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What is the driver circuit called?

2.108.1.16

Posted on December 9, 2016 at 12:06:48
Frihed89
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See the reference URL

 

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RE: What is the driver circuit called?, posted on December 9, 2016 at 13:51:06
Eli Duttman
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Looks like DC coupled, cascaded, common cathode 6SN7 sections.


Eli D.

 

Isn't 19.5K a bit high for a cathode resistor? NT, posted on December 9, 2016 at 13:58:03
Alpha Al
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nt

 

RE: Isn't 19.5K a bit high for a cathode resistor? NT, posted on December 9, 2016 at 14:14:50
Eli Duttman
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The heat associated with that value is the price you pay for DC coupling. Remember, the grid to cathode potential difference is what matters. Here, the grid is at a high positive potential, with respect to ground. In order for biasing to be correct, the cathode has to be at an even higher positive potential, with respect to ground.


Eli D.

 

Official term for this is: "6SN7 Eater", posted on December 9, 2016 at 14:40:38
Chip647
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Start-up will try to rip its little wings off until it is warmed up. It really needs a slow start. Replace the 5U4G with a 5AR4.

The 300B filament circuit of a simple voltage regulator with a big-ass electrolytic cap will sound very blahhh. It is in the signal path.

 

RE: Official term for this is: "6SN7 Eater", posted on December 9, 2016 at 15:25:47
Triode_Kingdom
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"Start-up will try to rip its little wings off until it is warmed up."

How so? Many designers use 6SN7s with SS rectifiers. They last for years.

"The 300B filament circuit of a simple voltage regulator with a big-ass electrolytic cap will sound very blahhh. It is in the signal path."

The cap at the output of the regulator is not in the signal path. Only the caps from each side of the filament to ground carry signal current. Not that this is a good thing, just sayin'...


 

RE: Official term for this is: "6SN7 Eater", posted on December 9, 2016 at 17:14:36
Eli Duttman
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The filament supply does SUCK. :>(( Proper current regulated filament "juice" costs some money, hence its absence.


Eli D.

 

The regulator cap is across the filament. It most definitely affects the signal., posted on December 9, 2016 at 17:37:58
Chip647
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But why use a 5U4 when the 5AR4 will help protect a DC coupled circuit?

And look at the 6SN7 heater circuit. Grounding one side of that "stout" filter circuit injects maximum noise into the heater. They could not afford a pair of 100 ohm resistors to fake a center tap? The whole thing is gross. I will bet it is at 7.5 volts "DC".

 

Can you describe the current path through the cap?, posted on December 9, 2016 at 18:57:14
Triode_Kingdom
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If you believe the cap is in the signal path, can you tell me where the differential signal voltage across the capacitor is coming from? Maybe draw the current flow?




 

RE: Can you describe the current path through the cap?, posted on December 10, 2016 at 06:42:46
Chip647
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The regulator cap bridges the 2 bypass caps.

This puts all three caps in play for each end of the filament. On one side of the filament the bypass will be the right side cathode bypass cap to ground as well as the regulator bypass cap in series with the other cathode bypass cap to ground. Same for the other side. Look at the parallel capacitor circuit that is created with the regulator cap.

 

RE: Can you describe the current path through the cap?, posted on December 10, 2016 at 07:59:31
Tre'
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Post withdrawn.

Tre'


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RE: Can you describe the current path through the cap?, posted on December 10, 2016 at 08:57:19
Triode_Kingdom
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You're merely describing how components are connected. You didn't say anything about voltage or current. Again, can you explain how a differential AC voltage is created across the cap?





 

RE: Can you describe the current path through the cap?, posted on December 10, 2016 at 13:56:35
Chip647
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Sorry to be dense. Possibly it would help if you could describe why the designer used a separate cap from each end of the filament. If there is no AC potential between the ends of the filament, one bypass cap would work perfectly, just chose a side to put it on.

 

RE: Can you describe the current path through the cap?, posted on December 10, 2016 at 14:50:48
Triode_Kingdom
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The 300B filament is equivalent to a 4.2 ohm resistor when hot. I can't think of any valid electrical reason to use two caps. It's just a habit with some designers, something that makes them comfortable. The point is, there's no differential voltage, and so no signal current through the parallel cap.

Here's a schematic from Andy Evans:










 

I love the Ron Coleman current regs, but we are talking about voltage regs, posted on December 10, 2016 at 17:39:23
Chip647
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There is a reason current regulation is much better than voltage regulation for the filament of a DHT output tube. For one, with current regulation there is no big honking electrolytic across the filament. Or do you argue that this does not matter sonically and the difference has to do with the configuration of the different chips?

What mechanism would cause the 5 volt DC to sound any different depending on the regulation type (current or voltage)?

 

RE: I love the Ron Coleman current regs, but we are talking about voltage regs, posted on December 10, 2016 at 21:39:44
Triode_Kingdom
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My take on this is that at least some of the sonic characteristics depend on the degree to which the supply isolates common mode signals at the filament from the mains. Supplies with 60Hz transformers are relatively poor in this regard, due to the capacitance between the windings.

As for the output capacitor, it's presence isn't dependent on whether the supply regulates current or voltage. Either type of regulator can use the cap or do without.

 

This is a good explanation from Rod, posted on December 11, 2016 at 08:13:11
Chip647
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I found it interesting

"Present a high impedance between the two ends of the filament. (Rod wasn't entirely sure why this should be, and have never found any reasonable explanation. But if you doubt whether it is true, just try current-driven heating. Then connect 1uF across the filament - all the wonderful sound suddenly disappears!)"

 

RE: This is a good explanation from Rod, posted on December 11, 2016 at 08:54:09
Tre'
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Using the schematic shown on the Bartola Valves website (with one side of the filament connected to the cathode resistor) I can see why all that is said would be true. The filament supply would be in the audio current path.

But I believe all of that changes when a hum pot (or a pair of resistors) is used to create a cathode connection that is at the center point of the filament and the filament supply is left (otherwise) floating.

Using the schematic below, can you describe the audio current path through the last cap of a floating DC filament supply connected to 1 and 4?






Tre'
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RE: This is a good explanation from Rod, posted on December 11, 2016 at 10:02:26
Triode_Kingdom
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"But if you doubt whether it is true, just try current-driven heating. Then connect 1uF across the filament - all the wonderful sound suddenly disappears!"

So says the person who's selling the product. And for the record, that's not an explanation. It's merely an unsubstantiated claim in an industry fraught with lies.

 

RE: This is a good explanation from Rod, posted on December 11, 2016 at 10:42:05
Chip647
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Not in that implementation. The one you posted has 25 ohms of the 925 ohm cathode resistor un-bypassed. The AN schematic bypasses these resistors with capacitors. The regulator capacitor is in series with these bypass caps. Therefore the regulator bypass cap makes a series circuit.

 

I have never heard of anyone implementing a Rod Coleman reg and being unhappy, posted on December 11, 2016 at 10:56:53
Chip647
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The site also has an article authored by Rod. I have used his regs quite a bit. he is a pretty bright guy. Nothing I have seen from him indicates that he is a huckster, just an engineer. I have been trying to figure out why the current reg is much better than the voltage reg. I have in my own tiny little brain aligned on the filament bypass cap as a bad guy.

You win, I am out of this thread.

 

Just to be clear..., posted on December 11, 2016 at 13:50:11
Triode_Kingdom
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My point was that the sonic benefit claimed here is unsubstantiated, and a conflict of interests exists. I did not accuse Mr. Coleman of wrongdoing.


 

Same conceptual problem, still wrong, posted on December 11, 2016 at 14:13:05
Triode_Kingdom
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"The regulator capacitor is in series with these bypass caps. Therefore the regulator bypass cap makes a series circuit."

You're describing the component connections again, but still not defining how or where the differential signal voltage necessary to create current through the cap would originate. The cap is in parallel with the filament. Do you believe that the AC signal voltage is different at the two ends of the filament?





 

RE: Same conceptual problem, still wrong, posted on December 11, 2016 at 16:09:10
Tre'
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Thanks TK.

I'm rethinking my DC supply.

Bridge-critical inductance choke-cap-resistor-cap.

The supply is "floating" with neither + or - grounded but....

The PT and choke are chassis mounted.

If I re-mount them, isolated from chassis, then the audio current flowing between the filament and ground through the stray capacitance of the filament PS will have no where to go.

While I'm at it, I think I'll replace the drop resistor with another choke just for good measure.

Tre'
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"Still Working the Problem"

 

Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 11, 2016 at 16:26:59
Chip647
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If AC is AC, then yes.

 

RE: Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 11, 2016 at 18:08:31
Tre'
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I was working from that perspective as well but, thanks to TK, I now realize that the filament is just a piece of wire and the audio current is not running between one end of it and the other end of it.

The audio current is running between the filament (as a whole) and ground (in these examples through the cathode bypass cap). (and continuing through the last cap of the B+ PS and the OPT and the plate of the tube and across the tube...... back to the filament)

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 11, 2016 at 18:20:02
Chip647
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It has been an interesting conversation. The filament is actually a piece of resistive wire, so it does act like a low value resistor.

From my experience, I have always had better results from a current reg than a voltage reg. If the cap has nothing to do with it, I guess that is fine. It does not make logical sense to me why there would not be some interaction given the ESR of the caps and the fact that we cannot build with perfect components.

 

Filament isolation, posted on December 11, 2016 at 20:32:58
Triode_Kingdom
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"If I re-mount them, isolated from chassis, then the audio current flowing between the filament and ground through the stray capacitance of the filament PS will have no where to go."

Also check the capacitance from primary to secondary. Some transformers perform very well in this respect, others are not so good. The Hammond 10V/4A transformers I initially used in my 211 SETs only measure about 50pF. If not for the residual hum, I would still be using them.


 

You have an active imagination, posted on December 11, 2016 at 20:42:20
Triode_Kingdom
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This discussion has absolutely nothing to do with current vs voltage regulators. You said the parallel capacitor is in the signal path. It isn't. Even after all the questions and responses in this thread, you have no idea why it would be. It's clearly a waste of time to bother you with facts. Really, your belief in this phenomenon has less basis than the earth being flat.



 

A little harsh TK, posted on December 12, 2016 at 05:17:57
Chip647
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Why bother discussing any thing on this site, I have attempted to answer your questions and have been polite.

I'm fully aware of the facts you have been trying to present. What I am trying to understand is the sonic impact of the cap across the filament. You have called me an idiot for implying that this part could have a sonic impact.

And before you try to say that a sonic impact is different than signal path, you are right, I will save you the time to call me an idiot again. I'm an idiot.

 

RE: A little harsh TK, posted on December 12, 2016 at 05:53:44
Triode_Kingdom
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OK, I retract "It's clearly a waste of time to bother you with facts." That wasn't called for. For the rest of it, I won't apologize for things I didn't say. I never commented on the sonic impact of the component, and I certainly didn't call you names. Let's get on with something else.














 

RE: Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 12, 2016 at 09:13:01
dave slagle
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I have to side with chip that the filament is indeed a piece of wire with resistance and therefore any AC signal through it will create a voltage difference due to ohms law. Following ohm up with kirchoff, one simply needs to pick any two nodes and apply ohms law to all of the possible paths between them in order to allocate the current through the various paths and assess the results.

In every case I can clearly see a current path through the filament supply. I also see different approaches giving varying levels of inclusion / exclusion but that just gets us into the judgement call of importance. The original audio note schematic with each side of the filament being bypassed to ground should reduce the signal current through the last cap but it cannot eliminate it completely.

A perfect (and extreme) example your logic is that if all of the grounds of a system are connected together with wire there can be no ground loops.

dave

 

RE: Can you describe the current path through the cap?, posted on December 12, 2016 at 09:43:41
dave slagle
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What is the best way to simulate the ac behavior of a filament? I conceptually view it as a tapped resistor with the AC signal being injected across its surface and in the most simple form a 300B filament can be looked at as a pair of 2 ohm series resistors with an AC signal being injected to their junction. If we accept this and look at it in spice it clearly shows that part of the AC signal current will traverse the filament supply. In the sim below I chose a current source with two different value caps across it to show this behavior and used a "traditional" DC bias scheme where one side of the filament is bypassed to ground.





If we accept the filament as tapped resistance model there can be no debate that AC does indeed traverse the last cap of a filament supply. Now lets look at the case where each side of the filament is referenced to ground through the same value cap.





Yea! we now have a fraction of a fetmoamp of current throught he cap so surely it must be out of the audio path for all but those golden eared marketers but maybe we should amp up the complexity of our filament model a bit and see what happens. Rather than that oh so symmetrical pair of two ohm series resistors, lets get radical and imagine the filament as 4 - 1 ohm series resistors and look at what happens when we look at a part other than the middle.





That cap is suddenly back in the picture and all of the above leaves me wiht two possibilities:

1- the last cap of the filament supply is most certainly in the signal path

2- my method for modeling the filament is fatally flawed.

dave




 

RE: A little harsh TK, posted on December 12, 2016 at 12:03:43
Chip647
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That's cool. I don't want you that retract anything. At some point I would like to understand how or if a filament bypass cap affects the signal. My logic is that if something affects the signal it is in some way in the signal path. I understand that from an engineering standpoint that is not technically a correct way to state that.
Thanks for your patience.

 

RE: Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 12, 2016 at 15:55:07
Tre'
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OK, but what's the amplitude of the voltage difference one end of the filament to the other end of the filament?

And how does a path provided by a cap across that filament affect the signal current flowing on toward ground and eventually back to the plate of the tube and then back to the filament?

And how does a cap across the filament cause there to be signal current flowing through the stray capacitance of any choke or power transformer to ground when there's going to be signal current flowing through those stray capacitances anyway?

And how does one keep signal current from flowing through the stray capacitances of the PS bits?

If there were chokes (not connected to the grounded chassis) in both legs of the supply would that be enough inductance to prevent signal current from flowing through the stray capacitance of the PT to ground?

And is the stray capacitance of the chokes and PT enough to cause audible problems to start with.

At this point all I have is questions, no answers.

I guess I need to, as you say, do the analysis and assess the results.

Thanks.

Tre'
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"Still Working the Problem"

 

RE: Do you believe that the AC signal voltage is different at the two ends of the filament? Yes, posted on December 12, 2016 at 18:34:17
dave slagle
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OK, but what's the amplitude of the voltage difference one end of the filament to the other end of the filament?

I like to look at it from a current loop perspective and I clearly see multiple possible current paths.

And how does a path provided by a cap across that filament affect the signal current flowing on toward ground and eventually back to the plate of the tube and then back to the filament?

Well, that depends if you believe that capacitors can have a "sound"

I didn't bring up any of the stray capacitance stuff so I don't have any comments.

dave

 

RE: Can you describe the current path through the cap?, posted on December 12, 2016 at 19:10:08
Tre'
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" I conceptually view it as a tapped resistor with the AC signal being injected across its surface and in the most simple form a 300B filament can be looked at as a pair of 2 ohm series resistors with an AC signal being injected to their junction. "

I think the AC signal is being injected equally at all points along the length of the filament.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Can you describe the current path through the cap?, posted on December 13, 2016 at 05:09:38
dave slagle
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the fact that the filament has resistance lets you easily figure out the impedance for current to complete its loop. given that the impedance of a typical last cap of a filament supply has a fraction of the impedance of the filament at audio frequencies, how can that filament cap not be a substantial part of the audio current loop?

dave

 

RE: Can you describe the current path through the cap?, posted on December 13, 2016 at 08:20:01
Tre'
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If the AC is being injected at the single point (the source) where the two 2 ohm resistors connect then how would here be a voltage differential across the ends of those resistors?

If the filament didn't need to be heated you could just tie those ends together and you would have one 1 ohm resistor.

For AC, a cap tying them together does just that.

There certainly is no AC voltage differential once the cap is in place.

No voltage differential, no current.

Or are you saying we have currents flowing in two paths across the cap between source and ground as shown below?





If so then the currents would need to be un-equal otherwise the net current across the cap would be zero and, in my mind, take the sonics (audible short comings) of the cap out of the picture.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Can you describe the current path through the cap?, posted on December 13, 2016 at 09:14:39
dave slagle
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look at my post above. I show exactly what you say if the signal all hits the middle point of the filament perfectly and acknowledge that the single exact midpoint reference is probably too simple of a model. Right below that I show a different point along the filament which gives current through the cap.

dave

 

RE: Can you describe the current path through the cap?, posted on December 13, 2016 at 09:43:26
Tre'
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Ok, I think I'm starting to understand.

If we have the above arrangement (and assuming a cap across the filament) can't we adjust the pot so the "symmetrical pair" model fits?

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Can you describe the current path through the cap?, posted on December 13, 2016 at 10:37:22
dave slagle
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Hey.

it doesn't appear so. For any given point across the filament there can be only one setting of the balance pot to prevent current through the cap and since there are infinite points across the filament, each would need its own setting of the "balance pot".

This whole concept is one of the arguments put forth for using a high impedance filament supply (either a CCS or LCL filtering) and it seems plausible to me.


dave

 

Wait a minute..., posted on December 13, 2016 at 13:02:16
Tre'
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Edit, I should preface this sentence with "looking at it your way"

For every 1-3 ohm model there is a 3-1 ohm model accruing at the same time and we're back to equal current.


Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

My Conceptual Model is Very Different, posted on December 13, 2016 at 20:27:35
Triode_Kingdom
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"What is the best way to simulate the ac behavior of a filament? I conceptually view it as a tapped resistor with the AC signal being injected across its surface"

But it is not tapped, nor is the AC signal applied at discrete points. Rather, the filament as a whole is modulated by the attraction of the electron cloud to the anode, an attraction that varies over time when an AC signal voltage is applied between grid and cathode. With this behavior in mind, i.e. a cathode acted upon identically everywhere along its length, with relative uniformity imposed by the dispersion of electrons into the cloud, I see no mechanism whereby one end of the filament might naturally accumulate a greater or lesser AC amplitude than the other. In all cases, I believe no differential signal voltage is created, therefore no differential signal current is encouraged to flow through the parallel capacitor.

Here's an experiment you can try at home...

Wire a 6.3V DHT as shown below. Use two 3V batteries (2-1.5V cells each) for convenience. Measure the current through the batteries. Kirchoff tells us they must be equal, because no current path exists other than the filament and the batteries. Now throw the B+ switch and measure the current through resistor Rk. Next, measure the current through each battery. You will find that the change in current through each battery is identical, proving that the two sides of the filament are impacted in an identical manner by the internal conduction of the tube and are contributing equally to the current into Rk. And, just in case you're tempted to think that the low resistance presented by the batteries across the filament forces this behavior, repeat with the addition of a resistor in series with each connection to the filament. The filament will continue to perform as if it were a single piece of wire with a signal source connected to its midpoint, and there will be no differential voltage or current.













 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 10:36:49
dave slagle
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Once the electrons exit the cloud and enter the conductor of the filament then Ohm and kirchoff are applied and I still see two distinct exit points with a resistance between them.

I fired up a 6A3 with a coleman reg and fed it with a 1Khz signal and was able to see that signal across the filament. I then placed a cap across the filament and the signal went away. I used fixed bias with one side of the filament grounded which should be the same as cathode bias with one sided referenced to ground.

I haven't thought or looked at the original reference of dual cathode resistors and caps from each end of the filament... maybe tomorrow.

This isn't all that different from the stranded wire discussion down the page where the point is made that the parallel stranded conductors are all at the same potential so no current will travel from strand to strand, yet in reality i'm sure it does but have no idea to what degree.

dave
dave

 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 11:12:03
Triode_Kingdom
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"I used fixed bias with one side of the filament grounded..."

That's not what we are discussing.






 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 11:24:13
dave slagle
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FIne... it is your contention that in a perfect world for this particular circuit absolutely 0 audio current will flow through the 470µf cap therefore making it inaudible.

It is my belief that the real world isn't that perfect and indeed there will be some current through the 470µ cap. How much and the audible effect of that current are up for debate. It seems the simplest way to test it would to use a current source and then add the capacitor on a switch and see if an audible change can be perceived.

On a slightly different note, why the two 150 ohm series resistors across the filaments?

dave




 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 12:47:16
Triode_Kingdom
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"It seems the simplest way to test it would to use a current source and then add the capacitor on a switch and see if an audible change can be perceived."

No, the simplest method would be the experiment I just described. Any imbalance created in the filament (as predicted by your conceptual model) can be easily observed and measured this way. If it's easier, use an isolated (ungrounded) SMPS across the filament for heating, and connect each end of the filament to ground with an identical value bias resistor.

I want to point out that this experiment creates a worst-case scenario, insofar as it applies a DC gradient across the filament. It's the gold-standard for defining how the heated filament really functions in the context of this discussion. If you can't measure an imbalance in this experiment, your model fails.

"On a slightly different note, why the two 150 ohm series resistors across the filaments?"

I don't know which resistors or schematic you're referring to.

 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 13:07:22
dave slagle
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If you connect each end of the filament to ground through an identical resistors the DC currents through each cannot be matched.

dave

 

RE: My Conceptual Model is Very Different, posted on December 14, 2016 at 13:51:34
Tre'
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"Once the electrons exit the cloud and enter the conductor of the filament..."

Is this a typo?

The electron flow is from filament to cloud.

Not that it makes any difference.


Tre'
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"Still Working the Problem"

 

Floating Supply, posted on December 14, 2016 at 14:20:26
Triode_Kingdom
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I think this is another instance in which the concept of a floating heater supply isn't being fully understood.

Below is another schematic that shows what I mean. It uses a 12V DC power supply to heat the filament of an 811A. Two equal value dropping resistors are used to provide 6.3V to the filament. The circuit creates voltages on each end of the filament that are perfectly balanced (equal and opposite) with respect to ground. Current through the two cathode resistors is also balanced.

According to your model, the application of B+ to this tube will create at least four measurable effects:

1. The negative end of the filament will be more attracted to the anode than the positive end. Therefore, it will convey a disproportionate increase in voltage to its bias resistor.

2. Anode current will no longer be evenly split between the two bias resistors.

3. Anode current will no longer equal the sum of the two currents in the bias resistors.

4. Differential voltage across the filament will no longer be exactly 6.3V.

It is of course my contention that none of this takes place. No differential voltage is created across the filament as a result of anode (signal) current. Both ends of the filament will always rise or fall equally. If you believe this is not true, I urge you to do the test.








 

RE: Floating Supply, posted on December 14, 2016 at 15:57:05
dave slagle
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In the setup you propose I still see one cathode resistor carrying more current than the other and assuming a "bias" of 10V one resistor will have 13.15V and the other 6.85V which across 200 ohms results in 66 and 34ma.

Can you please outline a test to show the AC behavior?

dave

 

RE: Floating Supply, posted on December 14, 2016 at 20:34:21
Triode_Kingdom
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"In the setup you propose I still see one cathode resistor carrying more current than the other and assuming a "bias" of 10V one resistor will have 13.15V and the other 6.85V which across 200 ohms results in 66 and 34ma."

Yes, you're correct. I was trying to create an analogy to the previous circuit, but without the batteries. I overlooked the fact that it creates a current imbalance in the cathode resistors. However, everything else is the same, and the physical phenomenon I was attempting to demonstrate hasn't changed.

I said:

"No differential voltage is created across the filament as a result of anode (signal) current. Both ends of the filament will always rise or fall equally."

This is the crux of the issue, and it can be proven with the circuit as-is.

 

RE: Floating Supply, posted on December 15, 2016 at 07:04:45
dave slagle
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OK... I see what your point is but I'm having a hard time figuring out how to measure it. A current probe would be great but trying to measure a tiny voltage across a floating series resistor is not an easy task.

back to the battery thing. In this case, given that the currents are out of phase with each other I would expect for 1A of filament current and 100ma of plate current to get roughly 1.050A of current through one battery and 0.950A of current through the other. It should be pretty easy to measure the currents through each with some precision but will the AC behavior follow this?

In my test lashup with unbypassed cathode resistors of 1500 ohms I was able to measure about 1mv of AC difference between the halves and when I do the math for the AC current I am only in the 1ma range so the differences I am looking for are probably in the microamp range. I know that upping the AC voltage swing and adding a load will get the AC currents up so any difference (or lack of) will be easier to see. It would sure be nice to simply have a current probe that measures microamps to clam on the wires to see what is going on but those things are expensive.

dave

 

RE: Floating Supply, posted on December 15, 2016 at 11:31:08
Triode_Kingdom
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"I'm having a hard time figuring out how to measure it."

The concept you're proposing would mean that the two ends of a DC-heated filament will not be elevated by the same amount when anode current flows. That's fundamental to the creation of a differential. If this is true, you only need to put a DVM across the heated filament, then apply anode voltage. Any differential created by the flow of anode current to the filament will be indicated by a change in the meter reading. I'll add that if it does this for DC anode current, it will also do it for AC.

Incidentally, I performed a similar experiment in the early '70s. I don't remember the details, but my position on this matter here in the forum is based on the conclusions I reached at that time. I'll have to eat a lot of crow if you prove me wrong.




 

RE: Floating Supply, posted on December 15, 2016 at 18:16:34
Tre'
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The heating current is not common to the bias current.

There is .01575 amps of heating current flowing across the two 200 ohm resistors in series but that current does not flow to or from ground nor does it flow across the vacuum to the plate.

The heating current only flows through the filament and the 'two resistors in series' which are in parallel to the filament.

There is a total of 100ma of bias current flowing through the two 200 ohm resistors. That current flows from ground to the filament and across the vacuum to the plate.

Each resistor has the same amount of bias current flowing through it and in the same direction, from ground to the filament.

Each resistor has the same amount of heating current flowing through it in one direction (negative to positive of the filament DC supply) and only appears to be flowing in opposite directions when looked at WRT the ground reference.

I don't think just measuring the voltages WRT ground when there are two different voltage sources, one where ground is part of the current path and one where ground is not part of the current path, gives the correct answer.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Floating Supply, posted on December 15, 2016 at 18:44:10
dave slagle
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OK... back to your original test with the two batteries. I used a dual bench supply in place of the batteries and set up the circuit as pictured with the measured results in the image. With the plate current applied the current increases through one resistor and decreases through the other but not in an exact 50-50 division which tells me one side is delivering more plate current than the other. The voltages on either half of the filaments also changed I presume due to the difference in current through each resistor.

I have no idea how this translates to AC behavior (I must have slet through that lecture in art school) so take this as a FWIW.

dave

 

RE: Floating Supply, posted on December 16, 2016 at 07:10:22
dave slagle
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The anode current is not represented in TK's sketch. A second supply of current would need to be added and if that current is indeed symmetrical then there will be no AC current or change in DC current across R4. A few posts down I built this circuit substituting I1 for a 6A3 filament and measured different currents through R3 and R5 where the above circuit gives identical currents.

Ultimately, I think we are trying to suss out how to build an accurate model for a DHT cathode. I have a filamentary DHT symbol for spice but no models to go with it.

dave

 

BTW, posted on December 16, 2016 at 07:28:53
Tre'
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I just added the filament chokes I got from you (30mH, 1.5 amp, 1.5ohms) to the filament end of my LCRC filament supply filter for my DIY 300b SET amps.


Sounds good!

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Floating Supply, posted on December 16, 2016 at 10:16:46
Triode_Kingdom
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Damn, I posted a fairly lengthy reply to this a couple hours ago, but it's not here. I'm sure I reloaded the page and checked for that, but it's just not on the forum. This will have to be a somewhat abbreviated version...

Your experiment seems to indicate that a differential was created across the filament as a result of anode current. If you have time, I'd like to see one small change. First, measuring across R1 and R2 should be accurate, but it might open the door for error. Can you try again by measuring directly across the filament? It might also be useful to confirm the relationship between anode current and filament differential (if such really exists) by applying a fixed anode voltage of maybe +300V, and varying the grid (rather than varying the anode voltage).

I'll probably duplicate the experiment myself this weekend. The only DHTs I have on had are 811s, so I'll need to cobble together a high voltage supply.

Thanks for taking time to do this. I'm hoping we can reach a consensus at some point. I'm pretty sure anode current doesn't create a filament differential, but if I'm wrong, it will mean revisiting several other fundamental concepts.


 

RE: Floating Supply, posted on December 16, 2016 at 19:43:15
dave slagle
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Hey,

I had to break the setup down today to do some real work. I am confident in the measurements across the resistors since I read them several times and the fact that they measured identically without the B+ applied (resistors measured within a few mOhms of each other) makes me trust the results.


I'd be very interested to contiunue this conversation and in the meantime I may try to search out some DHT spice models with a filament to see how that is modeled.

dave

 

RE: Floating Supply, posted on December 16, 2016 at 20:01:08
Triode_Kingdom
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I just finished my own experiment with this. I'll post results at the top of the forum in a new thread.

 

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