If the load impedance of the secondary 8-ohm and primary 3500 ohm electronic tubes; The secondary 4 ohm primary does not have a 3500 ohm electronic tube screen load impedance; Can an impedance switch be installed in the primary coil to maintain a constant load impedance of the output tube.
It was translated using translation software. I hope you can understand! thank you
It was translated using translation software. I hope you can understand! thank you
No I can’t. The impedance at the primary is reflected from the impedance at the secondary, which is reflected from the impedance at its load.
I understand what you're saying,
If I add a set of windings in the beginner level corresponding to a coil that outputs 4 ohms;
This way, the impedance reflected from the 4 ohm load to the primary remains constant at 3500 ohms.
3500:8 primary to secondary turns ratio of 22:1; By adding a primary winding and switching the switch, the turns ratio between the primary and secondary windings becomes 29.5:1,
In this way, whether it is a 4 ohm load or an 8 ohm load, the output power transistor always maintains a constant load impedance.
Is there such a design and production?
If I add a set of windings in the beginner level corresponding to a coil that outputs 4 ohms;
This way, the impedance reflected from the 4 ohm load to the primary remains constant at 3500 ohms.
3500:8 primary to secondary turns ratio of 22:1; By adding a primary winding and switching the switch, the turns ratio between the primary and secondary windings becomes 29.5:1,
In this way, whether it is a 4 ohm load or an 8 ohm load, the output power transistor always maintains a constant load impedance.
Is there such a design and production?
It might work in theory, but you should not do that.
1. The switch has to be high voltage rated. Don’t do that.
2. The plate of the output tube has to be connected all the time. If you toggle the switch, it momentarily disconnects the plate. That might damage the tube.
3. The bias point also changes after you toggle the switch.
1. The switch has to be high voltage rated. Don’t do that.
2. The plate of the output tube has to be connected all the time. If you toggle the switch, it momentarily disconnects the plate. That might damage the tube.
3. The bias point also changes after you toggle the switch.
Instead of 2 different primary windings with different turns, one way to make a 3500 Ohm primary work with 4 Ohm and 8 Ohm loads is to use 2 different turns numbers; usually they are connected in series to use the least turns possible.
Example: Secondary 100 turns to the 4 Ohm tap, plus 41 turns in series with the 100 turns, to make the 8 Ohm tap.
I have seen single ended that did use 2 taps for 3500 Ohms and 5000 Ohms. The windings were connected in series, with the total series turns for 5000 Ohms that are 1.41 times the turns for the 3500 Ohm primary tap.
That way, you could make it with only one secondary turns windings.
Example:
3500 : 8 Ohms
5000 : 8 Ohms
And, if you use all the primary turns (5000 tap), then 3500 : 4 Ohms
Example: Secondary 100 turns to the 4 Ohm tap, plus 41 turns in series with the 100 turns, to make the 8 Ohm tap.
I have seen single ended that did use 2 taps for 3500 Ohms and 5000 Ohms. The windings were connected in series, with the total series turns for 5000 Ohms that are 1.41 times the turns for the 3500 Ohm primary tap.
That way, you could make it with only one secondary turns windings.
Example:
3500 : 8 Ohms
5000 : 8 Ohms
And, if you use all the primary turns (5000 tap), then 3500 : 4 Ohms
For 3500 and 5000ohm primary the turnratio of 1.41 is wrong, it should be square root of 1.41=1.195
gorgon53,
Let me try and work the details out, and see If I am correct:
1.195 times the voltage in an Ohm tap.
Start with 4V in a 4 Ohm tap, terminated by 4 Ohms. Watts = (E squred)/R (4 squared)/4 = 4 watts
1.195 x 4 volts = 4.78V
Now, 4.78V on an 8 Ohm tap, terminated by 8 Ohms. Watts = (E squred)/R (4.78V squared)/8 = 2.856 Watts
Clearly, we did not give constant power into 4 Ohms versus 8 Ohms.
Regarding turns ratio: root applies to the Root of impedance, so you can start with impedance ration and get the turns ratio.
But the turns ration and voltage ratio are equal.
For constant Power:
There are 3dBV more in 8 Ohm tap, than dBv in 4 Ohms tap.
Proper load terminations: If the 4 Ohm tap and 4 Ohm load is 0dBV, then the 8 Ohm tap and 8 Ohm load is 3dBV.
0 dBV = 1 Volt.
dBV is according to 20log (Voltage measured/1 Volt)
Turns ratio is the same as the voltage ratio, Right?
Suppose there are 1.4x more total turns in an 8 Ohm tap, versus the total turns in 4 Ohms tap.
140 turns for 8 Ohms tap, requires 100 turns for 4 Ohm tap.
Turns: 140/100
Voltage: 1.40/1.00
20 log (1.4V/1.0V) = 2.92 dBV (close enough, the turns ration is 1.414213562 . . . to 1.0
3dBV is 1.41421.... x 1 Volt
3dBW is 2.0 times the wattage.
IRMC (I rest my case)
$0.04
Let me try and work the details out, and see If I am correct:
1.195 times the voltage in an Ohm tap.
Start with 4V in a 4 Ohm tap, terminated by 4 Ohms. Watts = (E squred)/R (4 squared)/4 = 4 watts
1.195 x 4 volts = 4.78V
Now, 4.78V on an 8 Ohm tap, terminated by 8 Ohms. Watts = (E squred)/R (4.78V squared)/8 = 2.856 Watts
Clearly, we did not give constant power into 4 Ohms versus 8 Ohms.
Regarding turns ratio: root applies to the Root of impedance, so you can start with impedance ration and get the turns ratio.
But the turns ration and voltage ratio are equal.
For constant Power:
There are 3dBV more in 8 Ohm tap, than dBv in 4 Ohms tap.
Proper load terminations: If the 4 Ohm tap and 4 Ohm load is 0dBV, then the 8 Ohm tap and 8 Ohm load is 3dBV.
0 dBV = 1 Volt.
dBV is according to 20log (Voltage measured/1 Volt)
Turns ratio is the same as the voltage ratio, Right?
Suppose there are 1.4x more total turns in an 8 Ohm tap, versus the total turns in 4 Ohms tap.
140 turns for 8 Ohms tap, requires 100 turns for 4 Ohm tap.
Turns: 140/100
Voltage: 1.40/1.00
20 log (1.4V/1.0V) = 2.92 dBV (close enough, the turns ration is 1.414213562 . . . to 1.0
3dBV is 1.41421.... x 1 Volt
3dBW is 2.0 times the wattage.
IRMC (I rest my case)
$0.04
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We are talking about a transformer having a primary 5000 ohm winding with a 3500 ohm tap. Say output is 4 ohm.
Winding and voltage ratio 5000/4 = 35.355
Winding and voltage ratio 3500/4 = 29.58
Primary winding and voltage ratio 5000 to 3500 tap = 35.355/29.58 = 1.195
If you want, you can use 5000/8 with a 3500/8 tap, but the full primary 5k to 3.5k tap ratio will still be 1.195 and not 1.41
No hard feelings, but IRMC also
Winding and voltage ratio 5000/4 = 35.355
Winding and voltage ratio 3500/4 = 29.58
Primary winding and voltage ratio 5000 to 3500 tap = 35.355/29.58 = 1.195
If you want, you can use 5000/8 with a 3500/8 tap, but the full primary 5k to 3.5k tap ratio will still be 1.195 and not 1.41
No hard feelings, but IRMC also
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Don’t say it. It’s been said too often in this thread. It’s meaningless. Output is measured in volts and amps. Ohms are reflected from the load. Terminology matters. And terms like dBv and ‘secondary 4 ohm primary’ have no place here.
This discussion is all very confused for something quite simple. If you have 3500 primary turns for a 4 ohm secondary, you need 3500x1.414=4949 primary turns for an 8 ohm secondary. But the better idea is to have a multi-tap secondary and leave the primary alone.
But if you put an 8R load on the secondary meant for 4R, you are REDUCING the load seen by the amp. This means slightly less power (valve amps don't quite behalf like transistor amps) but usually also lower THD.
If the amplifier doesn't mind the higher impedance on its output (LESS load) I would leave things as they are and use the whole transformer windings at all times. Some wonky designs might need the feedback network tweaking but the later good amps, eg any based on Mullard 5-20 should be very happy with no mods.
Hi, I have written an article some years ago, that explains this in a relatively easy way, see:
https://tubes.njunis.net/?p=117&lang=en
https://tubes.njunis.net/?p=117&lang=en
Post # 1 talks about the screen.
"The secondary 4 ohm primary does not have a 3500 ohm electronic tube screen load impedance"
The text was translated.
Is it lost in the translation?
Is it possible the plate was connected to a 5000 Ohm tap, and the screen was connected to the 3500 Ohm tap.
That would be Ultra Linear, at 70%.
Or is it possible that the desire was to have a multi tap primary, so that a switch can be used, allowing the tubes to always "see" 3500 Ohms, one switch position for an 8 Ohm load, and the other switch position for an 4 Ohm load?
What is the complete and correct translation for Post # 1?
"The secondary 4 ohm primary does not have a 3500 ohm electronic tube screen load impedance"
The text was translated.
Is it lost in the translation?
Is it possible the plate was connected to a 5000 Ohm tap, and the screen was connected to the 3500 Ohm tap.
That would be Ultra Linear, at 70%.
Or is it possible that the desire was to have a multi tap primary, so that a switch can be used, allowing the tubes to always "see" 3500 Ohms, one switch position for an 8 Ohm load, and the other switch position for an 4 Ohm load?
What is the complete and correct translation for Post # 1?
EJP, what you missed is, that my posts #6 and #8 where replies regarding post #5
So i tougth it would be a good idea to have that corrected so it would not mislead someone unfamiliar with simple things like that
IRMC, and would als very much appriciate not to receive any more "guidance" on what to say.
Read, try to understand what i said and what it refers to
Tnx
Clearly a 5k/3.5k transformers ilmpedance ratio is 1.41 and the total series turns for 5000 ohms should be 1.195 times the turns for the 3500 ohm primary (not 1.41)
So i tougth it would be a good idea to have that corrected so it would not mislead someone unfamiliar with simple things like that
IRMC, and would als very much appriciate not to receive any more "guidance" on what to say.
Read, try to understand what i said and what it refers to
Tnx
Yes.
The primary is designed to meet Saturation, HV & other requirements. Changing the secondary windings may lead to the Holy Smoke escaping and/or exciting Flash Bang stuff.
Multiple secondary taps would do the trick but my $0.02 is to leave things alone if using 8R speakers on a 4R optimised transformer. THD will be slightly better with good valve amp designs
Using 4R speaker son a 8R optimised winding is MUCH more dodgy so not recommended.
Gorgon53,
Correct. I was mistaken.
You said the turns for 5000 Ohms is 1.19 times the turns for 3500 Ohms.
Correct.
I was a little tired, and was thinking about the secondary with 8 and 4 Ohm taps. But that has 1.414 times the turns in the 8 Ohm tap, versus the turns in the 4 Ohm tap. The impedance ratio was 2:1.
It was not the 5000 Ohm to 3500 Ohm impedance ratio of 1.414, the impedances of this thread.
And that is how I made the error.
I will try and turn-over a new leaf; or perhaps turn-over a different number of transformer turns.
Correct. I was mistaken.
You said the turns for 5000 Ohms is 1.19 times the turns for 3500 Ohms.
Correct.
I was a little tired, and was thinking about the secondary with 8 and 4 Ohm taps. But that has 1.414 times the turns in the 8 Ohm tap, versus the turns in the 4 Ohm tap. The impedance ratio was 2:1.
It was not the 5000 Ohm to 3500 Ohm impedance ratio of 1.414, the impedances of this thread.
And that is how I made the error.
I will try and turn-over a new leaf; or perhaps turn-over a different number of transformer turns.
If the valve impedance is 3100 ohms that’s what you want in the transformer primary too, for optimal power transfer.
I have never understood this thread.
I have never understood this thread.
ejp,
Nice idea, For RF transmitters. Not nice for Audio.
Do you want low distortion, good damping factor, good frequency response in an Audio amplifier?
Then do not make the output tube plate impedance and the transformer primary impedance equal.
In RF, such as a 50 Ohm system, maximum power transfer is when output impedance is equal to the load impedance.
In RF, transmitter output impedance, transmission line impedance, and load impedance matching are Paramount.
Otherwise much of the power sent to the load are reflected back (consider that as a power loss at the load [not dissipated at the load]).
This is one place that the design characteristics of impedances for the best performance of RF systems,
Is very different than design characteristics of impedances, for the best performance of Audio systems.
The impedance ratios of source and load are different . . . RF versus Audio.
That is also why there is a lot of misunderstanding of speaker cables. Did you read any of those discussions/arguments?
Suppose the 8 Ohm tap of a tube amplifier has a damping factor of 10, the output impedance of the amplifier is 0.8 Ohms.
A speaker cable of 0.8 Ohms impedance is Not Practical.
And even if it was practical, the speaker load impedance is not 0.8 Ohms.
RF transmission line theory is great, but does Not apply to an Audio System.
Just my opinion . . . Prove me wrong, if it is wrong, it should be a simple task to write a bullet proof answer against what I said in this thread.
I am sorry if this starts another thread about speaker cables, if it does . . . OK
Nice idea, For RF transmitters. Not nice for Audio.
Do you want low distortion, good damping factor, good frequency response in an Audio amplifier?
Then do not make the output tube plate impedance and the transformer primary impedance equal.
In RF, such as a 50 Ohm system, maximum power transfer is when output impedance is equal to the load impedance.
In RF, transmitter output impedance, transmission line impedance, and load impedance matching are Paramount.
Otherwise much of the power sent to the load are reflected back (consider that as a power loss at the load [not dissipated at the load]).
This is one place that the design characteristics of impedances for the best performance of RF systems,
Is very different than design characteristics of impedances, for the best performance of Audio systems.
The impedance ratios of source and load are different . . . RF versus Audio.
That is also why there is a lot of misunderstanding of speaker cables. Did you read any of those discussions/arguments?
Suppose the 8 Ohm tap of a tube amplifier has a damping factor of 10, the output impedance of the amplifier is 0.8 Ohms.
A speaker cable of 0.8 Ohms impedance is Not Practical.
And even if it was practical, the speaker load impedance is not 0.8 Ohms.
RF transmission line theory is great, but does Not apply to an Audio System.
Just my opinion . . . Prove me wrong, if it is wrong, it should be a simple task to write a bullet proof answer against what I said in this thread.
I am sorry if this starts another thread about speaker cables, if it does . . . OK