Insulation between layers, multiple chambers, z winding.
Btw, what about a 1:1 transformer?
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OK, thanks. I have a lot to learn about this.
I would have to figure out if using a 1:1 transformer in my particular circuit would provide a low enough output impedance to be useful for my application. Also, 4:1 output transformers designed for DC current on the primary are already available (as are 1:1 interstage transformers of course), so if I don't decide to make it myself I could always buy it.
Speaking of making them, do you know of some website or something that shows what is needed in order to do this? Is it really cheaper in the long run? Thanks.
Making your own transformers is not cost effective in my experience. You need a coil winder, bobbins, end bells, E/I Cores, insulating material, ....
Then there is the time you spend winding, throwing away, winding again... to learn how to do it.
I'm winding my own transformers mainly as a learning experience, and because I want transformers that are not readily available.
For standard transformers, I usually buy from Edcor.
Then there is the time you spend winding, throwing away, winding again... to learn how to do it.
I'm winding my own transformers mainly as a learning experience, and because I want transformers that are not readily available.
For standard transformers, I usually buy from Edcor.
Thanks. You're probably right. I also want to be able to have an intelligent conversation with whoever makes these parts for me. So, I do want to learn more.
I looked at Edcor's website but did not see a plate choke mentioned. I see power chokes and various other transformers though. Thanks for the tip!
Surely!. That in the bottle neck in audio transformers designings. Followings Popilin concepts, you will get he lower capacitance in the windings. See also the PDF I posted some post above.
In respect to the frequency response, answer is YES, but I don't know their behaviour in respect to distortion in audio frequencies. Also, low µ triodes act better in this position, high and medium have higher plate resistances, making the capacitance trouble worse.
No, it is the impedance which makes the load to the tube, not the BEFM!
I wrote too quickly earlier. I intended to say triode-strapped VHF pentodes like the 6688, D3A, etc. These are high Gm tubes with low Rp.
If your tubes has low plate resistance in the selected operating point Q, then frequency response probably will be good.
OK, thanks.
I'm trying to visualize how the plate choke responds to AC signals from the plate. As I am trying to understand this, as the voltage on the plate changes, and it tries to draw more current through the choke, the choke reacts by producing a back emf that is in proportion to the degree of current change demanded by the tube. So, is it true that the instantaneous impedance changes throughout each AC cycle at the plate? In other words, does the plate see a larger instantaneous impedance the greater the voltage swing at the plate?
I'll try to be more clear.
Suppose you have the circuit in the figure below, in a previous post, or any other choke loaded triode or pentode.
The choke itself has a certain inductance that normally is specified in the data sheet, written in it, or always measurable by external means. Call it L [Hy].
Around it you have too: the resistive components in the circuit, load and plate resistance of the tube. Both in parallel from signal point of view Call them R[ohms]. And a capacitive component, derived from the winding intra-capacitance, wiring to core, and wiring to chassis, and tube and socket caps and.... Call, all of them in parallel, C [Farad].
Then, starting from 0Hz (DC) (to infinite) signal applied to the grid of the tube, response is almost null because of the low RDC, the winding DC resistance. Increasing a bit frequency, then there must be a first corner frequency f1 in which R is equal in magnitude to wl (2 pi f1 L) in which the first -3dB pole occurs. This is the usual half power point of the circuit. From this point, the frequency response will be flat at a maximum. At f1 phase will be 45° and inductive.
If you continue increasing the signal frequency, there must be a point in where L and C resonates. Then, the output will be at an absolute maximum, depending on the Q of the tank so formed, and reaching a resonant fr (peak) frequency.
Independently of this, there will be a second corner frequency f2, in which capacitive reactance (1/2 pi f C) will be equal to R, then from this point, a decaying -10db/dec to 0 db. This point is the maximum usable gain of the amplifier, and is the second -3dB corner (as can be seen in a Bode plot). Again, phase will be at 45°, this time capacitive.
Depending the frequency and the Q of the resonant point (1 / 2 pi sqrt (L C)) is, then the amplifier will be usable, or a complete disaster, and possibly self oscillating or several ringing in the response at the resonant frequency of the entire circuit. The higher the plate resistance (affecting R), the worse will be the low and the high extremes in the response.
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All depends what quality you want and if you are willing to learn all about transformers.
No budget transformers for me, but also Tango i will not buy because i think they are overpriced.
No budget transformers for me, but also Tango i will not buy because i think they are overpriced.
Yes, I understand about bandwidth, at least conceptually. I guess I didn't explain well enough. Let me try again.
I'm visualizing at the microscopic level changes in current at the plate of the tube as an AC signal is imposed on the grid when the plate is connected to a huge inductor. The inductor has energy stored in it from the DC supply. It resists changes in current by generating a back emf (as I understand it). So, is this resistance to change in current effectively a change in instantaneous impedance as seen by the plate? Does the plate experience a larger resistance to change in current the greater the size of the current swing? I know I'm mixing terms up a little here but I'm wondering if, as the plate current moves from the nominal DC value to some other value, and the inductor generates a back emf to compensate, does this look like a higher apparent impedance to the plate? I don't know how else to express what I'm trying to say, sorry.
Not only an inductor wired such a way has a BEFM. Any other, like a simple motor or ballast have a back electro motive forces which appears as response of changes in current superimposed from the external power souce. But you can ignore them, and be suffient with the impedance seen from the inductor terminals. It is the only you need to put it work.
This concept may be only important at very high di/dt (Rapid changes in current in too smaller periods of time) which may generate important and dangerous voltages spikes appearing in the circuit. They can arcing in tubes, sockets, and inductor's winding itself. This effect is positively used in electromagnetic fluorecent ballasts, in which, when the starter breaks the current flowing in the lamp and ballast wired in series, an important voltage transient is generated which helps trigger the lamp, and start lighting.
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No, you are confused, let me explain, when we have superimposed both AC and DC
Bac = (Uac x 10⁸) / (√2 π f S Np)
Hac = [4 √2 π Np i(AC)] / (9 l)
Bdc = [4 π μ Np i(DC)] / (9 l)
Hdc = [4 π Np i(DC)] / (9 l)
Hac = [4 √2 π Np i(AC)] / (9 l)
Bdc = [4 π μ Np i(DC)] / (9 l)
Hdc = [4 π Np i(DC)] / (9 l)
Then, there is implicit μ(AC) in Bac, and μ(DC) is explicit in Bdc.
As Bdc and Hdc are stationary fields, there is no hysteresis loop for them once steady state is reached, so we can consider μ(DC) as a constant, as we have an air gap, was defined μeff, and is understood that μeff takes the role of μ(DC).
Things are not that easy for μ(AC), do you remember the core nightmare? Did you read post#1?
Do you like assume that μ(AC) is a tensor? Do you like to work with incremental permeability?
My answer is NO, it is not necessary, as I said before μ(AC) is implicit in Bac, and of course it affects primary inductance, however has nothing to do with μeff.
In physics exist the so called linear superposition principle, and here was applied to decompose both fields (AC and DC), so one does not influence the other.
There are a lot of ways to measure magnetic permeability, in complex materials it can be done applying a Dirac δ function shape signal, but do not need that complication, if the manufacturer datasheet says "Typical DC Permeability Value μ=47000 @ 16000 Gauss" usually mean only this.
All derived equations are consistent, so the design goal is an air gap calculation (estimation if we don't know core datasheet)
lG = l (μ - μeff) / (μ μeff)
And after that, we can do a fine adjust (or verification), measuring Lp at lowest frequency and rated AC voltage, or different predictable values, e.g. 5 VRMS @ 50 Hz.With
μeff = Bdc(max) (9 l) / [4 π Np i(DC)]
Having said that, we can use derived equations and quiet sleep, let equations work for us.
You are guessing, if my German is not that wrong, datasheet on post#61 don't say anything about μ for that particular core.
You are still confusing μ, μ(AC), μ(DC), and μeff.
As Bac remains unchanged in the air gap, its only purpose is to limit μeff and consequently Bdc.
At maximum power always Bac reaches a maximum, if well below saturation region, the same must be for μ(AC), as μ(DC) remains constant, your defined effective μ also must reach its maximum. IMHO
As I said before, equations are consistent, and all design goals have been achieved.
Even more, is better to design a transformer for lowest possible frequency, because distortion reaches a maximum at Bac(max) due to magnetizing current also reaches its maximum, as it is a highly distorted current.
As a bonus, a transformer designed for 20Hz (more turns) will have lower distortion at 30Hz than the same transformer designed for 30Hz (less turns).
For that particular coil formers, here has a width of about 39.1mm to 39.8mm, that not differ too much with pictures on post#49, about 41mm, and lamination window is about 42.9mm
Pryde S.R.L - Productos
Here, modern magnet wire to about 1.25mm (I can assure up to there) has the same polyesterimide resin thickness, 0.025mm, at least that I use.
To wind 130 turns of 0.25mm magnet wire you need at least
131x0.30mm=39.3mm
To wind 34 turns of 1.05mm magnet wire you need at least
35x1.10mm=38.5mm
With my winding technique is easily achieved, if with yours don't, a few turns less, or a few thousandths of mm less diameter, will not affect too much the results, anyway, an experienced winder will know to do.
Master!
I didn't say anything different except that mu AC is not necessarily max at Bac max.
However even mu DC in reality is not precisely constant as function of DC current. It will decrease a little bit with increasing current until where it drops suddenly. In the example here with 2080 turns, with 0.3 mm gap, it will be about 340 with 80 mA and will go up to 350 decreasing the current to 70 mA. This is just for the records.... Measuring mu(DC) with a very small (10 mT) AC field is the only "easy" way I know. Such small perturbation won't change things. It's classical physics regime....
I usually measure values for mu because not all laminations are precisely what they are claimed to be. Didn't you know?
At max power Bac is max, the same is not true for mu(AC) as it depends on the material, its loop, where you are and the transformer parameters. For EI M6 at 1.6T you are not away from saturation at all and mu(AC) will be lower than that at 1T.
For average M6, Bdc=0.40-0.43T depending on the gap (0.4 to 0.3 nominal, respectively) with 2080 turns, 80 mA and 17.2 cm path length. With AC signal you can measure the inductance as function of signal (with 80 mA DC superimposed of course) and you will see that it initially increases, reaches a maximum and then starts to drop a long way before saturation. Verify where you get the max and see if this is the max Pout you assumed at 20Hz...
I don't think so.
For that particular coil formers, here has a width of about 39.1mm to 39.8mm, that not differ too much with pictures on post#49, about 41mm, and lamination window is about 42.9mm
Pryde S.R.L - Productos
Here, modern magnet wire to about 1.25mm (I can assure up to there) has the same polyesterimide resin thickness, 0.025mm, at least that I use.
To wind 130 turns of 0.25mm magnet wire you need at least
131x0.30mm=39.3mm
To wind 34 turns of 1.05mm magnet wire you need at least
35x1.10mm=38.5mm
With my winding technique is easily achieved, if with yours don't, a few turns less, or a few thousandths of mm less diameter, will not affect too much the results, anyway, an experienced winder will know to do.
Are you sure you are using the right wire or just guessing by adding 0.05 mm to the bare diameter?
I have found that insulation thickness increases a bit with increasing wire diameter. If your assumption were right a 0.16 mm wire would 0.21 mm for example. Instead mine is 0.199 (specified).
39.1 mm doesn't change things significantly. I don't believe you can put 130 turns turns over there with 0.3 mm wire.
Your technique is unimportant if you want to give an advice to another person as you cannot tell to do something without telling him HOW? You have to give him reasonable numbers. Still I don't believe a word of what you write about winding transformers. You have made so many mistakes until now....
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