G3YNH info: Maximally flat transmission bridge.

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.

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 50Ω, this implies an inductance of 35μH 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):

ηf =

Sensitivity at frequency f

Sensitivity at high frequencies

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

ηf =

| 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:

ηf =

| 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).

ηf =

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 ηf →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 ηf as an independent (input) variable. We start by squaring (1.4.5) and taking the reciprocal:

ηf²

= 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:

ηf²

- 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:

ηf²

- 1 =

Rh²

2XLi²

2

.

Rh²

2XLiRjk

2

.

Rik²

Rh²

This can be put into standard form:

Rh²

4XLi⁴

Rh²

4XLi²Rjk²

ηf²

- 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. :

XLi⁴

XLi² Rjk²

Rh² Rik²

ηf²

- 1

= 0

Hence a=1, b=1/Rjk² and c=-[4/(Rh²Rik²)][(1/ηf²)-1]
and the solution is:
(1/XLi)² = [ -b ±√(b²-4ac) ] / 2a
i.e.:

XLi²

-1

2Rjk²

± ½

Rjk⁴

Rh² Rik²

ηf²

- 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

4Rjk⁴

Rh² Rik²

ηf²

- 1

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

fη = 1

2π Li

-1

2Rjk²

4Rjk⁴

Rh² Rik²

ηf²

- 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:

fη =

(2√2)π Li ⁴√[ (1/ηf²)-1 ]

(1.4.7)

This expression is still a fair approximation to fη because it is intended that Rjk should be relatively large. More importantly however, it brings out the major influences, which are that fη 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/(4Rjk⁴) from the second square-root bracket, and then remove the factor 1/(2Rjk²) from the first square root bracket. This operation (noting that 2/√2=√2 ) gives:

fη = Rjk

(√2)π Li

-1 +

1 +

16 Rjk⁴ [(1/ηf²)-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 R0⁴ N⁸ [(1/ηf²)-1]

(N-1)⁴ Rh² Rik²

(1.5.1)

If the transformer has more than one turn on the primary, fη 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 fη 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 N⁴ [R0N²-(N-1)Rik]² [(1/ηf²)-1]

(N-1)⁴ Rik⁴

Which can be rearranged to give:

[(1/ηf²)-1]16 N⁴ 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:

fη =

(√2)π AL(N-1) √{ -1 + √[1 + [(1/ηf²)-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/√2. When ηf =1/√2, (1/ηf²)-1=1. Putting this into equation (1.4.7), and allocating the symbol fx to the phase-crossover frequency, we get:
fx = Rh / { (2√2)π Li }
where, from equation (1.1.5);
Rh = √(2 Li / Ch )
Hence:
fx = 1/ [ 2π √(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 Tanφ = 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: Tanφ = -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:

Tanφ =

XLi

XCh

Rjk

+ 1 +

XCh

XLi

The quadrature point occurs when Tanφ→∞, i.e., when the denominator of the expression above goes to zero. Hence, at the quadrature point:

Rjk

+ 1 +

XCh

XLi

= 0

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

Hence, expanding the reactances and rearranging:

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

(1.6.1)

and the phase angle can be calculated using:

φ = Arctan

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 = 50Ω as usual), and the total resistive secondary load (Rik) for which 50Ω 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 = 50Ω

Rik = 50Ω

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)
fη = R0 / [ (√2)π AL(N-1) √{ -1 + √[1 + [(1/ηf²)-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/ { 2π √[ 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 50Ω.
     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) Ω
or, taking the average of the asymmetric uncertainty:
Rk = 3283 ±1250 Ω
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 800Ω 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 and Rk. This puts a value on the resistor Rj; and also, if we care to specify the maximum power to be transmitted through the bridge, allows us to calculate the resistor power ratings.
     When calculating resistive losses, it is sensible to do so in the worst case. This occurs when all of the reactances have vanished from the system, i.e., in the high-frequency limit of the prototype model when XLi→∞ and XCh→0. We will start by allocating the symbol P0 to the maximum transmitted power; i.e., P0 is the power dissipated in the main load resistance R0 when the generator is operating at the design maximum output level and the load is purely resistive. We obtain the voltage appearing across R0 thus:
V = √(P0 R0)
It is often convenient to adopt P0=100W, firstly because this is a reasonable design criterion for high-sensitivity transmission bridges, and secondly, because all of the power losses calculated from it are then in %. When P0=100W and R0=50Ω, V=70.7 V RMS.
     From equation (1.1.1), the voltage across the current transformer secondary winding when all reactances have disappeared is:
Vi = V Rik / (N R0)
Hence the maximum power dissipated in (say) Rh is:
Ph = Vi² / Rh
i.e.:
Ph = P0 Rik² / ( N² R0 Rh )
and so on.

On the basis of the discussion above, the spreadsheet calculation can be extended as follows:

Additional input parameters:	k (or Rk) , P0
Formulae:	See mxf_AL67a.ods for actual implementation
Rk = 50 k / (1-k)	Nominal core loss resistance.
Rj = Rk Rjk / ( Rk - Rjk )	Secondary direct shunt resistance
Pk = P0 Rik² / ( N² R0 Rk )	Nominal core loss
Pj = P0 Rik² / ( N² R0 Rj )	Power in Rj
Ph = P0 Rik² / ( N² R0 Rh )	Power in Rh
Pik = Ph + Pj + Pk	Total power in current sampling network.

1.9 Trial voltage-sampling networks:
The voltage-sampling network parameters are essentially scaled versions of the current-network parameters, but there is a compromise involved in the choice of the scaling ratio. The problem is that if we make the upper voltage-sampling resistance R2 too small, the power dissipation in the network will be excessive; but if we make it too large, the compensation inductance Lv will become impractically large and the stray capacitance across R2 will make a significant contribution to the upper arm impedance.
When all of the reactances in the system disappear, the voltage across the voltage-sampling network is:
V' = V + Vii = V [1 + Rik / ( R0 N² ) ]
but
V = √(P0 R0)
hence
V' = [1 + Rik / ( R0 N² ) ] √(P0 R0)
The voltage across R2 is:
V2 = V' R2 / ( R1 + R2 )
i.e.,
V2 = [1 + Rik / ( R0 N² ) ] [√(P0 R0)] R2 / ( R1 + R2 )
The power in R2 is:
P2 = V2² / R2
i.e.,
P2 = P0 R0 [1 + Rik / ( R0 N² ) ]² R2 / ( R1 + R2 )²
Similarly:
P1 = P0 R0 [1 + Rik / ( R0 N² ) ]² R1 / ( R1 + R2 )²

Thus the calculations can be extended to evaluate voltage-sampling networks:

Additional input parameter: R2	Equation
Formulae: (See mxf_AL63.ods for implementation).
R1 = Rh R2 / ( N R0 )	(1.3.10)
P1 = P0 R0 [1 + Rik / ( R0 N² ) ]² R1 / ( R1 + R2 )²
P2 = P0 R0 [1 + Rik / ( R0 N² ) ]² R2 / ( R1 + R2 )²
P1+2 = P1 + P2
Lv = Li R1 / Rh	(1.3.10)
Cv = Ch Rh / R1	(1.3.10)

In the spreadsheet mxf_AL63.ods, the row for N=12 has been singled out for special attention. The value of Rik has been adjusted to make R0=50Ω for that case. The AL value has also been reduced to make the 1% drop-off point occur at 1.6MHz; producing the information that the <1% LF drop-off criterion can be met by using a core with AL=63nH or greater.
It was found that, with R2=2.2KΩ, the maximum power dissipated in the voltage sampling network will be about 2% of the transmitted power. The corresponding value for Lv is 33μH. This is a large inductance, but the drawback is not so great as in the case of the current transformer. The point is that a large transformer propagation-delay (which manifests itself as self-capacitance) requires invasive neutralisation arrangements; whereas the parallel capacitance of the lower voltage-sampling network can be balanced-out by placing capacitance across the upper voltage-sampling arm. Indeed, there may be situations in which it will be necessary to place additional capacitance across the lower network in order to offset the strays across the upper network.

1.10 Multi-turn primaries:
As was mentioned during the derivation of equation (1.5.1), the expression for the drop-off criterion fη must be multiplied by 1/Np² if the current-transformer has more than one turn on the primary. In that case also, the secondary inductance is given by:
Li = AL Ns²
and
N = Ns / Np
For the sake of generality, the ability to evaluate bridges which have an arbitrary number of primary turns has been included in the spreadsheet mxf_AL62.ods. Note that a bridge with a multi-turn primary will have a high insertion impedance and must either have a very large Lv or dissipate a large percentage of the transmitted power in the voltage sampling network. Such bridges however, have high detector sensitivity and so may be used with low-power generators.

Additional input parameter: Np	Equation
Formulae: (See mxf_AL62.ods for implementation).
N = Ns / Np
Li = AL Ns²
fη = R0 / [ (√2)π ALNp²(N-1) √{ -1 + √[1 + [(1/ηf²)-1] 16 U² (U-1)² ] } ]	(1.5.2) / Np²

1.11 Overboost:
The detector amplitude vs. frequency response for the maximally-flat bridge can be plotted using equation (1.4.5). It is rather more interesting however, to plot a version of the response function in which Ch is allowed to vary independently. This allows us to explore the effect of component tolerances, and to see whether there is any advantage in deviating from the exact maximally-flat condition.
The required response function is given by dividing equation (1.4.4) by equation (1.4.2):

ηf = (1+Rh/Rjk)

(1+Rh/Rjk)² +

XCh²

XLi²

XCh²

Rjk²

2XCh

XLi

Rh²

XLi²

(1.11.1)

This function is shown plotted below for a candidate bridge which has been adjusted to have its 1% drop-off point at 1.6MHz when the maximally-flat condition is imposed (see: mxf_sim_AL63.ods ).

The graph shows that when the boost capacitance is reduced below the value required for maximal flatness, an overboost occurs. Specifically, for the case examined, reducing Ch by 15% moves the -1% point to 1.1MHz, and gives a response of +1% at 1.7MHz. Broadly, this tells us that there is considerable latitude in the choice of Ch, and it is best to err on the low side. Whatever the choice of Ch however, Cv must always be kept in the correct proportion to it in order to maintain the frequency-independence of the balance condition.

1.12 SPICE simulation:
The circuit files listed below can be opened using the SPICE simulator program LTspice / SwitcherCAD III available free from the Linear Technology website. LTspice can export a pspice compatible netlist suitable for other simulation programs if so desired (but since LTspice is better than many commercial offerings, there may not be much point in doing so).

Current sampling network	mxflt_is.asc
Voltage sampling network	mxflt_vs.asc

The simulations confirm the theory developed above.

2. Type 2 Maximally flat networks:

>>>

3. High-frequency RVS model:
To make a bridge which will balance correctly over the entire 4¼-octave short-wave spectrum, it is always necessary to include some kind of high-frequency compensation. In particular; we must do something about the effective secondary parallel capacitance of the current-transformer, and we must take the parasitic capacitances of the upper and lower voltage-sampling networks into account. All of these capacitances will have a significant effect at high frequencies and so must be put into the model; but the analytical problem can be simplified by noting that they are all of the order of a few pF. It follows that any low-frequency response-tailoring scheme (involving capacitances of several nF) will only be affected at about the 0.1% level by such tiny capacitances. Hence we can devise our HF neutralisation arrangements by reference to the conventional RVS bridge; i.e., we can short-out any boost capacitors and forget them during this part of the analysis.
Shown below is an equivalent circuit for the RVS bridge with parasitic capacitances included. The presence of a Faraday-shield is implied by the absence of a stray capacitance from the through-line to the detector port. The circuit includes a neutralising capacitor Cn placed in parallel with the load resistance. This capacitor should be considered to represent a generic current-transformer HF phase-neutralisation scheme; i.e., neutralisation can be accomplished in various ways (see A6-4 section 18) but, in terms of their effect on the balance condition, all such techniques are equivalent to the inclusion of Cn.

Resistive Voltage-Sampling (RVS) bridge with parasitic capacitances and neutralising capacitor (Cn).

Cn is included in advance of any analysis because it is obvious by inspection that there can be no frequency-independent solutions for the balance condition in the absence of a neutralisation network. This point can be understood by considering the four principal impedances shown in grey boxes. These, subject to transformation in the case of the primary load impedance, are analogous to the four impedance-arms of a Christie-Wheatstone bridge. The voltage-sampling network consists of an RLC network and an RC network. The current-transformer is an RLC network, and so the primary load must behave as an RC network if all of the frequency factors are to drop-out of the balance relationships.

In order to find the balance conditions, we can use much the same procedure as was employed in section 1.3. This involves writing expressions for Vi and Vv (both derived from the same reference voltage, V) and equating them when the through-line is terminated in the target load resistance R0. The resulting expression is then rearranged to get all of the parallel impedances into reciprocal form, so that each may be expanded into a series of admittances.
In this case, our generic neutralisation method places a capacitance (actual or virtual) in parallel with the load, so that the total load impedance at balance is complex, i.e.:
Z0n = R0 // jXCn
Hence, by the ampere-turns rule and because I=V/Z0n, the output of the current transformer is:
Vi = V Zi / (Z0n N)
where
Zi = Rik // jXLi // jXCi

The output of the voltage-sampling potential-divider is:
Vv = V' Z1 / (Z1 + Z2) = V' (Z1 // Z2) / Z2
where
V' = V + Vii = V [1 + Zi / ( Z0n N² )]
Z1 = R1 // jXLv // jXC1
and
Z2 = R2 // jXC2

To balance the bridge, we set Vv = Vi and cancel V. Thus, in compact form:
[1 + Zi / ( Z0n N² )] (Z1 // Z2) / Z2 = Zi / (Z0n N)
Now, taking the reciprocal, and moving [1+Zi/(Z0nN²)] to the right-hand side:
Z2 / (Z1 // Z2) = [1 + Zi / ( Z0n N² )] (Z0n N) / Zi
Multiplying-out the right-hand side gives:
Z2 / (Z1 // Z2) = ( Z0n N / Zi ) + 1/N
Expanding 1/(Z1//Z2) and 1/Zi gives:

= N Z0n

Rik

jXLi

jXCi

Multiplying Z2 into the bracket gives the left hand side as (Z2/Z1)+1. Then subtracting 1 from both sides gives:

= N Z0n

Rik

jXLi

jXCi

- 1

We now need to expand Z1 and Z2. This is best accomplished by moving Z2 to the right-hand side, where it becomes an admittance. Also a certain proliferation of brackets is prevented by multiplying NZ0n into the bracket on the right-hand side before we do so:

NZ0n

Rik

NZ0n

jXLi

NZ0n

jXCi

- 1

Expanding Z1 and Z2 gives:

jXLv

jXC1

jXC2

NZ0n

Rik

NZ0n

jXLi

NZ0n

jXCi

- 1

And multiplying-out the brackets (noting that (1/N)-1 = -(N-1)/N and that j²=-1):

jXLv

jXC1

NZ0n

RikR2

NZ0n

jXLiR2

NZ0n

jXCiR2

N-1

NR2

NZ0n

jXC2Rik

NZ0n

XLiXC2

NZ0n

XCiXC2

N-1

jNXC2

The full expansion requires the substitution Z0n=R0 // jXCn. This is best accomplished by moving all of the terms with Z0n as a factor to one side of the equation, then dividing both sides by 1/Z0n. Thus (also recalling that XLXC=-L/C):

jXLv

jXC1

N-1

NR2

N-1

jNXC2

jXCn

RikR2

jXLiR2

jXCiR2

jXC2Rik

NC2

XCiXC2

Multiplying-out the left-hand side gives the final expansion, where all terms have dimensions of [1/Ω²]:

R0R1

jR0XLv

jR0XC1

N-1

NR0R2

N-1

jNR0XC2

jR1XCn

XC1XCn

N-1

jNR2XCn

N-1

NXC2XCn

(3.1)

RikR2

jXLiR2

jXCiR2

jXC2Rik

NC2

XCiXC2

The real part of this expression corresponds to the in-phase balance condition, and the imaginary part corresponds to the quadrature balance condition. Equating the reals gives:

R0R1

N-1

NR0R2

XC1XCn

N-1

NXC2XCn

RikR2

NC2

XCiXC2

(3.2)

From this, we can see that frequency-independence of the in-phase balance conditon is obtained when:

XC1XCn

N-1

NXC2XCn

XCiXC2

= 0

i.e., using XC=-1/(2πfC):

C1Cn

(N-1)C2Cn

N CiC2

= 0

Dividing throughout by C2 gives:

+ 1 -

= N Ci

(3.3)

Equating the imaginaries in (3.1) gives:

R0XLv

R0XC1

N-1

NR0XC2

R1XCn

N-1

NR2XCn

XLiR2

XCiR2

XC2Rik

(3.4)

But from the requirements for low-frequency balance given earlier as equation (1.3.10):
Lv = Li R2 / (N R0)
Hence the terms 1/(R0XLv) and N/(XLiR2) cancel. This leaves only terms involving capacitive susceptance, and so using XC=-1/(2πfC) and cancelling -2πf throughout:

(N-1)C2

NR0

(N-1)Cn

NR2

NCi

NC2

Rik

Muttiplying throughout by R2 and regrouping gives:

C1R2

+ C2R2

(N-1)

NR0

Rik

+ Cn

+ 1 -

N Ci

(3.5)

Although, on the diagram above, only the neutralisation capacitor is marked as adjustable, any of the stray capacitances can be padded to a higher value if necessary. Hence, in (3.3) and (3.5) it appears that we have two simultaneous equations with four adjustable parameters (C1, C2, Ci, Cn). This means that there are either an infinite number of ways in which high-frequency balance tracking can be accomplished; or, we are missing some crucial piece of information. The latter is, of course, the case; and in deducing the solution we eliminate the paradox. In the limit where XLv is very large (i.e. at high frequencies), the voltage-sampling network may be considered as a resistive potential-divider in parallel with a capacitive potential-divider. It is possible to make separate resistive and capacitive dividers which give exactly the same off-load output voltage, the only difference being the output impedances. If the outputs of such a pair of networks are connected together, there will be no changes in the output voltages. If the division ratios of the two networks are not the same however, then the combined output voltage will vary with frequency. Thus, in order to obtain a flat high-frequency response, we must impose the condition:
R2 / R1 = XC2 / XC1
i.e.:

R2 / R1 = C1 / C2

(3.6)

Substituting this into (3.3) gives:

+ 1 -

= N Ci

(3.7)

A check of the reasoning used in the derivation above can be had by substituting (3.7) into (3.5). The result, after cancellation is:

+ C2

(N-1)

NR0

Rik

= 0

Multiplying throughout by R0 and rearranging gives:

N R0

Rik

- 1

This is the transformer constant, introduced earlier as equation (1.3.12). Hence we can re-write (1.3.12) with this supplementary information:

N R0

Rik

- 1

(3.8)
Transformer constant

This result confirms the deduction (3.6) and the logical consistency of the working. Also, we can use it to substitute for R2 / R1 in equation (3.7) to give:

Ci / Cn = R0 / Rik

(3.9)

Now returning to equation (3.2); notice that when the high-frequency balance condition (3.3) is applied, equation (3.2) reduces to:

R0R1

N-1

NR0R2

RikR2

NC2

Multiply throughout by R0R2 and rearranging gives:

+ 1 -

NR0

Rik

R0R2Cn

NR0R2C2

Which, applying the cancellation given by equation (3.8) and then dividing throughout by R0R2 leaves us with:

NC2

This is an auxiliary balance condition, which links the frequency response of the voltage-sampling network to that of the current-sampling network. It supplements the collected balance relationships given earlier as equation (1.3.10).

N R0

N C2

(3.9)

From (3.9) we obtain the additional relationship:

Cn / C2 = R2 / R0

(3.10)

Hence, rather than being free to choose the various network capacitances, we find that their relative values must be constrained if we are to get the bridge to balance at high frequencies. If only one of the capacitances is determined by practical considerations (such as the desire to keep them all as small as possible) then all of the others are prescribed. This situation is remarkably different from that encountered when designing capacitive voltage-sampling (CVS, Douma) bridges, because the strays across the voltage-sampling network merely contribute to wanted capacitances in that case. It means that the RVS bridge, although conceptually simple in its prototype (low-frequency model) form, is actually the hardest to get right.
The high-frequency balance considerations have been added to the spreadsheet calculation mxf_hf_AL63.ods. A little experimentation with the numbers confirms that the capacitance across the upper voltage-sampling arm (C2) is always the smallest of the set, and that R2 must not be too large if the others are to be kept within reasonable bounds.

>>
We must place capacitance across the Itr sec to balance the freq response of the IS network

>

>>>>>
writing in progress.
>>>>

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Ch6 Contents

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