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----- Original Message ----
From: "crabtreejr@..." <crabtreejr@...>
To: loopantennas@...
Sent: Saturday, May 5, 2007 9:09:04 PM
Subject: Re: [loopantennas] Wellbrook ALA1530













Hello Nigel, Paul et al



AFAIK no schematics for the Wellbrook have ever been published. However the

harware content which was discussed here, and other information, leads me to

the conclusion that the amplifier is a push pull grounded base design, as Paul

suggests.



It is not that difficult to work out a possible configuration, as follows:



1. Basic physics of a small loop



The basic physics of a small receiving loop is well understood (see eg

Jasik's Antenna Engineering Handbook, 1st ed, 1961, p 6-2)



The unloaded voltage induced in a loop antena =

2.Pi.A.N.E.Sin( theta) / Lambda

where

Pi = 3.14

A = the area of the loop

N = the number of turns

E = the electric field strength

theta = the angle measured from the axis of the loop

lambda = the wavelength



Let us simplify this, assuming that N = 1, and theta = 90 degrees. Let us

also replace Lambda by c/f where c = the speed of light, and f = the frequency.

Then

V = 2.Pi.A.E.f / c



It can readily seen that for a constant field strength E, that the voltage

ouput is proportional to frequency. There is a simple method to overcome this

difficulty.



The untuned receiving loop can be considered as a voltage source in series

with the radiation resistance (very small, ie a small fraction of an ohm), the

loss resistance (probably somewhat larger), and the inductance of the loop,

which is the largest component. Take for example a (Wellbrook) loop with a

diameter of 1 meter, and a tube diameter of 25mm. Then the the inductance (L) is

ca. 2.33uH, with an impedance of ca. 14.6 ohms at 1MHz.



If one terminates the loop in an impedance in a resistance (R) much lower

than the inductance, then the current output is determined by the inductance, and

equals

V / (2.Pi.f.L) = (A.E) / (c.L)

which is frequency independent



Then the power output from the loop is i.i.R

= (A.A.E.E.R) / (c.c.L.L)

which is again frequency independent.



As the frequency goes down, the ouput goes down for constant E, with the -3dB

frequency equalling

1 / 2.Pi.L.R



2. Loop termination resistance for a ALA1530



To determine an approximate value for R, it is necessary to look at the

performance figures of a Wellbrook loop. Fortunately, there are performance

figures for an ALA1530 on the Wellbrook web site. Looking at the numbers the -3dB

frequency is ca. 550kHz, which would give an R of ca. 8.1 ohms.



It is that simple - terminate a one meter loop with a resistance of ca. 8

ohms, and you will get approximately the frequency response performance of an

ALA1530. Clearly the response falls away at very low and very high frequencies,

but that is probably due to various transformer effects.



3. Amplifier gain for a ALA1530



We must now consider the amplifier gain. Assuming a unity antenna factor, ie

one volt out for a field strength of 1 Volt per meter, then the gain required

is

power out / power in

Assuming the ouptut load is 50 ohms, then the required gain

= (c.c.L.L) / (A.A.R.50)



Sticking in the numbers which we already have, I calculated that for a unity

antenna factor, that the gain required would be ca. 33dB. The actual antenna

factor is ca. -7dB, and so the required gain is ca. 26dB.



To recap, we know know that to duplicate a ALA1530, we would need an

amplifier with an input impedance of ca. 8 ohms, and a gain of ca. 26dB.



4. Amplifier configuration



Assuming that the ALA1530 uses two transistors in its push-pull amplifier,

then the only way to do is through a common base amplifier with a transformer

coupled input and output. A Norton amplifier would not have sufficient gain.

It should be remembered that the input resistance of a Norton amplifier depends

upon the output load, which we do not want in this case.



The small signal input resistance of a common base amplifier is ca. 26mV

divided by the emitter current. To get some good linearity, a high quiescent

current is essential. Removing the heat from the transistors has to be

considered.



Let us assume that pretty much all of the stated 120mA supply current goes to

the two transistors, ie ca. 50-60mA each. Then the input resistance of each

transistor at 60mA bias is ca. 0.43 ohms. The two are effectively in series

and so the total input resistance is ca. 0.86 ohms.



The required turns ratio of the input transformer is SQRT( 8 / 0.86) = 3.04.

This is conveniently very close to 3.0, which is what you would wind. This

means that the transformer can be wound with a trifilar winding.



The amplifer gain effectively is approximately the output load resistance

divided by the input resistance. We need a gain of ca. 26dB, ie ca. 400.



Assume that the gain is 27dB, ie 500. The the required output resistance is

500 x 0.86 = 430 ohms. Assuming that the transformed output load is 50 ohms,

this would give a turns ratio on the output transformer of 2.93, ie very close

to 3.0, which could also easily be wound with a trifilar winding.



My conlusion is that a push-pull amplifier with the transistors each biassed

at ca. 50-60mA, and a 3:1 step down transformer on the input and output would

work give approximately the measured performance of the ALA1530. This should

be a fairly straightforward amplifier, if some attention is paid to the

detail. 2N5109 transistors should work quite well. There are, of course, other

choices.



5. Observations



Many people have wondered how the broadband matching was done in the ALA1530.

It is simply done by having an amplifier with a constant input impedance

lower than the inductive reactance of the loop at frequencies for which a

constant gain is desired.



Many people have thought that the amplifier must in some way be exotic. I

suggest that it is actually fairly simple, and that this might be the reason for

potting it.



Later versions of the ALA1530 have an amplifier with better intermodulation

performance. They also have a higher quiescent current, and a higher gain.

This suggests to me that there are more transistors, probably in a two stage

push-pull amplifier.



There was a broadband loop available in the UK before the Wellbrook ALA1530

was introduced. It was designed by Edward Forster, and sold by his company -

Phase Track Ltd, which was based in Reading. His patent - GB2235337 - gives

some useful background on broadband loop antennas and their amplifiers.



Finally, broadband loop antenas are a compromise solution. At lower

frequencies, the radiation resistance is very low, and hence they are very

inefiicient. This means that you will not be able to hear signals near the noise floor.

Of course, they do appear to offer better noise rejection than an active rod

antenna, unless you are very careful (eg see some of Dallas Lankford's recent

writings). They are also convenient, and can offer some directionality.



HTH and 73



John KC0G



In a message dated 5/5/07 12:28:19 PM Central Daylight Time,

nonlinear@rogers. com writes:

Dear Nigel

forgot that getting old, I remember that now, I remember seeing the pictures

but was fuzzy on whether for sure bipolars were used

I believe it was not known from the autopsy whether it was a Norton design

or not

otherwise it is likely a push pull grounded base design for low noise and

wide bandwidth

what do you think?

was there a schematic ever published for the Wellbrook

best

Paul

----- Original Message ----

From: "gandalfg8@aol. com" <gandalfg8@aol. com>

To: loopantennas@ yahoogroups. com

Sent: Saturday, May 5, 2007 12:36:40 PM

Subject: Re: [loopantennas] Wellbrook ALA1530

In a message dated 05/05/2007 17:28:55 GMT Daylight Time,

nonlinear@rogers. com writes:

highest gain and lowest noise more likely with jfets

ie more likely in the design than bipolar although not necessarily true if

one considers say the David Norton noiseless design with bipolar

however the Norton design is limited to smaller gains

------------ --------- --------- --------- --------- ---------

"Is more likely" is a bit different to the earlier "must have".

I was just intrigued as to your reasoning, considering that the hardware

content of a dismembered 1530 was discussed here in some detail only a few

weeks

ago and the active devices were obviously bipolar.

regards

Nigel

GM8PZR

** **** **** ***

See what's free at . com.

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