Hi Gents,
Re above from data Opt is 2.5/3.5k 10 watt, max ip 100 ma.
Question if I run this opt with 300B at 80 ma 425v for targetted 8 watt output
would core saturation set in ?
Thanks
Re above from data Opt is 2.5/3.5k 10 watt, max ip 100 ma.
Question if I run this opt with 300B at 80 ma 425v for targetted 8 watt output
would core saturation set in ?
Thanks
Depends of power AND frequency.
This OPT was used in Sun Audio amplifiers, mainly with 2A3, less often with 300B.
It has -relatively- small footprint (cut core?), so I would be surprised if it will be at 20Hz more Watts capable without rising distortion.
p.s.
http://www.wiredstate.com/forum/viewtopic.php?f=11&t=18684
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There will be less saturation if you use the 2.5k primary tap; instead of the 3.5k primary tap.
But,
The damping factor of a 300B driving 2.5k primary, is slightly less than 3.6
The damping factor of a 300B driving a 3.5k primary is slightly less than 5.0
The damping factor for the most part is due to the primary impedance / plate resistance, rp.
Given an rp of 700 Ohms . . .
The best the damping factor can be is 3.6 or 5.0, but they are both lower than that because the DCR of the primary is in series with the 700 Ohms; and the DCR of the secondary is in series with the load (resistor or loudspeaker).
An electric bass guitar goes down to about 40Hz (nowhere near 20Hz).
Build it and see what you get. Have fun listening.
But,
The damping factor of a 300B driving 2.5k primary, is slightly less than 3.6
The damping factor of a 300B driving a 3.5k primary is slightly less than 5.0
The damping factor for the most part is due to the primary impedance / plate resistance, rp.
Given an rp of 700 Ohms . . .
The best the damping factor can be is 3.6 or 5.0, but they are both lower than that because the DCR of the primary is in series with the 700 Ohms; and the DCR of the secondary is in series with the load (resistor or loudspeaker).
An electric bass guitar goes down to about 40Hz (nowhere near 20Hz).
Build it and see what you get. Have fun listening.
After a couple hours of use even big cores saturate, mainly if the speaker are 4Ω for this reason before the transistor era all speakers were 16Ω, It was a time when there was still no planned obsolescence.
FullRangeMan,
Decades ago, when several well known brands of loudspeaker manufacturers used Edge Wound Ribbon Voice Coils, 16 Ohms was quite common . . .
I believe one reason was the higher efficiency of those types of loudspeaker "motor".
Please explain how: "a couple hours of use even big cores saturate".
Decades ago, when several well known brands of loudspeaker manufacturers used Edge Wound Ribbon Voice Coils, 16 Ohms was quite common . . .
I believe one reason was the higher efficiency of those types of loudspeaker "motor".
Please explain how: "a couple hours of use even big cores saturate".
FullRangeMan,
Before the transistor era in Europe impedances of 3, 5, 7, and 8 Ohm were very common while 16 Ohm was not common at all.
Like 6A3sUMMER, I like to see you explain/proof your statement about even big cores saturating after a couple of hours, and why this effect would mainly occur with 4 Ohm loudspeakers. I never heard of that effect. So unless you explain/prove it, I don't believe it to be true.
Before the transistor era in Europe impedances of 3, 5, 7, and 8 Ohm were very common while 16 Ohm was not common at all.
Like 6A3sUMMER, I like to see you explain/proof your statement about even big cores saturating after a couple of hours, and why this effect would mainly occur with 4 Ohm loudspeakers. I never heard of that effect. So unless you explain/prove it, I don't believe it to be true.
Assuming we have total flux density factor of 1 distributed in 50% for DC magnetizing current to 50% for AC swing.
0.5 times flux density for 100mA DC magnetization current and 0.5 times for AC swing for a fixed frequency to give 10W
That means at 80mA DC you'll be using 80/100 * 0.5 = 0.4 factor DC flux density. It decreases linearly with decreasing current.
That permits you the chance to have a 0.6 factor (increase) of AC flux density.
AC flux density is linear with voltage, but power increases in a square way. With a 0.6 flux density factor you can basically increase primary swing to 0.6/0.5 = 1.2 times increase in voltage and 1.2^2=1.44 times increase in power, so the 10W max power should become 14.4W.
This power calculation is takes from the POV of voltage swing. You need to keep up with the equivalent current swing as well. For example a quiescent current of 80mA translates to 56,7mA RMS, which ( 0.0567^2 * 3500 ) translates into 11,25W of output power. You will be basically limited by your quiescent current.
What about frequency? That can make a huge difference.
I have seen much troubles with 2500Ω transformers at low frequencies because the tube is running out of current.
I have seen much troubles with 2500Ω transformers at low frequencies because the tube is running out of current.
Of course. However I see no way determining the flux density at a frequency and amplitude without knowledge on primary turns amount and core.
If you need to get a specific number of Watts output from a tube amplifier, be sure to calculate the output transformer's insertion loss.
That way, you know how many Watts the tube has to deliver to the output transformer's primary.
Example, With an output transformer that has:
A 3500 Ohm primary and a DCR of 175 Ohms
An 8 Ohm secondary and a DCR of 0.4 Ohms
About 0.45 dB is lost in the primary's DCR
About 0.45 dB is lost in the secondary's DCR
The output transformer's insertion loss will be about 0.9 dB (even if there are no other losses in the transformer).
-0.9 dB power loss = 1 / 1.23
If the output tube delivers about 9.8 Watts to the primary, 9.8 / 1.23
Then the secondary will deliver about 8 Watts to an 8 Ohm load.
(8 Watts from the tube = 6.5 Watts at the secondary).
For 80mA plate current, saturation will occur earlier on the 3.5k tap, and later on the 2.5k tap.
However, the inductance is higher on the 3.5k tap, versus on the 2.5k tap.
Try both taps, easy to change.
A 300B at 300Vp-k and 60mA has an rp of 700 Ohms. p-k = plate to cathode (filament = "cathode").
A 2A3 at 250Vp-k and 60mA has an rp of 800 Ohms.
A lower rp drives an inductance easier than a higher rp.
All Generalizations Have Exceptions.
Happy designing, building, and listening!
That way, you know how many Watts the tube has to deliver to the output transformer's primary.
Example, With an output transformer that has:
A 3500 Ohm primary and a DCR of 175 Ohms
An 8 Ohm secondary and a DCR of 0.4 Ohms
About 0.45 dB is lost in the primary's DCR
About 0.45 dB is lost in the secondary's DCR
The output transformer's insertion loss will be about 0.9 dB (even if there are no other losses in the transformer).
-0.9 dB power loss = 1 / 1.23
If the output tube delivers about 9.8 Watts to the primary, 9.8 / 1.23
Then the secondary will deliver about 8 Watts to an 8 Ohm load.
(8 Watts from the tube = 6.5 Watts at the secondary).
For 80mA plate current, saturation will occur earlier on the 3.5k tap, and later on the 2.5k tap.
However, the inductance is higher on the 3.5k tap, versus on the 2.5k tap.
Try both taps, easy to change.
A 300B at 300Vp-k and 60mA has an rp of 700 Ohms. p-k = plate to cathode (filament = "cathode").
A 2A3 at 250Vp-k and 60mA has an rp of 800 Ohms.
A lower rp drives an inductance easier than a higher rp.
All Generalizations Have Exceptions.
Happy designing, building, and listening!
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Uhhh...
A transformer 3500/8 Ohm with 175Ohm primary and 0,4 secondary has a loss of 0,43dB and 91% efficiency. Luckely less bad 🙂
If there is 9,8W from the tubes, secondary is 8,9W (only for the copper losses)
A transformer 3500/8 Ohm with 175Ohm primary and 0,4 secondary has a loss of 0,43dB and 91% efficiency. Luckely less bad 🙂
If there is 9,8W from the tubes, secondary is 8,9W (only for the copper losses)
btw the copperlosses have also impact at the distortion of the transformer. Keep the losses as low as possible i prefer at least below 0,25dB for a real top transformer.
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tubes4all,
It is 'as bad' as I said:
A two resistor series string measures 3500 Ohms. (Simulating a primary winding impedance)
The top resistor is 175 Ohms, and the bottom resistor is 3325 Ohms: 175 + 3325 = 3500 (primary winding impedance)
1 Volt is applied to the top of the 175 Ohm resistor, and the other end of 1 Volt is returned to the bottom of the 3325 Ohm resistor.
The voltage across the 3325 Ohm resistor is:
1 Volt x (3325 / (3325 + 175)) = 1 Volt x (3325 / 3500) = 0.95 Volt
The two resistor divider loss to the junction of the 175 Ohm and 3325 Ohm is:
20 x Log (0.95) = - 0.4455 dB
A two resistor series string measures 8 Ohms. (Simulating a secondary winding impedance)
8 Ohm - 0.4 Ohm = a Voltage divider of 0.4 Ohm and 7.6 Ohm.
1 Volt x (7.6 / (7.6 + 0.4)) = 1 Volt x (7.6 / 8) = 0.95 Volt
20 x (0.95) = - 0.4455 dB
A loss of 0.4455 plus another loss of 0.4455 = a loss of 0.891 dB
0.891 dB loss; not the 0.43 dB loss you calculated.
1 / (10^x), where x = (0.891 / 10)) = 0.8145% power efficiency
The efficiency is 0.8145%; not the 91% efficiency you calculated.
10 Watts x 0.8145 = 8.145 Watts
I often make mistakes in my calculations, please show me if my above calculations are incorrect in any way.
Thank You, tubes4all!
It is 'as bad' as I said:
A two resistor series string measures 3500 Ohms. (Simulating a primary winding impedance)
The top resistor is 175 Ohms, and the bottom resistor is 3325 Ohms: 175 + 3325 = 3500 (primary winding impedance)
1 Volt is applied to the top of the 175 Ohm resistor, and the other end of 1 Volt is returned to the bottom of the 3325 Ohm resistor.
The voltage across the 3325 Ohm resistor is:
1 Volt x (3325 / (3325 + 175)) = 1 Volt x (3325 / 3500) = 0.95 Volt
The two resistor divider loss to the junction of the 175 Ohm and 3325 Ohm is:
20 x Log (0.95) = - 0.4455 dB
A two resistor series string measures 8 Ohms. (Simulating a secondary winding impedance)
8 Ohm - 0.4 Ohm = a Voltage divider of 0.4 Ohm and 7.6 Ohm.
1 Volt x (7.6 / (7.6 + 0.4)) = 1 Volt x (7.6 / 8) = 0.95 Volt
20 x (0.95) = - 0.4455 dB
A loss of 0.4455 plus another loss of 0.4455 = a loss of 0.891 dB
0.891 dB loss; not the 0.43 dB loss you calculated.
1 / (10^x), where x = (0.891 / 10)) = 0.8145% power efficiency
The efficiency is 0.8145%; not the 91% efficiency you calculated.
10 Watts x 0.8145 = 8.145 Watts
I often make mistakes in my calculations, please show me if my above calculations are incorrect in any way.
Thank You, tubes4all!
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If i take this transformer as an example the losses i calculated are exactly the same as this datasheet
http://www.monolithmagnetics.com/si...0B single ended output transformer prelim.pdf
You can use your formules and check if your calculations are correct as well.
http://www.monolithmagnetics.com/si...0B single ended output transformer prelim.pdf
You can use your formules and check if your calculations are correct as well.
A Challenge for all of you Simulation Experts out there, please prove that my below calculations are wrong.
Thanks to any and all of you!
tubes4all,
Monolith Magnetics is well represented and respected in the industry.
Yes. And all of us do make mistakes.
However, you did not dis-prove my Post # 14 calculations.
I based it on a voltage divider principle, that is as old as electronics.
If you measure the primary impedance (secondary loaded) the impedance is the sum of the winding impedance in series with the primary DCR.
If you carefully measure the primary impedance (with the secondary un-loaded), you get a much higher impedance, it is the inductive reactance in parallel with the distributed capacitance (careful, you may be near resonance of the two), that is in series with the DCR. But at such high impedance, the DCR essentially drops out of the equation.
Using a $50,000 Vector Network Analyzer like I did, I measure the low frequency inductive reactance and calculate the inductance. Then I measure the high frequency distributed capacitive reactance, and calculate the capacitance.
Then I can calculate to predict the impedance at mid frequency from the parallel inductance and capacitance (and measure the impedance at mid frequency to check if the calculation is correct). At that high impedance, the series DCR essentially drops out of the equation when compared to the high impedance of the un-loaded parallel inductance and capacitance.
Note: many of the output transformers that I tested had un-loaded resonance in the range of 500Hz to 2kHz.
http://www.monolithmagnetics.com/si...0B single ended output transformer prelim.pdf
The Monolith 3300 Ohm primary data sheet says:
Primary DCR 106 Ohms
8 Ohm Secondary DCR 0.254 Ohms
Voltage divider action of primary (3300 - 106) / 3300 = 0.967878788;
20 x Log of that = - 0.2836 dB
Voltage divider action of secondary (8 - 0.254) / 8 = 0.96825;
20 x Log of that = - 0.2802 dB
Total loss of primary 0.2836 and secondary 0.2802 = - 0.5638dB
Thanks to any and all of you!
tubes4all,
Monolith Magnetics is well represented and respected in the industry.
Yes. And all of us do make mistakes.
However, you did not dis-prove my Post # 14 calculations.
I based it on a voltage divider principle, that is as old as electronics.
If you measure the primary impedance (secondary loaded) the impedance is the sum of the winding impedance in series with the primary DCR.
If you carefully measure the primary impedance (with the secondary un-loaded), you get a much higher impedance, it is the inductive reactance in parallel with the distributed capacitance (careful, you may be near resonance of the two), that is in series with the DCR. But at such high impedance, the DCR essentially drops out of the equation.
Using a $50,000 Vector Network Analyzer like I did, I measure the low frequency inductive reactance and calculate the inductance. Then I measure the high frequency distributed capacitive reactance, and calculate the capacitance.
Then I can calculate to predict the impedance at mid frequency from the parallel inductance and capacitance (and measure the impedance at mid frequency to check if the calculation is correct). At that high impedance, the series DCR essentially drops out of the equation when compared to the high impedance of the un-loaded parallel inductance and capacitance.
Note: many of the output transformers that I tested had un-loaded resonance in the range of 500Hz to 2kHz.
http://www.monolithmagnetics.com/si...0B single ended output transformer prelim.pdf
The Monolith 3300 Ohm primary data sheet says:
Primary DCR 106 Ohms
8 Ohm Secondary DCR 0.254 Ohms
Voltage divider action of primary (3300 - 106) / 3300 = 0.967878788;
20 x Log of that = - 0.2836 dB
Voltage divider action of secondary (8 - 0.254) / 8 = 0.96825;
20 x Log of that = - 0.2802 dB
Total loss of primary 0.2836 and secondary 0.2802 = - 0.5638dB
Did you see that MonolithMagnetics give a coppeloss 0,27dB ? Or they* are wrong or you are wrong.
Maybe you use another definition of copperloss?
* also a brand as Tango use the same calculations as Monolith Magnetics
The easiest way to understand transformer Rdc is imagining a cable with series resistance, contributing to power losses.
The secondary has its resistance reflected to the primary times the impedance ratio. The sum of the reflected secondary Rdc + and the primary Rdc is your overall Rdc value.
As a rule of thumb for my transformers, I avoid OPTs losses bigger than 20:1 Impedance/Rdc ratio. I keep my projects between 40:1 and 20:1.
The secondary has its resistance reflected to the primary times the impedance ratio. The sum of the reflected secondary Rdc + and the primary Rdc is your overall Rdc value.
As a rule of thumb for my transformers, I avoid OPTs losses bigger than 20:1 Impedance/Rdc ratio. I keep my projects between 40:1 and 20:1.
So between 0,34 and 0,22dB.
I prefer the lower value, remember it dosn't act as a normal resistor.