tube plate is still very much positive but has been 'level shifted' down (ie removing DC offset caused by quiescent voltage) so that it will start to swing negative at the grid of the next stage at some point.

At least that is what would appear to happen in my brain. Feel free to poke any holes in my theories though

[jmaf]

This is precisely the idea behind the negative voltage rail, in that "Negative voltages in tubes only serve to control the electron flow". I'm not actually making the tube swing negative per se, but the voltage at the grid to appear as an AC sine wave with a DC offset of 0 volts (because of the zener dropping the quiescent grid voltage to 0). The output at the tube plate is still very much positive but has been 'level shifted' down (ie removing DC offset caused by quiescent voltage) so that it will start to swing negative at the grid of the next stage at some point.

At least that is what would appear to happen in my brain. Feel free to poke any holes in my theories though

I see where you're coming from. Let's simplify things, instead of boring you with a ton of theory, just draw the next stage where you're coupling this to. It'll probably make things clearer. Remember, the next grid cannot be positive with respect to its cathode.

[kg]

I see where you're coming from. Let's simplify things, instead of boring you with a ton of theory, just draw the next stage where you're coupling this to. It'll probably make things clearer. Remember, the next grid cannot be positive with respect to its cathode.

why can't it?

especially if directly coupled?

[J M Fahey]

but what else am I gonna do with my 2 week holiday

Well, I might post here a dozen suggestions, but most would be unprintable following Forum rules .... although definitely FUN.
Back to electronics, what you are trying to do is a drifting and biasing nightmare.
Imagine you have 160V at a tube's plate, which is coupled into the next one's grid.
You think that if you used a 160V Zener you would heve 0V DC at said grid, but full AC voltage swing.
1) You won't, even under "ideal" conditions, because said grid will have a resistor to ground.
Zeners *need* some current passing through them to "Zene", you will have positive voltage through the grid resistor.
2) Under *real* conditions, plate voltage will drift or simply won't sit exactly at the same value day after day.
If it goes up just 1 V , which in a regular amp stage would change nothing, you will have 1V positive at the grid.
If it goes down 1V, it won't "Zene" and become an open circuit.
3) *If* you are willing to add real beefy negative voltages, servos and the like, don't waste your time, just build this which does what you want and then some:

It's brilliantly simple for what it does, and yes, uses level-shifting NE2 neon lamps (the 1952 version of Zeners). You could use modern Zeners there.
You will need +300V and -300V supplies, although the output can only swing +/-50V.
Or you can use an industry standard TL072

Imagine

[jmaf]

why can't it?

especially if directly coupled?

He can. It'll give us something more concrete to discuss upon, with voltages and values to discuss, not just a ton of theory

PS. I just reread your question. Why can't "it" (the grid) be positive, or why can't "it" be done?

[kg]

why can't the grid go positive?

[exclamationmark]

It's brilliantly simple for what it does, and yes, uses level-shifting NE2 neon lamps (the 1952 version of Zeners). You could use modern Zeners there.
You will need +300V and -300V supplies, although the output can only swing +/-50V.

I forgot to mention (somewhat importantly!) that the whole point of the zener diodes was to maintain the 'full' output swing, or at least that equivalent to a capacitively coupled stage. In the case of the above circuit I would rather much just stick a grid stopper between two stages, bias accordingly and be done with it, which basically gives me the same thing. The obvious disadvantage being that since quiescent voltage is always applied to the grid of the next stage, the bias will need to take this into account, and will reduce the voltage swing to perhaps half of what it could be. For instance if the quiescent grid/plate voltage is at 160v and we want to bias it at -3v, there would be a 163V voltage drop across the cathode, reducing the anode to cathode voltage and hence voltage swing.

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. Perhaps I could live with the lower output swing as only one or two stages are intended to be driven hard and I could just DC couple these ones... hmm...

[jmaf]

why can't the grid go positive?

Because it's much closer to the cathode than any other part of the tube. When the grid goes positive it pulls in electron current(in conventional flow it means "the grid conducts"), it in fact becomes a diode to ground from zero volts on up. But the grid isn't designed for electron flow, it's just a voltage barrier, in fact it's a very delicate structure. 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....