current and cathode bias SE

[tubeswell]

In a choke input filter, the same 1.22 derating factor would presumably hold. There may also be significant iron losses from the AC flux component, that would make it run even hotter and need further derating.

But you can't buy a current production choke designed to work in a choke input filter. They're all meant for the usual CLC arrangement. A good choke input filter really needs a swinging choke optimized for the particular application, anyway.

This is a CLC filter (albeit used as a whole-of-supply filter).

Screen current in a SE amp most certainly does change under load. Any non-linearity in the tube curves causes the average current to change under drive (RDH4 calls this "rectification") and the curve of screen current vs. drive is very non-linear, because the operating point goes into the region of high screen current on the negative excursions of plate voltage.

I assume that when plate current dips, screen current goes up, which would keep the tube current 'up', according to the concept that total (tube) current through a pentode is constant. But I did wonder whether, in 'real-world' conditions, the low-plate/high screen current part of the signal cycle of a Class A biased pentode under signal conditions would still overall add up to less current than the hi-plate/low-screen current part (because of the effect of the screen grid resistors and the supply resistor between the screen and plate/OT supply nodes), so that the total tube current demanded under drive would go through the supply in 'pulses' (i.e.: with peaks and troughs of current)? Is this what you are saying (albeit in a more intelligent way?)?

If so, then would not the charging and discharging function of the output capacitor on a CLC filter act to help reduce the effect of such (demanded) current pulses?

(Apologies for the lambbrainededness of my inquiry)

[lowell]

Hey all. Just want to report that my prototype is sounding great. Very classic tweed sort of tone and I've dialed it in for a good squishy overdrive tone. It is a single-ended amp with 1 6V6 into a 5k resistive load. I have currently been using my Peavey Bandit power-amp-in as a power amp. I guess I'm writing this to say that indeed a single ended amp can sound "saggy" and doesn't have to be "hard" like most 5F1 amps tend to sound.

[Wilder Amplification]

Hey all. Just want to report that my prototype is sounding great. Very classic tweed sort of tone and I've dialed it in for a good squishy overdrive tone. It is a single-ended amp with 1 6V6 into a 5k resistive load. I have currently been using my Peavey Bandit power-amp-in as a power amp. I guess I'm writing this to say that indeed a single ended amp can sound "saggy" and doesn't have to be "hard" like most 5F1 amps tend to sound.

But this does not mean that that "squish" you refer to (i.e. "compression) is coming from the power supply itself, which was the whole point of this thread in the first place.

However...I'm sure Steve Kimock has you believing otherwise.

[lowell]

In fact I measured the idle B+ current vs full signal current and there is only a small difference (don't have figure in front of me) so in fact you're mostly right about B+ sag. However more importantly what I'm trying to say is sag/compression need not be achieved by using a push-pull output OR more importantly a tube rectifier!! The amp is doing that "thing" as good as any 5E3 I've built.

[pdf64]

What does the cathode voltage do (idle cf full signal)? Pete.

[lowell]

Vk reduces by about 2-3v from 17v to 13v If I remember correctly. Which I was surprised by because I thought that as a tube is overdriven that Vk typically rises and "cools" the bias.

[MWJB]

For a given resistor value, a lower cathode voltage equates to cooler bias.

[Wilder Amplification]

For a given resistor value, a lower cathode voltage equates to cooler bias.

+1. Just as resistance to water flow creates pressure, resistance to current flow generates a voltage drop across said resistance. That voltage is indicative of the current flowing through the resistance (simple Ohm's Law). If the voltage drop across a fixed resistance is decreasing, this indicates that current flow through the resistance is also decreasing as well.

[kg]

if your cathode current DROPS under full output then it sounds like you're overdriving the output tube to cutoff during part of the cycle.

technically it's no longer running class A.

[MWJB]

Ha ha, true, but it is still 'running' & probably sounds great!

Class distinction is only realy valid under clean operation, all bets are off under overdrive, plus many, many SE guitar amps are biased closer to cut off than saturation even at idle, so are not symmetrically biased/ideal class A anyway...they're just cheap, entry level guitar amps with one power tube, normally cathode biased.

SF Champs buck the trend, but when many folk work out what they are actually dissipating, they go and rebias them cooler anyway.

[lowell]

Yes it's true that it's not true Class A anymore, however I would still call it a Class A amp being that it's single ended and "mostly" clean when the volume is down. Would you not?

[lowell]

For a given resistor value, a lower cathode voltage equates to cooler bias.

I don't understand this. From my understanding, if Vk is reduced then there is more current through the tube because Vgk is less thus allowing more current to flow. What am I missing here? Secondly if this is indeed colder bias, the zener on the cathode trick would not work... right?? It's my understanding that a zener who's value is just above quiescent Vk essentially fixed biases the stage when the cathode voltage RISES as the stage clips and the bias cools.

[MWJB]

Yes, if the value of Vk is reduced (but in your case it is not, it is constant, it is the voltage across it that changes), then current rises. But, if the cathode resistor remains at, let's say, 470ohms - as the voltage accross that 470 resistor drops, it is a sign that current is doing the same thing.

17/470= 36mA

13/470 = 28mA

[kg]

Yes it's true that it's not true Class A anymore, however I would still call it a Class A amp being that it's single ended and "mostly" clean when the volume is down. Would you not?

i'm just throwing that out there as an explanation as to why average current would drop as the tube is driven harder.

it's the only explanation. were the tube NOT cutting off, the avg current would absolutely without question get larger.

[Wilder Amplification]

i'm just throwing that out there as an explanation as to why average current would drop as the tube is driven harder.

it's the only explanation. were the tube NOT cutting off, the avg current would absolutely without question get larger.

Actually, if it were symmetrically biased, average current would stay the same and so would the voltage drop across Rk. However, I 100% agree that if it's biased in favor of the cold side, it will hit cutoff before it hits saturation, in which case average current DOES drop when overdriven, which explains the decreased Vk drop across Rk.

Lowell...as MWJB stated your cathode resistor isn't variable. It's a fixed value. Only the current flow through it is changing. This in turn changes the voltage drop by the same percentage as the change in current through the resistor. This is to state that if you have...say...arbitrarily speaking...a 50% change in current flow through a resistance, you will ALSO have a corresponding 50% change in the voltage drop across the resistor.

If you had say a 270R cathode resistor and 75mA flowing through it, you would have 20.25V across the cathode resistor -

E = I x R where -

E = Electromotive Force in Volts
I = Current in Amps
R = Resistance in Ohm's

0.075 Amps (75mA) x 270 Ohms = 20.25 Volts

Now if current through the same resistor increased to 150mA, you would have -

0.150 Amps (150mA) x 270 Ohms = 40.5 Volts

When current flow increased through the same resistance value, so did the voltage drop across the resistance by the same percentage.

Now if our current flow decreased to 25mA -

0.025 Amps (25mA) x 270 Ohms = 6.75 Volts

And as current flow decreased through the same resistance value, so did the voltage drop across the resistance...again by the same percentage.

Welcome to Ohm's Law.