로그인회원등록 장바구니주문조회운영자메일  상품 검색

게시판 검색           주요 링크 안내

 기술 연구
 ㆍ작성자 DHTsound ㆍ첨부#1 va012002ramsey.pdf (120KB) (Down:20)
 ㆍ추천: 0  ㆍ조회: 13237     파라피드 연구
Mathematical Derivation of the Parafeed Output Stage

I did this and to my surprise I noticed that the thing was resonating at certain values of choke & output transformer inductances and coupling capacitor values.  Sure enough when I dug through my trusty old Radiotron Designers Handbook 4,  there it was on page 519 a small table of values,  with the added cryptic note that "these values are sufficiently high to avoid resonance at an audible frequency." Really!

Here is the table, which is for a parafeed capacitor coupled interstage transformer. As the drive valve works in Class-A this should be similar to the output stage type, this one uses a resistor for the anode load instead of a choke.

 L 10H 20H 30H 50H 100H 150H C 4uF 2uF 2uF 1uF 0.5uF 0.5uF

As you can see there is a reciprocal relationship between the capacitor and transformer values,  so you can see that there is some kind of "factor" involved here.

I could find no statements explaining this "factor," mind you I do not have access to the copious reference material listed in RDH4.  Neither could I find a mathematical proof of the parafeed output stage in RDH4 nor anywhere else on the Net.

So I thought that I would ask those esteemed Bottleheads on the Bottlehead forum if they knew of any such proof.   Quite a substantial dialogue ensued, runs to twenty odd pages,  but the short answer was nobody else had a mathematical proof either.  Most advocated what I would call a "suck it and see" way of doing things and that is great,  however it did not appeal to the engineer in me.

Some of the things that came out of the forum discussion was that:

• As a general rule,  the coupling capacitor should be a reasonable size,  bigger than a couple of uF, heading up towards ten or so.

• The output transformer inductance had to be substantially larger than that of the plate choke.

• The value of the load reflected through the transformer was critical to overall performance.

Overall the level of discussion was excellent,  but it did not meet my needs to understand they why of the matter, rather that the it just is that was presented to me. Also, speaking personally, I find if I can understand the math/logic/schematics, it all falls into place so much quicker. So I started this quest to find the mathematical derivation of the parafeed output stage, and to try and determine the factors that make one successful or not.

This article is split into two parts, the first derives a mathematical statement for the parafeed output stage--this is my own work.  The second part goes on to describe a model that I created in Visual Basic for Excel.  Once I got over the hurdles of trying to work out what methods and object classes meant, I managed to develop a fairly sophisticated modeling tool.  I am going to look at setting up a web page with this on it.  I have always hesitated to do this in the past as I had nothing original to say.

Word of warning, I have taken some liberties with the nomenclature used, please forgive me.  Also, I only talk of impedances in general terms, X for a single item such as a capacitor,  and Z for more complex things such as inductors and resistors in series or parallel. My lecturers would be appalled at this, however it makes the document easier to read (well at least I think so.)  I did this as mucking around with complex impedances, with real and imaginary terms, is hard work and all it does is get in the way of what we are trying to do.

So please read on, it is fairly logical--well it is to me

Part One,  Mathematical derivation of the Parafeed output stage.

The parafeed output is shown in its AC form below (Figure 1).  The short circuit from the top of the plate choke to ground represents the ac short that is provided by the dc power supply output capacitors and low impedance of the PSU. (FWIW in the Pspice model that I made,  whether it was a short circuit or high value capacitor,  it made very little difference to the net result). Figure 1. AC equivalent of the Parafeed output

As can be seen, the output transformer is effectively in parallel to the choke via the coupling capacitor.

The above can now be redrawn to remove the output transformer. Figure 2.

In Figure 2, the voltage generator has a series resistance with it to simulate the plate resistance of the valve. The output transformer has been changed to a choke with the reflected output impedance shown in parallel to it. Also, I have omitted the leakage inductance (in effect I am using the low frequency transformer model, leakage inductance is only a problem at high frequencies,) The primary winding resistance can be included within the impedance equation for the primary (i.e., Ro+jwLo). The reflected secondary inductance can also be ‘lumped’ in with the primary inductance, the relationship between these two is the turns ratio.

Ei is the input voltage to the output arrangement, which when you look at it, is a filter.

The circuit can be further simplified by considering the reflected impedance of the secondary of the transformer and the transformer as one impedance (figure 3.) Figure 3.

From the above diagram we can see that the output voltage across ZL is derived as follows: Now we have to work out what Ei is as a portion of the generator voltage.  This is best done by again redrawing the circuit,  I promise you that this is the last time!! Figure 4.

From Figure 4, we can derive Ei as a portion of the generator voltage: Now putting all of this together, the following happens: The equivalent Z impedance is made up of the coupling capacitor in series with the output transformer in parallel with the plate choke,  this is represented in equation 4, below: Substituting (4) into (3) gives rise to the following: Multiplying the top and bottom with the denominator of (4) turns up the following: As you can see there are common terms to both the numerator and the denominator which further reduces the equation as follows: One further bit of manipulation comes up with the following: There is just one step to go,  that is to combine the left and central elements of the denominator The three bits of the denominator show the important elements of the parafeed circuit.

The "1," thankfully, tells us that provided nothing odd is going on with the reactive elements,  there is going to be less out of this circuit that what is put in.

The middle bit, expressed here: shows us that the relationship between the coupling capacitor and the output transformer is reciprocal (sound familiar?) which is what the table shows. It also shows that where the impedance of the choke is much greater than the generator resistance,  which is really the plate resistance, that the element above boils down to

Xc/ZL

So we have to be careful at low frequencies where the choke impedance is low compared to the generator resistance.

The third bit, see below, shows the relationship between the plate choke and output transformer inductance. Basically they are in parallel, which is self evident. What we have here, however, is their admittance, or the inverse of impedance.  Now remember that the whole equation here is upside down, so the parallel impedance has to be as large as possible, which makes the admittance small,  but magnifies it again when "turned upside down again" to give the final result.

This explains why the output transformer inductance (for inductance begets impedance, wL) has to be much greater than the choke.  With any parallel circuit,  the total Z is always less than the smallest Z in the circuit.  It can come close to the smallest Z only if the other parallel elements are much larger than the smallest.

Resonances

From the above we have shown that the rules of thumb spelled out to me in the forum chat are on the money.  The question that now arises is, what of the resonances?

If we go back to the equation below,  it can be seen that the output of the parafeed is going to be less than one as long as the sum of the two reactive elements in the denominator stays positive.  As this is a passive circuit normally this must be the case most of the time.  However there is a chance that they will sum to a negative number, in that case the output will be greater than one.  It is just a matter of finding what combination of frequency and component values that will cause this to happen. Well I did not fancy my chances at trying to use the above to determine output resonant frequencies. I would have been dealing with an equation with nine variables...no thanks. So I built myself a model. The first one I tried was in Pspice.  This was reasonably successful,  however,  I could not find a way of sweeping through capacitor, choke and transformer values to see what effect a change in either of these would have on the output.  That is where the second part of this article comes into play (I only have the demo version of ORCAD V8.0 to play with, and it does not support the above variable choke requirements, or if it does, I don’t know how to do it, which is quite likely!)

Part Two,  the Model

Before I go on,  here is a picture of the finished beast. Figure 5. Screen Shot

As you can see it runs in Excel.  I did this for two reasons, the first is the charting capability, the second is that it can also run Visual Basic.  This was necessary to create the routines for the complex impedance calculations

The graph on the left side of the screen is the result of putting in the numbers on the table on the right. The graphs can be printed off on their own.  As you can see, this is the output of the circuit, with a 100L output transformer, 20L choke, load of 7500 ohms, with a range of capacitors from 1 to 10 uF. It is possible to just as easily create phase response graphs.

There are four static values, all resistances:

• The generator resistance, Rg
• The choke winding resistance
• The output transformer primary winding resistance

Leaving the resistances static like this works ok, the effect of the winding resistances is negligible,  whereas the generator and load resistances have a major affect,  more of this later.

There are four variable values:

• Frequency sweep,  defined by Max, Min and Step values
• Capacitor range,  defined by Max, Min and Step values
• Output transformer inductance range, defined by Max, Min and Step values
• Choke inductance range, defined by Max, Min and Step values

when a particular component is swept, the max value of the remaining components are used as a static value.  I tried to generate 3D graphs by sweeping two parameters,  but is was difficult to see anything on the chart.  This will take further work to make graphs with any meaningful content. I may look into this later on as an add-on.

Internal workings

The model used to create the graphs is exactly the same as that shown in part one. To keep thing simple, I used equation 3, as this easier to work with.

I used complex numbers throughout and only converted to the absolute levels and phase angles at the end.  To do this I had to create a custom data type that comprised of two elements,  a real value and an imaginary one.  I also created customised functions to add, multiply,  divide (by using the complex conjugate) etc.

In order to ensure that the outcome was correct,  I did a couple of models longhand to make sure that my software was accurate,  this took some time and I found a few mistakes along the way.

What also first attracted me to excel was that it supports complex numbers directly in the spreadsheet. My first stab at doing this modeling was in the standard spreadsheet format,  however it was looking really ugly!!!

The results

I have used the following fixed values for the resistances,

• Rg = 700ohms

• Choke and transformer winding resistances = 350ohms

I have varied the output resistance later on,  to show the effect of the loading resistance.  All of the other parameters are shown in each chart. I was particularly interested in the 811A as a single ended unit, just love those anode caps! A good plate resistance for this thing is about 7500 ohms,  so I have used that.

Here is a graph showing the effect of keeping the load at 7K5,  the choke at 20L,  the output at 100L and sweeping the capacitor from 2 to 20uF. Figure 6. Output Graph, Capacitor Varies

Looks quite spiffy don’t you think? The blue line that droops down is the 1uF value, the rest as you can see cause resonance. What this is saying is that for this particular arrangement, keep the coupling capacitor low in value. Which is what it said in RDH4.

The next graph shows the same arrangement, but this time keeping the capacitor at 1uF, swinging the output transformer from 20 to 120 henries. Figure 7. Output Graph, Transformer Inductance Varies

In this example, it would seem that having too low an inductance is a real problem,  the big hump  there is with a 20L output transformer, so as you can see, and this is what the good folks in the forum told us, bigger is certainly better.

In the next graph,  I keep the output transformer at 100 henries,  the cap at 1uF,  and vary the choke between 2 and 40 henries. Figure 8. Output Graph, Chock Inductance Varies

What you can see here is that there is no real benefit in going for a huge inductance in the choke,  in fact it is neigh on impossible to get a plate choke of 40 henries with a substantial dc current flowing though it anyway.

So you can see, by a process of elimination,  we have come up with an output configure of a 1uF capacitor with a 10L choke,  and a 100L output transformer (this last item should be easier to do as there is no DC on the primary winding)

The graph of this arrangement is shown below. Figure 9. Output Graph, Model

Pretty smooth don’t ya think?

Just for interest sake, here is the phase response of the above. Figure 10. Phase Response of Model

This should not come as a big surprise,  as the sudden jump is where the capacitor starts to fight against the inductances of the choke and transformer.

One last graph--remember  I said that the value of the load resistance is critical?  Here is the graph that swept the output transformer value repeated, this time with a smaller load of 5000 ohms. Figure 11. Model using RL of 5K

See how the peaks have fallen? This is due to the heavier loading placed on the output by the lower impedance load.  It still shows however that a higher inductance transformer is required.

Conclusion

I hope that you have found this interesting,  I think that I have developed a tool that shows the why, and more importantly, a tool that allows the selection of the right components to ensure that the output is what it should be.

All I have to do now is to build one, but that is another article.

 HOME 쇼핑안내 회사소개 회사위치 제휴안내 운영자메일 사이트맵 