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Tube Experts: Analysis Requested (R.G. et. al.)

12/11/1999 11:56 PM
Joe Fuzz	Tube Experts: Analysis Requested (R.G. et. al.) I've been looking over the Tube Driver out at Jack's site and I'm stumped. Specifically, it's the tube section highlighted in the picture below. (Note: picture may take a second to load.) It's the 470k resistor connected to the grid/base of tube 2. Won't this hold the grid at +Vcc? It looks like, at best, the grid will only see the negative half of the AC wave. Your help/opinions appreciated!

12/11/1999 11:56 PM

Joe Fuzz

Tube Experts: Analysis Requested (R.G. et. al.)

I've been looking over the Tube Driver out at Jack's site and I'm stumped. Specifically, it's the tube section highlighted in the picture below. (Note: picture may take a second to load.)

It's the 470k resistor connected to the grid/base of tube 2. Won't this hold the grid at +Vcc? It looks like, at best, the grid will only see the negative half of the AC wave.

Your help/opinions appreciated!

12/12/1999 7:24 PM
AMZ	Re: Tube Driver Analysis Requested >>I've been looking over the Tube Driver out at Jack's site and I'm stumped. In the patent, it states that the pull-up resistor increases the input impedance, provides comparable output Z and allows more headroom before clipping (1.2v rms vs. 0.7v for a convential bias). I haven't analyzed it beyond this but it must work to some degree since I have had mail from many people who built the circuit and it worked well for them. regards, Jack

12/12/1999 7:24 PM

AMZ

Re: Tube Driver Analysis Requested

>>I've been looking over the Tube Driver out at Jack's site and I'm stumped.

In the patent, it states that the pull-up resistor increases the input impedance, provides comparable output Z and allows more headroom before clipping (1.2v rms vs. 0.7v for a convential bias). I haven't analyzed it beyond this but it must work to some degree since I have had mail from many people who built the circuit and it worked well for them.

regards, Jack

12/12/1999 9:46 PM

Joe Fuzz

Hi, Jack. Nice to hear from you. Hope your site redesign goes speedy and well.

quote:
"In the patent..."

I thought you designed this one! I thought that was your patent number.

quote:
"I haven't analyzed it beyond this but it must work to some degree since I have had mail from many people who built the circuit and it worked well for them."

I've built it -- it does work. In fact, it's a lot of fun. But, as I believe Aron pointed out in another thread, it varies a lot based on what tube you've got in there. I've tried 3 different tubes and gotten 3 different tones.

My question was an electronics one. If you look at the 470k resistor, it has 2 possible DC paths -- to the right leading to Tube 2's grid or to the left leading to the .047uF cap. Now, the cap will block any DC current to the left. And I *thought* that I_grid=0 for a tube which wouldn't allow any DC current to flow to the right. And if I_470k=0, then V_470k=0 so V_grid=V_CC.

Right?

12/12/1999 11:45 PM
AMZ	>>I thought that was your patent number. No, that's the patent number from Butler's original Tube Driver circuit. >>I thought that I_grid=0 for a tube which wouldn't allow any DC current to flow to the right. (As I understand it): Tubes aren't perfect devices and there is always some grid leakage. The tiny amount of current drawn across the signal grid resistor creates a positive voltage and secondary emission of electrons are attracted to the grid. Since the grid requires virtually zero current to perform its job, this space charge creates enough negative bias to allow the triode to operate. Since different tubes leak different amounts there may be a wide variation in performance from tube to tube in this circuit. regards, Jack

12/12/1999 11:45 PM

AMZ

>>I thought that was your patent number.

No, that's the patent number from Butler's original Tube Driver circuit.

>>I *thought* that I_grid=0 for a tube which wouldn't allow any DC current to flow to the right.

(As I understand it): Tubes aren't perfect devices and there is always some grid leakage. The tiny amount of current drawn across the signal grid resistor creates a positive voltage and secondary emission of electrons are attracted to the grid. Since the grid requires virtually zero current to perform its job, this space charge creates enough negative bias to allow the triode to operate.

Since different tubes leak different amounts there may be a wide variation in performance from tube to tube in this circuit.

regards, Jack

12/13/1999 11:23 AM
GFR	Altough it's been patented, it's an old technique called grid blocking. I've found it in a 60's book by Millman. With no signal applied, the gk diode is directly biased and the grid is near 0V. Grid current flows. As the grid is conducting, the input impedance is very low, not higher as the patent says. When you apply AC, the coupling cap is charged by the initial peak and then the grid is biased at a negative voltage (-Vpeak). Then it's a "normal" tube stage with high input impedance, until the cap discharges. Remember the famous blocking distortion that happens in Fenders (grid current flows, tube goes into cutoff and amp farts)? This is the same effect, but when the signal charges the cap it biases the tube OK instead of leading it to cutoff. While you generally want to avoid blocking, this design relies on it to work properly. This remembers me of the thread "One man's fart..."

12/13/1999 11:23 AM

GFR

Altough it's been patented, it's an old technique called grid blocking. I've found it in a 60's book by Millman.

With no signal applied, the gk diode is directly biased and the grid is near 0V. Grid current flows. As the grid is conducting, the input impedance is very low, not higher as the patent says.

When you apply AC, the coupling cap is charged by the initial peak and then the grid is biased at a negative voltage (-Vpeak). Then it's a "normal" tube stage with high input impedance, until the cap discharges.

Remember the famous blocking distortion that happens in Fenders (grid current flows, tube goes into cutoff and amp farts)? This is the same effect, but when the signal charges the cap it biases the tube OK instead of leading it to cutoff.

While you generally want to avoid blocking, this design relies on it to work properly. This remembers me of the thread "One man's fart..."

12/13/1999 7:21 PM
R.G.	Re: Tube Experts: Analysis Requested (R.G. et. al.) The second tube is operating "in clamp" as they used to say in the Golden Age. Tying the grid to Vcc through a high impedance doesn't pull the grid to Vcc because the positive grid voltage allows grid current to flow and the impedance of the grid is lower in that case, much lower than the 470K or so that ties the grid to Vcc. This connection is pretty much a constant grid current connection. It offers some biasing advantages for large signal conditions. The grid is effectively "clamped" to the cathode with a Vgk of approximately zero for all signals that aren't large enough to change the grid current from the resistor to Vcc. Larger signals can then be handled without sudden discontinuities as would happen if the grid were normally biased. When a tube is operated in clamp, the input impedance is much lower, and discontinuities happen on inputs large enough to run the grid back in to negative Vgk territory. It's an unusual connection, but it is a valid one. The file http://www.eden.com/~keen/mandt.htm (which I still need to get moved to geofex) has the following exerpt from Millman and Taub's "Digital and Switching Circuits" with some explanation: =========================================================== The grid volt-ampere characteristics of the 5965 tube are given in Fig. 6-33. At a given plate voltage the grid circuit behaves as a diode. By analogy with the definition of the dynamic plate resistance, the dynamic grid resistance r is given by dV/dIg, where V, and IG are the instantaneous values of grid voltage and current, respectively. The static grid resistance rg is defined as the ratio VG/IG. From Fig. 6-33 it appears that the difference in values between the static and dynamic resistances is not great, except possibly for small grid voltages. Furthermore, the value of the grid resistance ra is not a sensitive function of plate voltage. From Fig. 6-33 we find that for the 5965 tube, 250 ohms is a reasonable value for rG. For other tubes, the grid resistance may be much more variable than indicated above. For example, for a 12AU7 the static rG has values ranging from about 500 to 1,500 ohms, depending upon the values of grid and plate voltages A Clamped Grid If, as in Fig. 6-34, the grid leak is tied to the Vpp supply instead of to the cathode, then the grid-to-cathode voltage will approach nominal zero for values of R which are large compared with rG. For example, if R, = 1 M and Vpp = 300 V, then the grid current will be approximately 300 uA. From Fig. 6-31, we find that the grid voltage corresponding to this grid current is about -0.05 V. (If we assume that the value of rG = 250 ohms is valid at low grid voltages, then the calculated value of VG is 0.3 X 0.25 +0.075 V.) In many pulse circuits it is common to use this connection of the grid leak to a high positive voltage. Under such circumstances, where the grid is held at the cathode voltage because of the flow of grid current, we shall refer to the grid as being clamped to the cathode. Alternatively, the tube is said to be in clamp. Variability of Characteristics If the grid voltage is made a few volts negative, then the grid current reverses. This negative current is caused by the positive ions which are attracted to the grid. Since the positive-ion current comes from the residual gas in the "vacuum" tube, it is very variable from tube to tube, and is usually a small fraction of a microampere. Negative grid current can also result from thermionic or photoelectric emission from the grid.

12/13/1999 7:21 PM

R.G.

Re: Tube Experts: Analysis Requested (R.G. et. al.)

The second tube is operating "in clamp" as they used to say in the Golden Age. Tying the grid to Vcc through a high impedance doesn't pull the grid to Vcc because the positive grid voltage allows grid current to flow and the impedance of the grid is lower in that case, much lower than the 470K or so that ties the grid to Vcc. This connection is pretty much a constant grid *current* connection. It offers some biasing advantages for large signal conditions. The grid is effectively "clamped" to the cathode with a Vgk of approximately zero for all signals that aren't large enough to change the grid current from the resistor to Vcc. Larger signals can then be handled without sudden discontinuities as would happen if the grid were normally biased.

When a tube is operated in clamp, the input impedance is much lower, and discontinuities happen on inputs large enough to run the grid back in to negative Vgk territory. It's an unusual connection, but it is a valid one.

The file http://www.eden.com/~keen/mandt.htm (which I still need to get moved to geofex) has the following exerpt from Millman and Taub's "Digital and Switching Circuits" with some explanation:
===========================================================
The grid volt-ampere characteristics of the 5965 tube are given in Fig. 6-33. At a given plate voltage the grid circuit behaves as a diode. By analogy with the definition of
the dynamic plate resistance, the dynamic grid resistance r is given by dV/dIg, where V, and IG are the instantaneous values of grid voltage and current, respectively.
The static grid resistance rg is defined as the ratio VG/IG. From Fig. 6-33 it appears that the difference in values between the static and dynamic resistances is not
great, except possibly for small grid voltages. Furthermore, the value of the grid resistance ra is not a sensitive function of plate voltage. From Fig. 6-33 we find that for
the 5965 tube, 250 ohms is a reasonable value for rG. For other tubes, the grid resistance may be much more variable than indicated above. For example, for a 12AU7
the static rG has values ranging from about 500 to 1,500 ohms, depending upon the values of grid and plate voltages

A Clamped Grid
If, as in Fig. 6-34, the grid leak is tied to the Vpp supply instead of to the cathode, then the grid-to-cathode voltage will approach nominal zero for values of R which are
large compared with rG. For example, if R, = 1 M and Vpp = 300 V, then the grid current will be approximately 300 uA. From Fig. 6-31, we find that the grid voltage
corresponding to this grid current is about -0.05 V. (If we assume that the value of rG = 250 ohms is valid at low grid voltages, then the calculated value of VG is 0.3 X
0.25 +0.075 V.) In many pulse circuits it is common to use this connection of the grid leak to a high positive voltage. Under such circumstances, where the grid is held at
the cathode voltage because of the flow of grid current, we shall refer to the grid as being clamped to the cathode. Alternatively, the tube is said to be in clamp.

Variability of Characteristics If the grid voltage is made a few volts negative, then the grid current reverses. This negative current is caused by the positive ions which
are attracted to the grid. Since the positive-ion current comes from the residual gas in the "vacuum" tube, it is very variable from tube to tube, and is usually a small
fraction of a microampere. Negative grid current can also result from thermionic or photoelectric emission from the grid.

12/13/1999 9:59 PM

Joe Fuzz

Hi, R.G. I hope your site update goes smoothly!

quote:
"...because the positive grid voltage allows grid current to flow..."

This was not my assumption. I assumed that I_grid=0 at all times, like the gate of a JFET.

quote:
"Larger signals can then be handled without sudden discontinuities as would happen if the grid were normally biased."

This is good news as the next step for me is to re-design the opamp-preamp section to handle larger inputs before clipping.

Speaking of the "Golden Age," I've been talking to some of the folks at work who used tubes way back when and they all swear the circuit won't work with only 12V. When I tell them it does and sounds pretty good, they all ask "What are you doing with it?" I tell them. They respond, "Why in the hell would you want to do that!?" Obviously not guitar players.

Thanks for the insight. Follow up question: would there be an optimum DC bias point at the plate=collector=drain of tube 2? Is it Vcc/2 or should it be a couple of volts either way (i.e. force it into clipping on one side)? I think I measured Vcc/2 + 2V for one of the 12A~7's I tried.

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