Triode amplifier stage - design tutorial

[gtudoran]

Hello guys,

I got a lot of help from this forum so, i was thinking to give smth back. I've made a document that is very usefull for thoes who are useing valves (tubes).
In this article we will talk about triode amplification stages because is the most used in guitar amplification. I know there is a very large article about this from Valve Wizard but i will try to make this one a little more like a crush course, sthm more practical and smth to help you to build triode gain stages for any given triode and @ any anode voltage you like.

So lets us begin.
PART I - Theory Consideration

The most important thing in this is the Ia / Ua graph, this will show you the dependency of Ia function of Ua.

We will begin with all the things we know.Anode voltage depends on the PSU that you use and we will consider this 200v, the next step is the tube that you gonna use, and for this we will take 12AX7 (the most common one), so we have the HV and the tube type.

The guitar amplifier stage is called: Rezistive Load Amplifier Stage and the main caracteristic is the Ra (Anode Rezistor). This rezistor has 2 main functions: regulates the current that will pass trough the tube and has a role in the overall gain of the stage (a bigger rezistor will increse the gain but will cut the high frq.
So we could say that a good value for Ra is between 2-5 multiply Ri => Ra = (2 ... 5) x Ri where Ri is the internal tube rezistance (you will take Ri from the datasheet, if is not given then we could calculate it from the Ia/Ua graph).
The next thing is the Rg  or grid rezistor witch could be Rg~ 0.5 ... 1M, when the next amplification stage is a power stage  Ra should not exced the maximum value for the given power tube.
The next stept is the coupling capacitor and we will note it Cg. The value for this could be calculated with this formula: Cg = 159 / Fi*Rg where Fi is the limit for low freq. and Rg is the grid rezistor for the next stage. For practical use Cg will be in the interval 0.022 ... 0.05 uF.
The next step is very important and is the step where you decide how clean the signal will be at the  output of the stage, is the point where you will decide how negative will be the grid, and the general formula for this will be: Ug = Uin + (0.5 .. 1.5v) where Uin is the voltage that you expect from the input (guitar pickups, Ipod etc etc you get the point).
So, from here we could compute the next important thing: Rk or the cathode rezistor. Rk= (-Ug / Ia)*10^3 (the value that you optain is in OHM remember that and do not make mistakes ) Where Ug is the first grid voltage and Ia is the tube current in mA. For amplifier stages the Rk is bypassed by a capacitor witch is chosen by the low freq. that you whant / expect to pass. Bigger capacitor = more bottom-end. There is also a formula for that: Ck >= (1...2)* 10^3 / Fi*Rk where Fi is the low freq. that you expect and the Rk is the cathode rezistor in Kohm.

The part II of this tutorial is a practical design of a amplifier stage (hope will be ready this night).

Any feedback will pe appreciated (please excuse any mistakes in my english)

Best regards,
Gabriel Tudoran

[frequencycentral]

Very usefully broken down, credit to you for taking the time. Looking forward to the next installment........

[gtudoran]

So dear friends i've returned with the 2nd part of this tutorial. In tis part i will try to

explain how to make a practical triode stage amplifier going from a graph and somme other

things that we know.

First of all we will put on the paper the things that we know:
- Ua - anode voltage is one on them and we will consider it 60v (i already done the math for

this voltage)
- Ug1- the grid voltage that we expect to be remember that the grid voltage is eq. with the

Uin voltage + (0.5...1.5) so for this experiment we will consider Ug1=-1V (0.1+0.9). It is a

good voltage to start from
- For 6n16b in the datasheet we don't have the Ri (internal tube resistance), that is no

problem bc we can aprox. it (remember 10% is a very good aproximation and it will give you a

very good point to start from)
   So to calculate Ri we do this math:
   - we take our anode voltage 60v and our input voltage -1v and we will se on the

graph that thoes 2 values corespond to a current of aprox. 3.4mA (we will not need that

value right now) on the graph there is a red point that we mark it with A. We will make a

line that will touch the Vg curve in point A, after that we will make a paralel line with

the this one that will go trough the orig. of the graph (point 0). We will take a arbitrary

voltage (for this example we will take the anode voltage 60) and see what current we will

get on the graph:aprox 8.9mA. The only thing that we have to do now is to take put thoes

data in a formula like this: Ri = Ua / Ia => Ri= 60v / 0.0089A = 6741ohm =6.7k and this is

our internal tube rezistance +/- 10%. The hard part is gone for now .
   If we have the Ri we cand calculate the Ra witch is Ra=(2...5) x Ri we will take

2xRi = 14k (15k for the nearest value).
   After this step we can plot the lod line: First of all we will calculate the Ia for

the tube with the load resistor of 15k: so Ia=Ua/Ra => Ia=60v/15000ohm=0.004A (4mA), so we

will plot a line from 4mA to 60v (the green line).
   We will see that our tube in a staindby state will have a anode voltage of 41v and a

current going trough of aprox. 1.2mA.
   The next step is to see how much will be our amplification factor, for this we will

take the difference from the Ua (in our case is 41v and 20v hint. take a look at the 2

vertical green lines that interset the Ua axe, we took the voltage for -1v and for 0v Vg)

and divide them by the difference between the to Vg = -1 and 0 in absolute value (this is

1). So we have u= 41-20/1 =>u=21 (this is the amplification factor witch is verry much the

samme as the one form datasheet (u=25+/-5 in DS).
   The next step is to find out the Rk resistance, the formula for this is Rk= (-Ug /

Ia)* 10^3 => Rk= 1*10^3/1.2mA => Rk=833 ohm (from practice anything from 600 to 1k will do)
   Ck for the decouplig capacitor (the one parallel with Rk anything from 0.68 to 10uF

will do depending on how much bass response do you need to have) there is a method to

calculate this Ck >= ((1...2)*10^3) / Fi*Rk where Fi is the max low freq in Hz and Rk is the

cathode rezistor in kohm.
   Rg for the first stage will dictate the input impedance and a good value is 1M. The

coupling cappacitor from one stage to another from practice is 0.22uF, the formula to

calculate this is: Cg=159/Fi*Rg where Fi minimum low freq and Rg is the grid rezistor of the

next stage. In practice anything from 0.022uF and 0.05uF will assure you the minimum frq.

distorsion.
   And here we are at the end with a nice stage of amplification:
   Ra= 15k
   Ua= 60v
   Rg= 1M
   Rk=833ohm
   Cg (for the next stage) = 0.022uF
   Ck=anything from 0.68 to 10uF
   
   For the next stage we will need to know how much signal will pass so we have the

formula: Uout = K * Uin
    K = u / 1+ Ri/Ra + Ri/Rg from this we will have: K= 21 / 1+ 6.7/14 + 6.7/1000 => K=

21/1.4767 = 14.22 so our Uout from the first stage will be K * Uin = 14.22 * 1 ~ 14.22Vpp

   That is all guys... hope this will help you a little on the designs with tubes. I

will answer at any questions  as far as my knowlage goes.
   As a last sentence: pls. excuse my english mistakes.

Best regards,
Gabriel Tudoran
Analog Sound 

Graph: