Output Transformer Saturation Surpise

[nickb]

A transformer manufacturer can obtain the desired magnetising inductance for an OT, either with a lower reluctance core (more iron) and fewer primary turns, or more turns on a higher reluctance core (less iron). The former would tolerate a higher magnetising current before saturation occurs.

But I will read that chapter as I did get the book. I'll try to do some more tests under better controlled conditons and see if we can get some improved data, when I get a chance.

[Malcolm Irving]

Nick: I’m wondering if that single-ended OT you are looking at, either does not have an air-gapped core, or has an inadequate air-gap for the application? That might explain the high inductance figures quoted by the manufacturer and a propensity to saturate when dc is present?

[trobbins]

I would expect that the change in magnetising peak current level with excitation voltage would relate to the level of B-H entering in to the saturation curve region. Even though the test was done with AC only excursion, it could be related back to an SE BH excursion with respect to the voltage-frequency being applied and presuming a DC biased mid-point.

A similar, but probably impractical test, would be to move the idle bias current point in the amp as you suggest, but move it higher and with a lowered signal level. Moving it lower as you suggest should avoid excursion in to the saturation curve region and hence lower distortion of the waveform.

But I think the drooping incremental inductance when pushing in to the saturation side of the BH curve can be seen from a few viewpoints, especially if the viewpoint is more about an averaged impedance or reactance over a whole cycle.

[nickb]

I would expect that the change in magnetising peak current level with excitation voltage would relate to the level of B-H entering in to the saturation curve region. Even though the test was done with AC only excursion, it could be related back to an SE BH excursion with respect to the voltage-frequency being applied and presuming a DC biased mid-point.

A similar, but probably impractical test, would be to move the idle bias current point in the amp as you suggest, but move it higher and with a lowered signal level. Moving it lower as you suggest should avoid excursion in to the saturation curve region and hence lower distortion of the waveform.

But I think the drooping incremental inductance when pushing in to the saturation side of the BH curve can be seen from a few viewpoints, especially if the viewpoint is more about an averaged impedance or reactance over a whole cycle.

I planned to build a little constant current test jig but decided that, as you suggest, a simple test with a variac will be revealing enough so here are the results plotting Vrms applied at 50Hz with Ipk in mA.



At the bias point of 40mA it looks like the core has started to move into saturation. I didn't take measurements below 170V. In retrospect I wish I had as it would have made the shape of the curve clearer.

Nick: I’m wondering if that single-ended OT you are looking at, either does not have an air-gapped core, or has an inadequate air-gap for the application? That might explain the high inductance figures quoted by the manufacturer and a propensity to saturate when dc is present?

I read the chapter. Indeed yes the Danbury transformer is a quite different animal to the Hammond one. I think you are correct about the Danbury one being gapped and the Hammond being non-gapped.

The 1750C I'm testing is an older one with a 7K primary. I see they have updated the design to an 8k with a 5K tap and specify 50mA DC max. The inductance is high so I guess it's still not gapped. I got identical results comparing my old 1750C with the fitted Fender OPT so I guess the new one is not so similar. Shame. I wonder why they did that and confusingly for everyone kept the same part number?

I should have captured the current & voltage waveforms so I could differentiate the current and so calculate the working inductance since v = L (di/dt). I didn't think of it until afterwards. Too late now.

[Malcolm Irving]

...
At the bias point of 40mA it looks like the core has started to move into saturation ...

It does seem strange for single-ended applications. In a typical SE OT design, saturation occurs when magnetising current reaches about twice the intended quiescent dc.

EDIT: I suppose some saturation at 50Hz is acceptable, since the guitar only goes down to 82Hz.
Also, I shouldn't have compared 40mA at 50Hz with 40mA of dc. With sinusoidal voltage supply the core will begin to saturate at the peaks of the waveform, and will progressively saturate for a greater proportion of the waveform as the voltage is turned up.

[nickb]

It does seem strange for single-ended applications. In a typical SE OT design, saturation occurs when magnetising current reaches about twice the intended quiescent dc.

Quite. I checked the current probe calibration too. I note the slope is still fairly linear so the core is some way from fully saturated. I should have cranked the volts up higher but caution got the better of me

[Malcolm Irving]

When can a single-ended pentode saturate the output transformer?

Let’s say we have a single-ended pentode output stage with idle DC current of 50mA and a resistive load (referred to the primary of the OT) of 5k. The output transformer will typically be designed so that the 50mA dc sits halfway between zero magnetisation and saturation. So the OT will start to saturate at 100mA instantaneous magnetising current.

For this analysis the OT (as seen at the primary) is very accurately represented by a 5k resistive load (referred from a resistive load on the secondary) in parallel with a magnetising inductance. The OT primary current is then given by:

Ip = Ir + Im, where Ip = primary current, Ir = current into the referred resistive load, and Im = current into the magnetising inductance

The transformer will start to saturate at any instant when Im reaches 100mA. The 50mA quiescent dc is always present and is part of Im. If the pentode is producing a sinusoidal output, there is an additional sinusoidal current, call it Im(ac), through the magnetising inductance, and this would need to reach 50mA peak to reach saturation.

Im = Im(dc) + Im(ac), where Im(dc) = 50mA in this example

The same voltage is across the referred resistive load and the magnetising inductance, and therefore the ratio of peak values of Ir and Im(ac) is given by:

Im(ac) peak / Ir peak = Rload/(2.pi.f.Lm), where Rload is the resistive load referred to the primary (5k in this example), Lm is the magnetising inductance, and 2.pi.f.Lm is the magnetising reactance

Im(ac) is lagging Ir by 90 degrees, so we have to use vector addition to add a sinusoidal Im(ac) to a sinusoidal Ir.
We can take a few examples at different frequencies, and find out how much Ip the pentode must deliver, for Im to be at the point of saturation.

Case 1 (the -3dB point)
At this point 2.pi.f.Lm = Rload
and the peak values of Ir will equal that of Im(ac).
The peak value of Ip (at saturation) will then be sqrt(50*50+50*50) + 50 = 121 mA.
This is the instantaneous peak current that will have to be delivered through the primary winding by the pentode.

Case 2 (twice the frequency of case 1)
At this point, the reactance of Lm has doubled. If Im(ac) is still reaching saturation, Ir peak must now be doubled.
In this case the peak value of Ip (at saturation) will be sqrt(50*50+100*100) + 50 = 162 mA

Case 3 (half the frequency of case 1)
The magnetising reactance is now half that of case 1. Ir peak will now be half the value of Im(ac) peak.
The peak value required from the pentode is now sqrt(50x50+25x25) + 50 = 106mA

Can the pentode delver the currents required to reach saturation in the above cases? This can only be found out by looking at the plate curve for Vg1=0. This shows the highest current that the pentode can deliver without requiring significant grid current. Most SE class A output stages are designed with the resistive load line passing through this plate curve at, or near, the ‘knee’. With a bias point of 50mA, the intersection with the Vg1=0 plate curve will typically be at about 100mA.
To achieve saturation in the three example frequencies considered above, this current of 100mA must be increased to 121mA, 162mA and 106mA respectively. Once ac magnetizing current becomes significant, the resistive load line is expanded in the vertical direction (on the plate characteristics graph) by the ‘magnetising inductance load ellipse’. Whether this can be accommodated within the upper limit imposed by the Vg1=0 plate curve will depend on (among other things) how flat the plate curve is and whether the resistive load line was designed above or below the knee. Certainly, 106mA would seem to be achievable. True pentodes, such as EL34, have more ‘curved’ plate characteristics compared with beam tetrodes, such as 6V6, and would seem to have a greater chance or providing the necessary current for saturation.

[Malcolm Irving]

With a loudspeaker load, the load line is already elliptical at most frequencies, due to the reactive nature of the speaker. If the speaker is behaving inductively, this reduces the likelihood of OT saturation, because there is less ‘headroom’ for the magnetising current (to add to the load current) before the pentode’s current is limited by the Vg1=0 plate curve.

For quite a range of frequencies a loudspeaker's impedance is capacitive (a mechanical effect - not electrical). I think this might increase the 'headroom' and make saturation more possible, as the capacitive load current will be at the 'bottom' of its ellipse while the magnetising current is at the 'top' of its ellipse.

[nickb]

Can the pentode delver the currents required to reach saturation in the above cases?

Easily, I think. I just grabbed a 6V6 off the shelf and plotted at the working voltages of this amp viz Vg2=345V:




If you look back at the OP you can see the current peaks at 130mA. I wasn't driving it anywhere near as hard as I could.

[Malcolm Irving]

If I have the right figures, I think your amp has a bias current of 40mA, a B+ of 350V, and a referred load impedance of 8.75k. That would put a load line through the vertical axis at 80mA, roughly going through the knee of the blue curve.

I assume the red curve is the Vg1=0 curve, and the nominal load line is well below the knee for that one.

[nickb]

If I have the right figures, I think your amp has a bias current of 40mA, a B+ of 350V, and a referred load impedance of 8.75k. That would put a load line through the vertical axis at 80mA, roughly going through the knee of the blue curve.

I assume the red curve is the Vg1=0 curve, and the nominal load line is well below the knee for that one.

Near enough, it's 345V and 45mA.

The curves are plotted at Vg=0 to -25V in 5V steps. I hid the Vg legend on the plot as it was messy, forgetting that no-one else would know.

[benito_red]

I spent a lot of time building 5F1 clones. I was never satisfy with their sound. The real truth is 5F1 is a crap project with a even more crappy output transformer, and a crappy layout (original Fender 5F1 layout is defenetively a crappy amp), in fact with a decent layout and ground connections you can greatly reduce the noise. And is absolutely unfair to claim 5W of output power for a SE 6V6 while will never be possible to get more of 2,5W without to melt the 6V6. If you want to build a better amp that will outclass Fender or Victoria Champs and costing you less you MUST:
1) discard the 5F1 circuit and adopt 5F2-a (a simple tone control is important)
2) get a bigger airgapped transformer (it must have central column with about 625 square mm, similar the one used by VHT Special6
3) set an on/off switch for the negative feed-back
4) use at least a good 10" speaker with proper open back cabinet
5) adopt a star ground layout
the difference will be as night and day, it will exhibit much better sine shapes even at low end, and nobody listening such an amp will belive it's only a 2.5w amp

[olddawg]

Or you can spend $80 on CL and buy a used Epiphone Valve Jr that actually sounds good and don't have to go through the trouble.

[Mike Sulzer]

I spent a lot of time building 5F1 clones. I was never satisfy with their sound. The real truth is 5F1 is a crap project with a even more crappy output transformer, and a crappy layout (original Fender 5F1 layout is defenetively a crappy amp), in fact with a decent layout and ground connections you can greatly reduce the noise. And is absolutely unfair to claim 5W of output power for a SE 6V6 while will never be possible to get more of 2,5W without to melt the 6V6. If you want to build a better amp that will outclass Fender or Victoria Champs and costing you less you MUST:
1) discard the 5F1 circuit and adopt 5F2-a (a simple tone control is important)
2) get a bigger airgapped transformer (it must have central column with about 625 square mm, similar the one used by VHT Special6
3) set an on/off switch for the negative feed-back
4) use at least a good 10" speaker with proper open back cabinet
5) adopt a star ground layout
the difference will be as night and day, it will exhibit much better sine shapes even at low end, and nobody listening such an amp will belive it's only a 2.5w amp

There are some complicated questions regarding the transformer. I do not know if using a gap is the most efficient way to make a transformer for a single ended amplifier, but the gap is not essential for a good transformer. To see this, consider designing a transformer, same specifications, except specify a minimum operating frequency half of what we need and ignore the dc bias current through its primary. In the design of such a transformer the B field must be kept below some value that would cause the core to saturate. The relationship between the maximum field that is produced and the voltage across the transformer is given by Faraday's law of magnetic induction, which takes the following form here (eq. 22-4b, Radio Frequency Electronics, Jon Hagen):

Bmax = Vmax/( N*omega*(core cross section) ) < (values that causes core saturation)

Now we look at how Bmax changes when the dc bias current is introduced for SE operation. Since the flux no longer changes between positive and negative values, but rather varies between zero and twice that value, Bmax is twice as big as shown above. But we can make it stay at the original value by making the lowest frequency (omega) twice as high, that is, the value we originally wanted. This is one way to make an SE transformer without a gap. Not necessarily the best.

But what results in the cheapest good (or bad) transformer? The gap effectively lowers the permeability of the core, allowing more voltage before the maximum B field is reached, but it also lowers the inductance, which raises the minimum operating frequency. I do not think you can answer the question without going through the whole design process, which requires finding the best compromise in a coupled set of inequalities for Bmax, the losses, and the minimum inductance.

[bob p]

Over the years Fender used a wide variety of iron on the different SE champ circuits. Mercury makes reproductions of each one,which can make choosing one for an old build a bewildering proposition. A long long time ago I bought every Champ transformer that they offered and hooked them up on a bare chassis to a breadboarded 5F1 circuit, using a giant ohmite rotary switch that I had cannibalized out of a resistor decade box to switch between them. This allowed me to just twist the switch to change transformers in real time and compare the effects of different types of transformers on the circuit.

I'm not sure that I'd agree that you want an air gapped SE transformer with a champ. While it gives a more HiFi experience, IMO that's not what gives the circuit it's best tone. Those itty bitty transformers to a lot to shape the tone by cleaning up the top end. In some respects that makes the princeton type tone control unnecessary.

When this thread first came up I looked on my PC to find a snapshot of my transformer switching rig, but alas, it's gone. I had posted the photo here years ago, but that pic seems to have been lost in the Great Image Purge.