PPP OPT clarification

[Miles Prower]

How about an illustration to make this clearer? Actual Design PP Loadline.

As for OPTs, I used a 4K4 : 8R PP OPT. As for the impedances involved, this xfmr matches 8 ohms to an 1100 ohm load. 1100R : 8R corresponds to a voltage transformation of 11.73 : 1. If you want to make a push-pull OPT, then tou need two symmetrical halves for the primary, or a voltage ratio of 23.46. Doubling the voltage ratio increases the impedance transformation by a factor of four, therefore: 8(23.46)^2= 4403, or ~4400 : 8 across the whole primary. Each tube connected to the primary still sees the 1100R as its load since the primary center tap is at AC ground. Thus, when doing the loadline, use 1100R as the load resistance.

The DC loadline is red. This is used to determine the static Q-point. Since there's 350V at the plates, and the 6BQ6 has a Pd= 12W, 25mA of static DC gives a Pd= 8.75W -- comfortably under the max plate dissipation rating. This loadline intersects on the -30V grid line, so this establishes the Q-point.

The dynamic loadline (blue line) starts at the (350, 0) point. The justification for this being that the OPT primary responds to differential current only. With no signal, each VT draws the same current, and there is zero differential current, therefore no output, and no DC core magnetization, which is why PP OPTs don't require an air gap to avoid core saturation. This line ends at the (0, 318.2) point since: I= 350/1100= 318.2mA. Since this line starts at (350, 0) it's considered to be the "Class B" loadline. If the 6BQ6s were actually to be biased to Ip= 0 at no signal, then they'd be operating in Class B.

Having drawn the loadline, the max current is read where it intersects the Vgk= 0 line, since this design is to be a Class AB1 that doesn't pull grid current. This is 262.2mA, and this is a peak current. Since one 6BQ6 will pull that current at max no-clip signal while the other one is way into plate current cutoff, that will also be the total primary current, and so the output power becomes:

Irms= Ip / sqrt(2)
Irms= 262.2E-3 / sqrt(2)= 0.185Arms
Po= Rl * Irms^2
Po= (1100)(0.185^2)= 37.65W (Not including the ineviatble core and copper losses in the OPT)

Since the DC rail supplies two VTs, the total rail current is twice the peak plate current, or 524.4mAp. The average DC current then becomes:

Iave= Ip / pi
Iave= (0.5244)(3.414)= 0.167A

The DC input power: (350)(0.167)= 58.45W

Plate dissipation: Pd= (Pdc - Po) / 2
Pd= (58.45 - 37.65) / 2= 10.4W per plate, and we're still comfortably within specs.

To see just how deep into Class AB we are, add the Class A loadline. Since Class A means a duty cycle of 360 deg, the resulting primary load will be 2200R since we basically have two "generators" in series feeding the primary. Start at the intersection of the DC current and voltage lines. Since that means dropping 350V, the change in current for a change of 350V becomes: 350 / 2200= 159.1mA. This delta current is added to the Q-point current of 25mA, so the Class A line hits the current axis at (0, 184.1).

The intersection of the Class A and Class B loadlines is the transition point. Read the volatge there as: 297.14V. The Class A power becomes:

Vp= 350 - 297.14= 52.86Vp
Vrms= Vp / sqrt(2)= 52.86 / sqrt(2)= 37.28Vrms

Po= V^2 / R= (37.28^2) / 1100= 1.26W per VT. Since two VTs contribute to this, the total Class A power is: 2.52W.

So for the first two-and-a-half watts out of ~37W, we're in Class A, so this is operating deep into Class AB. Crossover distortion could be a concern here, but it turned out to not be much of a problem. Since the 6BQ6 is a TV horizontal deflection type, it can be biased quite a bit "hotter" in operation since TV HDs were rated very conservatively, due to the demanding nature of the intended operation. I've run them as hot as 65mA of plate current per tube without red-plating them, so they are a good deal tougher than comparable audio finals, such as a 6V6.

If you were going to parallel up, you'd have the same voltages, but twice the current. This would give you a load of 1100 / n. For two 6BQ6s per phase, Rl= 550R, and you'd need a 2200 : 8 OPT.

[Miles Prower]

One other indicator of performance is estimating harmonic distortion. From the Fourier series, to a first approximation, the h3 content is given by determining the plate current at the "half way" point. For a Vi= 30Vp, this would be at Vi= 15Vp. If one 6BQ6 has an instantaneous Vgk= -15V, the other one is well into plate current cutoff at Vgk= -45V. This makes it easier since there is no need to construct a composite characteristic for the Vgk= -15V line. The Class B loadline hits the Vgk= -15V line at 109.29mA (green line). The third order harmonic current is given by: I0.5= (Imax - 2I0.5) / 3

I0.5= (262.2 - 2*109.29) / 3= 14.54mA

The fundamental current is given by: If= 2(Imax + I0.5) / 3

If= 2(262.2 + 109.29) / 3= 247.66mA

The estimated THD= 100(I0.5 / If)

THD= 100(14.54 / 247.66)= 5.87%

Not real good, but not real bad either. Of course, it's a rough estimate since there may be higher order harmonics, and the assumption is perfect balance that perfectly cancels even order harmonics (not likely!) and all the higher orders of the Fourier series are ignored in favour of a simple graphical solution. Also, different VTs have different harmonic "personalities", and a low estimate doesn't mean good sound if the harmonics produced are all high order ones. The third isn't so bad since it's musically related, producing "edge" or "brightness". Besides, applying gNFB will improve linearity, damping, and reduce distortion arising from the front end and OPT nonlinearities.

In actual open loop operation, the third measured more like 3.0% with just a trace of a fifth and higher. Didn't sound bad running open loop, and gNFB improves the sonics quite a bit in practice.

[PeteN]

Chris: Yes, it's a (dual) triode, so I was operating under the assumption that instead of aiming for the 'knee', one goes for a tangent to the Pa max as you said, though thanks for your continued clarifications!

Miles: wow! Took a while to read, but in the process I took a handwritten copy as well as saving the page verbatim *taps head* Thankyou so much, can't have been a quick thing to type up.

Finally some formulas/rigidly defined methods! Yessssss! This is more like it

Thanks again, I owe you all a beer if you ever come to Cardiff. (offer valid at the students union only though, I need to start saving for transformers...)

[Sheldon]

Very nice tutorial Miles.

Sheldon