G3YNH info: Maximally flat transmission bridge.

TX to Ae

Ch6 Contents

A6.5

Part 2

A6.6: Resistive-Voltage-Sampling and Maximally-Flat
Transmission Bridges.

Inherent temperature compensation

Flat detector frequency-response

1. Prototype maximally-flat bridge.
1.1. Maximal flatness. IS network.
1.2. Maximal flatness. VS network.
1.3 Balance conditions.
1.4 LF drop-off.
1.5 Candidate transformers.
1.6 Quadrature point.
1.7 Preliminary design considerations

1.8 Losses in the current network.
1.9 Trial voltage-sampling networks.
1.10 Multi-turn primaries.
1.11 Overboost.
2. .
3. High-frequency RVS model.

Part 2 >>>.

Abstract:

>>>>> Work in progress

Introduction:
A problem with the design of conventional RF broadband current-transformers lies in reconciling the choice of coupling factor with the need to maintain good sensitivity at low-frequencies and good phase performance at high-frequencies. This is a particular drawback when trying to design accurate passive directional power-meters or return-loss analysers, because the low-frequency roll-off in detector sensitivity makes a nonsense of any attempt at indicator calibration. It was shown in section 6.2-10 that, in order to obtain amplitude flatness within 1% using a conventional transformer, it is necessary for the secondary reactance to be 7 times greater than the secondary load resistance at the lowest operating frequency. For a lower limit of 1.6MHz and a load resistance of 50W, this implies an inductance of 35mH which, in HF radio engineering terms, is large and in some cases impractical. The difficulty becomes apparent when it is observed that a large inductance can be obtained in one of three ways: by using a large number of secondary turns; by using a transformer core with a large magnetic path area; or by using high-permeability core material. None of these options is attractive. Using many turns or a large-area core implies a large conductor length; with attendant problems of propagation delay and winding resistance. Using many turns also implies a low coupling factor and hence low sensitivity. Resorting to high-permeability materials (which are more properly intended for EMC filtering and other non-critical applications) results in high core losses, strong dispersion effects (i.e., frequency-dependent inductance variation), and a huge temperature-coefficient of inductance. It starts to look as though good low-frequency performance is a lost cause unless frequency measurement and digital signal processing is included in the design; but there is a passive solution; the maximally-flat current-transformer network, which can achieve a flat output without the need for a large secondary inductance.
The theory of the maximally-flat current-transformer was introduced in Appendix 6.2 and confirmed experimentally in Appendix 6.3. The circuit is that of a conventional current-transformer network with an additional capacitor placed in series with the secondary winding. The capacitor is chosen in relation to the other circuit parameters so that the network becomes a maximally-flat second-order high-pass filter, the effect being to steepen the low-frequency skirt and give an almost-constant in-band amplitude response. The inductance requirement for an in-band amplitude flatness within 2% is reduced by a factor of about 4 by inclusion of this LF-boost capacitor.
In order to use the maximally-flat current-transformer as part of a transmission bridge; it is necessary to devise a companion maximally-flat voltage-sampling network. The point is to tailor the frequency response of the voltage network so that it tracks that of the current-sampling network in both magnitude and phase. Unfortunately, it is not possible to do so by modifying the conventional capacitive potential divider, because the counterpart of the boost capacitor is then a negative resistance. This means that a resistive potential-divider is mandatory if a solution is to be found using passive components. Resistive voltage-sampling (RVS) bridges are unsuitable for monitoring KiloWatt transmitters because the divider will typically absorb 1 or 2% of the input power; but it is important to consider the merits of a maximally-flat bridge in the context of an appropriate application. The point is to make an instrument capable of producing accurate return-loss or SWR measurements; in which case the power-level at which the reading is made is a matter of choice, and is preferably low. One virtue of the maximally-flat current-sampling network is that it requires a low number of turns on the transformer and therefore allows a high coupling factor. Hence it is ideal for sensitive bridges, i.e., bridges which give an off-balance output of several Volts when the transmitted power is in the 10 to 100W range.

1. Prototype maximally-flat bridge:

RF current-transformer.

¬ Theoretical prototype for a transmission bridge with a maximally-flat detector frequency-response (low-frequency model - does not include parasitic capacitances).

The first thing to notice about the prototype circuit shown above is that it is a low-frequency model. There is no component to represent propagation delay and other effects which mimic transformer secondary parallel capacitance; there is no component to represent the self-capacitance of the LF compensation coil Lv; and it is assumed that the resistor R2 has no capacitance. Such liberties can be taken at this stage of the analysis because the boost capacitances Ch and Cv are relatively large, and their reactances become correspondingly small at high-frequencies. Thus the bridge degenerates into a conventional RVS arrangement in the upper reaches of its frequency range, allowing the HF neutralising requirements to be deduced by reference to a simplified model.
The current-transformer is shown with a Faraday shield. This is done, not because a shield is necessary (see A6.4-19 and A6.5), but because it simplifies the circuit analysis. The shield adds stray capacitance from the secondary winding to ground; which, although not desirable, can be lumped with the self-capacitance of Lv in the high-frequency regime. Omitting the shield introduces stray capacitance from the through-line to the secondary winding, which complicates the model considerably; and by reducing the average capacitance per unit length of the through-line, gives rise to a mis-match which manifests itself as an increase in the apparent secondary parallel capacitance of the transformer. Notice also that the shield is earthed on the generator side of the transformer. This means that the capacitive current associated with the load-side part of the shield flows twice through the core, forward on the centre conductor and back on the shield, so that the overall effect on the transformer output is zero.
The current-transformer network differs from that described in Appendix 6.2 by the inclusion of an extra resistance Rjk. The reason for the addition is that the Thévenin-equivalent voltage and current sampling networks must be topologically equivalent if they are to have identical phase and magnitude characteristics. In the Thévenin equivalent circuit for the voltage-sampling network, R2 is in parallel with Lv. Hence we need a resistance in parallel with Li in order to achieve a frequency-independent solution for the bridge balance condition. Notice that this resistance has a double subscript. This is because it will be resolved eventually into two resistances in parallel: Rj, an actual resistor; and Rk, a resistance to represent the transformer losses.

1.1 Condition for maximal flatness - current sampling network:
Referring to the circuit diagram given above; we can write an expression for the voltage appearing across the current-transformer secondary (Vi) by first noting that I=V/Z0 and then applying the ampere-turns rule:
Vi = V Zi / (N Z0) (1.1.1)
where Zi represents the total load across the secondary winding, i.e.:
Zi = Rjk // jXLi // ( Rh + jXCh )
(where " // " means "in parallel with"), and N is the transformer turns ratio (N = Ns / Np). Generally, the number of primary turns Np=1 for a current-transformer, but there is nothing in principle to prevent the use of several turns of thin coaxial cable.
The high-pass filtered output Vi' is obtained from Vi via a potential divider composed of Ch and Rh, thus:

Vi'

[ Rjk // jXLi // ( Rh + jXCh ) ]

N Z0

( Rh + jXCh )

(1.1.2)

We now need to find the parameter relationship which gives the maximally-flat in-band magnitude response. Since the circuit is a linear network, we can start by dividing both sides of the expression by V to give the response function in dimensionless form. Inverting the function then allows the parallel combination of impedances to be represented as a series of admittances. Thus:

Vi'

N Z0

Rjk

jXLi

( Rh + jXCh )

Multiplying out, and regrouping the terms into reals and imaginaries (noting that 1/j = -j), gives:

Vi'

N Z0

Rjk

XCh

XLi

+ 1

+ j

XCh

Rjk

XLi

(1.1.3)

To find the reciprocal magnitude response, we take the magnitude of the expression above:

Vi'

N |Z0|

Rjk

XCh

XLi

+ 1

2

.

XCh

Rjk

XLi

2

.

Multiplying out gives:

Vi'

N |Z0|

Rh²

Rjk²

XCh²

XLi²

+1+

2Rh

Rjk

2XCh

XLi

2RhXCh

RjkXLi

XCh²

Rjk²

Rh²

XLi²

2RhXCh

RjkXLi

which can be rearranged:

Vi'

N |Z0|

Rjk

+ 1

2

.

XCh²

XLi²

XCh²

Rjk²

2XCh

XLi

Rh²

XLi²

(1.1.4)

The four right-most terms in this expression are frequency-dependent. Of these however, the first two, XCh²/XLi² and XCh²/Rjk² will be small within the passband because XCh² diminishes rapidly above the cutoff frequency, XLi² increases rapidly, and Rjk² will be relatively large. This leaves us to consider the last two terms, which can be placed on a common denominator thus:

2XCh

XLi

Rh²

XLi²

2XCh XLi + Rh²

XLi²

Rh² - 2 Li / Ch

XLi²

The frequency dependence of the in-band magnitude response can therefore be minimised choosing the circuit parameters so that:
Rh² - 2 Li / Ch = 0
Hence the boost capacitance can be calculated from the expression:

Ch = 2 Li / Rh²

(1.1.5)

An alternative version of this expression, which allows XCh to be eliminated from the network response function once the condition for maximal flatness has been imposed, is:

XCh = - Rh² / 2 XLi

(1.1.6)

1.2 Condition for maximal flatness - voltage sampling network:
Referring to the circuit diagram given above; the voltage VV is derived from an ordinary potential divider and can be written:
Vv = V' Z1 / (R2 + Z1)
i.e.:

jXLv // (R1 + jXCv)

R2 + [ jXLv // (R1 + jXCv) ]

Multiplying numerator and denominator by R2 gives:

R2 // jXLv // (R1 + jXCv)

The high-pass filtered output VV' is derived from VV via another potential divider, i.e.;
Vv' = Vv R1 / (R1 + jXCv)
Hence:

Vv'

R2 // jXLv // (R1 + jXCv)

(R1 + jXCv)

(1.2.1)

This expression is exactly analogous to equation (1.1.2). Hence, by inspection, the condition for maximal flatness of the voltage-sampling network is:

Cv = 2 Lv / R1²

(1.2.2)

This can be re-stated in reactance form as before:

XCv = - R1² / 2 XLv

(1.2.3)

1.3 Bridge balance conditions:
When the bridge is balanced, the arbitrary load impedance Z0 is replaced by the target load resistance R0. In that condition, the outputs of the current and voltage sampling networks must be equal at all frequencies insofar as the model provides an accurate description of the physical circuit. Using equation (1.1.2), we can write a condensed form of the dimensionless current transfer function as:

Vi'

N Z0

( Rh + jXCh )

(1.3.1)

Where: Zi = Rjk // jXLi // ( Rh + jXCh )
We can also write a condensed form of the voltage transfer function (1.2.1):

Vv'

(R2 // Z1)

(R1 + jXCv)

Where Z1 = jXLv // (R1 + jXCv)
But notice here that V' is not the same as V. There will be a voltage drop across the transformer primary given by:
Vii = I Zi / N²
i.e., the impedance looking into the current transformer primary will be the secondary load impedance divided by the square of the turns ratio. Vii can be expressed in terms of V by using the substitution:
I = V / Z0
Hence:
V' = V + Vii = V [ 1 + Zi / ( Z0 N² )]
Hence, the dimensionless voltage transfer function using the load voltage V as the reference is:

Vv'

(R2 // Z1)

(R1 + jXCv)

1 +

Z0 N²

(1.3.2)

The voltage and current transfer functions, both using V as the reference level, become equal when Z0 = R0. So too do their reciprocals, with the advantage that parallel impedances become sums of admittances. Hence, equating the reciprocals of equations (1.3.1) and (1.3.2), and moving the primary voltage-drop correction to the current-network side:

(R2 // Z1)

(R1 + jXCv)

1 +

R0 N²

N R0

( Rh + jXCh )

This, noting that 1/(a//b) = (1/a)+(1/b), can be rearranged as follows:

+ 1

(R1 + jXCv)

N R0

( Rh + jXCh )

(1.3.3)

Now notice that as f®¥, both XCv and XCh vanish, in which case the expression degenerates into the balance condition for a conventional RVS bridge. Observe also that the finite values of Ch and Cv are a matter of free choice; i.e., we do not have to impose the maximal flatness condition, and the amount of response-shaping can be abitrary within the constraints imposed by the circuit topology. The corollary is that the the frequency tracking of the boost networks, although essential for balance tracking, must also be accomplished regardless of balance considerations. It follows that we must impose the condition:
(R1 + jXCv) / R1 = ( Rh + jXCh ) / Rh
so that both boost networks have the same frequency response. Hence.:
XCv / R1 = XCh / Rh
which corresponds to a fixed capacitance ratio:

Cv / Ch = Rh / R1

(1.3.4)

This condition implies that when Vv'=Vi', then Vv=Vi, which means that the overall form of the balance condition for the maximally flat bridge is the same as for the RVS bridge at all frequencies, i.e., substituting (1.3.4) into (1.3.3):

+ 1

N R0

(1.3.5)

Now expanding the admittances 1/Z1 and 1/Zi we get:

jXLv

R1 + jXCv

+ 1

N R0

Rjk

N R0

jXLi

N R0

Rh + jXCh

(1.3.6)

Every complex expression can be arranged so that it has terms which are purely real and terms which are purely imaginary. When that is done, it can be treated as two separate equalities: that between the reals; and that between the imaginaries. Terms with a complex denominator can be separated by multiplying numerator and denominator by the complex-conjugate of the denominator. Thus equation (1.3.6) can be re-written:

jXLv

R2 (R1 - jXCv)

R1² + XCv²

+ 1

N R0

Rjk

N R0

jXLi

N R0 (Rh - jXCh)

Rh² + XCh²

and the real part is:

R1 R2

R1² + XCv²

+ 1

N R0

Rjk

N Rh R0

Rh² + XCh²

We can make a further distinction by noting that the expression above has terms which are frequency-dependent and terms which are frequency-independent. It can only be true at all frequencies if the sum of the frequency-independent terms on the left hand side is equal to the sum of the frequency-independent terms on the right hand side (and the same applies to the frequency-dependent terms). Hence we can deduce the requirement
( N R0 / Rjk ) + 1/N = 1
i.e.:

Rjk = R0 N² / (N-1)

(1.3.7)

In fact, the reason why Rjk was put into the model was so that this equality could be obtained. The only solution for Rjk®¥ occurs when N=1; i.e., by inspection of (1.3.6), when the 1/N term on the right-hand side cancels the 1 on the left-hand side.

Using (1.3.7) in (1.3.6), the balance condition now simplifies to:

jXLv

R1 + jXCv

N R0

jXLi

N R0

Rh + jXCh

(1.3.8)

This relationship must remain true in the limit of infinite frequency; i.e., when XL®¥ and XC®0, hence:
R2 / R1 = N R0 / Rh (1.3.9)
It can also be seen, by inspection of (1.3.8), that a frequency independent solution exists only when:
R2 / Lv = N R0 / Li
i.e.,
Lv / Li = R2 / ( N R0 )
and only when:
( R1 + jXCv ) / R2 = / ( Rh + jXCh ) / ( N R0 )
But we already know from (1.3.9) that R1/R2=Rh/(NR0). Hence:
XCv / R2 = XCh / ( N R0 )
i.e.:
Ch / Cv = R2 / ( N R0 )
Hence, collecting the various relationships:

N R0

(1.3.10)

A further important balance relationship comes from equation (1.3.6) in the limit where XL®¥ and XC®0:

+ 1 = N R0

Rjk

(1.3.11)

Now let us define a resistance Rik to represent the parallel combination of Rh and Rjk; i.e.:
Rik = ( Rh // Rjk ).
Substituting this into (1.3.11) and subtracting 1 from each side gives:gives:

N R0

Rik

- 1

(1.3.12)
Transformer constant

This is the principal voltage-sampling ratio or 'transformer constant'. It appears explicitly in the analysis of the conventional RVS bridge, in which the boost capacitors are shorted-out and the secondary load (including core losses) has degenerated into a single resistance. It will be required in section 3, where we will use the simplified RVS model as a basis for the high-frequency analysis,

1.4 Low-frequency drop-off :
Equations (1.3.7) and (1.3.10) tell us how to determine component values in the event of arbitrary choices of (say) N, R2, Rh and Li; but merely having the ability to balance the bridge does not constitute a proper design procedure. We now need to specify the permissible degree of detector sensitivity drop-off at the lowest frequency of operation, and use it to determine the required amount of transformer secondary inductance. To that end, we can start by defining a low-frequency drop-off factor (i.e., the relative magnitude response):

hf =

Sensitivity at frequency f

Sensitivity at high frequencies

Where sensitivity is defined as:
|Vdet| = |Vv' - Vi'|
Hence:

hf =

| Vv'(f) - Vi'(f) |

| Vv'(¥) - Vi'(¥) |

But the voltage and current sampling networks have the same frequency response. Therefore, for a given degree of mismatch at the load port, Vv' will remain in constant proportion to Vi' regardless of frequency. This means that we can evoke a complex constant, g say, (where g is a function of Z0) which allows is to write the magnitude response by reference either to the current-sampling network output, or to the voltage-sampling network output, but without the need for both. This assertion can be proved by examining equation (1.3.1) and noting that the point in establishing the balance condition is to arrange matters so that:
Vv' = V { Zi Rh / [ N ( Rh + jXCh )] }(1 / R0)
and
Vi' = V { Zi Rh / [ N ( Rh + jXCh )] }(1 / Z0)
so that when Z0®R0, Vv'-Vi' = 0
Hence if we define:
g = R0 / Z0
we get:

hf =

| g Vi'(f) - Vi'(f) |

| g Vi'(¥) - Vi'(¥) |

| Vi'(f) ( g - 1) |

| Vi'(¥) ( g - 1) |

| Vi'(f) |

| Vi'(¥) |

(1.4.1)

The reciprocal of the relative magnitude of the current-sampling network output was given earlier as equation (1.1.4). Identifying Vi' as Vi'(f), this becomes:

Vi'(f)

N |Z0|

Rjk

+ 1

2

.

XCh²

XLi²

XCh²

Rjk²

2XCh

XLi

Rh²

XLi²

(1.4.2)

where the frequency dependence becomes explicit upon expansion of the reactances. If we apply the condition for maximal flatness (1.1.6), this simplifies to:

Vi'(f)

N |Z0|

Rjk

+ 1

2

.

XCh²

XLi²

XCh²

Rjk²

(1.4.3)

And if we let f®¥, so that XL®¥ and XC®0, we get:

Vi'(¥)

N |Z0|

Rjk

+ 1

2

.

This can be simplified by taking the square-root of the square and noting that
(Rh+Rjk)/RhRjk=1/(Rh//Rjk). Thus:

Vi'(¥)

N |Z0|

Rh // Rjk

(1.4.4)

This, of course, is an expression for the relative output of an ideal current-transformer (infinite secondary reactance), where Rh // Rjk is the secondary load resistance. Now, to obtain an expression for the drop-off factor (1.4.1) (and noting that we are dealing with reciprocal transfer functions), we divide equation (1.4.4) by equation (1.4.3).

hf =

Rh // Rjk

Rjk

+ 1

2

.

XCh²

XLi²

XCh²

Rjk²

(1.4.5)

This tells us that the load impedance Z0 makes no difference to the frequency response. The turns ratio of the current transformer (N) does make a difference however, even though it does not appear explicitly; firstly, because the number of secondary turns (Ns) dictates Li once the transformer core has been selected; and secondly, because N determines Rjk according to equation (1.3.7).
Rjk = R0 N² / (N-1)
Notice also that:
1 + Rh / Rjk = Rh / ( Rh // Rjk )
so that hf ®1 as f®¥.
What we require for design purposes however, is to be able to specify a drop-off factor and find the frequency at which it occurs. This will enable us to adjust the circuit parameters (particularly Li) until the specified maximum drop-off is achieved at or below the minimum required working frequency. This entails solving (1.4.5) for f, with hf as an independent (input) variable. We start by squaring (1.4.5) and taking the reciprocal:

hf²

= 1 +

XCh²

XLi²

XCh²

Rjk²

Rh // Rjk

2

.

It will simplify matters from now on if we use the substitution:
Rik = ( Rh // Rjk ).
Where Rik (introduced earlier) represents the total resistive load on the transformer in the high frequency limit. Thus:

hf²

- 1 =

Rik²

Rh²

XCh²

XLi²

XCh²

Rjk²

The number of variables can also be reduced by using the substitution (1.1.6):
XCh = - Rh² / 2 XLi
Thus:

hf²

- 1 =

Rh²

2XLi²

2

.

Rh²

2XLiRjk

2

.

Rik²

Rh²

This can be put into standard form:

Rh²

4XLi

Rh²

4XLi²Rjk²

hf²

- 1

Rik²

= 0

which shows that it is a quadratic equation in (1/XLi)². In this case, the process of solving it will be assisted by multiplying throughout by 4/Rh². :

XLi

XLi² Rjk²

Rh² Rik²

hf²

- 1

= 0

Hence a=1, b=1/Rjk² and c=-[4/(Rh²Rik²)][(1/hf²)-1]
and the solution is:
(1/XLi)² = [ -b ±Ö(b²-4ac) ] / 2a
i.e.:

XLi²

-1

2Rjk²

± ½

Rjk

Rh² Rik²

hf²

- 1

This has two solutions; but 1/XLi² is positive, and the only way in which a positive right-hand side can be obtained is by taking the positive square root. Hence (also multiplying ½ into the square-root bracket):

XLi²

-1

2Rjk²

4Rjk

Rh² Rik²

hf²

- 1

Now let us identify
XLi = 2pfh Li
where fh is the lower frequency limit at which the output has diminished by a factor of hf . Hence:

fh = 1

2p Li

-1

2Rjk²

4Rjk

Rh² Rik²

hf²

- 1

(1.4.6)

The equation above tells us that the lower frequency limit is inversely proportional to the transformer secondary inductance.
Now, recalling that Rik = ( Rh // Rjk ), notice that when Rjk®¥, Rik®Rh. In that limit, equation (1.4.6) reduces to:

fh =

(2Ö2)p Li ÖÖ[ (1/hf²)-1 ]

(1.4.7)

(where ÖÖ indicates a quartic root). This expression is still a fair approximation to fh because it is intended that Rjk should be relatively large. More importantly however, it brings out the major influences, which are that fh can be reduced either by increasing Li or by reducing Rh.
Given that Rjk is finite however, equation (1.24) falls into its most convenient form when we forcibly remove the factor 1/(4Rjk

) from the second square-root bracket, and then remove the factor 1/(2Rjk²) from the first square root bracket. This operation (noting that 2/Ö2=Ö2 ) gives:

fh = Rjk

(Ö2)p Li

-1 +

1 +

16 Rjk

[(1/hf²)-1]

Rh² Rik²

(1.4.8)

where:
Rik = ( Rh // Rjk ) and Rjk = R0 N² / (N-1)

1.5 Evaluating candidate transformers:
The secondary inductance of the current transformer is derived from the number of turns according to the expression:
Li = AL Ns²
where AL is the inductance factor of the core (which can be taken from the manufacturer's data sheet or, preferably, measured), and Ns is the number of secondary turns. Also, from equation (1.3.7), we have
Rjk = R0 N² / (N-1)
We can use these substitutions in (1.4.8) to obtain an expression which can be used to determine the number of turns which must be wound on a given transformer core in order to achieve a desired low-frequency limit. For a transformer with a 1-turn primary, where Ns=N, we get:

16 R0

[(1/hf²)-1]

(N-1)

Rh² Rik²

(1.5.1)

If the transformer has more than one turn on the primary, fh is multiplied by a factor N²/Ns²=(1/Np)². This may make it look as though adding more turns to the primary is beneficial, but in fact, increasing Np reduces N and causes fh to increase slightly overall. Hence there is probably no advantage in using a multi-turn primary unless very high sensitivity (very low N) is required.
Equation (1.5.1) can be made more tractable for calculation purposes by using the identity:
Rik = ( Rh // Rjk )
i.e., if we adopt Rik (the total resistive load on the current-transformer) as a principal design parameter, we can eliminate Rh using:
1/Rh = (1/Rik) - (1/Rjk)
Substituting for Rjk using (1.3.7) we get:
1/Rh = (1/Rik) - [ (N-1) / (R0 N²) ]
which can be put on a common denominator:
1/Rh = [ R0 N² - (N-1) Rik ] / ( Rik R0 N² )
Using this substitution in (1.5.1) gives:

16R0² N

[R0N²-(N-1)Rik]² [(1/hf²)-1]

(N-1)

Rik

Which can be rearranged to give:

[(1/hf²)-1]16 N

R0²

(N-1)² Rik²

N² R0

(N-1)Rik

- 1

2

.

Noting the recursive nature of the last term in the square-root bracket, this can be put into a form suitable for two-step calculation:

fh =

(Ö2)p AL(N-1) Ö{ -1 + Ö[1 + [(1/hf²)-1] 16 U² (U-1)² ] }

(1.5.2)

Where: U = N² R0 / [ (N-1)Rik ] = Rjk / Rik = 1 + Rjk / Rh

1.6 Quadrature point:
Since the maximally-flat voltage and current sampling networks are second-order high-pass filters, their outputs at frequencies below the working range can be phase-shifted (relative to the generator) by more than +90°. Hence, at frequencies below the point at which +90° quadrature occurs, any phase analysis carried out will require that 180° is added to the result returned by an inverse-tangent function-call (see Appendix 6.2).
For the special case when Rjk®¥, the quadrature point coincides with the lower -3dB point. This is easy to demonstrate by noting that the relative voltage output of a network at the -3dB point is 1/Ö2. When hf =1/Ö2, (1/hf²)-1=1. Putting this into equation (1.4.7), and allocating the symbol fx to the phase-crossover frequency, we get:
fx = Rh / { (2Ö2)p Li }
where, from equation (1.1.5);
Rh = Ö(2 Li / Ch )
Hence:
fx = 1/ [ 2p Ö(LiCh) ]

For the general case when Rjk is finite, the quadrature-point frequency can be determined from the reciprocal current-transformer response function:

Vi'

N Z0

Rjk

XCh

XLi

+ 1

+ j

XCh

Rjk

XLi

derived earlier as
equation (1.1.3)

The phase angle of a phasor in the form a+jb is given by Tanf = b/a. Here however, we have the reciprocal of a relative voltage, i.e. a phasor in the form:
1 / ( a + jb ) = ( a - jb ) / ( a² + b² )
Hence the phase tangent in this case is given by: Tanf = -b/a. Notice also that (1.1.3) is not quite in the a+jb form because the main load impedance Z0 is complex. Since the tangent is a ratio however, Z0 is cancelled-out; which tells us that the network phase response (as distinct from the actual phase of the output) is not affected by the load. Thus the phase response is given by:

Tanf =

XLi

XCh

Rjk

+ 1 +

XCh

XLi

The quadrature point occurs when Tanf®¥, i.e., when the denominator of the expression above goes to zero. Hence, at the quadrature point:

Rjk

+ 1 +

XCh

XLi

= 0

Where XLi = 2pfx Li
and XCh = -1/( 2pfx Ch )

Hence, expanding the reactances and rearranging:

fx = 1/ { 2p Ö[ Li Ch (1 + Rh/Rjk) ] }

(1.6.1)

and the phase angle can be calculated using:

f = Tan

XLi

XCh

Rjk

+ 1 +

XCh

XLi

+ n 180

[degrees]

Where n=0 when f > fx, and n=1 when f < fx.

1.7 Preliminary design calculations:
Shown below is a snapshot of a spreadsheet calculation ( mxf_AL67.ods ) which determines the number of turns which must be wound on a transformer core having an AL of 67nH in order to meet various low-frequency drop-off criteria (a single-turn primary winding is assumed). The AL value was selected on the basis that the Amidon FT50-61 core has a published AL of 68.8nH, but small toroidal transformers typically have a leakage inductance of about 1 to 2%; so the coupled secondary inductance of the transformer will be about 0.98 of the total inductance. Since the published AL has a tolerance of ±25%, there is no point in resorting to decimal places. It is, of course, best to give AL as 0.98 of an actual measurement, but the published value has to suffice until the experimental work begins.
     The only other input parameters required for the part of the calculation shown are the bridge target load resistance (R0 = 50W as usual), and the total resistive secondary load (Rik) for which 50W is also a reasonable starting value.
     The calculation tells us that for cores in the middle of the tolerance range, and assuming a working frequency range of 1.6MHz and above; the LF drop-off can be kept within 1% by using a 12-turn transformer (with a small downward adjustment of Rik), or within 2% by using an 11-turn transformer, or within 5% by using a 10-turn transformer.
     Notice incidentally, that the -3dB frequency is slightly different from the quadrature frequency (fx) due to the finite value of Rjk.

Input parameters:

R0 = 50W

Rik = 50W

AL = 67 nH / turn²

Formulae used (see open document spreadsheet file mxf_AL67.ods )	Equation
Li = AL N²
U = N² R0 / [ (N-1)Rik ]	(1.5.2)
fh = R0 / [ (Ö2)p AL(N-1) Ö{ -1 + Ö[1 + [(1/hf²)-1] 16 U² (U-1)² ] } ]	(1.5.2)
Rjk = R0 N² / (N-1)	(1.3.7)
Rh = Rjk Rik / ( Rjk - Rik )
Ch = 2 Li / Rh²	(1.1.5)
fx = 1/ { 2p Ö[ Li Ch (1 + Rh/Rjk) ] }	(1.6.1)

1.8 Losses in the current-sampling network:
It was mentioned earlier, that the resistance Rjk directly in parallel with the current-transformer secondary winding was given a double subscript because it is a combination of resistive losses in the transformer core (Rk) and an actual resistor to make up the difference (Rj); i.e.:
Rjk = Rj // Rk
This means that we need to make an estimate for Rk in order to determine Rj.
     In Appendix 6.1, a formula relating the transformer parallel loss-resistance to the transfer efficiency factor was given as:
Rk = k Ri / (1-k)
where Ri is the load on the transformer during the efficiency measurement, and k is the factor by which the output voltage falls short of that of an ideal transformer. It was found that for transformers wound on type FT50-61 toroids, with 8 to 12 turns on the secondary and a 1-turn primary, k was about 0.965 ±0.01 at 30MHz when Ri was 50W.
     For the purpose of designing ordinary transmission bridges, it is sufficient to assume that k does not depend on frequency and adopt a value equivalent to the worst case. For the maximally-flat bridge however, Rjk is a low-frequency balance-tracking parameter (albeit a minor one) and so we need to adopt a k-value which is appropriate for the region in which the high-pass network does its work. This, presuming that we are designing with the HF spectrum in mind, is somewhere around 2 to 4 MHz.
     Shown below is the graph of complex permeability for type 61 ferrite:

From the graph we can see that the material is entering a dispersion region on moving upwards through the HF spectrum; which means that we can expect greater efficiency at 3MHz than at 30MHz. It is extremely difficult to convert this information into an accurate value for k, but an educated guess which will not be far from the mark says that k at 3MHz will be somewhere between 0.98 and 0.99. It is reasonable therefore to model the maximally-flat bridge on the basis that k=0.985 ±0.005, an assumption that has the virtue of placing the upper edge of the 99.7% confidence interval at 1. Taking k = 0.985 ±0.005, we get:
Rk = 50 k / (1-k) = 3283 (+1667, -833) W
or, taking the average of the asymmetric uncertainty:
Rk = 3283 ±1250 W
This estimate is, of course, crude; but there are various reasons for supposing that the uncertainty will not matter. Firstly, in the spreadsheet calculation discussed above, it was found that there was very little difference between the -3dB point and the phase crossover frequency. This means that Rjk has only a small effect on the frequency response. Secondly, Rk is a lot larger than the various calculated values for Rjk (600 to 800W for viable designs); which means that Rj will be only a little greater than Rjk. The overall uncertainty of an asymmetric parallel combination is weighted towards the uncertainty in the lowest-value component. Hence, it appears doubtful that there will be any need to adjust Rj on test. For those who have the facility to measure the impedance of the secondary winding in the 1.6 to 4MHz region moreover; it is possible to obtain a fair estimate for Rk directly and so refine the value for Rj.

Although the estimation procedure given above is rough, it nevertheless allows us to split the transformer load into its three resistive components Rh, Rj<