Direct Coupling with Zener Diodes

[exclamationmark]

Ok, got it. The pot must be connected as shown, but the wiper goes to the bias node, providing a adjustable voltage for the grid. The wiper should have a cap to gnd so the voltage reference stays low impedent, unless it is relatively low ohmic. With the voltage tapped from the wiper, you see what happens in the middle...

Edit: forgot some... The Ac coupled circuit will work fine. But the DC coupled one will not. The DC current going from the preceding stage thru the resistors will swamp the LED bias voltage. If you get the voltage right it's b/c the resistor values are such that you end up losing too much signal. What's the point? If those zeners you got function ok at low currents, well half a mA or so aint gonna mess up the stage that much. Since you are going hybrid (mixxing SS and tube) how about using a CCS for the plate load? That way you can have large plate resistance, and still have the voltage drop you want over load.


That's a brilliant idea. I've only tried this with a LED/diode from cathode to gnd and the tube's cathode current is the only current going thru the LED. 2-4mA isn't a lot for a steady LED voltage, maybe that's why I've never liked that scheme...Gotta try that CCS trick! When the current thru the LED can be much higher than cathode current, the LED should indeed provide a constant voltage and signal swing will not affect it.
Thanks man, learned something new today
Check circuit here, is that what you mean?

Yeah this (the cathode LED CCS) is how the hi-fi guys commonly tend to do it. Although I suppose guitarists find it more rock 'n roll to have it operating near its knee doing who knows what without the CCS.

I figured the reason why circuit #2 didn't work without ridiculously large (100k+) resistors for R1 + R2 was the current swamping. Circuit simming on LTSpice, it seems to work perfectly given that the resistors are of sufficient size (I find it funny to watch SPICE attempt to model grid-current limiting, though, haha!).

You're right about just making the CCS lower current, it's just that I don't exactly trust these 50uA zeners that much. I'd much rather put something like 10mA through some 'normal' zeners to get reliable bias. As I said, the resistive circuit eats a little more than half the signal which is fine by me (because I end up throwing most of it out anyway with 12ax7's!) but one major advantage is that I can 'tune' the voltage supply to provide the bias whenever it feels like drifting (I guess the same could be said by placing a pot at the cathode...).

Gonna start measuring some of my existing circuits to figure out how bad plate drift actually is from start up to steady state. If the drift isn't too bad, a I can stick a pot across a single 0.6 volt schottky diode in the B- supply to 'fine tune' the bias for that stage. Of course I could also get a proper voltage regulator IC for the B- supply... I suppose I'm not really using zeners to level shift in the slightest anymore!

I'd actually really like to try putting a CCS as the plate load one day. What exactly happens if you jam a 20mA CCS on top of a tube and it's only capable of drawing a certain amount of current (that's nowhere near 20mA)? This might be fun to research...

Edit: Ignore that last part, I found that the load line is simply a horizontal line... But uhh... what happens when the B+ supply is reached. Won't the clipping be of a different character though? I guess I would have to see (hear?) how this sounds for myself (unless anyone can tell me otherwise )

Double Edit: I assume it approximates a mu-follower arrangement except the CCS being solid state and all. I don't think a solid state CCS would appreciate being turned on and off rapidly at clipping however...

[J M Fahey]

KG posted datasheets of triodes operating with the grid positive so it's scientific fact that they continue to amplify whatever is showing up on the grid given a low enough driving impedance.

+1

[Chuck H]

Right. Aren't there operating classes where that is exactly the point??? Isn't that class B? Or any "class" indication that doesn't have a "1" suffix other than class A? I don't know for sure because I never build such amps. But I'm sure I've read a little about it.

[kg]

chuck: class and grid current have no direct relationship... no reason you can't have class A2!

however, grid current and cathode/self bias are mutually exclusive.

[jmaf]

however, grid current and cathode/self bias are mutually exclusive.

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.

Both tubes are cathode biased in that dc-coupled pair.

[J M Fahey]

The second triode you mention, the cathode follower, has around 160V (positive respect to ground, by the way) on its grid.
I would hardly call that "cathode biased".
Otherwise its cathode would sit only a few volts above ground, instead of around 160V. (Let me check the , grasp! ... datasheet)
The cathode resistor you see there (around 100K to 47K depending on amp model) is *not* repeat *not* a "bias" resistor but a "load" resistor.
Wonder if you know the difference.
To make it easier to you:
*if* you need to have, say, 150V above ground on a cathode, and you want to *cathode* bias that triode, you can use a cathode biasing resistor (duh) , plus the load resistor.
Of course, said triode's grid must be DC referenced to the corresponding end of the bias resistor, and it will need an input signal coupling capacitor.
In the direct coupled pair you can save 3 parts= 1 capacitor and 2 resistors.
Since an image is worth 1000 words:

[kg]

Both tubes are cathode biased in that dc-coupled pair.

do you really consider a directly coupled cathode follower to be cathode biased?

[jmaf]

I guess this is why they invented algebra...

B = +B voltage for this circuit
Rp1 = Plate resistor 1
rp1 = Plate resistance 1
Rk1 = Cathode resistor 1
Vp1 = Voltage at plate 1

Rk2 = Cathode resistor 2
rp2 = Plate resistance 2
Vp2 = Voltage at plate 2
Gm2 = Gm1 = Transconductance for both tubes

For numerical comparisons, assume a 12AX7 and Rk2 = Rp1 = 100k OHMs

Solving for tube 1 at DC.

Ik1 = B / (rp1 + Rp1 + Rk1)

Solving for tube 2 at DC.

Ik2 = B / (rp2 + Rk2)

We know Rp1 = Rp2 (the matched 100k resistors I mentioned). We also know rp1 = rp2 (matched triode sections, we assume).

We see Ik2 > Ik1 because Rp1 + Rk1 > Rk2 (triode one has a cathode resistor more than triode two). So Ik2 > Ik1. All this is in my original post which originated much controversy.

At DC we have:

B = (Ik2 * rp2) + (Ik2 * Rk2)
B = (Ik1 * Rk1) + (Ik1 * rp1) + (Ik1 * Rp1)

When is the grid negative?

Vp1 = our 2nd triode grid voltage directly coupled from triode one plate voltage
Vk2 = our 2nd triode cathode voltage

Vg2 = DC Bias = Vp1 - Vk2

If the grid is to be negative, then Vp1 < Vk2. Let's see when this happens.

Bias = B - (Rp1 * Ik1) - B - rp2 * Ik2 (Voltage of plate 1 minus voltage of cathode 2.)
Bias = Rp1 * Ik1 - rp2 * Ik2

For Bias to be negative, we need Rp1 * Ik1 < rp2 * Ik2

rp2 is fixed, it is given by the triode datasheet. We know Ik2 > Ik1 from above, so to have negative bias we need rp2 > Rp1

That is, for the second grid to be negative, we need DC plate resistance of triode 2 to be higher than first plate load resistor Rp1.

Unless I messed up on the algebra, that proves grid 2 is negative at DC, as the 12AX7 plate resistance is less than 100k ohms.

We know the DC operating point of grid two is negative, let's check the AC swing.

DC Vp1 = B - Ik1 * Rp1

So that's our maximum theoretical peak positive swing : B - Ik1 * Rp1

For the second grid to go positive, we need that swing to be larger than the swing on that triode's cathode, that is :

B - Ik1 * Rp1 > Gm2 * (B - Ik1 * Rp1) * Rk2

That is, the maximum swing from plate one must be larger than the second tube's transconductance * swing from plate one applied to the grid * Rk2 which will give us the voltage tube two develops with that swing.

B - Ik1 * Rp1 > Gm2 * (B - Ik1 * Rp1) * Rk2
1 > Gm2 * Rk2
or
Rk2 < 1 / Gm2

For a Sovtek LPS 12AX7, transconductance is 0.0017 A/V so

Rk2 < 588 OHMs

For tube two grid to swing above Bias DC voltage you'd need a 588 ohms cathode resistor.

Unless I've messed up real bad, I've proven both that grid two is always negative and that Rp1 must be equal to Rk2, which is what I claimed on my post about matching the resistors and that Fender/Marshall dc-coupled pairs never drive the grid positive. I appreciate any corrections.

[Steve Conner]

Where do you show that Rp1 must be equal to Rk2? I don't see that anywhere in the math, the only conclusion you have is Rk2 > 588.

You are also making the mistake of doing a large signal analysis with small signal parameters. (If we are driving the circuit hard, as we do in guitar amps, then at the positive peak, tube 1 is cut off, so its gm is zero and its rp infinite. Therefore it vanishes from the analysis, tube 2's grid is simply connected to B+ by tube 1's plate load resistor.)

[jmaf]

Where do you show that Rp1 must be equal to Rk2? I don't see that anywhere in the math, the only conclusion you have is Rk2 > 588.

We can only conclude that Ik2 is bigger than Ik1 if we assume the two resistors to be identical.

About the big/small signal analysis, I'll give it some thought, thank you.

EDIT

The line

We know Rp1 = Rp2 (the matched 100k resistors I mentioned). We also know rp1 = rp2 (matched triode sections, we assume).

Becomes

We know Rp1 = Rk2 (the matched 100k resistors I mentioned). We also know rp1 = rp2 (matched triode sections, we assume).

[kg]

jmaf, i don't know what you're trying to prove here, other than the notion that you are right and i am wrong, which is not going to happen.

as stephen said, one only has to consider what happens if the plate loaded stage cuts off.

no math necessary... TALK IT OUT IN YOUR HEAD.

first plate = second grid = b+.

put another way, forget all the rest of the circuit. just connect the grid of a cathode follower directly to the b+ rail through a 47k-100k resistor, and see what happens. change the cathode load resistor to different values if you want.

IF that grid node does NOT reach b+ then you've got grid current flowing. PERIOD!

[jmaf]

jmaf, i don't know what you're trying to prove here, other than the notion that you are right and i am wrong, which is not going to happen.

Sure, Mr. Feynman.

I just wanted to post the above so anyone can test what you said. Sure you're right and I am wrong. As long as I had posted my reasoning for what I said so anyone who can add and subtract can test. Engineering is a science, not politics.

EDIT: Richard Feynman was a very humble, very gracious scientist, someone who would discuss 1st grade math and still act like it was totally new to him as long as he could contribute to another person's knowledge. Richard Feynman is very different from what we've seen on this discussion, unfortunately I must say.

[redelephant]

jmaf. You are also forgetting that the tube's rp and gm changes as the ik changes. For instance, when the CF swings down and has lower ik the gm goes down and rp goes up, when the CF swings up, it's vice versa. (For a 12AX7, at vp=250V and ik=0.5mA (swinging down) the gm=0.9mA/V and rp=115kohms, the other way (swinging up) vp=100V and ik=2mA the gm=2.0mA/V and rp=55kohms).
But numbers mean you must do the algebra, I stopped using algebra when I figured out how to read tube datasheets and graphs. Not to diss math, by all means, I wish I used it more just to keep it up.

Look at the graphs of the 12AX7, and as it swings up towards b+, it pretty much needs a positive grid if you want more than 2mA with a vp of 100volts.

No matter how you chose Rp1 and Rk2 values, the CF will start to reach a point where it's vp is too low for the amount of ik it is asked to pass to the load. Besides, you've neglected the AC coupled load, which is significant b/c in use, there is signal, and the loading is quite substancial.
So in real circuits, there is no logic to match Rp1 and Rk2.

[imaradiostar]

Why not just build a test circuit with a moving coil meter inserted between the plate of the first and the grid of the second? Feed the circuit a sine wave and as the signal goes up the meter will move, showing that the grid current is drawn only on a portion of each cycle. When no grid current is drawn (lower signals) the positive and negative halves will average in the inertia of the meter, giving no movement.

There's nothing like an experiement to prove your theory. Of course, you might find for some plate voltages and Rk2 values that grid current is drawn with no input signal at all!

jamie

[redelephant]

Yeah, he could do that, or just use a multimeter and measure over the grid-stopper if he uses those.