Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: gontek on April 10, 2025, 05:30:26 am
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Hi, I'm trying to calculate perfect transformer for my project,
As I know (from Rob Robinette calculator) OT impedance = PlateVoltage^2/PlateDissipation,
but should I get idle or max PlateDissipation?
And, why typical tranformer for 50 El34 has 3.4K?
if we, let's say, have: 420 B+ and 25W PD ? it should be aroud 7k
sorry for silly question...
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50 EL34 !! You mean 50 watts amp with EL34 ?
No need to break your head with calculation , see transformer manufacturer recommendations.
Hammond;
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50 EL34 !! You mean 50 watts amp with EL34 ?
No need to break your head with calculation , see transformer manufacturer recommendations.
Hammond;
thx, yep, I meant 50W,
my project uses pcl86 tubes, they are rated 9W plate dissipation, and with my, not typical B+ =~316V, there is no dedicated transformer
why 1750P & 1750Q impedance differs so much? If jcm800 & jtm45 plate voltages seems to be close
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Why 50 watts (and why are talking about EL34) if you use 9 watts tubes ?
Will you use 6 tubes ?
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Why 50 watts and why are talking EL34 if you use 9 watts tubes ?
Will you use 6 tubes ?
no, I'm just trying to resolve impedance calculations on real amp,
why jcm800 uses 3.2K, instead of ~7K like JTM45?
Is it just bias difference?
Is formula (B+*B+)/PlateDissipation correct?
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Read answer no 3
https://ampgarage.com/forum/viewtopic.php?t=36600
and here ;
https://el34world.com/Forum/index.php?topic=26516.0
https://theultimatetone.com/thread-33.html
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May this be of some help ?
(https://i.imgur.com/z5cajh0.png)
Note the relation from B+ Voltage -- G2 Voltage and Zout
Franco
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Is formula (B+*B+)/PlateDissipation correct?
It is a simplification, a ball-park figure, an approximation, not a precise design formula.
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thank you for your patience,
I see, maybe, my question is not so stupid,
I've got a great idea, I will order Hammond 125E, which has a lot of different ratios, and will experiment :)
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thank you for your patience,
I see, maybe, my question is not so stupid,
I've got a great idea, I will order Hammond 125E, which has a lot of different ratios, and will experiment :)
Référence: 125E
Audio transformer, universal push-pull tube output, 15 watts
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I'm just trying to resolve impedance calculations on real amp,
why jcm800 uses 3.2K, instead of ~7K like JTM45?
Ohm's Law: Resistance = Volts / Current
Or ---------> Impedance = Volts / Current
Or ----> Impedance is a ratio of (AC) Volts to (AC) Current
Forget "Idle Bias" & "Idle Plate Dissipation" as those are mostly irrelevant to the output transformer.
The Output Transformer primary Impedance is about AC Volts and AC Current present when we play the amp.
What AC Volts do we need across each Primary Impedance to develop 50 Watts?
Volts = √(Power x Impedance)
JTM45: Volts = √(50w x 7kΩ) = 591.6v AC ----> ~418v Peak per side
JCM800: Volts = √(50w x 3.4kΩ) = 412v.3 AC ----> ~291.5v Peak per side
With "Volts" lower and "Impedance" lower for the JCM800 output transformer, then how did "Current" change?
Ohm's Law
Current = Volts / Impedance
JTM45: Current = 591.6v AC / 7kΩ = 84.5mA ----> 119.5mA Peak
JCM800: Current = 412v.3 AC / 3.4kΩ = 121.3mA ----> ~171.5mA Peak
For the same power output, the JCM800 must pull more Current through the output transformer while the voltage-drop developed across the Primary Impedance is smaller. The Ratio of Volts to Current changed.
Where this matters is output tube screen voltage, and how hot the tube runs when driven.
The screen voltage needs to be high enough to support the peak plate current pulled by the tube.
The higher peak plate current means the tube runs hotter when driven with audio.
The hotter operation of the tube (with high peak plate current) when driven means its Idle Bias needs to be cooler, so it spends a little longer in cut-off to cool the plate. This usually is not accounted for in the simple Bias Calculators that work on idle plate dissipation.
Using lower or higher output transformer Primary Impedance can also change how the output tube distorts.
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First of all:
Va max for PCL86 is 250V.
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First of all:
Va max for PCL86 is 250V.
Very low !
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Hi, I'm trying to calculate perfect transformer for my project,
As I know (from Rob Robinette calculator) OT impedance = PlateVoltage^2/PlateDissipation
my project uses pcl86 tubes, they are rated 9W plate dissipation, and with my, not typical B+ =~316V, there is no dedicated transformer
First of all:
Va max for PCL86 is 250V.
no, I'm just trying to resolve impedance calculations on real amp
Whether you know it or not you are asking, "How do I design an Output Section?"
Since we're probably not gonna learn Electronic Design in a day, copy an existing plan:
Hammond AO-44 (https://www.captain-foldback.com/Hammond_sub/schematics/AO44.jpg)
PCL86 (https://frank.pocnet.net/sheets/030/p/PCL86.pdf) = ECL86 (https://frank.pocnet.net/sheets/010/e/ECL86.pdf) with a different heater
Data Sheet says "250v max" but Hammond had 310v at their plate. Nobody makes "new PCL86" so whatever you get your hands on will be able to tolerate the same.
Page 4 of the linked ECL86 data sheet shows load impedance for push-pull ECL86 pentodes at 8-9kΩ. You can assume anything 7-10kΩ will be "close enough."
The Hammond AO-44 schematic has notes at the output transformer that can be used to calculate a similar result.
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First of all:
Va max for PCL86 is 250V.
Wut?
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First of all:
Va max for PCL86 is 250V.
This notes 300V?
https://tube-data.com/sheets/010/p/PCL86.pdf
First of all though might rather be the heater supply:- how to provide 13.3V?
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HBP, links not working.
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HBP, links not working.
This one ? It work for me
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I have from the Philips Datasheet oft he PCL86:
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I have from the Philips Datasheet oft he PCL86:
That's the 1961 datasheet, up-rated to 300V in 1968.