## Choke loading Anodes.

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #61

Sorry for delay Teech, I've been wrapped up in work (:shock:) all today and break times have been spent sorting out skate-related stuff.

Homework will be done tonight when I get home.

Back in a jiffy.

DTB

Homework will be done tonight when I get home.

Back in a jiffy.

DTB

*"The fat bourgeois and his doppelganger"*

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #63

Sorry I'm late Sir, the dog ate my Bus and my homework was late.

Right done it (I think).

I've had to do this by hand not on the 'poot as the 76 data sheet is in .pdf format and I couldn't work out a way of drawing onto a pdf, I know there's a way to do it but don't tell me now.... I'm counting

First up I plotted and drew on the 20K loadline and the first suggested OP.

For the 2nd OP... 100V at 2.5mA I worked out that with a 350V HT I need to drop 250V to get the anode at 100V (without a calculator too...get me!!). So, if I want to drop 250V @ 2.5mA Ohms law says 250V/0.0025A = 100,000R = 100K ohm resistor.

OK, next step was to plot the 2nd OP on the curves = 100 V (on the x axis) and 2.5mA (on the y axis) gives me a spot, marked with a blob.

Next point, valve totally shut-off = no current flowing = nothing dropped across resistor = Ht = 350v on the anode at 0mA. Thats another plot marked (same as other OP1 of course as its the same HT and no current being pulled).

Last point requires ohms law. Valve fully conducting, 350V being dropped across Ra (note capital 'R'), ohms law says I=V/R = 350/100,000 = 0.0035A = 3.5mA

I'm sure all that above is correct.

To make up for being late.... get this .......I've worked out that with OP1 I'm dropping 500mW across that 20K resistor so I'd spec a 1W resistor in that position.

With OP2 I'm dropping 0.625W so would probably go for a 2W there.

Sorry the pic is a bit scribbly but I think its right.

DTB

Right done it (I think).

I've had to do this by hand not on the 'poot as the 76 data sheet is in .pdf format and I couldn't work out a way of drawing onto a pdf, I know there's a way to do it but don't tell me now.... I'm counting

First up I plotted and drew on the 20K loadline and the first suggested OP.

For the 2nd OP... 100V at 2.5mA I worked out that with a 350V HT I need to drop 250V to get the anode at 100V (without a calculator too...get me!!). So, if I want to drop 250V @ 2.5mA Ohms law says 250V/0.0025A = 100,000R = 100K ohm resistor.

OK, next step was to plot the 2nd OP on the curves = 100 V (on the x axis) and 2.5mA (on the y axis) gives me a spot, marked with a blob.

Next point, valve totally shut-off = no current flowing = nothing dropped across resistor = Ht = 350v on the anode at 0mA. Thats another plot marked (same as other OP1 of course as its the same HT and no current being pulled).

Last point requires ohms law. Valve fully conducting, 350V being dropped across Ra (note capital 'R'), ohms law says I=V/R = 350/100,000 = 0.0035A = 3.5mA

I'm sure all that above is correct.

To make up for being late.... get this .......I've worked out that with OP1 I'm dropping 500mW across that 20K resistor so I'd spec a 1W resistor in that position.

With OP2 I'm dropping 0.625W so would probably go for a 2W there.

Sorry the pic is a bit scribbly but I think its right.

DTB

*"The fat bourgeois and his doppelganger"*

### #64

Very good Mr Dave. Spot on.

You now almost have enough to build two amplifier stage susing a 76. But the question is which line is best?

Well, they both have their advantages. the 20k one can swing a bigger output voltage. if you imagine the signal on the grid can drive the grin voltage above and below its center point. For the moment assume that bad things happen if you pass the 0v grid line. So looking at the graph, for the 20k line, the grid can go +14v (ish) to about 0v. And -14v to about -28v. So it can handle a 28v p/p signal. The 100k one has more problems, it can only go about 5v positive, so even though it can go a lot further negative, if we assume that the signal we want to amplify is symmetric, that limits us to 20v p/p input signal.

SO the 20k can produce the most output, so is it best?

Maybe, maybe not, remember the anode resistance, and Mr I's rule of 3 times ra as a load. Well, 20k is only about 2 terms the ra. And if we look at the grid lines as the load line passes through them we see that they are closer together when going negative than going positive, this means that the output from the 20k load will be distorted, in this case, more distorted than the 100k.

So it depends on what we want max voltage swing, or lower distortion.

Of course there are any number of load lines you might what to try, and if you build the amp, you can pick the point you want, draw a line, build the amp and have a listen. Hours of interest.

Now your next job is to work out the gain of the two amps. To do that, move the same distance along each line from the center point. Maybe pick a distance that the gap in one direction of the grid lines near the point. So for wach load, find what 4v movement on the grid is, and draw a point on each line on either side of the op point. Then drop a vertical line to the bottom scale and find what the distance between the places the lines cross the axis. Mesasure that in volts, and that will show the voltage out for a 0 to +4v signal and a 0 to -4v signal. Add then together, and this will be the voltage out from a 8v p/p signal. So divide the output voltage by the input (8v) and that will give you the gain of the amp for that load resistance. Then see how that compares to the Âµ of the valve.

You now almost have enough to build two amplifier stage susing a 76. But the question is which line is best?

Well, they both have their advantages. the 20k one can swing a bigger output voltage. if you imagine the signal on the grid can drive the grin voltage above and below its center point. For the moment assume that bad things happen if you pass the 0v grid line. So looking at the graph, for the 20k line, the grid can go +14v (ish) to about 0v. And -14v to about -28v. So it can handle a 28v p/p signal. The 100k one has more problems, it can only go about 5v positive, so even though it can go a lot further negative, if we assume that the signal we want to amplify is symmetric, that limits us to 20v p/p input signal.

SO the 20k can produce the most output, so is it best?

Maybe, maybe not, remember the anode resistance, and Mr I's rule of 3 times ra as a load. Well, 20k is only about 2 terms the ra. And if we look at the grid lines as the load line passes through them we see that they are closer together when going negative than going positive, this means that the output from the 20k load will be distorted, in this case, more distorted than the 100k.

So it depends on what we want max voltage swing, or lower distortion.

Of course there are any number of load lines you might what to try, and if you build the amp, you can pick the point you want, draw a line, build the amp and have a listen. Hours of interest.

Now your next job is to work out the gain of the two amps. To do that, move the same distance along each line from the center point. Maybe pick a distance that the gap in one direction of the grid lines near the point. So for wach load, find what 4v movement on the grid is, and draw a point on each line on either side of the op point. Then drop a vertical line to the bottom scale and find what the distance between the places the lines cross the axis. Mesasure that in volts, and that will show the voltage out for a 0 to +4v signal and a 0 to -4v signal. Add then together, and this will be the voltage out from a 8v p/p signal. So divide the output voltage by the input (8v) and that will give you the gain of the amp for that load resistance. Then see how that compares to the Âµ of the valve.

Little known fact, coherent thought can destructively interfere with itself leaving no thought at all, that’s why I prefer incoherent thought.

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #65

I've got a problem here Nick, you say that the 100K loadline can only swing 5v positive and that we're assuming our signal is symmetrical....wouldn't that mean that it'd be a 10V p/p i/p signal max? 5Volt +ve and 5V -ve?The 100k one has more problems, it can only go about 5v positive, so even though it can go a lot further negative, if we assume that the signal we want to amplify is symmetric, that limits us to 20v p/p input signal.

DTB

*"The fat bourgeois and his doppelganger"*

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #67

Phew, I thought I was missing summat major! thanks for continuing with the schooling BTW.Nick wrote:Yep, you are right, sorry, its been a long day.

OK, I've think I've done this right. I had to re-do my pictures, too many lines all in black.

http://i195.photobucket.com/albums/z109 ... adline.jpg

Now then, I'm estimating this but I think for an 8V P/P i/p I'm achieving approx 95 Volt o/p. Therefore 95/8 = 11.875... for both loadlines, TBH I was expecting the 100K loadline to be low gain but low distortion and the 20K to be high gain high distortion. The BIG break through tonight for me though is visually seeing how the loadlines bunch together the more -ve the 20K loadline goes. Now THAT is a penny dropper, a ruddy great 50p one at that

A u of 11.875, not a zillion miles way from the claimed 13.8, Oooooo!

As regards the scenario gain Vs distortion... oooo, given that this is a 76 'pre-amp' that'd be strapped in front of an existing power amp I'd choose low gain and low distortion.

DTB

*"The fat bourgeois and his doppelganger"*

### #68

Yep, good stuff, though the 20k line looks like 10v p/p. You need to use the 4v distance from the 20k line, not the 4v one from the 100k line, you might notice the two distances are not the same.

The big thing as you have seen is you need to see how the grid lines intersect the load line, and partially ignore what he lines do off the load line. Thats what I meant before about the grid lines becoming straight not as important as the gaps between them.

So next step, check the gain again, then try and draw a tangent to the grid lines at the op point and determine the ra.

The big thing as you have seen is you need to see how the grid lines intersect the load line, and partially ignore what he lines do off the load line. Thats what I meant before about the grid lines becoming straight not as important as the gaps between them.

So next step, check the gain again, then try and draw a tangent to the grid lines at the op point and determine the ra.

Little known fact, coherent thought can destructively interfere with itself leaving no thought at all, that’s why I prefer incoherent thought.

### #69

And its worth pointing out that the 76 is one of the most linear valves you are going to find. Try the same on just about any other triode and you will see what I mean.he BIG break through tonight for me though is visually seeing how the loadlines bunch together the more -ve the 20K loadline goes. Now THAT is a penny dropper, a ruddy great 50p one at that

Little known fact, coherent thought can destructively interfere with itself leaving no thought at all, that’s why I prefer incoherent thought.

### #70

Dave, I think you've already sussed this, but just to push the point home (hope you don't mind me butting in) another sum you can do is look at the anode voltage swing in each direction to get an idea of distortion level.

Taking the 20K load line first, I reckon that moving the grid input from -13.5 to -27 volts would move the anode from 250 to about 345 volts, so about 95 volts. Swinging the grid the other way from -13.5 up to 0 would move the anode from 250 down to about 115 volts, so about 135 volts. So our symmetric +13.5/-13.5 volt input signal has become a very non-symmetric -135/+95 output signal (also note the signal has been inverted). This non-symmetry would be a pretty good measure of how much second harmonic distortion would be generated. 135 is 42% bigger than 95

Doing the same for the 100K line gives about -160/+140.

160 is only 14% bigger than 140, so much less distortion.

In both cases the distortion will pretty much scale with signal amplitude, so if this stage was only swinging a few volts, distortion would be much less.

Taking the 20K load line first, I reckon that moving the grid input from -13.5 to -27 volts would move the anode from 250 to about 345 volts, so about 95 volts. Swinging the grid the other way from -13.5 up to 0 would move the anode from 250 down to about 115 volts, so about 135 volts. So our symmetric +13.5/-13.5 volt input signal has become a very non-symmetric -135/+95 output signal (also note the signal has been inverted). This non-symmetry would be a pretty good measure of how much second harmonic distortion would be generated. 135 is 42% bigger than 95

Doing the same for the 100K line gives about -160/+140.

160 is only 14% bigger than 140, so much less distortion.

In both cases the distortion will pretty much scale with signal amplitude, so if this stage was only swinging a few volts, distortion would be much less.

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #71

Sorry Nick, I'm not with you here, can you scribble what you mean onto the last jpeg I uploaded.Nick wrote:Yep, good stuff, though the 20k line looks like 10v p/p.You need to use the 4v distance from the 20k line, not the 4v one from the 100kline, you might notice the two distances are not the same.

I was trying to guess +4V and -4V (8V p/p) using the distance between the gridlines as a ref. i.e the distance between gridlines closest the OP = 4V which on my ruler is about 12mm. So I scored a line at + 12mm either side of the OP.

I think I've got the wrong end of the stick but maybe you could illustrate what it should show if you get time,

Sorry about the upcock.,

DTB

*"The fat bourgeois and his doppelganger"*

### #72

I have drawn in thick black the distance between the lines, a 4v distance along the 20k line. Then take the same distance and go up and down (red and green) along the 20k line, and I get to a place thats short of where you got to.

I think you used the distance along the 100k line on the 20k line (or at least I think you may be doing that).

I think you used the distance along the 100k line on the 20k line (or at least I think you may be doing that).

### #73

I'll have a go, hope I'm not making things worse.

Taking the 20K loadline, that load line represents the range of possible states at the anode. So lets assume you had set the valve up at the 250V, 5mA op point with a 20K load. Then assume you measure the voltage and current at the plate - all the possible combinations of current and voltage that you might measure will all lie along that load line. Exactly where you would be along the load line would be determined by the grid voltage. So with a grid voltage of -13,5V you would be at the op point. If you change the grid volts to -8V, then to find out the conditions at the anode, you move along the 20K load line until it intersects with the -8V grid line.

If you specifically want to move +4 volts on the grid from the op point, you have to draw a new grid line. You have to guesstimate where a -9.5V gridline would be if they put one on. You could actually draw -9, -10 and -11V lines in between the -8 and -12V lines, and then draw one in between the -9 and -10V lines, that would be your guesstimated -9.5V line. Where it intersects with the loadline is the +4v point.

Similarly if you want to find the conditions at -4v from the op point, you have to draw in a guesstimated -17.5v grid line. Where that intersects with the loadline is the -4v point.

Taking the 20K loadline, that load line represents the range of possible states at the anode. So lets assume you had set the valve up at the 250V, 5mA op point with a 20K load. Then assume you measure the voltage and current at the plate - all the possible combinations of current and voltage that you might measure will all lie along that load line. Exactly where you would be along the load line would be determined by the grid voltage. So with a grid voltage of -13,5V you would be at the op point. If you change the grid volts to -8V, then to find out the conditions at the anode, you move along the 20K load line until it intersects with the -8V grid line.

If you specifically want to move +4 volts on the grid from the op point, you have to draw a new grid line. You have to guesstimate where a -9.5V gridline would be if they put one on. You could actually draw -9, -10 and -11V lines in between the -8 and -12V lines, and then draw one in between the -9 and -10V lines, that would be your guesstimated -9.5V line. Where it intersects with the loadline is the +4v point.

Similarly if you want to find the conditions at -4v from the op point, you have to draw in a guesstimated -17.5v grid line. Where that intersects with the loadline is the -4v point.

Last edited by Max N on Fri Jul 17, 2009 1:40 pm, edited 1 time in total.

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #74

Ah! I see, I've redone it on my big printed off version, I think I'm measuring approx 75-80V P/P swing.

So lets say, 78V/8V = mu of 9.75?

DTB

So lets say, 78V/8V = mu of 9.75?

DTB

*"The fat bourgeois and his doppelganger"*

- Dave the bass
- Amstrad Tower of Power
**Posts:**11709**Joined:**Tue May 22, 2007 4:36 pm**Location:**NW Kent, Darn Sarf innit.

### #75

Yeah, gotcha, cheers Max.

2 teachers, this is gonna cost a lot in apples!

DTB

2 teachers, this is gonna cost a lot in apples!

DTB

*"The fat bourgeois and his doppelganger"*