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-   -   Understanding OPT primary Z and load matching. (http://www.diyaudio.com/forums/tubes-valves/39725-understanding-opt-primary-z-load-matching.html)

percy 20th August 2004 07:38 PM

Understanding OPT primary Z and load matching.
 
First of all, when a manufacturer specifies the impedance of the primary winding, what is the frequency at which it is being specified ? Is there a standard ? I could not find this information for the big names in transformers like Lundahl, James,... .

Now since each side of a transformer is nothing but an inductor, I would assume that it never presents a constant impedance(load) to the output tube. If a transformer's primary Z is rated at say 6K then I believe I have to interpret it as the transformer presents a load of 6K to the output tube at one frequency only. The load - that is the impedance of the primary - would however be very different at different frequencies, especially at the ends of the audio frequency spectrum(20hz - 20Khz). Thats a really WIDE range of load the output tube will 'see'!

Is my understanding correct ? If it is then what is considered a good match ?

Thanks!

Sch3mat1c 20th August 2004 08:21 PM

Impedance [ratio] depends on the turns ratio. It gets decoupled a bit at higher frequencies due to leakage inductance and shunted due to parasitic capacitance, while at LF it gets shunted by inductance, but for the most part 50-20,000Hz (for a reasonable piece of iron) it holds.

Tim

percy 20th August 2004 09:00 PM

Ok. so as long as the secondary has a constant load the tube will see a load of the rated primary Z over almost the entire range ?

But I guess what I am still not able to clear from my mind is the visualization of the primary winding as an 'inductor in series' with the plate. What effect does that have on the AC(audio) signal flowing through it ?

Sch3mat1c 20th August 2004 09:33 PM

The customary basic transformer model is this:

Start with an ideal transformer. This responds to 0Hz (while pradoxically conducting DC with zero resistance ;) ) and transforms it perfectly based on the turns ratio = voltage ratio. No voltage or frequency limit.
Now put an ideal inductor in parallel with the primary. This is the primary inductance, which allows say <5Hz to be counted as DC (the operating current) and passed to the tube. You can just as well put this on the secondary (minding its value based on impedance ratio = turns ratio squared), it doesn't matter.
Next, you need a perfect resistance in series, because the wire isn't perfect, and a series inductance as well for leakage inductance (a matter of coupling factor). Then a capacitor in parallel across the terminals to represent capacitance between the turns and layers of wire. The result looks like: (two pins for input) > parallel capacitor (i.e. B+ to plate) > series resistor and inductor > parallel inductor > ideal transformer.

Now all becomes clear. Primary inductance is the primary winding itself acting like a choke, a simple fact of being wire wound around a core and creating a magnetic field. This is represented by the inductor in parallel with the ideal transformer.

The effect of it on AC only applies to low frequencies, as its reactance is astronomical compared to the load impedance at any middle or upper frequency. Typically, when Xl = Rl is considered cutoff frequency. For triode amplifiers, it can be Xl = Rp instead (Rp is typically 3 times lower than Rl so it can be an important distinction).

Saturation is a different LF limit which depends on magnetic field strength, and thus the relation of frequency vs. voltage.

Tim

jlsem 21st August 2004 12:30 AM

Quote:

Ok. so as long as the secondary has a constant load the tube will see a load of the rated primary Z over almost the entire range ?
The impedence indicated by the manufacturer is what the tube will see if a resistor of a constant value is connected across the secondary , i.e. an 8ohm resistor on the 8ohm tap. A better question is At what frequency is an interstage transformer rated for a given primary impedence, seeing as there is no secondary load in class A1?

Speakers, of course , do not present a constant load so the tube will see different impedences at different frequencies, up to three times the rated load as the frequency approaches Fr. This helps delay clipping somewhat at lower frequencies in actual practice.

Quote:

But I guess what I am still not able to clear from my mind is the visualization of the primary winding as an 'inductor in series' with the plate. What effect does that have on the AC(audio) signal flowing through it ?
The value of the primary inductance determines the shape of the AC load line which is always elliptical when the tube sees an inductive load. If the primary inductance is too low the path of the load line will approach the shape of circle at low frequencies, thus dipping into areas of operation where distortion is high and the tube is at cut-off. Hence the high distortion and poor low frequency response of cheap output transformers which may be physically large enough, but have a low primary inductance.

You can also see how the higher impendence of a speaker driver at low frequencies presents a more horizontal axis for an elliptical AC load line and prevents its path from dipping too near cut-off.

John


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