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Elevated Heater Question

I am trying to understand where to elevate a DC regulated heater supply with respect to the HT DC ground.

In one classic circuit I have seen is a 12AX7A with one of the sections as a plate follower that feeds the second section, which is a cathode follower.

If B+ = 250 VDC and the DC voltages at each cathode are about <10 VDC for the plate follower section and about 190 VDC at the cathode of the cathode follower section, how does one go about setting the DC elevation point for the heaters?

The 12AX7A tube manual states that the:

Heater Negative (with respect to the cathode) Maximum Voltage = 200 Volts.
Heater Positive (with respect to the cathode) Maximum Voltage = 100 Volts.

If +40 VDC is selected as the desired elevation point above the cathode voltage, that would seem to imply that two different supplies elevated to different values are required, which is obviously not possible with a single 12AX7A.

How do you reconcile this?
 
I don't think you actually mean "plate follower." It sounds as though you really mean "common cathode voltage amplifier."

In any case, if you're using two different tubes for those two sections, yes, you can use two different heater supplies, each elevated to something above the cathode of its respective stage. If you're using the same tube for both sections (and one should note that using a 12AX7 this way is poor design practice, it's a lousy cathode follower), then you need to pick a voltage that will cause the Vhk(max) ratings of each section to not be exceeded. For the example you give, a +40V elevation will keep things within ratings, but certainly the heater is well negative of the cathode of the cathode follower stage.
 
Here is the example circuit from LTSpice:

[IMGHTTPDEAD]http://www.mdbq.net/diyaudio/12ax7.jpg[/IMGHTTPDEAD]

Are you saying it is poor design to put both of these stages in one bottle?

In this circuit LTSpice shows a cathode voltage of 1.5 VDC and 127 VDC respectively for a 250 VDC B+.

So, the trick is to simply elevate the LT supply to 41.5 VDC and the -85.5 VDC between the heater and second cathode is acceptable because it satisfies the Heater Negative specification of 200 VDC?
 
So, the trick is to simply elevate the LT supply to 41.5 VDC and the -85.5 VDC between the heater and second cathode is acceptable because it satisfies the Heater Negative specification of 200 VDC?

Yes- the 41.5V isn't very critical as long as Vhk(max) is satisfied and the heater is positive with respect to the cathode of the voltage amplifier. The poor design practice is the use of a 12AX7 as a cathode follower- for that, you want a tube with high transconductance and that can swing some current. The 12AX7 is neither.

The thing to be careful of in a circuit like this is to think through what happens at power up and power down, not just the static voltages after warmup.
 
Yes- the 41.5V isn't very critical as long as Vhk(max) is satisfied and the heater is positive with respect to the cathode of the voltage amplifier.

In the second stage the heater is negative with respect to the cathode voltage. Did you mean that the magnitude of difference is less than Heater Negative Specification?

The poor design practice is the use of a 12AX7 as a cathode follower- for that, you want a tube with high transconductance and that can swing some current. The 12AX7 is neither.

Ah, that is interesting because I see this done in so many circuits.

The thing to be careful of in a circuit like this is to think through what happens at power up and power down, not just the static voltages after warmup.

Assuming the LT and HT supplies are both turning on and off at the same basic rate, is the issue more to do with the time it takes for the heaters to reach a steady-state temperature?
 
Two problems with the circuit shown in post 3:
- during startup, when the HT supply is present but the cathodes are still cold, the grid of the second triode is at +250V while its cathode is at 0V. This is a lot of voltage across a small gap. The solution is to add a signal diode or miniature neon lamp.
- the CF is likely to run near grid current which will load the first stage and cause distortion. The 12AX7 does not like sinking much anode curent with low anode voltage. The high mu means that the window of normal use between grid cutoff and grid current is unusually narrow.

There is the separate issue of the 12AX7 making a poor CF, but this would depend on what it is being asked to drive.
 
I am trying to understand where to elevate a DC regulated heater supply with respect to the HT DC ground.

In one classic circuit I have seen is a 12AX7A with one of the sections as a plate follower that feeds the second section, which is a cathode follower.

If B+ = 250 VDC and the DC voltages at each cathode are about <10 VDC for the plate follower section and about 190 VDC at the cathode of the cathode follower section, how does one go about setting the DC elevation point for the heaters?

The 12AX7A tube manual states that the:

Heater Negative (with respect to the cathode) Maximum Voltage = 200 Volts.
Heater Positive (with respect to the cathode) Maximum Voltage = 100 Volts.

If +40 VDC is selected as the desired elevation point above the cathode voltage, that would seem to imply that two different supplies elevated to different values are required, which is obviously not possible with a single 12AX7A.

How do you reconcile this?

This will just about work as is with one side of the heaters connected to ground. The heaters at -190V relative to the CF cathode, just within the 200V spec. The heaters are less than 10V below the other cathode, again within the 200V spec. However, if there is any significant signal level, the CF cathode could rise well about +200V and take it outside the 200V spec. This os one reason heater elevation is used.

If you elevated the heaters to +75V relative to ground then they would be -190 +75 = -115 relative to the CF cathode - well within the 200V spec and with plenty of headroom for large signals. They would also be +75 - a bit above the other cathode, again well within the 100V spec. The other advantage of the heaters being +ve with respect to the first cathode is that the diode formed by the heaters and the cathode is now reverse biased which eliminates one form of hum pickup.

Not that the values for heater breakdown vary from manufacturer to manufacturer and from data sheet to data sheet even for the same tube.

Cheers

Ian
 
Two problems with the circuit shown in post 3:
- during startup, when the HT supply is present but the cathodes are still cold, the grid of the second triode is at +250V while its cathode is at 0V. This is a lot of voltage across a small gap. The solution is to add a signal diode or miniature neon lamp.

Good point.
I do not remember the ionization potential for Ne, but as in my country valves are expensive and difficult to obtain, as a general rule I use a delay of about 30 seconds to warm heaters, and after that, a soft start for B+.
Also I would put some garlic, if that helps. :D
 
Good point.
I do not remember the ionization potential for Ne, but as in my country valves are expensive and difficult to obtain, as a general rule I use a delay of about 30 seconds to warm heaters, and after that, a soft start for B+.
Also I would put some garlic, if that helps. :D

90V or so for a typical neon lamp. Connected between grid and cathode, it should protect the tube quite nicely.

Chris
 
90V or so for a typical neon lamp. Connected between grid and cathode, it should protect the tube quite nicely.

Chris

I would not be so sure


Ion Bombardment
The vacuum in a valve is not perfect, so there are gas molecules randomly floating between the anode and cathode.
If an electron should be accelerated towards the anode from the cathode, there is always a chance that it will collide with a gas molecule and have sufficient energy to remove an electron from that molecule, rendering it positively charged and attracted to a lower potential such as the cathode.
These ions can have a significant momentum when it strikes the cathode.

Case 1: Vak = 0, Hot Cathode
In metals, at normal temperature, the conduction band is essentially filled of electrons only up to the Fermi energy EF, to extract an electron from the metal is therefore necessary to give a starting energy ei, but at high temperatures the occupation of electronic states extends above EF.
If the temperature is high enough, some electrons reach energies greater than EF + ei, and escape from the metal.
If the temperature increases further, the Fermi energy level is widely exceeded and electrons have enough energy to collide with a gas molecule, and have enough energy to remove an electron from that molecule, rendering it positively charged and attracted to a lower potential such as the cathode.
The force acting on the ions, is due to the electric field created by the charge distribution between the cathode and anode, ie. the electron cloud.
F = e E
For simplicity we assume that ions are fermions, then from the Fermi-Dirac distribution, some ions after interacting with the electron cloud, hit the cathode with enough momentum to be absorbed into its surface.
This phenomenon is called "Cathode Poisoning"

Case 2: Vak = B+, Cathode warms from cold
The cathode starts to warm up to a certain temperature barely above the Fermi energy level.
For simplicity we assume that only one electron is emitted.
This electron is accelerated toward the anode and assuming that collides with at least one gas molecule, and has enough energy to produce an ion.
Now there is no electron cloud, then the force exerted on the ion, is due to the electric field created by the potential difference between cathode and anode.
F = e E = - e grad (φ)
The ion hit the cathode with enough momentum to produce sputtering in the cathode.
This phenomenon is called "Cathode Stripping"


As it stands, we are in case 2, with the aggravation of having to 90 V or so between grid and cathode.
If I were the cathode, would be praying... :D
 
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I would not be so sure


Ion Bombardment
The vacuum in a valve is not perfect, so there are gas molecules randomly floating between the anode and cathode.
If an electron should be accelerated towards the anode from the cathode, there is always a chance that it will collide with a gas molecule and have sufficient energy to remove an electron from that molecule, rendering it positively charged and attracted to a lower potential such as the cathode.
These ions can have a significant momentum when it strikes the cathode.

Case 1: Vak = 0, Hot Cathode
In metals, at normal temperature, the conduction band is essentially filled of electrons only up to the Fermi energy EF, to extract an electron from the metal is therefore necessary to give a starting energy ei, but at high temperatures the occupation of electronic states extends above EF.
If the temperature is high enough, some electrons reach energies greater than EF + ei, and escape from the metal.
If the temperature increases further, the Fermi energy level is widely exceeded and electrons have enough energy to collide with a gas molecule, and have enough energy to remove an electron from that molecule, rendering it positively charged and attracted to a lower potential such as the cathode.
The force acting on the ions, is due to the electric field created by the charge distribution between the cathode and anode, ie. the electron cloud.
F = e E
For simplicity we assume that ions are fermions, then from the Fermi-Dirac distribution, some ions after interacting with the electron cloud, hit the cathode with enough momentum to be absorbed into its surface.
This phenomenon is called "Cathode Poisoning"

Case 2: Vak = B+, Cathode warms from cold
The cathode starts to warm up to a certain temperature barely above the Fermi energy level.
For simplicity we assume that only one electron is emitted.
This electron is accelerated toward the anode and assuming that collides with at least one gas molecule, and has enough energy to produce an ion.
Now there is no electron cloud, then the force exerted on the ion, is due to the electric field created by the potential difference between cathode and anode.
F = e E = - e grad (φ)
The ion hit the cathode with enough momentum to produce sputtering in the cathode.
This phenomenon is called "Cathode Stripping"


As it stands, we are in case 2, with the aggravation of having to 90 V or so between grid and cathode.
If I were the cathode, would be praying... :D

Okay. Let's assume we add a delayed B+ start of about 30 seconds after filaments are turned on. That is what i did with my power amp.
 
The neon goes from grid to cathode in the CF.

The issue with Vgk is not the voltage, but the field: volts/m. During warm up the anode will be at 250V, so if there are any free electrons (e.g. due to photoemission) then they will already be reaching 250eV energy so can already produce ions. We don't worry about 250V on the anode of a cold valve causing trouble. Well, some people might worry but they shouldn't.

The issue with Vgk is that the gap is so small that the electric field becomes large. This can cause direct ionisation if the valve is gassy. I suppose in theory it could also cause direct damage to the cathode surface. Once a neon strikes the voltage drops to somewhere around 40-50V and that is fine.

It could be argued that frame grid valves such as ECC88 might need a smaller voltage as the grid-cathode spacing is much smaller. People who are worried can use a diode instead. Any ordinary low capacitance signal diode will do. In normal operation it is reverse biased and bootstrapped so does not affect the circuit.