Thanks. I didn't know this !
For the case of the first transformer, taps are 1W, 2W, 4W. So, for this case does that apply? I mean 1W x 4 = 4W, but 4W tap is not in the center. Am I wrong?
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Short answer is ohms law.
You know the voltage they are suppose to work with (100V) and you know the power (4W)
The current is P/V=0.04W and resistance is V^2/P= 2500 ohms
Assuming an efficiency of 100%, the power is the same at both sides: primary and secondary.
So, with 8 ohms you have V=(RP)^(1/2) = 5.65 V.
The voltage ratio of the transformer is 17.67.
Now do the same for all other taps and you have the impedances. Excel says it is:
Considering 66 Turns for the secondary of 8 ohms gives N=SQRT(Z_1/8)*66=
Or
The centre tap is where the impedance is 1/4 of the highest impedance value. Here it is the 2 Watt tap?.
Correct me if I am wrong.
With the 10W transformer, the 2.5W tap is 1/4 of the power, but also 1/4 of the max. impedance. Here they differ.
Cheers
You know the voltage they are suppose to work with (100V) and you know the power (4W)
The current is P/V=0.04W and resistance is V^2/P= 2500 ohms
Assuming an efficiency of 100%, the power is the same at both sides: primary and secondary.
So, with 8 ohms you have V=(RP)^(1/2) = 5.65 V.
The voltage ratio of the transformer is 17.67.
Now do the same for all other taps and you have the impedances. Excel says it is:
| Voltage | 100 | 100 | 100 | 100 |
| Wattage | 4 | 2 | 1 | 0.5 |
| Current | 0.04 | 0.02 | 0.01 | 0.005 |
| Sec | 8 | 8 | 8 | 8 |
| Sec Curr. | 0.70710678 | 0.5 | 0.35355339 | 0.25 |
| Sec. Volt. | 5.65685425 | 4 | 2.82842712 | 2 |
| Ratio | 17.6776695 | 25 | 35.3553391 | 50 |
| Imp Pr. | 2500 | 5000 | 10000 | 20000 |
Considering 66 Turns for the secondary of 8 ohms gives N=SQRT(Z_1/8)*66=
| Cumulative Turns | 1166.72619 | 1650 | 2333.45238 | 3300 |
| Turns | 1166.72619 | 483.273811 | 683.452378 | 966.547622 |
Correct me if I am wrong.
With the 10W transformer, the 2.5W tap is 1/4 of the power, but also 1/4 of the max. impedance. Here they differ.
Cheers
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Hi guys,
I think I made a mistake when building my amp.
I've connected the heaters of the 6N17B and 6N16B in series, since the specs are the same, but the voltages are very different.
The heater of the 6N17B has a voltage of 5.3V and the 6N16B has a voltage of 6.7V.
I guess, if I want to do this, I have to measure the resistance of each valve and use the ones that have the closest value. But as the current is 0.4A, even 1Ω resistance would give a difference of 0.4V.
Is there a way to use them in series, or do I have to just apply 6V to each one of them? Since the power adapter used is 12V DC, I could use either a 7806 or a resistive divider. The problem with the 7806 is that it will need a heatsink, which I would like to totally avoid.
Any ideas?
Cheers,
Pedro
I think I made a mistake when building my amp.
I've connected the heaters of the 6N17B and 6N16B in series, since the specs are the same, but the voltages are very different.
The heater of the 6N17B has a voltage of 5.3V and the 6N16B has a voltage of 6.7V.
I guess, if I want to do this, I have to measure the resistance of each valve and use the ones that have the closest value. But as the current is 0.4A, even 1Ω resistance would give a difference of 0.4V.
Is there a way to use them in series, or do I have to just apply 6V to each one of them? Since the power adapter used is 12V DC, I could use either a 7806 or a resistive divider. The problem with the 7806 is that it will need a heatsink, which I would like to totally avoid.
Any ideas?
Cheers,
Pedro
If you intend to connect tube heaters in series, you'd have to check the individual heater currents, not the voltages. Both nominally are the same for your tubes. So you need to connect a resistor in parallel with the higher voltage tube with a resistance that yields equal voltages for both.
Best regards!
Best regards!
These tubes should have the same heater current and voltage. I never took a closer look at the current/voltage.
The datasheet says, they have same heater voltage and current. Current can vary 10%, though.
You can try the resistor to balance them, as Kay said.
You can also use a 1 ohm resistor to check the actual current of the series string, to help identifying the right resistor.
The datasheet says, they have same heater voltage and current. Current can vary 10%, though.
You can try the resistor to balance them, as Kay said.
You can also use a 1 ohm resistor to check the actual current of the series string, to help identifying the right resistor.
Hi,
I am not sure I fully agree with this.
The voltage is fixed and if both heaters are connected in series, they have to have the same current flowing through them. The only difference is the resistance of each filament, which will cause a difference in voltage.
I have 12V connected to one side of the 6N16N and the other side connected to the 6N17B, which in turn connects to ground.
For the 6N17B I've measure a resistance of 6.4 ohm while for the 6N16B I've measured a resistance of 8ohm.
So, using the resistor divider equation:
V_6N17B = R_6N17B / (R_6N17B + R_6N17B) * 12 -> V_6N17B = 5.3V
V_6N16B = 12 - 5.3 = 6.7V
and that matches exactly what I am measuring.
As you said, I can add a resistor in parallel with 8 ohms to obtain 6.4 ohms (or a 1.6 ohm in series with 6.4 ohm to obtain 8ohm).
So I can obtain 1.5ohm using two 3 Ohm resistors in parallel (close enough to 1.6 ohm)
To obtain 6.4 ohm out of an 8 ohm resistor I will need a resistor of 0.7 ohm, which I can achieve using 4 x 3 ohm in parallel (0.75 ohms).
On top of this, I will have the resistor's tolerance that will make these values deviate from the calculated ones. At least they will be closer than what I have now.
Does this make sense ?
Cheers,
Pedro
Hi guys,
I came across an approach for designing an inverterless push-pull amp that differs from the traditional one (post #112).
Instead of grounding the grid of one of the valves, the output (anode) of one will be fed into the grid of the other, making the signals actually out-of-phase.
Has anyone tried the approach before ? This one, should work as a "proper" pp output stage, where the traditional will only work in class A and give half the power.
Cheers,
Pedro
http://electronbunker.ca/eb/TubeAudio_01a.html
I came across an approach for designing an inverterless push-pull amp that differs from the traditional one (post #112).
Instead of grounding the grid of one of the valves, the output (anode) of one will be fed into the grid of the other, making the signals actually out-of-phase.
Has anyone tried the approach before ? This one, should work as a "proper" pp output stage, where the traditional will only work in class A and give half the power.
Cheers,
Pedro
http://electronbunker.ca/eb/TubeAudio_01a.html
I have not tried this one yet, but would be something interesting to simulate.
On the other hand, this seems very much like the self-split approach. What happens in class B? Where is the signal coming from if the tube where the feedback comes from is in cut-off? I might be missing something, though.
On the other hand, this seems very much like the self-split approach. What happens in class B? Where is the signal coming from if the tube where the feedback comes from is in cut-off? I might be missing something, though.
Hi,
Great observation Thomas and thanks Kay for clarifying this.
Question:
When looking at feedback loops, I've noticed that the loop only connects to one of the taps of the output transformer and not to each one of them.
If I have a transformer with 4 and 8 ohm taps and the feedback resistor is connected to the 4 ohm tap, what will happen if I use the 8 ohm tap ?
Cheers,
Pedro
Great observation Thomas and thanks Kay for clarifying this.
Question:
When looking at feedback loops, I've noticed that the loop only connects to one of the taps of the output transformer and not to each one of them.
If I have a transformer with 4 and 8 ohm taps and the feedback resistor is connected to the 4 ohm tap, what will happen if I use the 8 ohm tap ?
Cheers,
Pedro
Hi,
I have just connected the amp to a spectrum analyser and the reslt was 50Hz noise and many of its harmonics.
Please find, below screenshots and a file with the sound of 50Hz noise and, later, at max volume oscillation.
From my personal dropbox:
https://www.dropbox.com/scl/fi/gdl0...tion.mp3?rlkey=xv86p4re7cqkfr2wuu84us11n&dl=0
Cheers,
Pedro
I have just connected the amp to a spectrum analyser and the reslt was 50Hz noise and many of its harmonics.
Please find, below screenshots and a file with the sound of 50Hz noise and, later, at max volume oscillation.
From my personal dropbox:
https://www.dropbox.com/scl/fi/gdl0...tion.mp3?rlkey=xv86p4re7cqkfr2wuu84us11n&dl=0
Cheers,
Pedro