tubes4all,
I do not have any of the test equipment you mentioned.
Please post a schematic on this thread, of your favorite amplifier that you designed.
Do not be afraid that someone will steal your circuit.
Or is the favorite amplifier that you own, a commercial product. If so, then do not post the schematic.
Have fun!
I have posted some of my amplifier designs on this forum (actual built and working amplifiers).
I will be posting more of my amplifier designs on this forum (actual built and working amplifiers).
I hope somebody copies and builds those designs.
No fear here.
I do not have any of the test equipment you mentioned.
Please post a schematic on this thread, of your favorite amplifier that you designed.
Do not be afraid that someone will steal your circuit.
Or is the favorite amplifier that you own, a commercial product. If so, then do not post the schematic.
Have fun!
I have posted some of my amplifier designs on this forum (actual built and working amplifiers).
I will be posting more of my amplifier designs on this forum (actual built and working amplifiers).
I hope somebody copies and builds those designs.
No fear here.
You get what you paid for. The windings must be all over the place? p.s. you should run a feedback loop to this wide bandwidth amp
GNF loves a good transformer@6A3sUMMER
Thanks, so there’s a 1 dB loss on the full range due to DCR.
39,5 H are 4963 Ohm at 20 Hz, so:
20 log (1980/3110) = -3.9 dB
105 mH are 4.98 Ohm at 20 Hz, so:
20 log (5.68 / 8.7) = -3.7 dB
Totally -7.6 dB at 20 Hz compared to midband, -8.6 dB considering input power. Is it correct?
At 20 kHz 7.6 nF are 1047 Ohm, so:
20 log (783/3110) = -12 dB
Here comes what the producer said: the capacitance applies to only a small part of the winding. How this will affect the calculations? I think I will need to consider in parallel to a part of the winding, not the full impedance.
Indeed many on this forum use this transformer and are happy with it.
Doing the calculation considering EG only 1/3 of the windings in parallel to this capacity, I’ll get:
1047 || (3110/3) = 521 Ohm
So the primary would become: 3110/3 + 521 + 3110/3 = 2594 Ohm
20 log (2594/3110) = -1.57 dB at 20 kHz compared to midband, -2.57 dB considering input power.
So I guess that capacity will be in parallel with something like 1/3 to 1/2 of the winding.
Am I doing the calculations right?
How those numbers are influenced by output tube(s) rp?
Thanks
Roberto
Thanks, so there’s a 1 dB loss on the full range due to DCR.
39,5 H are 4963 Ohm at 20 Hz, so:
20 log (1980/3110) = -3.9 dB
105 mH are 4.98 Ohm at 20 Hz, so:
20 log (5.68 / 8.7) = -3.7 dB
Totally -7.6 dB at 20 Hz compared to midband, -8.6 dB considering input power. Is it correct?
At 20 kHz 7.6 nF are 1047 Ohm, so:
20 log (783/3110) = -12 dB
Here comes what the producer said: the capacitance applies to only a small part of the winding. How this will affect the calculations? I think I will need to consider in parallel to a part of the winding, not the full impedance.
Indeed many on this forum use this transformer and are happy with it.
Doing the calculation considering EG only 1/3 of the windings in parallel to this capacity, I’ll get:
1047 || (3110/3) = 521 Ohm
So the primary would become: 3110/3 + 521 + 3110/3 = 2594 Ohm
20 log (2594/3110) = -1.57 dB at 20 kHz compared to midband, -2.57 dB considering input power.
So I guess that capacity will be in parallel with something like 1/3 to 1/2 of the winding.
Am I doing the calculations right?
How those numbers are influenced by output tube(s) rp?
Thanks
Roberto
Hi,
I agree; on my side I have always posted diagrams and dozen of test lab ( also with photo)
I am lucky because in addition on my two AP + other stuff ( three tube tester + Sofia + R&S) I have access at the Audioreview lab with three AP (+ others) where in some case there are automatic test procedures written inside the lab.
This make a great differences respect the the thousands of simulation I see whitout a complete set of real test lab so they stay in the virtual world.
And I can't understand some guys tha insist to speak without any real data to discuss, just one square wave at 80mV (just an example
)to Zintolo.
Do you have a minum of test set or not?
Try to use a good sound card and ARTA (with LIMP) or REW software. Also Spectralab is interesting but is costs.
So you can chek life what happen; the numbers helps but the real life can be (is ) different
Walter
“ And I can't understand some guys tha insist to speak without any real data to discuss, just one square wave at 80mV (just an example ) “
Yes Walter, that is quite understandable if you don't understand the physics of a transformer. People helped you with the Radiotron designers manual and Partridge's articles, but it didn't help, you ignored the theory.
And because you don't understand transformers, you don't understand the value of measurements, and bragging about expensive equipment doesn't help either.
) “ Yes Walter, that is quite understandable if you don't understand the physics of a transformer. People helped you with the Radiotron designers manual and Partridge's articles, but it didn't help, you ignored the theory.
And because you don't understand transformers, you don't understand the value of measurements, and bragging about expensive equipment doesn't help either.
zintolo,
The -1 dB loss from the DCRs of the transformer applies at 1kHz, I have not seen significant losses at that frequency other than due to DCRs.
There can be additional losses at low and high frequencies.
Examples: The primary inductance at low frequencies, and the leakage reactance at high frequencies (in addition to the effective distributed capacitance, however it may be distributed, if it is only over part of the windings, that makes it very complex to find the total effect).
Perhaps that uneven distribution of capacitance is due to the method of toroid windings.
I would expect that if 7.6nF is in parallel with only part of the primary windings, the resonance of LC might be of some concern.
Suppose it is in the middle 1/3 of the windings. That is like L, series LC, L. What kind of primary impedance does that present versus frequency.
I am not sure I follow some of your calculations.
As you said, 39.5H is 4964 Ohms at 20Hz. That is correct.
The 300B @ 300V plate to filament, and 60mA plate current has a plate impedance, rp, of 700 Ohms.
At 20Hz, 700 Ohms rp drives the Complex impedance of L (4964 Ohms) that is in parallel with the reflected load impedance to the primary (5,000 Ohms).
At 1 kHz, rp of 700 Ohms drives 5k Ohms. 300B u = 3.85
Gain at 1kHz = 3.85 (5000/5700) = 3.377
If at some low frequency, the transformer complex impedance presents 2500 Ohms, then . . .
Gain at 1kHz = 3.85 (2500/3200) = 3.01
Comparing the response of 1kHz to the low frequency, we get 20 log (3.01/3.377) = -0.999 dB (-1dB)
If the transformer primary at 20Hz only presents 2500 Ohm, then the response is only -1 dB from the output at 1kHz, where the primary presents 5000 Ohm.
The insertion loss from DCR is a constant (-1dB).
Frequency response is relative to 1kHz, so at the -3dB low frequency, and -3dB high frequency, they are 1/2 power versus what they are at 1kHz.
By the way, 105mH at 20 Hz is 13.2 Ohms . . . it is Not 4.98 Ohms as you stated.
The -1 dB loss from the DCRs of the transformer applies at 1kHz, I have not seen significant losses at that frequency other than due to DCRs.
There can be additional losses at low and high frequencies.
Examples: The primary inductance at low frequencies, and the leakage reactance at high frequencies (in addition to the effective distributed capacitance, however it may be distributed, if it is only over part of the windings, that makes it very complex to find the total effect).
Perhaps that uneven distribution of capacitance is due to the method of toroid windings.
I would expect that if 7.6nF is in parallel with only part of the primary windings, the resonance of LC might be of some concern.
Suppose it is in the middle 1/3 of the windings. That is like L, series LC, L. What kind of primary impedance does that present versus frequency.
I am not sure I follow some of your calculations.
As you said, 39.5H is 4964 Ohms at 20Hz. That is correct.
The 300B @ 300V plate to filament, and 60mA plate current has a plate impedance, rp, of 700 Ohms.
At 20Hz, 700 Ohms rp drives the Complex impedance of L (4964 Ohms) that is in parallel with the reflected load impedance to the primary (5,000 Ohms).
At 1 kHz, rp of 700 Ohms drives 5k Ohms. 300B u = 3.85
Gain at 1kHz = 3.85 (5000/5700) = 3.377
If at some low frequency, the transformer complex impedance presents 2500 Ohms, then . . .
Gain at 1kHz = 3.85 (2500/3200) = 3.01
Comparing the response of 1kHz to the low frequency, we get 20 log (3.01/3.377) = -0.999 dB (-1dB)
If the transformer primary at 20Hz only presents 2500 Ohm, then the response is only -1 dB from the output at 1kHz, where the primary presents 5000 Ohm.
The insertion loss from DCR is a constant (-1dB).
Frequency response is relative to 1kHz, so at the -3dB low frequency, and -3dB high frequency, they are 1/2 power versus what they are at 1kHz.
By the way, 105mH at 20 Hz is 13.2 Ohms . . . it is Not 4.98 Ohms as you stated.
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From https://electronicsclub.info/impedance.htm
Inductive Reactance, XL
Inductive reactance, XL is small at low frequencies and large at high frequencies. For steady DC (frequency zero), XL is zero (no opposition), which means that inductors pass DC but block high frequency AC.
Inductive Reactance, XL = 2pifL
XL = reactance in ohms (ohm)
f = frequency in hertz (Hz)
L = inductance in henrys (H)
Input impedance (ZIN) is the impedance 'seen' by anything connected to the input of a circuit or device (such as an amplifer). It is the combined effect of all the resistance, capacitance and inductance connected to the input inside the circuit or device.
Check this also: https://www.aikenamps.com/index.php/output-transformers-explained
In tube design, you print the tube curves for your desired B+ and Bias, or find the closest, if you are a pro you use a tube tracer and get the curves for the desired B+, then you extrapolate the loading by the transformer and try to optimize the lowest distortion (the most symetrical) gain line around the bias point and try to maximize the power/bandwidth. The less Induction, the more power you can obtain from your output tube because it can swing more voltages. The higher B+ the higher power it can also swing, your primary limit is that the higher the B+ the less bias to stay inside maximum heat dissipation, and the higher bias the more linear the curves become. The designer must use his visual senses and build simulations, draw many load lines until it is perfect.
The less induction, the more power but also higher Zout, and less power in the bass region. Transformers cannot pass square waves into the sub 100hz region. The high bandwidth over 20 Khz is only important if you are using GNF because it needs this to stabilize as a headroom, the feedback loop effectiveness decreases as the maximum bandwidth decrease, also you need to limit the bandwidth of the feedback loop, at low end and top frequencies.
Inductive Reactance, XL
Inductive reactance, XL is small at low frequencies and large at high frequencies. For steady DC (frequency zero), XL is zero (no opposition), which means that inductors pass DC but block high frequency AC.
Inductive Reactance, XL = 2pifL
XL = reactance in ohms (ohm)
f = frequency in hertz (Hz)
L = inductance in henrys (H)
Input impedance (ZIN) is the impedance 'seen' by anything connected to the input of a circuit or device (such as an amplifer). It is the combined effect of all the resistance, capacitance and inductance connected to the input inside the circuit or device.
Check this also: https://www.aikenamps.com/index.php/output-transformers-explained
In tube design, you print the tube curves for your desired B+ and Bias, or find the closest, if you are a pro you use a tube tracer and get the curves for the desired B+, then you extrapolate the loading by the transformer and try to optimize the lowest distortion (the most symetrical) gain line around the bias point and try to maximize the power/bandwidth. The less Induction, the more power you can obtain from your output tube because it can swing more voltages. The higher B+ the higher power it can also swing, your primary limit is that the higher the B+ the less bias to stay inside maximum heat dissipation, and the higher bias the more linear the curves become. The designer must use his visual senses and build simulations, draw many load lines until it is perfect.
The less induction, the more power but also higher Zout, and less power in the bass region. Transformers cannot pass square waves into the sub 100hz region. The high bandwidth over 20 Khz is only important if you are using GNF because it needs this to stabilize as a headroom, the feedback loop effectiveness decreases as the maximum bandwidth decrease, also you need to limit the bandwidth of the feedback loop, at low end and top frequencies.
gabdx,
You said:
"In tube design, you print the tube curves for your desired B+ and Bias, or find the closest, if you are a pro you use a tube tracer and get the curves for the desired B+, then you extrapolate the loading by the transformer and try to optimize the lowest distortion (the most symetrical) gain line around the bias point and try to maximize the power/bandwidth. The less Induction, the more power you can obtain from your output tube because it can swing more voltages. The higher B+ the higher power it can also swing, your primary limit is that the higher the B+ the less bias to stay inside maximum heat dissipation, and the higher bias the more linear the curves become. The designer must use his visual senses and build simulations, draw many load lines until it is perfect.
The less induction, the more power but also higher Zout, and less power in the bass region. Transformers cannot pass square waves into the sub 100hz region. The high bandwidth over 20 Khz is only important if you are using GNF because it needs this to stabilize as a headroom, the feedback loop effectiveness decreases as the maximum bandwidth decrease, also you need to limit the bandwidth of the feedback loop, at low end and top frequencies."
My Generalizations, about output stages that do not have any negative feedback:
Tube output stage design usually involves a compromise of one or more factors.
distortion, damping factor, frequency response, etc.
To drive inductive reactance that is low impedance at low frequencies, you need the plate impedance, rp, to be low.
I am confused by your statement that high bias is better. For A1 operation, bias is a negative voltage. Did you mean, high current, instead of high bias?
With high bias volts (high negative volts), the current goes down, but the plate impedance, rp, goes UP.
(more difficult to drive inductive reactance at low frequencies).
The lower the plate impedance, rp, versus the primary impedance RL (ratio), the higher the damping factor will be.
And, within the limited power resulting from that ratio of rp to RL, the lower the distortion will be.
Also, with the low plate impedance, rp, to RL ratio, the wider the frequency response will be.
Need more power? But at the same time you want high damping factor, low distortion, and wider frequency range . . .
Then pick a more powerful output tube, and a bigger output transformer.
There is no Perfect load line for the lower power tube to get all of that.
Sorry.
Without global negative feedback, versus With global negative feedback:
If your output transformer is saturating, global negative feedback will only make it clip harder (even more saturated).
Can not be fixed; Except by getting a bigger output transformer (or lowering the volume control - less output power).
How many music signals look like a 100Hz square wave? (except all full clipping electric guitars do have square waves; but the guitar amp output transformer will probably have downward sloped square waves, and guitar speakers that are presented with a pure electrical square wave will not make the acoustic wave square anyway).
Compromises exist.
Just relax, and Enjoy the Music!
Just my opinions
You said:
"In tube design, you print the tube curves for your desired B+ and Bias, or find the closest, if you are a pro you use a tube tracer and get the curves for the desired B+, then you extrapolate the loading by the transformer and try to optimize the lowest distortion (the most symetrical) gain line around the bias point and try to maximize the power/bandwidth. The less Induction, the more power you can obtain from your output tube because it can swing more voltages. The higher B+ the higher power it can also swing, your primary limit is that the higher the B+ the less bias to stay inside maximum heat dissipation, and the higher bias the more linear the curves become. The designer must use his visual senses and build simulations, draw many load lines until it is perfect.
The less induction, the more power but also higher Zout, and less power in the bass region. Transformers cannot pass square waves into the sub 100hz region. The high bandwidth over 20 Khz is only important if you are using GNF because it needs this to stabilize as a headroom, the feedback loop effectiveness decreases as the maximum bandwidth decrease, also you need to limit the bandwidth of the feedback loop, at low end and top frequencies."
My Generalizations, about output stages that do not have any negative feedback:
Tube output stage design usually involves a compromise of one or more factors.
distortion, damping factor, frequency response, etc.
To drive inductive reactance that is low impedance at low frequencies, you need the plate impedance, rp, to be low.
I am confused by your statement that high bias is better. For A1 operation, bias is a negative voltage. Did you mean, high current, instead of high bias?
With high bias volts (high negative volts), the current goes down, but the plate impedance, rp, goes UP.
(more difficult to drive inductive reactance at low frequencies).
The lower the plate impedance, rp, versus the primary impedance RL (ratio), the higher the damping factor will be.
And, within the limited power resulting from that ratio of rp to RL, the lower the distortion will be.
Also, with the low plate impedance, rp, to RL ratio, the wider the frequency response will be.
Need more power? But at the same time you want high damping factor, low distortion, and wider frequency range . . .
Then pick a more powerful output tube, and a bigger output transformer.
There is no Perfect load line for the lower power tube to get all of that.
Sorry.
Without global negative feedback, versus With global negative feedback:
If your output transformer is saturating, global negative feedback will only make it clip harder (even more saturated).
Can not be fixed; Except by getting a bigger output transformer (or lowering the volume control - less output power).
How many music signals look like a 100Hz square wave? (except all full clipping electric guitars do have square waves; but the guitar amp output transformer will probably have downward sloped square waves, and guitar speakers that are presented with a pure electrical square wave will not make the acoustic wave square anyway).
Compromises exist.
Just relax, and Enjoy the Music!
Just my opinions
Last edited:
Yes, high bias = high current bias
THD rises at the same power the more Inductance your transformer/load line increase, it reduces THD at lower powers and extend the low end response, lowers Z out, but it loses a lot of power at 1khz,
You should (IMO) prioritize cathode feedback, bootstrapping?, GNF to lower the Z out, and choose the transformer with the most power transfer (which is the less input resistance), it is a big sacrifice to draw a load line with a high RL
I don't agree that GNF makes a properly designed tube amp clip harder, a no feedback design will clip hard as the tubes are saturated, your transformer should never saturate before the output tubes, with GNF the THD curve is steeper, audibly if properly designed the GNF should sound better at same clipping/non-clipping powers. When designing the tube amp you take in account that the GNF will try to saturate the OT by restoring the flat response bellow 60hz, lower value capacitors in specific coupling areas should prevent a low signal boost to reenter the amplifier and saturate the OT or cause motor boating.
Transistor can do low frequency square waves that is because it is a current coupling, it has almost 0 ohm at low frequency.
In SET 2xRp, of the tube = ideal RL of the transformer which maximize that tube power output, it is a very well known formula...
In PP, there is no such thing because it is designed to go into class AB and B
IOW A 1500 ohm transformer would maximize the output power of the 300B, technically 2500 to 5K is used
THD rises at the same power the more Inductance your transformer/load line increase, it reduces THD at lower powers and extend the low end response, lowers Z out, but it loses a lot of power at 1khz,
You should (IMO) prioritize cathode feedback, bootstrapping?, GNF to lower the Z out, and choose the transformer with the most power transfer (which is the less input resistance), it is a big sacrifice to draw a load line with a high RL
I don't agree that GNF makes a properly designed tube amp clip harder, a no feedback design will clip hard as the tubes are saturated, your transformer should never saturate before the output tubes, with GNF the THD curve is steeper, audibly if properly designed the GNF should sound better at same clipping/non-clipping powers. When designing the tube amp you take in account that the GNF will try to saturate the OT by restoring the flat response bellow 60hz, lower value capacitors in specific coupling areas should prevent a low signal boost to reenter the amplifier and saturate the OT or cause motor boating.
Transistor can do low frequency square waves that is because it is a current coupling, it has almost 0 ohm at low frequency.
In SET 2xRp, of the tube = ideal RL of the transformer which maximize that tube power output, it is a very well known formula...
In PP, there is no such thing because it is designed to go into class AB and B
IOW A 1500 ohm transformer would maximize the output power of the 300B, technically 2500 to 5K is used
@6A3sUMMER thank you. On your last post calculations were half with 3k primary impedance and half with 5k, but it is clear.
Let me resume with the 3k primary impedance to see if I got it right:
300B μ = 3.85
300B rp = 700 Ohm
Primary impedance = 3000 Ohm
Primary Rdc = 110 Ohm
Secondary Rdc = 0.7 Ohm
Primary inductance = 39.5 H
Secondary inductance = 105 mH
Assuming a constant 8 Ohm load on the secondary, the midband primary impedance will be:
110 + (8.0 + 0.7) x (3000/8) = 3373 Ohm
That means the 300B works at a real 3k load when the speaker has an impedance of 7 Ohm.
So midband gain is:
3.85 x 3373 / (3373+700) = 3.19
And midband DF is:
3373 / 700 = 4.8
Then as we were saying 39.5 H are 4963 Ohm at 20 Hz, so the impedance becomes:
110 + (4963 || 3263) = 2079 Ohm
Gain at 20 Hz is:
3.85 x 2079 / 2779 = 2.88
so, compared to midband, it's:
20 log (2.88 / 3.19) = -0.89 dB
DF at 20 Hz is:
2079 / 700 = 2.97
If those are right, I will then go to the 20 kHz data.
Let me resume with the 3k primary impedance to see if I got it right:
300B μ = 3.85
300B rp = 700 Ohm
Primary impedance = 3000 Ohm
Primary Rdc = 110 Ohm
Secondary Rdc = 0.7 Ohm
Primary inductance = 39.5 H
Secondary inductance = 105 mH
Assuming a constant 8 Ohm load on the secondary, the midband primary impedance will be:
110 + (8.0 + 0.7) x (3000/8) = 3373 Ohm
That means the 300B works at a real 3k load when the speaker has an impedance of 7 Ohm.
So midband gain is:
3.85 x 3373 / (3373+700) = 3.19
And midband DF is:
3373 / 700 = 4.8
Then as we were saying 39.5 H are 4963 Ohm at 20 Hz, so the impedance becomes:
110 + (4963 || 3263) = 2079 Ohm
Gain at 20 Hz is:
3.85 x 2079 / 2779 = 2.88
so, compared to midband, it's:
20 log (2.88 / 3.19) = -0.89 dB
DF at 20 Hz is:
2079 / 700 = 2.97
If those are right, I will then go to the 20 kHz data.
zintolo,
My Post # 671 did not mention a 3k primary.
Where did you get the 3k primary?
Perhaps you were referring to what I said:
"If at some low frequency, the transformer complex impedance presents 2500 Ohms, then . . .
Gain at 1kHz = 3.85 (2500/3200) = 3.01"
The number above, "3200" is simply: 2500 primary + 700 rp = 3200.
Any way, it is very late here, I am tired, and I have to finish reading, and then get up early.
My Post # 671 did not mention a 3k primary.
Where did you get the 3k primary?
Perhaps you were referring to what I said:
"If at some low frequency, the transformer complex impedance presents 2500 Ohms, then . . .
Gain at 1kHz = 3.85 (2500/3200) = 3.01"
The number above, "3200" is simply: 2500 primary + 700 rp = 3200.
Any way, it is very late here, I am tired, and I have to finish reading, and then get up early.
@6A3sUMMER
our conversation and examples are based since a few posts on the output transformer I linked at post #626 that is 3k.
You replied to my request using that transformer as reference for calculations, helping me in that.
That's why the last post using 5k was confusing.
our conversation and examples are based since a few posts on the output transformer I linked at post #626 that is 3k.
You replied to my request using that transformer as reference for calculations, helping me in that.
That's why the last post using 5k was confusing.
gabdx,
If your Hi Fi Stereo music amplifier clips, whether it is a tube or a transistor, it is not Hi Fi (turn the volume down, or get a more powerful amplifier).
This thread is Hi Fi Stereo, Tubes / Valves; It is not an Instruments & Amps Guitar Amplifier Thread.
The very last page of the Heathkit W-5M manual has a very educational curve about saturation with the presence of global negative feedback . . .
you should look at it, if you have not seen it before.
If I had a little more time, I would teach about the Even Order Harmonics of a Square Wave that has anything other than exactly 50% Up, and 50% down time (it is not all Odd Harmonics then). And how the Even Order Harmonics eventually have more power than the odd harmonics at very high frequencies.
And, I would teach about what a square wave looks like when you:
A. Either completely remove one, and only one, Odd Harmonic
Or,
B. Reverse the phase of one, and only one, Odd Harmonic
Surprise: You see that one and only one Odd Harmonic riding on top of the previously flat top and flat bottom of the square wave.
And I could mention a lot more time and frequency domain "oddities".
We all have our favorite amplifier circuits and topologies.
There are lots of amplifiers topologies that work and sound very good . . . If they are implemented properly.
The original Tacoma Narrows Bridge is an example of a design that was not properly implemented.
Do you know of a commercial tube amplifier with an output transformer that will not saturate when playing back Telarc's definitive recording of the 1812 Overture (a real canon that has 6Hz fundamental frequency at high volume)?
If you know of one please list the manufacturer and model number.
I am talking about a reasonable playback volume, so you can feel the effect.
Next, Please name the manufacturer and model of the loudspeaker that can faithfully play that same Telarc recording of the 1812 Overture.
Just Saying
If your Hi Fi Stereo music amplifier clips, whether it is a tube or a transistor, it is not Hi Fi (turn the volume down, or get a more powerful amplifier).
This thread is Hi Fi Stereo, Tubes / Valves; It is not an Instruments & Amps Guitar Amplifier Thread.
The very last page of the Heathkit W-5M manual has a very educational curve about saturation with the presence of global negative feedback . . .
you should look at it, if you have not seen it before.
If I had a little more time, I would teach about the Even Order Harmonics of a Square Wave that has anything other than exactly 50% Up, and 50% down time (it is not all Odd Harmonics then). And how the Even Order Harmonics eventually have more power than the odd harmonics at very high frequencies.
And, I would teach about what a square wave looks like when you:
A. Either completely remove one, and only one, Odd Harmonic
Or,
B. Reverse the phase of one, and only one, Odd Harmonic
Surprise: You see that one and only one Odd Harmonic riding on top of the previously flat top and flat bottom of the square wave.
And I could mention a lot more time and frequency domain "oddities".
We all have our favorite amplifier circuits and topologies.
There are lots of amplifiers topologies that work and sound very good . . . If they are implemented properly.
The original Tacoma Narrows Bridge is an example of a design that was not properly implemented.
Do you know of a commercial tube amplifier with an output transformer that will not saturate when playing back Telarc's definitive recording of the 1812 Overture (a real canon that has 6Hz fundamental frequency at high volume)?
If you know of one please list the manufacturer and model number.
I am talking about a reasonable playback volume, so you can feel the effect.
Next, Please name the manufacturer and model of the loudspeaker that can faithfully play that same Telarc recording of the 1812 Overture.
Just Saying
Last edited:
@6A3sUMMER
you don't have to say sorry, it's me that I have to say thank you for your time and help.
So, back on track:
300B μ = 3.85
300B rp = 700 Ohm
Primary impedance = 3000 Ohm
Primary Rdc = 110 Ohm
Secondary Rdc = 0.7 Ohm
Primary inductance = 39.5 H
Secondary inductance = 105 mH
Primary Leakage Inductance = 10.88 mH
Effective Primary Capacitance = 7.6 nF
Until now we ve calculated:
Midband primary impedance = 3373 Ohm
20 Hz primary impedance = 2079 Ohm
Now talking about 20 kHz:
7.6 nF at 20 kHz are 1047 Ohm
10.3 mH at 20 kHz are 1297 Ohm
8000 / 8 = 375
Considering the capacitance in parallel with 1/3 of the primary, we'll have:
8.7 x (375/3) + 1047 || [8.7 x (375/3)] + 8.7 x (375/3) = 1087.5 + 1047 || 1087.5 + 1087.5 = 2708 Ohm
Considering the capacitance in parallel with 1/2 of the primary, we'll have:
8.7 x (375/2) + 1047 || [8.7 x (375/2)] = 1631.2 + 1047 || 1631.2 = 2269 Ohm
Considering the capacitance in parallel with the full primary, we'll have:
1047 || [8.7 x (375)] = 793 Ohm
Then primary Rdc and Leakage inductance are in series with that, so the impedance at 20 kHz will be respectively:
2708 + 110 + 1297 Ohm = 4115 Ohm
2269 + 110 + 1297 Ohm = 3676 Ohm
793+ 110 + 1297 Ohm = 2200 Ohm
you don't have to say sorry, it's me that I have to say thank you for your time and help.
So, back on track:
300B μ = 3.85
300B rp = 700 Ohm
Primary impedance = 3000 Ohm
Primary Rdc = 110 Ohm
Secondary Rdc = 0.7 Ohm
Primary inductance = 39.5 H
Secondary inductance = 105 mH
Primary Leakage Inductance = 10.88 mH
Effective Primary Capacitance = 7.6 nF
Until now we ve calculated:
Midband primary impedance = 3373 Ohm
20 Hz primary impedance = 2079 Ohm
Now talking about 20 kHz:
7.6 nF at 20 kHz are 1047 Ohm
10.3 mH at 20 kHz are 1297 Ohm
8000 / 8 = 375
Considering the capacitance in parallel with 1/3 of the primary, we'll have:
8.7 x (375/3) + 1047 || [8.7 x (375/3)] + 8.7 x (375/3) = 1087.5 + 1047 || 1087.5 + 1087.5 = 2708 Ohm
Considering the capacitance in parallel with 1/2 of the primary, we'll have:
8.7 x (375/2) + 1047 || [8.7 x (375/2)] = 1631.2 + 1047 || 1631.2 = 2269 Ohm
Considering the capacitance in parallel with the full primary, we'll have:
1047 || [8.7 x (375)] = 793 Ohm
Then primary Rdc and Leakage inductance are in series with that, so the impedance at 20 kHz will be respectively:
2708 + 110 + 1297 Ohm = 4115 Ohm
2269 + 110 + 1297 Ohm = 3676 Ohm
793+ 110 + 1297 Ohm = 2200 Ohm