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|6th March 2010, 03:13 PM||#1|
Join Date: Dec 2003
design a preamp, so easy....
I posted this elsewhere in a number of parts, its all my own theory,
I hope its understandable, if you don't know, persevere and read, I think its pretty good, but undertstand there were some disagreements with people on other matters, hence some of the writing, don't let that get in the way.
I hope its comprehensible, please critique, but I thought it was so good I had to share it here.
this tutorial will cover selection of
working out cathode resistor
and working out anode resistor
based upon load line.
I believe its foolproof.
1/ take your selected preamp valve
it should have a mu of less than 20 as we don't require much for a preamp
I have selected ecc82.
2/ For a preamp, we require it not to overload when presented with the strongest source.
cd has an output say of 2 volts rms
peak to peak this is 2.82 volts
if we halve that, we have our minimum bias point
that makes 1.4 volts, so we select anywhere between 1.5-2volts for our bias point.
3/ go to our ecc82 curve. I will select 1.5 volts
place a dot on the 1.5 volt line.
read across to the left hand line.
This is the current our valve will run at.
We find a dot on the 1.5 volt will read across and give a current of circa 3.6 milliamps.
cathode resistor is thus ohms law, V/I and is 1.5/3.6 ma =470 ohms
4/ now to our load line.
to do this, we select our psu voltage our valve will run at.
I am choosing about 150-160 volts.
draw a line from our 1.5 volt dot on the grid line to 150 volts, and then upto the left hand axis.
this is our load line.
our load resistor value calculates easily.
Its defined as ohms law, V/I.
150 volts, our psu volts, divided by the point at which our load line passing through our bias point of 1.5 volts goes on and meets the vertical line.
if you do the maths, 150/7 millamps or so, we have a load resistor of 22k ohms.
you can do this with any valve for a preamp
simply select 1.5-2 volts grid, read off horizonally the current.
cathode resistor is V/ current.
then draw a line to selected psu volts on the bottom axis from the bias point selected.
continue that line upto the vertical axis.
divide your psu volts by the current that your line touches the axis.
and you get load resistor value.
how to design a simple preamp that will work perfectly.
I hope that's of some help to those wondering about load lines.
you two, there are people on here confused as hell about what a load line is and how to design a simple valve stage.
my example has distilled that to 4 simple steps.
1/ pick your 2 volt grid line
2/ read across your current, your cathode resistor is ohms law, V/I.
3/ run a ruler from that 2 volt dot to your chosen HT volts, in this case, 150,
4/ your anode resistor value is ohms law, 150 volts divided by current at where that big line crosses the current axis, about 7 milliamps, yielding a resistor of 22kohms.
you have a fully working valve stage.
2 ohms law calculations.
you can add a cathode follower direct coupled on the end with the same load resistor as the anode stage, 22kohms, direct couple it.
2 preamps, choose any valve you want, choose the 2volt grid line.
simple as that.
people were begging for a simple explanation of load lines, load lines do not tell you how to design, they want to know how to design a simple valve stage.
they will not find simpler than this.
2 applications of ohms law, V/I to work out 2 resistors.
add a 0.1uf cap and 1MEG resistor on in and out and away you go.
do this for any valve, any current, and load resistor, any psu volts if the valve will take it.
just set your bias to 1.5 volts min, and draw 2 lines.
this does, if you know it, fine, its for people who wish to.
I am sure the people who this will hopefully help will not want me to delete this in disgust.
Or am I throwing pearls to pigs?
Philip, you should know better, perhaps choose an ecc88, and build a preamp using this, and then judge it, or go and design one if you know all of this.
did I say it was the 'best' amplifier?
did I say it makes others obselete?
its the easiest guide you will ever find to get a perfect working valve stage. I said nothing else.
I can knock up a fully working design now and guantee it will work, if you can't do that, or design one from scratch, I suggest you learn carefully about what's here,
this is pure gold.
you will not see this anywhere at all.
honestly, your like the ****, you have a great man in your midst and you don't have the ears to hear or the eyes to see.
this will make a foolproof preamp that anyone can design.
I don't think I have ever seen it written any easier way that that.
oops the attach went to the top, here it is for an ecc88
note the 2 volt grid line, place a dot, read off the current on the left, about 5 milliamps
cathode resistor is 2/5 milliamps = 200 ohms?
take a ruler, from the dot, choose your ht, 150 volts, draw a line from the dot to 150 volts, that's your HT on the horizontal scale.
continue tthat to the left current vertical axis, we get about 10ma, so your anode resistor is
ohms law, V/I which is 150/10ma =15kohms
a perfect working valve stage.
hope that's easy.
my credibility does not matter, you can submit this to any valve genius, and he will say that this is a foolproof way to design a working valve stage that no one has ever made this simple
your listening once again to the hearsay of others and prejudging, philip, wihtout reading and learning and seeing the elegance of this.
if you wanted to design an atomic bomb, you would ask einstein, not a back street cobbler.
in fact this is so good, I have a good mind to remove it, as people that diss this are not worthy of reading it.
|6th March 2010, 03:14 PM||#2|
Join Date: Dec 2003
how to design an srpp preamp in easy steps.
* look at your chosen valve data sheet, look at the max current the valve can take
* pick any value between 1/10 and 1/2 that.
* choose a voltage between 1.5 and 2.5 volts
* divide that volts by your chosen current.
* that's your resistor values
* pick a power supply voltage between 150 and 300 volts, build your power supply to give this voltage, on load, ie accounting for 2x the current, ie 2 valves sucking current, if using 1 mains transformer, as the chokes, series resistors, or transformer secondary, will lose volts depending on their resistance. ( its acutally, volts lost = = 2x your chosen current x total series resistance of chokes, resistors and transformer secondary in the power supply.)
build your srpp preamp, referring to a schematic for what goes where
perfect doesn't imply the best in the world, it means it will work first time, the values are perfect for it, and not need any further design work.
so you can now go and design 3 preamps:
1/ a basic voltage amp
2/ add a cathode follower on the end for a 2 stager, using the same load resistor as the voltage amp
3/ a srpp amp.
ahh, nice to hear greg, I was expecting a shoot me down in flames post
how does it sound? it sounds like the best preamp in the world to me.
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.
you can even use this to design driver stages for your single ended output valve.
just remember output valves require and input voltage to the grid to drive them to full power, so your driver stage has to provide this, that's your load line, the peak positive and negative voltage, left and right, up and down the load line, and the output from your driver valve has to be twice that, that the output valve requires, that's what load lines are for.
ideally it needs to be in excess.
eg, say your 2a3 tube is biased for 50 milliamps.
it uses a 1k ohm cathode resistor.
the input voltage required is ohms law, V=IR, which is 50 volts.
we require a driver valve that can give 100 volts , ie 50 volts left on your load line, and 50 volts right, more than this as we wish for excess.
so we put our grid volts on our driver that will give more than 50 volts to the right to our power supply value.
and more than 50 volts left to the cutoff point on the left .
this is measured as the output volts on the horizontal or X axis of our selected driver tube.
that's what a load line is for, to give required volts to drive a following valve.
if you look at the 6922 graph, look at the 0 volt grid line.
it corresponds to 40 volts on the X axis.
our power supply is 150 volts
so at this psu, our 6922 can provide 150-40 =110 peak to peak voltage swing.
divide that by 2, we get 55 volts
our 6922 is man enough to drive a 2a3 valve as its in excess of 50 volts
simply put our bias point at half way on the x axis.
what's our cathode resistor?
I reckon its ohms law, V/I, and is about 2.5 volts grid divided by 5 milliamps = 500 ohms.
bingo a working 2a3 ampllifier
6922 valve, 150 volts psu, 500 ohm cathode resistor, 15k ohm load.
if we want an srpp driver stage we simply use 2x 500 ohm resistors, and it should work first time.
so 3 x 2a3 amplifiers yield their design secrets:
its really so easy.
* simple voltage amp
* anode into cathode follower amp
* srpp using 2x500 ohm, all the same as you would do a preamp.
hope that' s easy.
output valves also require current, but I can't recall the calculation here, basically an output valve grid is a capacitor and needs current to charge it. that must be supplied, but that's over complicating things.
I did this, as I wanted to try to find out how to distill all about designing an amp into easy steps, so I could understand exactly what I was doing, instead of having to goto the web and use designs, which may well be done by people who don't know what they are doing.
Its here, and anyone can use it to design their perfectly working preamp, simply choose any valve you desire, follow the steps, and it will work.
then you can tweak capacitors, perhaps even stick in a variable pot to alter the load resistor value, and the cathode resistor, to your hearts content.
one thing I don't understand is feedback, but I have a scope now, and am going to get a signal generator, when I figure that out, I will post all about feedback.
I think the easiest way is to stick a variable resistor in a feedback configuation and tweak it to see the effect on sound from zero to infinity.
let me know which bits you don't follow, this hasn't been composed just on the fly.
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.
* 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.
|7th March 2010, 09:06 AM||#3|
Join Date: Feb 2009
I read the first section and this occured to me: I believe loadline calculation should refer to the anticipated load and arrangement of grid curves (or dynamic characteristics graph) as well, rather than just assuming that the load is close to infinite and that any point along the loadline will do just fine as long as it is within confines of class A1. Other than that you got all the key parts covered
mod verb, transitive /mod/ to state that one is utterly clueless about the operation of device to be "modded" and into "fixing" things that are not broken; "My new amplifier sounds great so I want to mod it."
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