how to design a perfect preamp in 4 easy steps

[andrew ivimey]

'actually, of course I haven't built one yet, I am getting some parts, and am going to build a 2a3, and a few preamps, and poss a phonostage.'

groan

:p

[colin.hepburn]

Hi All
Too who it may concern
Can you pleas STOP THE DIGS at Ian pleas it doesn't matter if he has or Has’nt built an amp I don’t know of any teachers that build every thing they teach do you
Ian has laid out his tutorials in an easy step by step way Ian is willing to help others on here rather than grip at others which seems to be an ongoing pain in the ass on this forum
Pleas stop this and let Ian post his tutorials without the hassle
Thank you
Colin

[colin.hepburn]

HI all
Hi all this is me that’s typing this as most of the members here will be aware that I did post that I am dyslexic with numbers and letters some time ago which seemed to be fine with well all of you Here but this makes things like ohms law and basic maths a bit of a challenging thing to be fare I didn’t get maths a school building from a kit or with a schematic is fine and the help I have had from most of you hear and a cross the road is well very much been welcomed by me so pleas understand this I thank you all for that I think most tutorial’s take it for granted that you know ohms law yes its basic maths but this is where I F…k up the basics remember I am dyslexic and I didn’t get this formula stuff at school OK so V/I /R but it s the formula I get wrong or at lest mess-up
so to the don’t mind me line for righting this well I am the blues man and at now 52 years and just lost me mum to cancer back in June there so well we carry on don’t we
so pleas let Ian post his tutorials with out the hassles from OK lads here it is as far as I can see Ian has dun his home work here by laying out what he is explaining well plus the maths formula is laid our so I can vary the load lines and yes in my view Ian is far more for understanding of some one what is having difficulty with understanding this stuff than some but the some may not have even thought about it so that’s OK by me OK the 3 bottles of wines kicking in now OK it 5 am and still got the sen 650s on though me 12B4a valve headphone amp playing the blues bliss yes I built it so would like to leave all you artists here with this image pleas do as instructed on the image and expand it all the way up to 400% but don’t close you eye s looking at a white ish wall will do but before you post an image back on here have you designed it and drawn it out first does it work
OK thank you all
PS i have been saving up and buying bit so got lost do do this winter like wad phone 2/3
preamp
2a3 plus frugal horns so hope to be busy and getting some help from you all cheers

Colin

[Deco-Doctor]

Originally Posted by colin.hepburn

Where did you get the load line software?

Glass-Ware TubeCAD

http://www.glass-ware.com/GlassWare_Programs.htm

[colin.hepburn]

Ah Some handy looking software there
thanks for the link

[Ianm2]

A fascinating relation has revealed itself whilst I have been playing around with this, yet another revelation.

to get your cathode resistor, pick a nice point on the 2 volt grid line.

divide this 2 volts by your selected operating current, read off of the left hand axis.

now, read off your voltage at this dot, on the lower axis

use that, to divide by that same current, you get a resistance.

use this value as the load resistor, and as if by magic,

your power supply volts your loadline touches on the horizontal, will be magically double the volts of that point on the lower axis.

and the current your diagonal loadline touches will be double the current your dot is at.

and the loadline will pass through your dot.

and if you divide the power supply volts by the end current, you will get exactly the same value of resistor, ie your loadline, as if you did calculate it from the voltage the dot is at, divided by the current, that dot is at, too.

complete magic, and that really will get you a perfect amplifier stage.

A

[Ianm2]

ND,,,

• not only that, if you halve that load resistor,
  
• on the 0 volt grid line, you will get exactly twice the current that your dot is at, and the voltage will be exactly halve way in between on the bottom axis, too.
  
• there is something very odd going on here,
  
• to find this out, print out a ecc83 graph.
  
• put your dot at 1.5 volts grid, your current set at 1 milliamp to get a 1k5 cathode resistor.
  
• you will find this correstponds to 180 volts on the bottom axis, so your load resistor is 180 divided by 1 milliamp=180k ohms
  
• your power supply is double 180 volts, 360 volts, your loadline runs from 360 volts to double the 1 millamp point, 2 milliamps on the vertical axis.
  
• however, halve that resistor, and you will get a power supply of 270 volts, 270 volts divided by 90 k ohms = 3 milliamps, which is where your loadline draws from to,

and magically, the 0 volt gridline is at doulbe the operating current of 1 milliamp, 2 milliamps, and the bottom volts are 90 volts, the top one is 270 and half way is your dot, at 180 volts.

very odd indeed.

go check it out, it works for all valves.

so in summary, print out ecc83 graph or aka 12ax7

1/ choose 1.5 volts grid line, pick 1 milliamp

2/ read off current you place your dot at, 1ma

3/ cathode resistor is 1.5 volts divided by this current, 1k5 this is ohms law, R = V / I

4/ read off volts dot is at on horizontal, 180 volts

5/ power supply is double that voltage ie 360 volts

6/ load resistor value is that voltage dot is at, divided by that current. ie 180 volts / 1 ma or 180k this is ohms law, too

loadline draws from power supply diagonally to double the current dot is at. ie 360 volts to 2 ma

load resistor is also power supply volts divided by double the current dot is at, where it crosses the vertical current graph. ie 360 / 2 = 180k this is ohms law again

7/ half the load resistor. 90k

8 half power supply volts 90 volts, add this to the dot ( 180 plus 90 = 270 volts)

thats your new power supply, for your 90 k loadline, the loadline draws from that 270 volts to triple the current your dot is at. ie 3 ma

check, divide 270 by 3ma = 90k ohms. ohms law.

9/ 0 volt gridline occurs at double the current the dot is at. ie 2 ma.

pure magic.