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Ξεκιναω ενα θεματακι προς συζητηση για το εν λογω μοντελο της γνωστης εταιρειας otl ενισχυτικων.
Πηρα το εναυσμα να γραψω κατι, για καποιες ασαφειες που γραφωνται κατα καιρους .
Ποιο κατω θα δειτε τι γραφει καποιος ριβιουερ, αλλα να δουμε ποσο κοντα στην αληθεια βρισκεται ακομα και ο κατασκευαστης του μηχανηματος.
Παμε λοιπον.
The amplifier is built with only a single stage of gain and no feedback, quite unusual even for an OTL, the only one of this type I know of. To reduce distortion without feedback is not easy, so the amplifier circuit is fully differential throughout, resulting in even-order harmonic cancellation at every stage throughout the amp. It is also all-triode and class A for best linearity. There is only a single gain stage and, of course, there is no output transformer.
Πραγματι το κυκλωμα αποτελειται απο μονο ενα σταδιο ενισχυσης σε κασκοδικη τοπολογια με την γνωστη 6sn7 και στην πορεια αναλαμβανει ενας ακολουθος καθοδου με τον ιδιο τυπο λυχνιας να οδηγησει τις τελικες . Να πουμε επισης οτι η τοπολογια του συγκεκριμενου ενισχυτη ειναι κυκλοτρονικη.
Οπως γραφει ποιο πανω ο ριβιουερ ο ενισχυτης λειτουργει με χωρις αναδραση. Τι σημενει αυτο΄; οτι ο ενισχυτης με ενα απλο υπολογισμο με βαση τον τυπο των λυχνιων εξοδου που εχει και στην τπολογια που λειτουργει η αντιστασταση εξοδου του ειναι 10ωμ. !!!!!!Για να το κανω ποιο απλο εχει 8 διπλοτριοδες λυχνιες ανα καναλι [16 δηλαδη] με ανυπαρκτο συντελεστη αποσβεσης!!!!!!!!!!!!!
Να πουμε σε αυτο το σημειο οτι αν ο ενισχυτης λειτουργουσε σε τοπολογια Φουτερμαν με τον ιδιο αριθμο λυχνιων θα ηταν περιπου στο μισο!!!!
Αν , δε, εκανε και χρηση αναδρασης θα μπορουσε να παει ανετα σε επιπεδα ενισχυτου στερεας καταστασης της περασμενης 10ετιας!!!
Κοιτωντας λιγο στο νετ για περισσοτερες πληροφοριες επεσα πανω σε αυτο το πολυ ενδιαφερον αρθρο που κατα καποιο λογο επιβεβαιωνει τους προβληματισμους μου για τον συγκεκριμενο ενισχυτη . Κανει αναφορα για την αντισταση εξοδου του και την ισχυ του .
Speltz's Discovery #5
(not in chronological order)
In my opinion the output impedance of these OTLs is a moving target. I say this because I can see problems in applying the different measuring methods to the circlotron circuit. This means that I believe that none of the methods are completely correct. I will review my thoughts on the 1/2 power transfer method, my thoughts on the "voltage" method, and discuss a third method. Also, it is important to keep in mind that a huge double digit damping factor because of a decimal point output impedance is not to be assumed as the ultimate goal of a power amp. A woofer's step response is best when it is "critically dampened", and this usually happens with a few ohms of amplifier output impedance, not with something like 0.03 ohm that the solid state amps boast.
The output of an amplifier can be modeled as a perfect AC voltage source along with a series output resistor. In this simple model, the value of this resistor is by definition the specified output impedance. The "voltage" method of determining the value of this output resistor is to have the amplifier output a fixed voltage (say 1 Vrms @ 1khz) into no load. Next apply a load and reduce its value until the output is reduced to exactly half (0.5 Vrms). In this model the value of the load must be the value of the output resistor since the voltage is exactly split between the two resistors (output resistor and load resistor).
When this method is used on a M-60, a load of 11 ohms (nominal) will half the output voltage. So the voltage method results in what seems like a 11 ohm output impedance. But this is of course assuming that a perfect AC voltage source with a series resistor is an accurate model of the M-60.
Next I will try plugging in values for the above model and figure max power. For example, let us say that the voltage source can put out a maximum of 10 Vrms (before clipping, limited by its power supply rails). And let us make the series output resistor 10 ohms. These are easy round numbers. Now I start applying different loads and solve for output power into the load.
I found this interesting because the maximum power point is when the load resistor and the output resistors are exactly the same value. It does not matter what values of maximum voltage or series output resistor are chosen, the lower 1/2 power point will always be when the load resistor is at 17.4% of the series output resistor. So in this model the 1/2 power load resistance of 1.74 ohms is not an accurate representation of the actual (10 ohm) series output resistance.
So does this mean that since a M-60 puts out its maximum power into 14 to 16 ohms, that then its output impedance is about 15 ohms. No, I don't think that is true either.
Adding 3 dB of negative feedback will decrease the output impedance of the model to 5 ohms. Yet with this 1/2 power method applying any amount of negative feedback from the output will not change the result. It will still do only 2.5 watts into 10 ohms, and 1.25 watts into 1.74 ohms. How can this 1/2 power method be proper?
There is yet another way to measure output impedance of an amplifier, and that is to drive a signal into the amplifier's output and measure how much it moves. Obviously an amplifier with an output impedance of 0.00 ohms will not move at all. And the higher the output impedance is, the easier the signal driving backwards into the amp can "push" it around. I have the luxury of having a Hewlett-Packard spectrum analyzer on my work bench at work. It has a 600 ohm frequency sweeping output and a tracking input that can display linear or log scale amplitude. This can easily be set up to measure impedance of resistors, transformers, phone lines (my speciality), capacitors, inductors, amplifiers, etc. Its accuracy allows me to see the equivalent series resistant of electrolytic capacitors down to thousands of an ohm.
Using this method a stock M-60 shows a nominal 11 ohm output impedance which seems to support the "voltage" method. It also becomes very easy to see the output impedance rise and fall with decreasing and increasing bias current (respectively) and drop to 7 ohms with the -2dB F.B. installed.
Yet I am still not happy with simply declaring the output impedance of a M-60 as 11 ohms because the measurement is taken with the M-60 in a static (no audio) state. I believe that the output impedance in these OTL amps is a moving target, dynamically changing with the music, possibly running in the range between 3.4 ohms and 11 ohms.
I figure that since each half of the circlotron is biased at less than half of the current that it is capable of creating, ...........then each half of the circlotron can turn "ON" farther (harder) than it can turn "OFF". Or think of it this way, The sum of the current of the (+) phase plus the (-) phase is not always equal throughout the cycle of a single sine wave cycle. I will walk you through one cycle of a sine wave from a M-60 at a 60 watt output level into an 8 ohm resistive load:
1) Start at idle, 600mA + 600mA = |1.2A |
2) At positive peak, 3.9A + 0mA = | 3.9A |
3) Back through zero crossing, 600mA + 600mA = |1.2A |
4) Down to the negative peak, 0mA + 3.9A = | 3.9A |
5) And returning back to zero, 600mA + 600mA = | 1.2A |
Remembering that I could easily see how output impedance tracked bias current in a fairly linear fashion (more current = lower output impedance)with the HP analyzer, it is easy to imagine how the output impedance of these amps varies dynamically throughout the single cycle of a sine wave. The amp actually holds on to the load (speaker) harder the farther it tries to move it. In the above example there is 3.25 times the total current flowing at the top and bottom of the sine wave as there is through the middle. So there is possibly 3.25 times less output impedance at those points in time than when the amp is sitting in a static state.
So let us try this: 11 ohms divided by 3.25 = 3.4 ohms!
My final thoughts:
The M-60 is an extremely difficult amplifier to model. I believe the output impedance is dynamically changing in a range between about 11 and 3.4 ohms when driving an 8 ohm resistive load to 60 watts, and in a range between 11 and 2.3 ohms when driving a 4 ohm resistive load to 45 watts.
Paul Speltz
###
Ο συγκεκριμενος μηχανικος λεει οτι γινεται και χρηση αναδρασης 3db!
Πηρα το εναυσμα να γραψω κατι, για καποιες ασαφειες που γραφωνται κατα καιρους .
Ποιο κατω θα δειτε τι γραφει καποιος ριβιουερ, αλλα να δουμε ποσο κοντα στην αληθεια βρισκεται ακομα και ο κατασκευαστης του μηχανηματος.
Παμε λοιπον.
The amplifier is built with only a single stage of gain and no feedback, quite unusual even for an OTL, the only one of this type I know of. To reduce distortion without feedback is not easy, so the amplifier circuit is fully differential throughout, resulting in even-order harmonic cancellation at every stage throughout the amp. It is also all-triode and class A for best linearity. There is only a single gain stage and, of course, there is no output transformer.
Πραγματι το κυκλωμα αποτελειται απο μονο ενα σταδιο ενισχυσης σε κασκοδικη τοπολογια με την γνωστη 6sn7 και στην πορεια αναλαμβανει ενας ακολουθος καθοδου με τον ιδιο τυπο λυχνιας να οδηγησει τις τελικες . Να πουμε επισης οτι η τοπολογια του συγκεκριμενου ενισχυτη ειναι κυκλοτρονικη.
Οπως γραφει ποιο πανω ο ριβιουερ ο ενισχυτης λειτουργει με χωρις αναδραση. Τι σημενει αυτο΄; οτι ο ενισχυτης με ενα απλο υπολογισμο με βαση τον τυπο των λυχνιων εξοδου που εχει και στην τπολογια που λειτουργει η αντιστασταση εξοδου του ειναι 10ωμ. !!!!!!Για να το κανω ποιο απλο εχει 8 διπλοτριοδες λυχνιες ανα καναλι [16 δηλαδη] με ανυπαρκτο συντελεστη αποσβεσης!!!!!!!!!!!!!
Να πουμε σε αυτο το σημειο οτι αν ο ενισχυτης λειτουργουσε σε τοπολογια Φουτερμαν με τον ιδιο αριθμο λυχνιων θα ηταν περιπου στο μισο!!!!
Αν , δε, εκανε και χρηση αναδρασης θα μπορουσε να παει ανετα σε επιπεδα ενισχυτου στερεας καταστασης της περασμενης 10ετιας!!!
Κοιτωντας λιγο στο νετ για περισσοτερες πληροφοριες επεσα πανω σε αυτο το πολυ ενδιαφερον αρθρο που κατα καποιο λογο επιβεβαιωνει τους προβληματισμους μου για τον συγκεκριμενο ενισχυτη . Κανει αναφορα για την αντισταση εξοδου του και την ισχυ του .
Speltz's Discovery #5
(not in chronological order)
In my opinion the output impedance of these OTLs is a moving target. I say this because I can see problems in applying the different measuring methods to the circlotron circuit. This means that I believe that none of the methods are completely correct. I will review my thoughts on the 1/2 power transfer method, my thoughts on the "voltage" method, and discuss a third method. Also, it is important to keep in mind that a huge double digit damping factor because of a decimal point output impedance is not to be assumed as the ultimate goal of a power amp. A woofer's step response is best when it is "critically dampened", and this usually happens with a few ohms of amplifier output impedance, not with something like 0.03 ohm that the solid state amps boast.
The output of an amplifier can be modeled as a perfect AC voltage source along with a series output resistor. In this simple model, the value of this resistor is by definition the specified output impedance. The "voltage" method of determining the value of this output resistor is to have the amplifier output a fixed voltage (say 1 Vrms @ 1khz) into no load. Next apply a load and reduce its value until the output is reduced to exactly half (0.5 Vrms). In this model the value of the load must be the value of the output resistor since the voltage is exactly split between the two resistors (output resistor and load resistor).
When this method is used on a M-60, a load of 11 ohms (nominal) will half the output voltage. So the voltage method results in what seems like a 11 ohm output impedance. But this is of course assuming that a perfect AC voltage source with a series resistor is an accurate model of the M-60.
Next I will try plugging in values for the above model and figure max power. For example, let us say that the voltage source can put out a maximum of 10 Vrms (before clipping, limited by its power supply rails). And let us make the series output resistor 10 ohms. These are easy round numbers. Now I start applying different loads and solve for output power into the load.
I found this interesting because the maximum power point is when the load resistor and the output resistors are exactly the same value. It does not matter what values of maximum voltage or series output resistor are chosen, the lower 1/2 power point will always be when the load resistor is at 17.4% of the series output resistor. So in this model the 1/2 power load resistance of 1.74 ohms is not an accurate representation of the actual (10 ohm) series output resistance.
So does this mean that since a M-60 puts out its maximum power into 14 to 16 ohms, that then its output impedance is about 15 ohms. No, I don't think that is true either.
Adding 3 dB of negative feedback will decrease the output impedance of the model to 5 ohms. Yet with this 1/2 power method applying any amount of negative feedback from the output will not change the result. It will still do only 2.5 watts into 10 ohms, and 1.25 watts into 1.74 ohms. How can this 1/2 power method be proper?
There is yet another way to measure output impedance of an amplifier, and that is to drive a signal into the amplifier's output and measure how much it moves. Obviously an amplifier with an output impedance of 0.00 ohms will not move at all. And the higher the output impedance is, the easier the signal driving backwards into the amp can "push" it around. I have the luxury of having a Hewlett-Packard spectrum analyzer on my work bench at work. It has a 600 ohm frequency sweeping output and a tracking input that can display linear or log scale amplitude. This can easily be set up to measure impedance of resistors, transformers, phone lines (my speciality), capacitors, inductors, amplifiers, etc. Its accuracy allows me to see the equivalent series resistant of electrolytic capacitors down to thousands of an ohm.
Using this method a stock M-60 shows a nominal 11 ohm output impedance which seems to support the "voltage" method. It also becomes very easy to see the output impedance rise and fall with decreasing and increasing bias current (respectively) and drop to 7 ohms with the -2dB F.B. installed.
Yet I am still not happy with simply declaring the output impedance of a M-60 as 11 ohms because the measurement is taken with the M-60 in a static (no audio) state. I believe that the output impedance in these OTL amps is a moving target, dynamically changing with the music, possibly running in the range between 3.4 ohms and 11 ohms.
I figure that since each half of the circlotron is biased at less than half of the current that it is capable of creating, ...........then each half of the circlotron can turn "ON" farther (harder) than it can turn "OFF". Or think of it this way, The sum of the current of the (+) phase plus the (-) phase is not always equal throughout the cycle of a single sine wave cycle. I will walk you through one cycle of a sine wave from a M-60 at a 60 watt output level into an 8 ohm resistive load:
1) Start at idle, 600mA + 600mA = |1.2A |
2) At positive peak, 3.9A + 0mA = | 3.9A |
3) Back through zero crossing, 600mA + 600mA = |1.2A |
4) Down to the negative peak, 0mA + 3.9A = | 3.9A |
5) And returning back to zero, 600mA + 600mA = | 1.2A |
Remembering that I could easily see how output impedance tracked bias current in a fairly linear fashion (more current = lower output impedance)with the HP analyzer, it is easy to imagine how the output impedance of these amps varies dynamically throughout the single cycle of a sine wave. The amp actually holds on to the load (speaker) harder the farther it tries to move it. In the above example there is 3.25 times the total current flowing at the top and bottom of the sine wave as there is through the middle. So there is possibly 3.25 times less output impedance at those points in time than when the amp is sitting in a static state.
So let us try this: 11 ohms divided by 3.25 = 3.4 ohms!
My final thoughts:
The M-60 is an extremely difficult amplifier to model. I believe the output impedance is dynamically changing in a range between about 11 and 3.4 ohms when driving an 8 ohm resistive load to 60 watts, and in a range between 11 and 2.3 ohms when driving a 4 ohm resistive load to 45 watts.
Paul Speltz
###
Ο συγκεκριμενος μηχανικος λεει οτι γινεται και χρηση αναδρασης 3db!
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