Average Screen Current

[HotBluePlates]

I posted the following to another forum, and thought it might be useful to some folks here.  The goal is to answer, "What is the average screen current," so we can answer questions like, "What is my screen dissipation," and, "Do I need to change my screen resistor."  As always, "copy a known-good plan" still works but this is for folks who want to know why the known-good plan worked.


"Optimum" screen grid resistor for maximum power output is 0Ω. It's why many 1950s amps like this, and this, and this, and this had no screen resistors.

The screen voltage dictates the peak plate current the tube can manage, which is the point of the graph below, found on the top of Page 6 of the 6L6GC data sheet. When screen volts fall, the plate current the tube can manage also falls.​
​
​

​

​
The screen draws some current at idle, but much more when the tube is pushed momentarily (aka "dynamically") to higher plate current. That is the point of the dashed Screen Current Curves shown below and found on the bottom half of Page 7 of that same 6L6GC data sheet.​
​

​


The screen current curves show that when we drive to Control Grid (G1) to 0v to hit peak plate current, screen current rises from about 3.5mA (idle) to 40mA (peak plate current).​
​
When amp manufacturers started adding screen resistors to amps with 6L6s, 470Ω was a typical value. We see that here & here as an example.​
​
Those amps show around 440v on the screen. 3.5mA x 440v = 1.54 watts, unconditionally safe for 6L6GC. 40mA x 440v = 17.6 watts, too hot for the tube's 5 watt screen dissipation rating (page 1 of the 6L6GC data sheet linked earlier).​
​
The 470Ω resistor causes the screen voltage to drop: 470Ω x 40mA = ~19v. Alone, that's not enough to prevent the screen from over-dissipating. However that doesn't matter.​

[HotBluePlates]

What matters is the average screen current over the whole signal cycle.

     Half of the signal cycle, the screen current is at the idle value (or less) because that side of the Class AB push-pull stage is shut off. As a worst-case, let's assume it doesn't fall to zero and stays at 3.5mA.​
​
     To average the other points in time, we need to consider the grid-driving voltage of a Sine Wave, and the equal-angle changes of G1 voltage. Then we look at the curves provided on the data sheet to read off G2 (screen grid) current.​
​
     We will consider the screen current at 0º, 45º, 90º, 135º, 180º, 225º, 270º, 315º, and 360º. These currents will be summed, then divided by 9 to determine the average screen current.​
​
     The G1 voltage at Sin 45º and Sin 135º is 70.71% of the difference from -45v to -0v: -13.2v (Green Dot on the graph)​
​
     We assume the plate & screen voltage is steady at 400v, and the idle bias is -45v (a little hotter than what Fender shows on their schematics of a bit over -50v).​
​
     A loadline of ~3.6kΩ has been plotted as an example delivering almost 50w output, because we need this to determine the plate voltage for each point in the signal cycle. A Dashed Red line has been extended from the Plate Loadline to the Screen Curve for that G1 Voltage, at that Plate Voltage.​
​

After you run through all the steps, the average screen current is 11.33mA, so the average screen dissipation is 11.33mA x 400v = 4.985 watts. Perfectly within the screen dissipation rating.​
​
     Note that the real average screen current will be lower because some portion of the cycle between 180º and 360º will have 0mA of screen current. I simply didn't want to deal with the hassle of trying to estimate when that happens.​
​
     The Fender amps with 400+ volts also have a 470Ω screen resistor, which will shave screen dissipation some more.​

[shooter]

G1 voltage at Sin 45º and Sin 135º is 70.71% of the difference from -45v to -0v

talk like this kept me up late in my youth, probably nightmares now 

[DuaneOh]

This is very helpful as it confirms what I thought. I'm playing with an EL34 Class AB cathode biased design and have been using LTspice to look at loadlines, here is what I've got:

Va to K=342v
Vg2 to K=328v
Ia quiescent = 73ma
Ig2 quiescent = 13.5ma
r to r = 4k
Vg1 @ 0v = Ia @ 291ma
Vg1 @ 0v = Ig2 @ 83ma

I calculated average full power Ia as 291ma x .707 = 206ma
Ia @ 206ma = Vg1 @ -8.5v
I followed a line down from this point to Ig2 @ -8.5v = 38ma
So, 328v (Vg2) x 38ma = 12.5 watts     But I've got a 1k g2 resistor:
328v - 1k x 38ma = 290v x 38ma = 11 watts

EL34 data says Wg2 (Limiting) = 8 watts   Is this pushing it? Did I calculate this wrong?

I know that it will be somewhat less due to Vg2 being 290v at full power, plus power supply sag and that the EL34 will be in cutoff for a portion of the cycle. But as it gets pushed further into clipping the average Ia and Ig2 will rise.

Thanks for your help.

[DuaneOh]

I figured that the EL34 will be in cutoff for about 25% of the time, so 11 watts x .75 = 8.25 watts?

[HotBluePlates]

G1 voltage at Sin 45º and Sin 135º is 70.71% of the difference from -45v to -0v

talk like this kept me up late in my youth, probably nightmares now 

 

So the idea is we can calculate & know the behavior of the output tube when something regular & predictable (like a Sine wave) is applied as input to the tube.

Now while a sine wave voltage is applied to the tube grid, screen current is not a sine wave.  See below:

We must divide the G1 drive signal into equal-parts, and the easiest way to do that is in degrees of the signal cycle.  Consider that 0º, 180º and 360º are the same as what we have at idle.  90º is the positive-peak of the input signal, while 270º is the negative-peak of the input signal.

It would be nice if 45º that lies halfway between 0º and 90º were an input signal of 50% of the peak input voltage.  But it's not, because Sin 45º = 0.70711 or "70.71%" of the peak voltage at 90º.

This is very helpful as it confirms what I thought. I'm playing with an EL34 Class AB cathode biased design and have been using LTspice to look at loadlines, here is what I've got:

Ia quiescent = 73ma
Ig2 quiescent = 13.5ma
...
Vg1 @ 0v = Ia @ 291ma
Vg1 @ 0v = Ig2 @ 83ma

I calculated average full power Ia as 291ma x .707 = 206ma
Ia @ 206ma = Vg1 @ -8.5v
I followed a line down from this point to Ig2 @ -8.5v = 38ma
So, 328v (Vg2) x 38ma = 12.5 watts     But I've got a 1k g2 resistor:
328v - 1k x 38ma = 290v x 38ma = 11 watts

EL34 data says Wg2 (Limiting) = 8 watts   Is this pushing it? Did I calculate this wrong?

IMO, yes you calculated wrong.  If I used the same steps, my example above would have an average screen current of 20.5mA instead of the 11.33mA that is the correct answer.  IOW, your steps probably result in a too-large estimate of average screen current.

     - In my example, "70.71%" was for G1 grid voltage.  70.71% of peak plate current of 330mA is ~233mA, but 70.71% of G1 volts happens at a plate current of ~205mA (and somewhat lower screen current).

Looking at the Green Screen Current curve above, we see it rises from the idle amount to some peak during the positive-going part of the input signal.  The steps you used assumes screen current peaks again during the negative-going part of the signal cycle, when it is actually falling towards 0mA.  Instead, try the following:

     - Make a note of your idle bias voltage. (~29v based on your post?)

     - Draw the loadline.

     - Note the change from Idle Grid Bias Voltage to 0v in grid volts, and calculate the grid voltage 70.71% of the way towards 0v.  (If your case was "idle bias at 29v" then 29v x 0.7071 = 20.52v, and 29v - 20.52v = 8.48v)

     - Using the curves of of plate current (heavy lines) and screen current (dashed lines), make a note of screen current at 0º, 45º, 90º, 135º, 180º, 225º, 270º, 315º, 360º of G1 voltage.  If you look closely, some of these points are duplicates of others.

     - There are 9 total points where you annotated screen current.  Add up all those currents, then divide by 9.  The result is the "average screen current."

     - You are free to add more places to note screen current, and you will get a more accurate average-value.  Just remember that they must be equally-spaced over the entire signal cycle.

[DuaneOh]

So the idea is we can calculate & know the behavior of the output tube when something regular & predictable (like a Sine wave) is applied as input to the tube.

Now while a sine wave voltage is applied to the tube grid, screen current is not a sine wave.  See below:


We must divide the G1 drive signal into equal-parts, and the easiest way to do that is in degrees of the signal cycle.  Consider that 0º, 180º and 360º are the same as what we have at idle.  90º is the positive-peak of the input signal, while 270º is the negative-peak of the input signal.

It would be nice if 45º that lies halfway between 0º and 90º were an input signal of 50% of the peak input voltage.  But it's not, because Sin 45º = 0.70711 or "70.71%" of the peak voltage at 90º.

Ok, I think I got it. I've been trying to figure out if I can get away with a smaller g2 stopper so that if I use 6L6's, KT66's, etc., instead of EL34's I won't get as much screen compression. (worth it?) I did a spreadsheet with the values I got from the loadline.

The screen supply voltage is 342v but with a 1k g2 stopper at 13.5ma quiescent I used 328v for Vg2.



I hope that this looks correct enough.
Then I decided to add Rg2 as a variable to see how it affected the Vg2 and Wg2. Ig2 will also change so I am cheating a bit. This is with a 1k g2 resistor.



I changed Rg2 to 470 ohm.



There are variables not accounted for, but am I in the ballpark?

Thinking about the worst case for a guitar amp, full clipping where the output looks like a square wave, with a 1k g2 stopper I would have:
Wg2 = 83ma x 259v / 2 (50% duty cycle) = 10.75 watts
W-Rg2 = (83ma x 1000)² / 1000 / 2 = 3.45 watts

This is a rough approximation and would not happen very often or maybe not at all, I'm wondering what safety factor should be applied for heavy clipping?

Thanks for your help.

[Lectroid]

Mr. Plates,

I'm confused about the Bassman schematic you listed in your post.  There were no screen resistors, it's true.  But the screen node did come off after a 10K dropping resistor from the power amp plates node.  I'm pretty sure you know what you're doing, so can you explain why that dropping resistor would not be basically a screen resistor?

Thanks.

[PRR]

Different kind of "screen resistor". Due to the cap, it averages the current, does not act quickly on sudden peaks of current.

FWIW, I think 10K here is more a parlor-radio value. If you beat the amp the screen voltage will drop way off.

[HotBluePlates]

Ok, I think I got it. ...

I agree, looks like you nailed it!

... I've been trying to figure out if I can get away with a smaller g2 stopper so that if I use 6L6's, KT66's, etc., instead of EL34's I won't get as much screen compression. ...

6L6 and KT66 naturally have lower screen current, so the bigger resistor is not a problem at all.

EL34 needs the big resistor because it does not have the aligned-grids of the 6L6/KT66, and screen current peaks higher in use.  Hence a greater need to restrain screen dissipation.

... Thinking about the worst case for a guitar amp, full clipping where the output looks like a square wave, with a 1k g2 stopper I would have:
Wg2 = 83ma x 259v / 2 (50% duty cycle) = 10.75 watts
W-Rg2 = (83ma x 1000)² / 1000 / 2 = 3.45 watts

This is a rough approximation and would not happen very often or maybe not at all, I'm wondering what safety factor should be applied for heavy clipping?

If you assume the supply voltage will not sag/drop, then:

     8w / 0.083A = ~96 volts ---> allowable on the screen
     342v - 96v = 246 volts ----> to be dropped by the screen resistor
     246v / 0.083A = ~3kΩ ----> required screen resistor

But slugging the screen volts down to ~96v will clamp plate current & also screen current.  Therefore screen current will not get that high.  And along the way getting there, grid-drive from the phase inverter will be clamped by the EL34's G1 grid-current, and there will likely be bias-shift/grid-blocking that will tend to slide the tube towards a colder bias.  These things also mean less plate & screen current.

So what now?

     Take the info as "3kΩ is an upper limit, maybe 1kΩ is a lower limit."
     Test an actual amp driven very hard with 2 meters:  1 for screen volts (screen-to-cathode), another for screen current (maybe volts across screen resistor, maybe actual series ammeter).
     Start closer to 3kΩ, drive the amp hard, start backing off in the 1kΩ direction until you get the desired balance of screen dissipation & power output & low-compression/sag.

I'm confused about the Bassman schematic you listed in your post.  There were no screen resistors, it's true.  But the screen node did come off after a 10K dropping resistor from the power amp plates node.  ... can you explain why that dropping resistor would not be basically a screen resistor?

TL;DR  -  You're ignoring that filter cap.



There is a Resistor, and it ultimately connects to the Screen.  That's true.

Maybe the more-complete name of what we're discussing is the "Series Screen Dropping Resistor."  Its job is to prevent over-dissipation of the screen by shedding volts as screen current rises.

     Remember Ohm's Law:  Volts (across the Resistor) = Current x Resistance.

There's that 10kΩ resistor right before the 6L6 screens.... but there's also a 16µF filter cap after the resistor & directly connected to the 6L6 screens.

     The capacitor stores Charge (electrons) to build up to a voltage.  That cap supplies the brief peaks of screen current, and then the upstream power supply re-charges the cap.  Prior to some severe current demand that outstrips the ability of the power supply to re-charge the cap fast enough, screen volts stays unchanged.


So you can change the "power supply dropping/decoupling resistor" (10kΩ), and get a different steady-state DC Voltage.  But the "series screen dropping resistor" is used for brief voltage-changes to keep screen dissipation in check.

[Lectroid]

PRR, HBP,

Thank you for explaining that.  I knew that the 10K was in the power supply, more properly speaking, but you've both explained it so that I learned something beyond the question I asked.  Happens to me all the time here...

Thanks again.

[dwinstonwood]

Thank you HBP for sharing this knowledge! I've bookmarked it along with some of your other posts. I'll need to read and work through it a few times, but some of it is already making sense good sense.

[Lectroid]

Ditto.  I know it will take me some time to digest this, but it should be saved on the site somewhere.  This is a great tutorial. 

[DuaneOh]

Now that I’m starting to understand the interactions of all this, I ran a few more simulations.

If you assume the supply voltage will not sag/drop, then:

     8w / 0.083A = ~96 volts ---> allowable on the screen
     342v - 96v = 246 volts ----> to be dropped by the screen resistor
     246v / 0.083A = ~3kΩ ----> required screen resistor

The tube would be in cutoff for a portion of the signal so it would need to be averaged (RMS)? Looking at the loadline it is in cutoff for about 25% of the signal.

But slugging the screen volts down to ~96v will clamp plate current & also screen current.  Therefore screen current will not get that high.  And along the way getting there, grid-drive from the phase inverter will be clamped by the EL34's G1 grid-current, and there will likely be bias-shift/grid-blocking that will tend to slide the tube towards a colder bias.  These things also mean less plate & screen current.

From one simulation I found that at Vg1 = 0V, Vg2 will drop to 281V through the 1k g2 resistor instead of the calculated 259V. I believe this is because as Vg2 drops the loadline changes and g2 pulls less current. And Ig2 then pulls .062A, which makes sense.

The RMS values in the simulation are Vg2 = 320V and Ig2 = .036A when the signal is fully clipped which makes Wg2 = 11.5 watts. This is with a 1k g2 stopper. Does this make sense?

As examples I've been looking at the Matchless Chieftain, Clubman and BadCat HotCat. It's interesting that they can use 1k g2 stoppers and they run at higher voltages, but use 10watt+ resistors. I am guessing that they are hard on EL34's, and that contributes to their sound?

Looking at the Philips El34 data, Wg2 "Limiting" = 8W. How far can g2 be pushed and get "reasonable" life (2 years?) without catastrophic failure? This amp will have a built in attenuator so the EL34's will get worked.....

Thanks for all your help on this.

[HotBluePlates]

Thinking about the worst case for a guitar amp, full clipping where the output looks like a square wave, with a 1k g2 stopper I would have:
Wg2 = 83ma x 259v / 2 (50% duty cycle) = 10.75 watts
W-Rg2 = (83ma x 1000)² / 1000 / 2 = 3.45 watts

If you assume the supply voltage will not sag/drop, then:

     8w / 0.083A = ~96 volts ---> allowable on the screen
     342v - 96v = 246 volts ----> to be dropped by the screen resistor
     246v / 0.083A = ~3kΩ ----> required screen resistor

The tube would be in cutoff for a portion of the signal so it would need to be averaged (RMS)? Looking at the loadline it is in cutoff for about 25% of the signal.

I did not spend much time working through the example.  When you said "square wave" I assumed the G1 drive for the output tube was square-wave.  That was a mistake, as was not-averaging the screen current.  But unless/until you post a loadline I'm a little limited (Imgur.com is an excellent spot to upload the photo, then link here).

So I'll try again & assume your "75% On, 25% Off" estimate is accurate.

     - Off 25% of the time, so we will break the signal cycle into 4 parts.

     - If plate output is square-wave, then so is screen current (because it's a % of total cathode current).  So screen current is (as a very rough first approximation) 83mA for 75% of the signal cycle.

     - (83mA + 83mA + 83mA + 0mA) / 4 = 62.25mA

     - If the plates are pulling peak current 75% of the time, the B+ voltage will sag in a real amp.  But for worst-case we will assume it doesn't and that the screen voltage node of the power supply likewise stays at 346v.

Calculate the allowable voltage for the average screen current:

     - 8w / 62.25mA = 128.5v
     - 342v - 128.5v = 213.5v (to drop across the series screen resistor)
     - 213.5v / 0.06225A = ~3.4kΩ


Something is gonna give before the output tubes get there.  B+ volts will sag, screen volts will sag due to the series resistor & reduce plate output, G1 drive voltage will clamp, etc.

You can sim all you want, but you will find some factor (or several) overlooked.  Use the math to establish reasonable limits, then go test it on the bench.

[DuaneOh]

You can sim all you want, but you will find some (or several) factor overlooked.  Use the math to establish reasonable limits, then go test it on the bench.

I am definitely guilty of over thinking things, part of my OCD..... I'll get to building it soon, but I've got some other fish to fry.     Thanks for your help.

[pdf64]

I can see how a full power sine wave may have a 75% on, 25% cut off cycle. 
But with a square wave, wouldn’t it be more like 50% on, 50% cut off?

[DuaneOh]

I can see how a full power sine wave may have a 75% on, 25% cut off cycle. 
But with a square wave, wouldn’t it be more like 50% on, 50% cut off?

I was going off the loadine, It makes sense that it would be 50% - 50%.