No it's not. It is confused especially because there are two conceptual mistakes: 1) the secondary AC current cannot balance the primary AC flux (as stated earlier) and so you need headroom for the signal 2) the AC induction can be such that you can cross both B=0 and/or saturation, it depends on how the transformer is made and how it is used.
If you say that a DC current is causing saturation then you have no room left for the signal. Instead if you say a certain max DC current is recommended to get the rated power such current will be well below saturation.
Take the Lundahl LL1682/50 mA below:
http://www.lundahl.se/wp-content/uploads/datasheets/1682.pdf
It says that 50mA DC current is the current that will allow a maximum power of 8W @30Hz into a 5R load. Of course you can use less than 50 mA but you won't get 8W from any tube at reasonable distortion level.....
The 50 mA will generate 0.9T DC induction and 8W into 5R will generate 0.7T for a total induction of 1.6T. At 1.6T the core is still below saturation. The DC current that will cause saturation (i.e. 1.8T DC induction) is 100 mA. Exactly 2x.
The partition between DC induction of 0.9T and AC induction of 0.7T is the same for all Lundahl transformers.
Of course you can use the above transformer at 60 mA but the useful output power at 30Hz for 1.6T total induction will be about 5.5W into 5R.
The Transcendar of this thread is rated 90 mA max DC bias which is NOT the "max DC current before saturation" BUT it is the max DC current you can use for 10W output without saturation. Is this the same thing??
It is not clear down to what frequency as Euro21 who used it didn't tell how much DC current he used for his 300B.
Yes. The counter MMF from the secondary is what will stop the primary AC flux to increase any further. What secondary AC current will not do is balancing the primary induced AC flux. Otherwise the primary AC current will respond to it. So you cannot use the full headroom for DC current. The saturation (or near) DC current is that current taking all (or nearly all) the available headroom.
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I stand corrected on this as I missed a couple of posts. 1) Transcendar responded that 90 mA is a max current that will result in induction close to saturation. Euro21 wrote in another post later that he used 70 mA.
So it sure that at 70 mA DC current can be considered a 10W transformer as the power response is within 0.5 dB down to 40Hz. This not a bad figure being an affordable product.
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Not sure what you mean by that. There are three significant flux paths in a SE transformer:
(1) The main flux path, which links both the primary and the secondary. The flux in this path is driven by imbalance between primary and secondary MMFs.
(2) The primary leakage flux, which links only with the primary winding.
(3) The secondary leakage flux, which links only with the secondary winding.
(2) and (3) only have relatively small flux levels and cause leakage inductances.
Sorry that had to be B.
Forget (2) and (3). Yes it is point (1).
All started from this
It seems that you can operate the transformer at DC current just below saturation to get the rated power.
This statement is simply not true and was in response to the fact that Dual Triode wants to find the optimal use of his transformer (i.e. use the all the headroom available without reaching saturation or crossing zero and at same time getting the max output power).
Now taking again the datasheet of LL1682 above let's see if it can't go anywhere near saturation or zero.
Already used as it is it will saturate for 8W just below 30Hz.
This transformer has a ratio that results in 8.8K for the more typical 8R load and has very good insulation. Now I want to use a 211 but at lower voltage accepting less power than usual. The 211 datasheet says that at 750V/34mA I can get 5.6W with 8.8K.
What happens if I use the LL1682/50mA at 34 mA? The DC induction will be 0.61T and .99T will be available towards saturation. How much primary voltage will get you there? It's 283V (i.e. about 9W) which the 211 in the above condition cannot provide. However 0.62T in the other direction will now result in zero crossing. How much primary voltage is this? It's 177V (i.e. less than 3.6W).
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As the triode approaches cutoff
All,
This is why I attempted to leave the power triode out of my version of the Single End OPT discussion. The triode can only produce positive current. With increasingly negative grid voltage the decreasing current flow approaches “cutoff” and the output begins to clip. The triode output will become seriously nonlinear as the triode approaches cutoff.
The current output cannot go negative. It will not cross zero
I was off target bringing up zero crossing for Single End OPT’s.
DT
All,
This is why I attempted to leave the power triode out of my version of the Single End OPT discussion. The triode can only produce positive current. With increasingly negative grid voltage the decreasing current flow approaches “cutoff” and the output begins to clip. The triode output will become seriously nonlinear as the triode approaches cutoff.
The current output cannot go negative. It will not cross zero
I was off target bringing up zero crossing for Single End OPT’s.
DT
Yes it is.
I should be careful speaking for DF96, but I believe it is clear he did NOT say the secondary current balances the primary AC flux as you state. Nor do I say that. You need to study ampere-turn balance. What is being claimed is that ampere-turn balance does not contribute additional flux over and above what is generated by standing DC current plus AC magnetizing current. DF96 can reject or support this idea, but that is my understanding of his post, and is completely correct.
The watts output to the speaker is a result of the AC voltage applied to the speaker, assuming a simplified model that an amp is a voltage source. Based on this AC voltage applied to the complex load, we have an AC current that flows. This current obeys transformer law, being present by ampere-turns in both primary and secondary, and does not create additional flux in the transformer core.
However, the AC voltage applied to the primary winding does cause an 'extra' current to flow based on primary inductance of the transformer. This 'extra' current is present only in the primary, and is very much *smaller* than the load current. This is the magnetizing component, and is what results in the additional flux I suspect you are referrring to. In all cases it is always magnetizing current, and is only present in the primary winding. Certainly this will result in saturation at some level to some degree. I think calling is 'signal current' is a misnomer (believe it was called that somewhere) and might be contributing to the debate back and forth. I see 'signal current' as being impressed into the speaker, which is not the same as magnetizing current which contributes to flux.
However, the AC voltage applied to the primary winding does cause an 'extra' current to flow based on primary inductance of the transformer. This 'extra' current is present only in the primary, and is very much *smaller* than the load current. This is the magnetizing component, and is what results in the additional flux I suspect you are referrring to. In all cases it is always magnetizing current, and is only present in the primary winding. Certainly this will result in saturation at some level to some degree. I think calling is 'signal current' is a misnomer (believe it was called that somewhere) and might be contributing to the debate back and forth. I see 'signal current' as being impressed into the speaker, which is not the same as magnetizing current which contributes to flux.
Of course the triode current cannot be negative. But Bac>Bdc doesn't mean the triode current has to be negative.
In the example above if you apply 177V rms at the primary then Bac is 0.62T while Bdc is 0.61T. So Bac>Bdc. However 177V rms mean 500V peak-to-peak and this is about 57 mA peak-to-peak. This approx. +/- 28.5 mA peak around the DC current of 34 mA. It still is 5.5 mA above cut-off. The approximation is not bad considering the excellent linearity of the 211. So the triode current does not become negative and actually is still above cut-off.
Let's re-check the transformer working conditions @30Hz:
Bdc is proportional to Idc. So if 50mA results in 0.9T then 34 mA results in 0.612T. Bac is proportional to applied volts. If 200V rms produce 0.7T then to get 0.62T you need 177V rms. In fact 177V rms ruslts (neglecting losses) in 3.6W which is less than max power of 5.6W.
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I meant AC flux density (i.e. Bac) created applying signal at the primary. I have already corrected this in response to Malcom. The flux is only one. You cannot balance this. If Bac becomes too high you get saturation and becomes really easy as the frequency decreases. In fact most output transformers saturate well above 20Hz at the rated power.
Neglecting losses,
P = Ep² / Zp ==> Ep = √(P Zp) = Uac
Take this as a maximum value, then magnetic field B is given by
Bac(max) = (Uac x 10⁸) / (√2 π fo S Np)
I hate the word flux because all fields involved have defined their own flux, having said that, this magnetic field B is the maximum reached in transformer operation.
The magnetic field B has an associated magnetic field H
B = μ H
To avoid the use of tensors, we must treat magnetic permeability, μ, as a constant; this treatment is greatly justified in SE OPT, because the air gap in the core make μ quite constant over the entire magnetic hysteresis loop B vs H.
The magnetic field H is associated to a current called magnetizing current, Im, which is a fraction of the primary current, the value of this current depends on the transformer parameters and the frequency, and this current does not cause any additional flux, this current causes the whole flux (B = μ H), anyway, the term magnetizing current is not the best because the core is already magnetized by DC bias current.
Ok but ac current can increase and decrease respect to DC current. Flux takes the same shape of the applied excitation. Zero crossing means crossing one axis and there two defining a plane. Crossing the H axis (Bac>Bdc and opposite in sign) doesn't imply change of sign in excitation current. This already results in some more distortion. Of course it is worse if you run a loop around the origin (mulitple zero crossing)
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Nothing to thank, it is my pleasure, and I think that you are understanding quite well SE OPTs, just to avoid complaints and confusions, try to change “AC swing” by “AC voltage swing” in your previous posts.
In PP OPTs, when B=0 on the magnetic hysteresis curve B vs H, in the core the vector B change its sense of direction, from this → to this ← and vice versa, fortunately, in SE OPTs the current always flows in the same direction and zero crossing never happens.
And you are treating an SE OPT as though it is a PP OPT, or even worse…
Now you are treating an SE OPT as though it is a power transformer, in a PT, if you put a load on the secondary, the secondary AC current produces a magnetomotive force
F = Ns Is
By lenz's law, "If an induced current flows, its direction is always such that it will oppose the change which produced it" then, on the primary we have an opposed force
F = Np Ip
As Ip and Is are in opposed phase, the corresponding magnetic fields Hp and Hs are cancelled each other, the only "survivor" is the one corresponding to magnetizing current, H, which was already taken into account in B = μ H, with usual values of magnetic permeability (μ in the order of 10⁴), H and then magnetizing current will be quite small.
For a SE OPT, the magnetizing current could be quite high because the core has an air gap, even more at low frequencies.
Maybe you are misunderstanding how an SE OPT works.
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Impressive discussion.
Magnetics seem to be an art u find books predating the triode yet so difficult to grasp. I have tried and keep giving up. So I just wire up the amp and readjust my calculated bias to the actual bias while looking at tha scope. So much easier.
Fascinating subject tho so I may some day try understanding more than utter basics once again....
Magnetics seem to be an art u find books predating the triode yet so difficult to grasp. I have tried and keep giving up. So I just wire up the amp and readjust my calculated bias to the actual bias while looking at tha scope. So much easier.
Fascinating subject tho so I may some day try understanding more than utter basics once again....
True.45 said:
Not true. The max DC current is just that: max DC current. It will be set either by saturation or wire heating; probably the former, or both at the same point. It is not the max DC current recommended to get a particular power output - you have completely misunderstood what the datasheets say.
Yes - provided that you leave room for the magnetising current, not the total primary AC current. You don't seem to be able to distinguish between these two different concepts.
Yes. We may be using slightly different language, but I believe we are saying the same thing: a transformer (with secondary AC current) is not a choke. The whole point of a transformer is that secondary AC current roughly balances primary AC current (taking into account turns-ratio); in an ideal transformer they would exactly balance.zigzagflux said:
I don't see how we can have a meaningful discussion about details of audio transformer behaviour when some seem unable to grasp basic transformer action.
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