What are terminal 1 & 4 of 300B tube

[audiousername]

I've split the reply into two parts, so I'm at least reducing the chance of losing it all again...

Originally posted by Kofi Annan
On the other half of the triode, there's 98V at the cathode (according to the schematic) and there's a cathode resistor that is bypassed with a capactor. Is this a bias setting for the other half of the triode? If so, why is the capacitor there?

I initially went into this in great detail, but the concise version will have to do for now.

As you've said before, the plate voltage of the first stage is 90V. There is a 1K gridstopper resistor at the grid of the next stage, but no coupling capacitor, so the stages are DC coupled. Assuming class *1 operation (no grid current), the voltage at the grid is also 90V. Why? Because V=IR, if no current flows, no voltage is developed no matter what R is. From this, we find that the grid voltage wrt (wrt=with respect to) the cathode is -8V. The cathode resistor has been (I assume, haven't checked) sized to provide the required bias.

Now, on to the matter of the capacitor over "there". I'll use an example to quantitatively explain the reason why. (Component sizing and the like is for another day... Anyway, I'm not so fond of maths)

Let's say we have a triode, set up with a 500 ohm cathode resistor (which is not bypassed by a capacitor), and the standing current is 10mA at the idle conditions. We can find that the Ec is -5V.

Aside: Whoops, should have said this in the last post. Eb and Ec are wrt the cathode.

I'll apply a 1V positive pulse to the grid of the valve. One would expect this to change Ec from -5V to -4V. We will see if this is the case. As an easy conversion factor, I'll assume an effective transconductance of 1mA/V, so for our 1V pulse, current increases by 1mA. Ok, so now this 1mA causes a voltage to be developed across the cathode resistor equal to V=IR=1mA x 500 ohm = 0.5V. Uh-Oh.....

This means that at that instant, the grid voltage wrt to ground has changed from 0V to 1V, but the cathode voltage has changed from 5V to 5.5V!! so Ec has only gone from -5V to -4.5V (not -4V as expected).

Since the changes counteract each other, this is a form of NEGATIVE FEEDBACK Heresy! Bring out the pitchforks!

In this case, the feedback reduces gain and distortion, but increases output impedance from that stage.

So, if we don't want this feedback, we bypass the resistor with a suitably large capacitor such that at all frequencies of interest, the R||C (resistor in parallel with the capacitor) appears as a short circuit. Hence, no AC voltage is developed across it, and the feedback is removed. That is why the capacitor is there.

 

[audiousername]

I'll use an example to quantitatively explain the reason why. (Component sizing and the like is for another day... Anyway, I'm not so fond of maths)

Sorry, I meant qualitatively.Had to make another post since the edit time was over.

Anyway, doing a lot of replying to my own posts.... this time zone thing is annoying when trying to communicate with people on the other side of the world...

 

[Kofi Annan]

this time zone thing is annoying when trying to communicate with people on the other side of the world...

I agree. Please either move closer to the UN building or petition the Australian Parliament to adopt the US time schedule. This will make things much easier.

Sorry I didn't respond yesterday. The DIYAudio forms were down (or at least, not working for me).

There is a 1K gridstopper resistor at the grid of the next stage, but no coupling capacitor, so the stages are DC coupled.

OK-- when the two stages of the same triode are connected with a resistor, its considered DC coupling. Got it. In the memory. Not gonna lose it. Here's the stupid question: why would you couple it with a capacitor or by any other means (direct, no passive component, e.g.)?

Assuming class *1 operation (no grid current), the voltage at the grid is also 90V. Why? Because V=IR, if no current flows, no voltage is developed no matter what R is.

I think you covered this later, but shouldn't the grid voltage be zero? I think you're saying that the grid voltage wrt the cathode is 90V (-90V, maybe?), but I'm probably too stupid to get it. That's US government-backed public education for you.

This means that at that instant, the grid voltage wrt to ground has changed from 0V to 1V, but the cathode voltage has changed from 5V to 5.5V!! so Ec has only gone from -5V to -4.5V (not -4V as expected).

For true negative feedback, wouldn't the grid and cathode voltage have to increase and decrease in kind (1V grid, -1V cathode)? Yeah, I know, dumb question.

So, if we don't want this feedback, we bypass the resistor with a suitably large capacitor such that at all frequencies of interest, the R||C (resistor in parallel with the capacitor) appears as a short circuit.

I think I get this one. The capacitor will pass AC but block DC. The AC will go to ground and will never see the resistor and, so, can't have an effect on the feedback of the AC signal to the grid. The DC, however, will be grounded through the resistor.

We still need the resistor though for the proper bias, so its necessary to have it there for the tube topology to work. We just don't want the negative feedback it brings, so we send the AC through the back door with the capacitor.

Hail yeah.

Looks like the power stage is next and its capacitor coupled. I would imagine that this is to clean up the signal and void the DC before it reaches the power tube. The 220K to ground and the 1K to the grid are puzzling.

Hmmm.....

How did we arrive at the value for the coupling cap and what are the resistors doing there? Hey waiter! I didn't order any resistors with my capacitor coupling!

Seriously, you can stop answering questions at any time, but you risk losing out on the Big Money Prize. Should he go for it audience?!?

I'll buy you a Foster's should I ever find myself down under. If you come to the UN building in New York, I could buy you a stab wound or at least a mugging.

Thanks,
Kofi

 

[audiousername]

OK-- when the two stages of the same triode are connected with a resistor, its considered DC coupling. Got it. In the memory. Not gonna lose it. Here's the stupid question: why would you couple it with a capacitor or by any other means (direct, no passive component, e.g.)?

Well, you could probably connect the grid of the 2nd half of the 6SN7 to the plate of the 1st half directly, provided that the layout is such that high frequency garbage isn't picked up. About why you'd use capacitors to couple stages - basically so we can forget about the DC component and just concentrate on what we're interested in for music - AC.

I think you covered this later, but shouldn't the grid voltage be zero? I think you're saying that the grid voltage wrt the cathode is 90V (-90V, maybe?), but I'm probably too stupid to get it. That's US government-backed public education for you.

No, I don't mean that the grid voltage wrt the cathode is 90V.

The grid voltage wrt ground is 90V.

Hmm, Ok... Here goes. The valve doens't see anything apart from the things at its pins. It sees Eb and Ec, and from that a suitable Ib flows. (I hope you're ok with the abbreviations.) So, the valve doesn't know that its cathode is actually sitting at 98V higher than ground. To the valve, the cathode voltage is just what everything else is referenced to. So, when you measure the grid voltage as 90V, and the cathode voltage at 98V, Ec = grid voltage wrt ground - cathode voltage wrt ground= 90 - 98 = -8V. Similarly Eb = plate voltage wrt ground - cathode voltage wrt ground = 285 - 98V = 187V. If you had the 6SN7 curves there you'd be able to estimate the current flowing through the valve (Ib).

For reference
Eb = plate voltage wrt cathode
Ec = grid voltage wrt cathode
Ib = plate current (for class *1 triodes this also equals the cathode current)

For true negative feedback, wouldn't the grid and cathode voltage have to increase and decrease in kind (1V grid, -1V cathode)? Yeah, I know, dumb question.

Hmm, that would be positive feedback. Bad for the stability of the amp. In the example you gave (+1V grid and -1V cathode at the same time), the effective pulse applied to the grid is +2V. Why? Because as the grid increases voltage, the cathode simultaneously decreases, making the pulse applied to the grid appear larger. Hence, the feedback is positive, not negative - negative feedback always reduces gain. For negative feedback, the grid voltage increases, but the cathode voltage simultaneously increases, thus making the pulse applied to the grid appear smaller. So, if we want negative feedback, we yould have the grid change by +1V, and the cathode voltage increase as well (by say, +0.5V). Still confused? Remember that the valve only sees what is at its pins, and refers all voltages with respect to its cathode, and read it again.

I think I get this one. The capacitor will pass AC but block DC. The AC will go to ground and will never see the resistor and, so, can't have an effect on the feedback of the AC signal to the grid. The DC, however, will be grounded through the resistor.

We still need the resistor though for the proper bias, so its necessary to have it there for the tube topology to work. We just don't want the negative feedback it brings, so we send the AC through the back door with the capacitor.

Hail yeah.

Yep, I think you get it too.

Looks like the power stage is next and its capacitor coupled. I would imagine that this is to clean up the signal and void the DC before it reaches the power tube. The 220K to ground and the 1K to the grid are puzzling.

Clean up the signal? Um... not really. But what it does do is block the DC from reaching the grid of the output valve. Why would we want to do this? Why not direct couple like the two halves of the 6SN7 are? Well, some people do. But there are disadvantages.

It says on the site that Joseph Esmillia likes 300Bs at 80mA or so. So taking this 80mA, the 300B is running at around 70V at its cathode (Ec = -70V). If we wanted to DC couple the stages, we move the grid from 0V (ground potential) to 285V (DC potential of the previous stage's plate. To maintain Ec at -70V, the cathode now must be at 285V + 70V = 355V. The 300B is also running at around Eb = 355V, so the plate in this new DC coupled arrangement will be at 355V + 355V = 710V Considering that the cost of the power supply increases approximately proportionate to the square of the voltage, this will be expensive! Not only that, let's look at the cathode resistor. P=VI, so with 355V across it, and 80mA through it, the cathode resistor will be dissipating 355V * 0.08A = 28.4W at idle! (Though you can use this 'free' DC at the cathode of the output valve for other purposes, like usiing it to power your input/driver valves!)

To avoid this, the capacitor is used to block DC, allowing an output from the previous stage at +285V, but a ground potential input for the output stage.

Hmmm.....

How did we arrive at the value for the coupling cap and what are the resistors doing there? Hey waiter! I didn't order any resistors with my capacitor coupling!

Well, actually you did. Not ordering resistors with your capacitor coupling is like ordering dinner without a plate. I suppose you could do it, but it is not common. If we left the resistor out, there would be no voltage reference fot the output valve's grid at DC. It doesn't know whether it should be at 0V, 30V, 300V or 300B (well 'B' is the Russian symbol for volts) - sorry that was a cheap pun. But anyway, we have to refer the grid of the output valve to something, and ground is convenient - especially when we're doing cathode bias. Placing a resistor between the grid and ground pins the grid to ground potential (remember, the grid ideally does not source nor sink any current, and Ohm's law requires current in order to have voltage). So, could we use any old resistor?

Unfortunately, no. Sadly, we do not have ideal grids, and they sometimes do source or sink a little current - even when Ec is negative. So, too large a resistor will cause the grid to drift from ground level (V=IR), hence we will not exceed the maximum value stated in the valve datasheet. Another, perhaps less obvious reason, is the fact that too low a grid resistor decreases the input impedance of the output stage. (Since the grid is essentially an infinite impedance, the input impedance is simplified to the grid resistor). By the way, I didn't define grid resistor. It's that 220K thing between the grid of the output valve and ground. Too low an input impedance results in the gain of the previous stage falling, and generally increases distortion from the previous stage (by decreasing the effective value of the previous stage's plate load resistor). Um... perhaps that was a little complicated.

Ok, so how do we determine the grid resistor value. I basically would use the value recommended in the datasheet or find the closest thing in my junk pile. Now, how do we determine the required value of the coupling capacitor? Basically, we need to choose a low-frequency cutoff. What we need the capacitor for is to pass AC but block DC. But in between (i.e. low frequencies) the capacitor ceases to be an effective short and begins to block a little of our beloved AC. We need to decide when this happens, lets say 5Hz - some designers go lower, some higher. The frequency at which this happens is given by 1/(2*pi*R*C) where R is the grid resistor (in Ohms) and C is the coupling capacitor (in Farads). A little algebra gives us C = 1/(2*pi*R*F) where F is the -3dB point of this low-frequency rolloff.

Seriously, you can stop answering questions at any time, but you risk losing out on the Big Money Prize. Should he go for it audience?!?

I don't mind answering your questions. It's nice to be able to help I think I might have strayed into technobabble a bit here, so if you don't understand what I'm talking about, say which part and I'll try to be clearer.

I'll buy you a Foster's should I ever find myself down under. If you come to the UN building in New York, I could buy you a stab wound or at least a mugging.

LOL... no stab wounds or muggings please. They're available freely where I am

 

[Kofi Annan]

I needed to sleep on this last night. Lots of information here.

voltage at the grid is also 90V. Why? Because V=IR, if no current flows, no voltage is developed no matter what R is.

I think I'm starting to get this now. The thing is, why is there no current running from the plate, through the 1K resistor to the grid? You've probably let out a big sigh here, but unlike others on this board, electronics is very new to me. I am a Secretary General of the United Nations by trade, not an engineer.

So, when you measure the grid voltage as 90V, and the cathode voltage at 98V, Ec = grid voltage wrt ground - cathode voltage wrt ground= 90 - 98 = -8V. Similarly Eb = plate voltage wrt ground - cathode voltage wrt ground = 285 - 98V = 187V. If you had the 6SN7 curves there you'd be able to estimate the current flowing through the valve (Ib).

I get this. I really do. It makes sense that the grid (in a perfect world) will not have any current flowing through it. It simply needs to become more negative / positive wrt the cathode to attract / repel the free electrons.

I printed out the 6SN7(GT) curves and it looks like at Eb = 187V and Ec = -8V, the Ib = about 2.45mA. Sound right?

So, if we want negative feedback, we yould have the grid change by +1V, and the cathode voltage increase as well (by say, +0.5V). Still confused? Remember that the valve only sees what is at its pins, and refers all voltages with respect to its cathode, and read it again.

I read it again. And again. And again. And I think I get it. I need to see the components of the tube in a vacuum (one bad pun deserves another). The grid must see no increase in voltage, so by increasing the cathode voltage while the grid voltage decreases, the gap is only getting wider wrt the cathode voltage. See? Got it. I think.

However-- it seems that the grid WILL see and .5V change, as in your example. Are you saying that the small difference in voltage wrt cathode is close enough for negative feedback?

But what [the coupling capacitor] does do is block the DC from reaching the grid of the output valve.

Got that part, and I think I understand why we need to couple with the cap.

If we left the resistor out, there would be no voltage reference fot the output valve's grid at DC.

Whoa. OK-- the capacitor blocks DC (voltage) and passes AC (signal) to the grid of the 300B. This makes sense now. You've identified the 220K resistor as the "grid resistor", so why is it necessary to have the 1K resistor that's in series with the grid to ground? In this case, shouldn't we say that the grid resistor is 1.2K (or, if you prefer, 1K2)?

The answer probably lies in the reason that the 1K is not in series with the AC from the 6SN7 AND the ground simultaneouly, but again, I don't get why.

A little algebra gives us C = 1/(2*pi*R*F) where F is the -3dB point of this low-frequency rolloff.

Yeah! I LOVE formulas (formulae?)! It looks like the designer likes the idea of the -3dB rolloff as well since the .22uF cap works fine for this rolloff point.

I don't mind answering your questions.

That may change...

Looks like there's AC on the filaments, but I think the 300B is directly heated. I know that this will cause a hum and that the hum can be controlled with the balance pot across the filament pins, but why wouldn't we just rectify the AC and send DC to the filament? I think I have an idea about this one, but I'd like to hear it from you.

I'm betting that the 880R resistor tied to the ground lug of the pot is to bias the 300B and the cap is there to avoid negative feedback.

Couple of general questions also. I know that a no NFB tube design is preferred for SETs, but I'd like to know why. Also, it looks like he's feeding the B+ though the primary of the OPT before it gets to the plate of the 300B. I don't remember seeing this before-- I thought that the final signal went to the primary of the OPT? I may be wrong (likely), but if not, why is he feeding the B+ though the OPT primary?

Next round of questions: POWER STAGE!

Thanks a million,
Kofi

 

[andy2]

Don't you think the cap at the power supply for the output tube 300b is a little small?

I have a VTL amp with 120w ultranlinear/ 50w triode mode that has two hugh capacitors at 3mf (3300uf).

 

[audiousername]

Don't you think the cap at the power supply for the output tube 300b is a little small?

I have a VTL amp with 120w ultranlinear/ 50w triode mode that has two hugh capacitors at 3mf (3300uf).

3300uF is a VERY large amount of capacitance when talking about valve amplifiers, and I think it would kill any sensible thermionic rectifier with the pulse currents generated.

The required capacitance to keep the ripple voltage low is proprortional to the load current, so SS amps, say a Pass Labs type amp drawing 4A continuously need massive amounts of capacitance. In contrast, valves are intrinsically high impedance devices, and draw relatively small load currents (generally in the mA range), albeit at high voltages. So we don't really need that much capacitance.

Even if I could afford 3300uF of 500V capacitors, I wouldn't put them in my amp. The energy stored in a capacitor is equal to 0.5xCxV^2. so 3300uF charged to 400V gives 264 Joules Forgetting to discharge those caps when working on your amp may very well be the last thing you do....

 

[audiousername]

I think I'm starting to get this now. The thing is, why is there no current running from the plate, through the 1K resistor to the grid? You've probably let out a big sigh here, but unlike others on this board, electronics is very new to me. I am a Secretary General of the United Nations by trade, not an engineer.

As long as the grid remains negative wrt the cathode, it repels electrons, thus (ideally) electron current does not flow out of the grid. Conversely, electron current does not flow into the grid because it is not sufficiently hot enough (or suitably coated) to emit electrons into the vacuum within the valve.

Of course all of this is idealised, a small current will probably flow for a variety of reasons, but if this current becomes large enough for us to worry about, you probably have a bad valve.

BTW, I'm not an engineer either

I get this. I really do. It makes sense that the grid (in a perfect world) will not have any current flowing through it. It simply needs to become more negative / positive wrt the cathode to attract / repel the free electrons.

I printed out the 6SN7(GT) curves and it looks like at Eb = 187V and Ec = -8V, the Ib = about 2.45mA. Sound right?

Errgh.... Here is where I get confused You get it, but I don't.

Applying Ohm's law to the voltage given on the schematic gives I=V/R=98/23.5K=4.17mA, but looking at curves I have (actually for the 7N7, but the difference is supposed to be just the type of base), tells me that Ib should be ~3mA. Eeeek! I don't know what's going on. Either I've made a mistake somewhere (quite likely) or the voltages are out a little.

Help! I need some poor soul to check what's going on.

I read it again. And again. And again. And I think I get it. I need to see the components of the tube in a vacuum (one bad pun deserves another). The grid must see no increase in voltage, so by increasing the cathode voltage while the grid voltage decreases, the gap is only getting wider wrt the cathode voltage. See? Got it. I think.

However-- it seems that the grid WILL see and .5V change, as in your example. Are you saying that the small difference in voltage wrt cathode is close enough for negative feedback?

Yep, for zero negative feedback, the grid must see the full input signal. So any change in cathode voltage (no matter how small) which effectively subtracts from the input signal (therefore making it appear smaller) constitutes negative feedback. The negative feedback need not be total.

I get your bad pun, (sorry for that) but here's is a diagram of the components inside the vacuum anyway which I think is pretty neat

An externally hosted image should be here but it was not working when we last tested it.

(From http://www.r-type.org/static/basics.htm)

Not complicated things built on silicon which are difficult to picture, valves are basically little light bulbs with other bits of wire and metal in them

Whoa. OK-- the capacitor blocks DC (voltage) and passes AC (signal) to the grid of the 300B. This makes sense now. You've identified the 220K resistor as the "grid resistor", so why is it necessary to have the 1K resistor that's in series with the grid to ground? In this case, shouldn't we say that the grid resistor is 1.2K (or, if you prefer, 1K2)?

The answer probably lies in the reason that the 1K is not in series with the AC from the 6SN7 AND the ground simultaneouly, but again, I don't get why.

The grid resistor is what defines the DC potential of the grid. It need not necessarily go to ground (though this is common). The 1K resistor you're talking about is a gridstopper resistor, placed there in the hope of not allowing the valve to pick up RF rubbish. (The leads within the amplifier can act as little antennae, and well valves were used to detect radio waves for many years. 1950 was a bad year for valves worldwide.) You could probably ignore the 1K resistor if your layout is good enough. In terms of the audio-frequency operation of the circuit, its effect is usually negligible - i.e. it's nothing to worry about, pretend it's not there for now.

Looks like there's AC on the filaments, but I think the 300B is directly heated. I know that this will cause a hum and that the hum can be controlled with the balance pot across the filament pins, but why wouldn't we just rectify the AC and send DC to the filament? I think I have an idea about this one, but I'd like to hear it from you.

You could. Some people do. If I could afford a 300B (or wasn't so much of a fiscal retentive) I probably would use DC. This is 2005, not 1925. Hum is not acceptable. Many people don't for subjective reasons - "AC sounds better", and do the best they can to eliminate hum while using AC.

I'm betting that the 880R resistor tied to the ground lug of the pot is to bias the 300B and the cap is there to avoid negative feedback.

Yep. You're learning.

Couple of general questions also. I know that a no NFB tube design is preferred for SETs, but I'd like to know why.

Because no-NFB designs are simpler and easier to design and thus more difficult to stuff-up. Other reasons too (clipping behaviour, distortion characteristics, fashionable...) SET isn't really my cup of tea.

Also, it looks like he's feeding the B+ though the primary of the OPT before it gets to the plate of the 300B. I don't remember seeing this before--

Placing the OPT primary (the 'load') between B+ and the output valve's anode is the most common way to connect a single-ended output stage, and is called series feed.

I thought that the final signal went to the primary of the OPT? I may be wrong (likely), but if not, why is he feeding the B+ though the OPT primary?

The final signal is developed across the OPT primary (it must be for sound to come out of your speakers, good or bad ) In this case, the signal is developed between the 300B's anode and B+.

 

[Kofi Annan]

As long as the grid remains negative wrt the cathode, it repels electrons, thus (ideally) electron current does not flow out of the grid. Conversely, electron current does not flow into the grid because it is not sufficiently hot enough (or suitably coated) to emit electrons into the vacuum within the valve.

I think I get this. I really do.

BTW, I'm not an engineer either

I don't know if this is encouraging or discouraging.

Errgh.... Here is where I get confused You get it, but I don't.

No, no, no... I just looked at the plate curves. I don't get it... I'm just repeating what the picture told me. Maybe it has something to do with the intangibles, like unwanted capacitance? Just a guess. A wild, stupid guess.

The negative feedback need not be total.

Meaning that the increase in grid voltage need not EXACTLY match the increase in cathode voltage to have negative feedback, right?

here's is a diagram of the components inside the vacuum anyway which I think is pretty neat

You're right. This is pretty neat. I remember when girls, puck rock and sniffing glue was pretty neat. I'm getting old.

The 1K resistor you're talking about is a gridstopper resistor, placed there in the hope of not allowing the valve to pick up RF rubbish.

Yeah! Got this one! Got it. Any rationale for selecting the 1K value?

You could [use DC on the 300B filaments]. Some people do. If I could afford a 300B (or wasn't so much of a fiscal retentive) I probably would use DC. This is 2005, not 1925. Hum is not acceptable. Many people don't for subjective reasons - "AC sounds better", and do the best they can to eliminate hum while using AC.

I've heard that AC will allow an even distribution of the tube detritus (dust, etc.), while DC will spread the junk sealed inside the tube to a central location, shortening the life of the tube. I don't know if I explained this well, but it sounds like voodoo to me. Nothing against voodoo, you understand. Screamin' Jay Hawkins was a Voodoo Wildman and I love him like a crazy uncle.

Yep. You're learning.

Yeah! Take that, Dad! I can learn someting!

I think I've got a basic understanding of the circuit now. There's obviously some other questions that have yet to form, but I feel 1.2 billion percent more comfortable than I have before.

Now the power stage:

I've read that tubes are not choosy when it comes to an exact measure for the B+, but let's talk PSU filtering.

There's a rectifier tube and some electrolytics and chokes after the power supply. I know what the rectifier tube does (conversion of AC into DC), but its unclear to me how the AC flows into the required DC. If there's 380VAC on the plate and cathode of the GA37 rectifier, and the filament winding of the transformer provides the 5V 3A for the rectifier heater, then does the raw DC come back out of the filaments?

Also, how did JE determine the proper proper filtration for the DC? It looks like a CLC for the B+ and a CRC for the preamp plates. How do we determine what's best for PSU filtering? I mean, I can use the Duncan Amps PSU designer, if necessary, but why are the components where they are? That is, how do we determine the value for the choke, the caps, resistors, etc.?

The question mark key on my keyboard is almost worn out. Thanks again for being such a sport. I don't want to spoil the surprise, but next month, I am recommending that Australian Rules Football be the International Sport of Peace and Unity to the UN Governing Council.

Please keep that under your hat.

Seriously, thanks so much for putting up with all my asenine questions.

Kofi

 

[audiousername]

I don't know if this is encouraging or discouraging.

Neither do I.

I learnt most of what I know from reading. Yes, some electronics was part of the physics course at high school but that covered only the simple form of Ohm's Law. I was a little bit of a physics nerd at high school (ok, maybe a bit more than a little ), got interested in this stuff and tried to read what I could.

If you're interested in some other resources, the TubeCad Journal is an excellent online resource. But perhaps the best I've come across for the basics, and not so basic, is Morgan Jones' Valve Amplifiers (a book, and a large one at that). Sadly I recently had to return that to the library (again). I will buy a copy one day...

The Radiotron Designer's Handbook (by Fritz Langford-Smith, an Australian ) is probably the most respected text around on valve circuits, and is now available online (i.e. for free), though it can get a bit complicated.

No, no, no... I just looked at the plate curves. I don't get it... I'm just repeating what the picture told me. Maybe it has something to do with the intangibles, like unwanted capacitance? Just a guess. A wild, stupid guess.

Stray capacitances and the like do not affect the DC operation of the circuit, only AC (since all capacitors are ideally an infinite impedance to DC). So no.

I still do not know why the voltages are different. I need help

Meaning that the increase in grid voltage need not EXACTLY match the increase in cathode voltage to have negative feedback, right?

Yep.

I might say this right now, before things get carried away. This is just one form of negative feedback - there are many other ways of feeding back an inverted phase output to the input (which is the essential part of the negative feedback mechanism)

You're right. This is pretty neat. I remember when girls, puck rock and sniffing glue was pretty neat. I'm getting old.

I'm feeling old too. Did some long overdue cleaning out of my store room recently and found stuff from ten years ago. It seemed like yesterday when I put it in there. Old shopping bags with store names which closed years ago. This stuff is disturbing....

Yeah! Got this one! Got it. Any rationale for selecting the 1K value?

Umm... Optimal gridstopper resistor values are invariably determined by experiment. There are many vairables involved and basically a resistor which is about right is put in, and its effect measured. For those who don't have 'scopes, or the knowledge to use them, either forget about the gridstoppers entirely (and hope for no oscillation, or break-in from taxi firms), or make the gridstoppers so large that no RF rubbish can get in.

The rationale between gristoppers is that they form a low-pass filter between the grid-stop resistance and the Miller-amplified Cag (anode-grid capacitance) of the valve. In addition, the resistance provides damping to resonant stuff (the technical term ) that might be going on. I think that's how they work anyway...

I've heard that AC will allow an even distribution of the tube detritus (dust, etc.), while DC will spread the junk sealed inside the tube to a central location, shortening the life of the tube. I don't know if I explained this well, but it sounds like voodoo to me. Nothing against voodoo, you understand. Screamin' Jay Hawkins was a Voodoo Wildman and I love him like a crazy uncle.

There isn't supposed to be any detritus inside the vacuum. It's supposed to be an... er... vacuum? It sounds like voodoo to me as well. Ok there will probably be some rubbish that isn't supposed to be in there but there shouldn't be so much that we have to consider it. Actually on second thoughts, maybe this refers to lingering gas ions floating around somehow. I dunno...

I think what has been argued before is that DC heating means that the part of the filament that is more positive will tend to emit less electrons than the part which is more negative (since there is a greater voltage between the more negative part of the filament and the anode, compared with the more positive part and the anode.) Thus, it may be argued that part of the filament will be worn out prematurely, shortening life. AC heating eliminates this because the positive and negative parts swap 50 times a second in this part of the world (60 times a second in the US). I am yet to be convinced that this is of significance.

Yeah! Take that, Dad! I can learn someting!

I think I've got a basic understanding of the circuit now. There's obviously some other questions that have yet to form, but I feel 1.2 billion percent more comfortable than I have before.

That's great

Now the power stage:

I've read that tubes are not choosy when it comes to an exact measure for the B+, but let's talk PSU filtering.

Umm... for triodes, the current through the valve is not only dependent on Ec (also known as Vgk if there isn't enough terminology already), but also on Eb.

Mathematically, Ib is proportional to Ec^(3/2), but is also proportional to Eb^(3/2). So, increasing or decreasing B+ too much will also affect the current through the valve and other 'stuff'. For pentodes, Ib is approximately independent on Eb, but is proportional to Ec2^(3/2). It seems like we're not up to pentodes yet, so I'm not quite sure why I said that. It must be late...

There's a rectifier tube and some electrolytics and chokes after the power supply. I know what the rectifier tube does (conversion of AC into DC), but its unclear to me how the AC flows into the required DC. If there's 380VAC on the plate and cathode of the GA37 rectifier, and the filament winding of the transformer provides the 5V 3A for the rectifier heater, then does the raw DC come back out of the filaments?

The rectifier is a GZ37, not that this is particularly important for now.

Yes, the unfiltered B+ (which is DC with AC hum voltages superimposed on it) comes out of the filament. Being pedantic, the GZ37 is actually an indirectly heated rectifier with the heater internally connected to the cathode, so the unfiltered B+ actually comes out of the cathode and heater (there being no filament). That was an essentially meaningless sentence.

Also, how did JE determine the proper proper filtration for the DC? It looks like a CLC for the B+ and a CRC for the preamp plates. How do we determine what's best for PSU filtering? I mean, I can use the Duncan Amps PSU designer, if necessary, but why are the components where they are? That is, how do we determine the value for the choke, the caps, resistors, etc.?

We determine the value for the chokes and capacitors by looking at our junk box hoping there is something useful in there. Invariably, the junk box is full of the bad kind of junk and we check out the website of our favourite dealer in all things thermionic and magnetic to find what fits our budget. Then we use PSUDII to determine whether this produces acceptably low hum. If it doesn't the pocket is in for another beating.

What we look for in chokes is generally current rating (insulation strength, i.e. maximum DC voltage allowable beween the windings and the generally earthed core/frame will generally not be a problem with sane amplifiers). Capacitors need to be suitably voltage rated, and with adequate ripple current rating. Unfortunately maximum ripple current is not usually specified, so we must ignore it.

Seriously, there are rigourous ways to find out what you require, but then you'll just have to go with what is available. So it's easier just to work backwards from what you can get your hands on.

The question mark key on my keyboard is almost worn out. Thanks again for being such a sport. I don't want to spoil the surprise, but next month, I am recommending that Australian Rules Football be the International Sport of Peace and Unity to the UN Governing Council.

Please keep that under your hat.

Seriously, thanks so much for putting up with all my asenine questions.

No worries.

 

[andy2]

With respect to AC vs. DC heating of the filament, I had some background in working with IC design at Intel which has a similar issue with ac and dc current affecting the longevity of the metal conductor.

In IC design and physical layout, we had different rules regarding the size of the metal that carries either dc or ac current. It has to do with electron migration in one dirrection (DC) or in equal but opposite migration (AC). If electron migrates in one direction, it creates more "wear and tear" on the metal, as opposed to both direction which sort of even out the "wear and tear" but you know what I am trying to say.

So at Intel, the layout rules say that the width of the conductor that carries DC current has to be 3 or 4 times that of the metal that carries AC current to combat metal deterioration due to electron migration.

I don't know if this applies directly to tube filament but DC current does affect the life time of a metal.

 

[Bas Horneman]

I don't know if this applies directly to tube filament but DC current does affect the life time of a metal.

I've read that it does. Maybe not in that ratio...but DC does shorten the life of the filament.

But can be countered by reversing the polarity on your filament every 6 months or so.

 

[fdegrove]

Hi,

But can be countered by reversing the polarity on your filament every 6 months or so.

You could do that...
Don't know of anyone who actually does that but, yes....

The phenomenon that's being described as electron migration (that's what it is) is also known as "notching".

As far as longevity goes, the first element of a tube to die (actually it becomes exhausted) is the getter.

At then of the day I wouldn't worry too much about either possibilities, sometimes you just need to replace a dead tube and that's the end of that.

Cheers,

 

[Kofi Annan]

I learnt most of what I know from reading.

I learnt most of what I know from an Australian.

I was a little bit of a physics nerd at high school

A nerd??? On the DIYAudio boards???

If you're interested in some other resources, the TubeCad Journal is an excellent online resource. But perhaps the best I've come across for the basics, and not so basic, is Morgan Jones' Valve Amplifiers (a book, and a large one at that). Sadly I recently had to return that to the library (again). I will buy a copy one day...

I've been on TubeCad Journal quite a bit. The article on the grounded cathode amplifier is currently at my bedside where its been read and read again. Understanding it better with each read.

I also took a look at the Morgan Jones book at Amazon and it looks right up my alley. I'll be bagging that one soon.

Mathematically, Ib is proportional to Ec^(3/2), but is also proportional to Eb^(3/2). So, increasing or decreasing B+ too much will also affect the current through the valve and other 'stuff'.

Gotcha. I guess what I mean is that a tolerable range in PSU voltage for a solid state / op amp device is smaller than that of your basic tube. Anyway, that's the way it seems.

For pentodes, Ib is approximately independent on Eb, but is proportional to Ec2^(3/2). It seems like we're not up to pentodes yet, so I'm not quite sure why I said that. It must be late...

Ahhh... pentodes... You probably made a mistake by bringing this up.

Don't know much about these, but from what I've read, I can expect more distortion with these over triodes. I know that theres additional grids to stop the flow of stray electrons in the tube. I also know that they can be "triode strapped" but that's about where my knowledge ends. Actually, my knowledge ends significantly before then, I was just trying to sound impressive.

Bearing in mind that I will be ordering the Morgan Jones book relatively soon (first, gotta get a paper route for some extra cash) and you can refer me there, can you give me the basic idea as to why we might choose a pentode over a triode for a preamp tube?

Being pedantic, the GZ37 is actually an indirectly heated rectifier with the heater internally connected to the cathode, so the unfiltered B+ actually comes out of the cathode and heater (there being no filament). That was an essentially meaningless sentence.

It has plenty of meaning for me. So, when wiring the GZ37 for the 300B amp (I typed GA37 last night... tired fingers), you'd solder one wire from the 5V 3A winding to a fliament and the other to the filament and RCL network?

We determine the value for the chokes and capacitors by looking at our junk box hoping there is something useful in there.

I can deal with that. But we are also using the resistors to drop voltage to meet the requirements of the recipient tubes, which should be as easy as V=IR, right? Also, I think the PSU designer does a nice job with this. Anything to make it a bit easier.

Thanks again for all your help.

But can be countered by reversing the polarity on your filament every 6 months or so.

I'd just call one of the DHT Filament Polarity Reverser guys in the phone book. They don't work on Fridays, though-- got a good union. "Mr. Polarity" is my favorite-- "Turn Your Negative Into A Positive! Swappin' Filament DC since 1932!"

sometimes you just need to replace a dead tube and that's the end of that

That's what I'm talkin' about.

Thanks again.

Kofi

 

[audiousername]

A nerd??? On the DIYAudio boards???

LOL

I also took a look at the Morgan Jones book at Amazon and it looks right up my alley. I'll be bagging that one soon.

Yep, I think it would be great for you. Morgan also has a new book out, titled Building Valve Amplifiers. Unfortunately I couldn't borrow that from a library, so I had to buy it. When you finally reach the stage of building something, it is an excellent resource and is likely to save pulling your hair out looking for the source of that annoying hum! I think quite a few people who frequent these parts have those books, so if you have questions, at least we'll know what you're on about. Who knows, the author might even answer your question....

Gotcha. I guess what I mean is that a tolerable range in PSU voltage for a solid state / op amp device is smaller than that of your basic tube. Anyway, that's the way it seems.

Well, a 5V drop is not so significant when you have a 400V supply, but is very significant when you're on a 15V supply!

Ahhh... pentodes... You probably made a mistake by bringing this up.

Don't know much about these, but from what I've read, I can expect more distortion with these over triodes. I know that theres additional grids to stop the flow of stray electrons in the tube. I also know that they can be "triode strapped" but that's about where my knowledge ends. Actually, my knowledge ends significantly before then, I was just trying to sound impressive.

Bearing in mind that I will be ordering the Morgan Jones book relatively soon (first, gotta get a paper route for some extra cash) and you can refer me there, can you give me the basic idea as to why we might choose a pentode over a triode for a preamp tube?

I think I made a mistake bringing that up as well. A pentode has five elements (excluding the heater): the cathode, the control grid (g1), the screen grid (g2), the suppressor grid (g3) and the anode. Pentodes are notciably noisier than triodes, and their high plate resistance means that most of the power supply hum is coupled to the output. I would stay away from pentodes (in pentode mode) for most preamp applications. Of course you can triode-strap them (connect g2 to plate via small resistor), then you just basically have a weird triode (with three grids).

It has plenty of meaning for me. So, when wiring the GZ37 for the 300B amp (I typed GA37 last night... tired fingers), you'd solder one wire from the 5V 3A winding to a fliament and the other to the filament and RCL network?

No, you'd wire 5V across the heater pins (pins 2 and 8), and connect pin 8 to the input of your power supply filter (i.e. the caps, chokes and stuff).

I can deal with that. But we are also using the resistors to drop voltage to meet the requirements of the recipient tubes, which should be as easy as V=IR, right? Also, I think the PSU designer does a nice job with this. Anything to make it a bit easier.

Yep, V=IR, the great truth. You'll be using PSUD for the rest of the power supply, so just add that extra RC leg to your simulation.

Thanks again for all your help.

No worries

I'd just call one of the DHT Filament Polarity Reverser guys in the phone book. They don't work on Fridays, though-- got a good union. "Mr. Polarity" is my favorite-- "Turn Your Negative Into A Positive! Swappin' Filament DC since 1932!"

I'm not sure if they had the technology to rectify and smooth (low voltage/high current) heater supplies back in 1932, and if they did it would have probably been prohibitively expensive. Lead-acid accumulators would have been used for DC, and changing polarity would have been as easy as turning the battery around!

 

[audiousername]

At then of the day I wouldn't worry too much about either possibilities, sometimes you just need to replace a dead tube and that's the end of that.

Yep, all valves die eventually - usually power output ones first, which is a pain, because they're more expensive than the small signal ones. Reality is cruel sometimes...

Of course this is made all the less painful when you use cheap types that nobody else cares about

 

[audiousername]

In IC design and physical layout, we had different rules regarding the size of the metal that carries either dc or ac current. It has to do with electron migration in one dirrection (DC) or in equal but opposite migration (AC). If electron migrates in one direction, it creates more "wear and tear" on the metal, as opposed to both direction which sort of even out the "wear and tear" but you know what I am trying to say.

So at Intel, the layout rules say that the width of the conductor that carries DC current has to be 3 or 4 times that of the metal that carries AC current to combat metal deterioration due to electron migration.

I don't know if this applies directly to tube filament but DC current does affect the life time of a metal.

I'm not sure how significant this is when talking about valves, because I would imagine the wire which the heater is composed of would be significantly thicker than traces on an IC. No idea how big the current-carrying conductors are on an IC but I picked up a valve and checked the thickness of a 6.3V 300mA heater wire (the wire alone, not including the white insulation). It's easily visible to the naked eye and probably about the same thickness as hair, or maybe a bit thinner.

 

[fdegrove]

Hi,

and changing polarity would have been as easy as turning the battery around!

Errrrrr....That's actually the hard way...In casu the heavy way to go about it...

What surprises me a little is that no one has mentioned so far that there's an audible difference whether you connect the bias resistor of a DHT such as the 300B to either pin#1 or #4....
IOW it changes depending to which pin you return heater/cathode to ground.

Fellow member Guid Tent brought this to the attention here on this forum for the first time a few months ago while presenting a formidable DC supply for DHTs....

No idea if there's a similar preference with IDHT(ubes), never tried it.

Cheers,

 

[audiousername]

Hi Frank,

Errrrrr....That's actually the hard way...In casu the heavy way to go about it...

LOL I didn't think if that. But maybe after you've gone down to the local radio shop to recharge your A cells, you'd turn them around?

What surprises me a little is that no one has mentioned so far that there's an audible difference whether you connect the bias resistor of a DHT such as the 300B to either pin#1 or #4....
IOW it changes depending to which pin you return heater/cathode to ground.

Fellow member Guid Tent brought this to the attention here on this forum for the first time a few months ago while presenting a formidable DC supply for DHTs....

This is the first I've heard of that. Do you mean when using DC heating?

Actually, I can't test this out myself anyway. I don't own any 300Bs.

No idea if there's a similar preference with IDHT(ubes), never tried it.

I would doubt it, as the heater is much less involved in the signal-carrying than the filament is in a DHT, but you never know.

 

[fdegrove]

Hi,

Did a little search and I think I just found Guido's thread regarding the DHT heaters:

DHT DC HEATING

I would doubt it, as the heater is much less involved in the signal-carrying than the filament is in a DHT, but you never know.

True.
You may be surprised by the amount of signal you'd find there however...

Cheers,