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Old 25th April 2012, 03:49 AM   #1
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Default Help getting Imax and Eb approximations for a push-pull amp

Hi to all!

Im a electrical engineer student and I have made several tube amps for guitar using existing schematics and stuff, but now I want to design my own amplifier form 0.

I understand the concepts, the diferents classes of operation, biasing etc.. also the roles of components, maching a OT etc....

For this purpose I downloaded a book called "Designing vacuum tube amplifiers and related topics" by Charles R. Couch

The book explain step by step how to design a guitar amplifier, starting for the power stage

I realy like the aproach of this book, very simple and easy understanding, absolutly foolproof and great practical data



For the amp I want to design, will be a 6L6s (not CG) push pull AB1 amplifier of 25W

Refering to the book, I searched in the datashet for the graph "Plate current (Ib) in the (Y) axis and Plate voltage in the (X) axis", then locate the line Ib @ Ec1 = 0, then put a point in the middle of the curve (from the end of the "knee" of the curve to the end of the line), then get the Imax for that point in the graph, arround 180mA (with Ec2 = 250V)

Ok, so far so good.. then the next formula appears for making an estimation of the plate swing voltage:

Pout = 0.32 x Imax x Eb ; rearranging the terms: Eb = (3.125 x Pout) / Imax

Where Imax (obtained from the plate curves) = 0.18 and Pout = 25

This give me 434.0V of plate swing voltage, exceeding the datasheet maximun plate voltage rating of 360V...

Then for linearity concerns, looking at the same curve, is noted that the 6L6 start compressing when the plate voltage is below +/- 60V, so we need the plate swing voltage to be above 60V, this implies:

Eb = 434 + 60 = 494V..almost 500V of plate voltage...much higher than the maximun rating of 360V...

A iteration is made to calculate a new Imax from the same graph, now using 60% of 494V of plate voltage (Ib @ Ec1 = 0 implies this condition, control grid at 0V and a plate voltage 60% of normal operating voltage (anyway in my datasheet it only say Ec1 = 0)), this gives me arround 190mA



Then the quiescent current... Iq = Imax / pi

replacing Imax = 0.19 and pi = 3.14, then Iq = 60.4mA, this implies 30.2mA per tube...

At this point my desing is looking like:

Eb= 500V (seems to me a little high......for 25W....)
Ec2= 250V
Imax= 190mA
Iq= 60.4mA

Looking at the typicall operating conditions at the datasheet I found that the push-pull AB1 configuration for 26.5W is:

Plate voltage: 360V
Grid #2 voltage: 275V
Maximun signal plate current: 132mA
zero signal plate current: 88mA

In the book the tube used is a 6JN6, and all this makes perfect sense and very nice numbers...

but with I and my 6L6s seems to be like a tube fryer amp....

I realy liked this book, is very simple and and practical in concepts and desings, but I dont think is working for me... at least I believe that 500V of plate voltage is.... too high for just 25W of power...

Please any help or advise on this matter? I will really apreciate any input!

Saludos!
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Old 25th April 2012, 05:21 PM   #2
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I had the same question when I started with tubes. This is how it was explained to me.

The "design center maximum" ratings are for tubes at the "Quiescent" state, meaning no signal. It is perfectly acceptable for the voltage to swing above the rated maximum when a signal is applied. This goes for the plate power as well. Your loadline can pass into/through the maximum power curve, as long as your "Q" point does not exceed maximum power.

The best explanation is found in the RCA Radio Designers Handbook (1952), chapter 3. Do a search online, I found a PDF copy I could download.
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Old 4th June 2012, 09:32 PM   #3
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Quote:
Originally Posted by Anko deth View Post
At this point my desing is looking like:

Eb= 500V (seems to me a little high......for 25W....)
Ec2= 250V
Imax= 190mA
Iq= 60.4mA

Looking at the typicall operating conditions at the datasheet I found that the push-pull AB1 configuration for 26.5W is:

Plate voltage: 360V
Grid #2 voltage: 275V
Maximun signal plate current: 132mA
zero signal plate current: 88mA


Saludos!
Instead try following Charles' instruction near the end of the book using his spreadsheet, the results make more sense, I think... Take a look here.

Jaz
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