TX to Ae RF Bridges - -

6.3 Transmission Bridges: Part 1.

Revision in progress:
The author has recently developed a method for current-transformer bridge evaluation that gives accurate information using inexpensive test equipment. The text below will be revised in the light of new experimental results. The theory does not change substantially, but correction for component non-idealities can improve results considerably.

6-17. Current Transformer Bridges:
The relationship between the input and output of an electrical network is called its transfer function. Determination of the balance relationship for an impedance bridge using a current transformer now boils down to writing down the transfer functions for the voltage and current sampling networks and setting the them to be equal when the balance criterion has been met. Once again, this problem can be split into a mid-band analysis and a low-frequency analysis, and we will begin with the former.

Fig. 6-17#1

From the previous section (with a slight change of notation) the output of a loaded current transformer of infinite secondary inductance is given by the relationship:
 Vi = I Ri / Ni 6-17.1
where I is the current flowing between the generator and the load, and Ni is the number of turns in the secondary winding (adding the subscript 'i' to the current-transformer turns will allow us later to introduce inductive devices to the voltage sampling network). The relationship between the current I and the generator voltage V is:
 V = I (Zii + ZL) 6-17.2
where ZL is the load impedance; and Zii is the insertion impedance (i.e., input impedance) of the current transformer and is real if the secondary inductance is very large, i.e.:
 Zii = Rii = Ri / Ni² 6-17.3
Using expression (6-17.2) to eliminate I from expression (6-17.1), and changing Zii to Rii, we obtain:
 Vi = V Ri / [ Ni (Rii + ZL) ]
We can also write an alternative form for this expression by using (6-17.3) to eliminate Rii:
Vi = V Ri / Ni [( Ri / Ni² ) + ZL]
which rearranges to:
 Vi = V Ri / [( Ri / Ni ) + Ni ZL]

The ratio Vi/V is the dimensionless transfer function for the current sampling network, to which we will assign the symbol ηi (where η ("eta") is written in bold because the function it represents is complex). Hence:
 ηi = Vi / V = Ri / [ Ni (Rii + ZL) ]
When the bridge is balanced, ZL becomes equal to R0, the target or characteristic resistance of the system, and so we may define the transfer function at balance as:
 η0 = Ri / [ Ni (Rii + R0) ]
(here η is now real, and so is written un-bold). Also notice, that since the insertion resistance Rii is usually small in comparison to R0, we may often be inclined to use the approximation:
 η0 = Ri / (Ni R0)

In the bridge shown above, the voltage sampling network is a resistive potential divider. The transfer function for this network (assuming ideal resistors) is:
 ηv = Vv / V = R1 / (R1 + R2)
(Vv is in phase with V, and so their ratio is real). To obtain the mid-band balance condition for the bridge, all we have to do is equate η0v, hence:
 η0 = Ri / [ Ni (Rii + R0) ] = R1 / (R1 + R2)

Example: A prototype current transformer has a 10 turn secondary winding and a 50Ω secondary load resistor. This transformer has a nominal insertion resistance (input impedance) Rii=Ri/Ni²=0.5Ω. The target load resistance R0 is 50Ω.
The bridge balance condition is:
R1 / ( R1 + R2 ) = Ri / Ni (Rii + R0) = 50/[10(50+0.5)] = 50/505 = η0

There is, of course, an infinite range of resistor values that will satisfy the relationship
η0=R1/(R1+R2). We may attempt to tackle this problem by first observing that the ideal situation would be for the generator to be loaded with a resistance equal to R0 when the load impedance is R0. If there were no voltage sampling network, the generator would see a load of R0+Rii, and so we should explore the possibility of choosing the total resistance R1+R2 such that when it is placed in parallel with R0+Rii, the generator sees a load equal to R0, i.e. we want the condition:
 1R0 = 1(R1 + R2) + 1(Rii + R0)

which, by rearrangement, gives the solution:
 R1 + R2 = R0 (Rii + R0) / Rii

In the example given above, R0=50Ω and Rii=0.5Ω. Hence we may calculate that the ideal value for R1+R2 is 50×50.5/0.5=5050Ω. For the balance condition, we have already established that R1/(R1+R2) must be equal to 50/505, and so R1/5050=50/505, and R1=5050×50/505=500Ω. R2 is of course, 5050-R1=4550Ω.

While these values for R1 and R2 may be ideal for the generator however, they are not so good for swamping the effects of stray capacitance, and the higher value resistor will need to be of a very low-capacitance type if its impedance is not to be capacitive at 30MHz [see Compoments & Materials, section 2-7]. High values of resistance in the network also imply that the voltage sample may be susceptible to loading due to the input impedance of the detector; and while such loading will not affect the balance condition (because the detector draws no current when Vv=Vi), it is undesirable in situations where the degree of unbalance must be measured, or where current is drawn from the network by more than one circuit (e.g., dual bridges for forward and reflected power measurement). Partial compensation for a rising frequency-response of the network can be effected by placing a small trimmer capacitor across R1, and for a falling response by placing a very small trimmer capacitor across R2; but still 4550Ω is a very high resistance in radio applications, and it is unlikely that the perfect loading that justifies the choice will be achieved in practice. The calculation above is for a 10 turn transformer, and if the number of turns is increased, the 'ideal' voltage sampling network resistance rises accordingly. Consequently, while perfect loading is just about feasible when using a 1:10 transformer, it is not attainable with a 1:40 transformer. We might therefore consider reducing the total resistance of the voltage sampling network so that it is in the region of (say) 1KΩ, e.g., by choosing R1=100Ω, and R2=910Ω. In that case, at balance, the generator sees a load of 1010Ω in parallel with 50.5Ω, i.e., 48.1Ω; which corresponds to an SWR of 50/48.1=1.04:1 and is therefore well within the load tolerance of a normal HF radio transmitter. What will finally resolve the issue however, is not the niceties of perfect loading, but the power dissipation in the voltage sampling network.

If the bridge is designed to withstand a maximum power throughput of say 100W, then the transmitter output voltage at that power level is nominally 70.7V (assuming a design load resistance of 50Ω). If we use a voltage sampling network with a total resistance of 1010Ω, it will draw a maximum current of 70.7/1010=70mA. The power dissipated in R1 is then 0.07²×100=0.49W, and the power dissipated in R2 is 0.07²×910=4.46W; which immediately shows why designers are inclined to avoid using resistive potential dividers in bridges which are intended to be left in situ after the impedance matching step has been completed. If we increase the total resistance of the network to 5050Ω, then the maximum current is 14mA, the power dissipated in R1 is 0.014²×500=98mW, and the power in R2 is 0.014²×4550=0.89W. The use of resistive voltage dividers therefore involves a compromise between excessive power dissipation on the one hand, and non-ideal resistor behaviour and susceptibility to loading by the detector on the other. Consequently, using the scheme in conjunction with very high power generators is impractical. The approach is workable in low-power applications however, and in bridges which are not intended to be left in situ; and it is, if nothing else, theoretically illuminating when we turn our attentions to the low-frequency analysis.

6-18. Low-frequency Compensation:
"I get a perfect 1:1 SWR on 160m, but the transmitter refuses to give its full output."

There is a certain consensus in the amateur-radio literature, that there is no need to pay too much attention to the secondary inductance of current transformers. There is also a consensus in some quarters that it is very difficult to get radio stations to work well in the 160m (1.8 - 2MHz) band, and these two opinions are not entirely unrelated. The point is, that having installed an ATU and thrown away the cheapernet coax leading to the G5RV antenna, it sometimes transpires that the transmitter still refuses to give full output, and the MTBF (mean-time between failures) of the power-transistors is rather short; even though the working SWR is always apparently 1:1. The reason, of course, is that the impedance indicator (SWR bridge) is lying, and we are now in a position to show just how outrageously inaccurate an uncompensated bridge can be. Here we will base our analysis on bridges built using sensibly chosen transformer cores; but it does not take much imagination to work out what can happen if undocumented cores are used.
In [current transformers, section ?] we showed that the reactance of the transformer secondary winding at the lowest frequency of operation should be made at least equal to, and preferably larger than, the secondary load resistance. While this choice limits the low-frequency desensitisation to 3dB however, it does nothing to preserve the integrity of the balance condition. If a bridge is left un-compensated (which many are) it will still balance convincingly at low frequencies, but it will guide the user to mis-adjust the antenna tuner in a way that is particularly undesirable. We can explore the effect by comparing the mid-band current-transfer function at balance with the actual transfer function, and thereby obtain an expression for the load impedance that will be obtained. This derivation incidentally, is the same regardless of the type of voltage sampling network, i.e., it applies to all un-compensated current transformer bridges, not just the ones that use resistive potential dividers.
For the purpose of this analysis, we will use the approximation that the current transformer input impedance is small in comparison to the system load. This allows us to write the mid-band current transfer function at balance as:
η0 = Ri / Ni R0
At low frequencies, the transformer load will no-longer be Ri, but Ri in parallel with the secondary inductance, i.e., Zi=(jXLi//Ri), and due to this discrepancy, the bridge will no longer balance when the system load is equal to R0, but at some other impedance Zbal. The act of balancing the bridge by adjusting the ATU controls will ensure that the incorrectly chosen load impedance will be such as to bring the current analog into phase and magnitude coincidence with the voltage analog (no matter what voltage sampling method is used); and so the current transfer function will still be equal to η0. Hence:
η0 = (jXLi//Ri) / Ni Zbal
Putting these two equations together we have:
Ri / R0 = (jXLi//Ri) / Zbal
which contains the required relationship between transformer winding reactance and the load impedance foisted upon the bridge's hapless victims. Zbal can now be written explicitly, i.e.:
Zbal = R0 (jXLi//Ri) / Ri
and this can be expanded using the parallel-to-series transformation [AC Theory, Section 18]:
Zbal = R0 (Ri XLi² + j XLi Ri²) / ( Ri² + XLi² ) Ri
i.e.,
 Zbal = R0 (XLi² + j XLi Ri) / ( Ri² + XLi² )

The deviation of Zbal from R0 is shown in the Z-plane diagram below for various values of XLi expressed as a proportion of Ri. Zbal lies on a constant conductance circle corresponding to G=1/R0.

Fig. 6-18#1

Note that if we have XLi=Ri at 1.8MHz, the bridge will balance at 25+j25Ω instead of 50Ω (assuming R0=50Ω). The region between XLi=Ri/2 and XLi=4Ri can be regarded as fairly typical for uncompenstated bridges. The point at XLi=Ri/8 is approximately what happens when a transformer that should have been wound on an FT50-61 is wound on a T50-2 instead, but this is so bad (Zbal=0.8+j6.2Ω) that we might expect the operator to become suspicious (or notice the flames coming out of the power amplifier). Finally notice that even when XLi=16Ri the bridge is not perfect (Zbal=49.8+j3.1Ω). If we design for XLi=Ri at 1.8MHz, then XLi=16Ri at 28.8MHz, and so it transpires (presuming somewhat optimistically that there are no other parasitic or dispersive effects) that 'low-frequency' compensation will help to maintain correct bridge balance throughout the entire operating frequency range.

6-18a. Modified voltage sample.
From the foregoing discussion, we must conclude that an un-compensated current-transformer bridge is a pathological device that has the potential to turn a radio transmitter into a barbecue at low frequencies. We cannot necessarily solve this problem by increasing the transformer secondary inductance, because more inductance means a longer line-length, and this will cause large phase errors at high-frequencies. We also need to be able to control the turns-ratio rather than have it dictated by low-frequency considerations. Good design practice therefore requires the inclusion of some kind of LF compensation scheme.
It transpires that the situation for which we must compensate is extremely simple: A parasitic element that has the frequency law XLi=2πfLi and draws current in quadrature to the current analog appears in parallel with the current analog. The most straightforward remedy is homeopathic: we can place a device that has the same frequency law and phase relationship in parallel with the voltage analog. If the voltage sampling network is a resistive potential divider, then the device in question is an inductor. The modified bridge circuit is shown below:

Fig. 6-18#3.

Sticking to the approximation that the insertion impedance of the current transformer is small in comparison to the system load impedance ZL, the transfer function at balance for the current sampling network (now complex) is:
 η0 = (jXLi // Ri) / Ni R0
Our objective therefore, is to define the voltage sampling network transfer function (ηv=Vv/V) in such a way that it is equal to η0 when ZL=R0, regardless of frequency.
Now observe that the voltage sampling network consists of two impedance in series, one being R2 and the other being jXLv in parallel with R1. Let us therefore, for the time being, define Z1=(R1//jXLv), in which case the transfer function can be written:
ηv = Vv / V = Z1 / (Z1 + R2)
The process of balancing the bridge at all frequencies now becomes a matter of equating ηv=η0 and solving for Lv. If we expand all of the expressions for elements in parallel however, the amount of algebraic manipulation involved will quickly get out of hand, and so we will resort to a few mathematical tricks. The tricks in question are: latterly, to redefine the problem in terms of admittances, but firstly, to observe that the voltage-sample transfer-function is related to the expression for Z1 in parallel with R2:
Z1 // R2 = Z1 R2 / (Z1 + R2)
Hence:
ηv = (Z1 // R2) / R2
i.e.;
 ηv = (R1 // jXLv // R2) / R2
Now, the general expression for any number of impedances in parallel is:
Zpar = 1 / [ (1/Z1) + (1/Z2) + (1/Z3) + . . . . . + (1/Zn) ]
which may be verbalised as: "add the admittances and take the reciprocal". We may also observe that we may just as well equate the reciprocals of the voltage and current transfer functions as the transfer functions themselves. hence:
1/η0 = Ni R0 / (jXLi // Ri) = 1/ηv = R2 / (R1 // jXLv // R2)
Which gives:
 Ni R0 [ (1/jXLi) + (1/Ri) ] = R2 [ (1/R1) + (1/jXLv) + (1/R2) ]
The use of reciprocal transfer functions has caused the problem to separate directly into real and imaginary parts; or more to the point, it has fragmented the expression above into a frequency independent part and a frequency dependent part. The frequency independent part is:
Ni R0 / Ri = R2 [ (1/R1) + (1/R2) ] = (R1 + R2) / R1
which is equivalent to the earlier mid-band analysis in the approximation that the current transformer insertion impedance is small. The 'frequency dependent' part is:
Ni R0 / jXLi = R2 / jXLv
i.e.:
Ni R0 / j2πf Li = R2 / j2πf Lv
but the j2πf terms cancel (i.e., the compensating inductor has removed the frequency dependence), and the balance condition is satisfied when:
 Lv = Li R2 / Ni R0
Thus if Lv is calculated according to the relationship above; insofar as the inductances of the transformer and the compensating inductor do not vary with frequency, compensation for the shunting effect of the current transformer secondary inductance will be achieved throughout the entire frequency range.

Example: A bridge uses a current transformer that has a 12 turn secondary wound on an Amidon FT50-61 core. The AL value for the core is 68nH/turn², and so the secondary inductance is ALNi²=9.79μH. The current transformer has a 50Ω secondary load resistor and the system characteristic impedance R0 is also 50Ω. The nominal mid-frequency insertion impedance for the transformer is therefore 50/12²=0.35Ω, which will be neglected in the calculation but allowed-for by making the balance condition slightly adjustable. A resistive voltage sampling network will be used, with a total resistance of about 5KΩ to keep the dissipation in the resistors to around 1W when the generator is delivering 100W. The 'mid-band' balance condition is:
Ri / ( Ni R0 ) = R1 / (R1 + R2)
i.e.,
R1 / (R1 + R2) = 1/12
If we make R1+R2 exactly 5KΩ, then R1=5000/12=416.67Ω and R2 is 4583.33Ω, but these are horrible values and so we will make R1 a 430Ω 1% resistor. Now we observe that:
12 R1 = R1 + R2
i.e.,
R2 = 12 R1 - R1 = 11 R1
and so R2=4730Ω. This value will be very slightly low because we have neglected the current transformer insertion impedance, and so we will allow for some adjustment by constructing R2 from a 4.3KΩ 1% 1W resistor in series with a 1KΩ cermet trimmer.
The compensating inductance is:
Lv = Li R2 / Ni R0 = 9.79μH × 4730 / (12 × 50) = 77.2μH
If the inductor is wound on another FT50-61 core, the theoretical number of turns required is:
Nv = √(Lv / AL ) = 33.7 (i.e., 34).
Ideally, the value for Li used in calculating the compensating inductance should be determined by measurement at a frequency close to the lowest operating frequency of the bridge. Lv should then be made up to the required value on a laboratory component-measuring bridge. Both the transformer and the inductor should use the same type of core material, so that any dispersive effects (variation of AL with frequency) are cancelled. The practical circuit might look something like this:

Fig. 6-18#4.

Notice that a trimmer capacitor has been placed across the upper arm of the voltage sampling network. Its purpose is to compensate for the self-capacitance of the inductor and the capacitance to ground at the transformer secondary due to the presence of the Faraday shield. These paratitic capacitances will usually add up to something in the region of 4pF, and so the compensation trimmer, which has a value in inverse proportion to ratio of the two voltage sampling resistances, will be less than 1pF. It can be formed by placing two pieces of stiff wire in proximity and adjusting the spacing.

Another example of a bridge using a resistive voltage sampling network with a compensating inductor is given in ref [39].

[39] Measuring Rho - The Alternative to SWR, Bob Pearson, G4FHU, Rad Com, Feb 1998, p23-26.
(See also, by the same author: "How Big is a Bad SWR?" Rad Com, March 1993, p64-65, April 1993, p62-63.).
Article describes a dual-meter (forward and reflected signal) bridge with switch to enable the two meters to read the voltage and current samples directly, this permitting |Z| to be estimated as |V|/|I|. Uses toroidal current transformer with coax electrostatic screen, and resistive voltage sampling network with shunt inductor for LF compensation . Inductor and current transformer both wound on Philips RCC14/5 toroid, ferrite type 4C65 (AL=55).

6-19. The Douma - Buschbeck (capacitive voltage sampling) bridge:
The principal limitation of the resistive voltage sampling network is, of course, its power dissipation. The power loss can however be brought practically to zero by adopting a capacitive potential divider, as shown in the prototype circuit below. This configuration, with low-frequency compensation that will be discussed shortly, was submitted for US patent in 1953 by Tjiske Douma, of Sierra Electronic Corp. California, and patented in 1957 [USP 2808566]. Douma's principal inspiration was to build the lumped equivalent circuit of a fixed-position loop-probe reflectometer invented by Sontheimer and Korman in 1944 (the basis of the transmission-line bridge) [USP 2423416]. Ulrich Rohde (of Rohde & Schwarz), however, asserts that the configuration was invented in Germany by one Dr Buschbeck, sometime before 1949 [21]. The person to whom he refers is probably Werner Buschbeck of Telefunken, who was certainly working in this field in the late 1930s (see for example [USP 2165848] ). Many of the German Patent Office records were lost or destroyed at the end of the Second World War, and it is likely that Rohde's attribution is correct. Direct evidence of Buschbeck's priority is absent however, and the independence of Douma's invention is clear from his citation of prior art; and so here we will generally refer to the circuit as 'Douma's bridge'. The bridge in its uncompensated form is closely related to (and often confused with) a circuit known as the "Bruene Directional Wattmeter" (i.e., SWR bridge) [40][41][42] (which will be discussed in a later section).

Refs:
[21] "New approach to measuring SWR at high-frequencies", Ulrich L Rohde DJ2LR, Ham Radio, May 1979 p34-35.
Attributes the capacitive voltage sample / current transformer bridge to a German inventor, one Dr Buschbeck "more than 30 years ago" [i.e., before 1949], but draws Buschbeck's circuit without secondary loading of the current transformer [possibly an ommission].
[40] An Inside Picture of Directional Wattmeters, Warren B Bruene W0TTK [now W5OLY], QST April 1959 p24-28.
[41] In-Line RF Power Metering Doug DeMaw W1CER [now W1FB], QST Dec 1969 p11- 16.
Construction article for Bruene-type bridges for 3.5-30MHz.
[42] Ferromagnetic Core Design & Application Handbook, M F "Doug" DeMaw W1FB, 1st edition, 2nd printing 1996. MFJ publishing co. http://www.mfjenterprises.com/ .Bruene bridge: p94-95.

The trade-off with capacitor voltage sampling is that the capacitor network shifts the load phase angle from the generator's point of view. Douma offered a solution to this problem, that involves using a second voltage sampling network with a frequency dependent output to be summed with the existing voltage and current samples, allowing part of the load capacitance to be referred to the generator side of the bridge [USP 2734169] (i.e., Douma compensated for any generator-side capacitance by correcting the balance condition so that the generator would always see a resistive load - see section ?). This complicated solution seems to have found little favour however, and the usual approach is to keep the capacitance small; i.e., to see if it is possible to design a capacitive network with enough output to drive a detector over the desired frequency range without falling outside the generator's load tolerance limits.

We will begin, as usual, with the 'mid-band' analysis:

For a capacitive potential divider, the transfer function is:
ηv = Vv / V = jXC1 / (jXC1 + jXC2)
= (-j/2πfC1) / [(-j/2πfC1)+(-j/2πfC2)]
The -j/2πf factors in the numerator and denominator can be cancelled, giving:
ηv = Vv / V = (1/C1) / [(1/C1)+(1/C2)]
= (1/C1) / [(C2 + C1)/(C1 C2)]
which simplifies to:
 ηv = Vv / V = C2 / (C1 + C2 )

The capacitance seen by the generator is given to good approximation by the formula for capacitors in series, i.e.,
Cs = C1 C2 / (C1 + C2 ) = ηv C1

The tangent of the phase angle for a parallel combination of resistance and reactance is R/X. If, as recommended by Underhill & Lewis [43], we set the maximum (negative) value for the phase angle of the load seen by the generator to be about -7°, this immediately sets a maximum value for Cs. No great accuracy is required for this stipulation, and so we will neglect the insertion impedance of the current transformer in this case and give the rule of thumb:
Tan(-7°) ≥ R0 / XCs
i.e., for a 50Ω system:
-XCs ≥ 407Ω
The reactance of the capacitor network will of course be lowest at the upper frequency limit for the bridge, so if we set XCs=-407Ω at 29.7MHz, we obtain Cs=1/2πfXCs=13.2pF.

Ref:
[43] "Automatic Tuning of Antennae". M J Underhill [G3LHZ] and P A Lewis.
SERT Journal, Vol 8, Sept 1974, p183-184. Gives criteria for achieving 1.2:1 SWR, i.e., 45 ≤ R ≤ 56Ω, 17.5 ≤ G ≤ 22.5 mS, -7° ≤ φ ≤ +7°.

Now, if we assume the same balance condition as in the previous example, i.e., η0v=1/12, then C1=Csv=158.4pF and C2=Cs/(1-ηv)=14.4pF. We may therefore make C1 (say) two 82pF capacitors in parallel, and C2 a 3-30pF trimmer; and calibrate the system by adjusting C2 with the bridge connected to a 50Ω dummy-load resistor. In this case, by using preferred value capacitors, C1 has been nominally adjusted to 164pF and C2 becomes 164/11=14.9pF. Note however, that the reactance of C2 at 1.8MHz is nominally -5931Ω, and so we can expect the bridge sensitivity to be reduced by detector loading effects at the lowest end of the frequency range. The frequency independence of the uncompensated voltage sample however, will remain fairly good over the whole range; because the capacitive network will have absorbed most of the system strays during the calibration process.

LF Compensation:
The bridge of course, will be of dubious accuracy without low-frequency compensation, and so we must modify the voltage sampling network. In this case, the compensation does not require a shunt inductor, but surprisingly, a resistor. The reason is that the compensating element must draw current at +90° relative to the current flowing in the rest of the network; and since the network is capacitive, a resistor satisfies this criterion whereas an inductor would draw current in anti-phase.

>> Korman 1942 "Radio Frequency Wattmeter" US Pat. 2285211 used this method.

The compensated circuit is shown below:

Here we may write:
ηv = Vv / V = Z1 / (Z1 + jXC2)
Where Z1 = (Rv // jXC1)
Notice however, that the expression for Z1 in parallel with XC2 is:
(Z1 // jXC2) = Z1 jXC2 / (Z1 + jXC2)
Therefore:
ηv = (Z1 // jXC2) / jXC2
Hence:
ηv = (Rv // jXC1 // jXC2) / jXC2
Which, when inverted gives:
1/ηv = jXC2 [ (1/jXC1) + (1/jXC2) + (1/Rv) ]
In this case, Rv ends up in the imaginary (frequency dependent) part, i.e.,
1/ηv = XC2[(1/XC1)+(1/XC2)] + jXC2 / Rv
Now, by substituting XC=-1/2πfC (remember that capacitive reactance is negative) and re-arranging, we obtain:
1/ηv = [ (C1 + C2) / C2 ] -j/(2πf C2 Rv )
Which, since -j=1/j, can also be written:
 1/ηv = [ (C1 + C2) / C2 ] + 1/(j2πf C2 Rv )
The real part of this expression is, of course, the reciprocal of the transfer function from the mid-band analysis. The imaginary part is dimensionless (as a transfer function should be) because capacitance × resistance = time, and time is the reciprocal of frequency.
The reciprocal of the complex transfer function for the current transformer was derived in the previous section, and when the bridge is at balance it can be written:
1/η0 = Ni R0 [ (1/Ri) + (1/jXLi) ]
and so by substituting XLi=2πfLi and equating the imaginary parts of the reciprocal voltage and current transfer functions we obtain:
1/(j2πf C2 Rv ) = Ni R0 / j2πf Li
i.e.,
 Rv = Li / Ni R0 C2
If this condition is satisfied, and provided that the transformer secondary inductance does not change significantly with frequency; compensation for the inductive reactance of the current transformer secondary winding will be achieved throughout the entire frequency range.

If we return to our previous example, where Li=9.79μH, Ni=12, and C2 ended up as 164/11=14.9pF, then using the formula above, the compensating resistance is 1094.6Ω. This is within 0.5% of 1.1KΩ, and so may be constructed using a single 1.1KΩ resistor or by placing two 2.2KΩ resistors in parallel. Since the resistor provides a DC path for a diode detector, the RF choke can be omitted, but it is often still included in order to maximise the detector sensitivity (the resistance of the detector DC-return path is effectively in series with the meter). Notice that the choke is connected across the detector port, not across the voltage sampling network output as is the case in some designs. A choke of less-than infinite inductance connected across one of the sampling networks will upset the balance condition, whereas an imperfect choke loading the detector port will only affect the magnitude of off-balance meter readings.

Circuits that work in essentially the same manner as Douma's bridge are used in impedance (SWR) indicators and power-amplifier protection circuits, in HF radio equipment manufactured by Collins Radio, MFJ, Kenwood, and many other companies. From the foregoing calculations however, it can be seen that the detector output will be susceptible to loading at low frequencies, and the load seen by the generator will soon become unacceptably reactive if the upper frequency limit is extended (we will look at solutions to this problem in sections 6-?22 and 6-?23). The bridge however can be made suitable for monitoring high-power transmitters by the simple expedient of increasing the number of turns in the current transformer secondary to keep the power dissipation in the secondary load resistor to about 1 Watt (and adjusting the voltage sampling network components accordingly). Note that when using the bridge with high-power transmitters, it is advisable to carry out flux-density calculations [38].

Ref|:
[38] Amidon Associates Inc. (Technical data book) Jan 2000. Maximum flux density calculations and recommendations: p1-35.
Information is also available online from: www.amidoncorp.com .

Douma's bridge and variants of it are often implemented without the LF compensation resistor, and if this is the case, it is important to read the circuit documentation. Many so-called "HF" bridges are only designed to work from 3.5 to 30MHz [44] and whereas a Sontheimer-Korman (transmission-line) bridge [USP 2423416] will merely become grossly insensitive below its design minimum frequency, an un-compensated current-transformer bridge will become grossly inaccurate below its design minimum frequency. A further caveat arises when modifiying HF transceivers for general coverage, the point being that some crude modifications will remove transmit-inhibit signals for frequencies below 1.6MHz, in which case a power-amplifier protection circuit based on an uncompensated current transformer bridge will allow full-drive into highly unsuitable load impedances , and there is great danger that the output transistors will be destroyed.

Ref:
[44] After the Second World War, the 160m band remained unavailable to USA radio amateurs until 1955 (see AC6V's Amateur Radio History page). Even then, the band was primarily allocated to the Loran navigation system. Hence many amateur and commercial circuit designers simply ignored the existence of the 160m band, and to many operators, HF started at 3.5MHz. The problem with a current transformer bridge designed to work from 3.5MHz up, is that it still seems to work when you try it on 160m, and the user has no way of knowing that it is grossly innacurate. This is one source of the folklore relating to the difficulty in getting a radio station to work on 160m. The other is the myth that you need an 80m long back garden (by that reckoning, the BBC Radio 4 transmitter on 198 kHz would need a 757m long antenna, but it doesn't).

6-x. Generator power-factor correction:
In the Douma bridge circuit shown below, an inductor Lα has been placed in series with the generator output. This simple modification can provide an almost frequency-independent compensation for generator-side parallel capacitance.

From the necessary design considerations for the Douma bridge, we know that at high frequencies, the resistor Rv is chosen such that:
Rv >> XC1.
Also, when the bridge is balanced, no detector current is drawn from the junction of C1 and C2. Hence, at high frequencies, to a good approximation, the voltage sampling network subjects the transmission line to a parallel capacitance that we may compute from the series capacitor formula as:
Cα = C1C2/(C1+C2)
When the bridge is balanced, the load is equal to R0. If we neglect (temporarily) the insertion impedance of the current transformer and start with Lα=0, the effect of Cα is to move the impedance seen by the generator away from R0, the path taken being a clockwise rotation around a circle of constant conductance (see diagram right). A finite value for Lα, on the other hand, will move the resultant impedance vertically in the Z-plane, along a line of constant resistance. Hence a value of Lα can be chosen that will restore the generator phase angle to zero.

The resistive load seen by the generator by virtue of the presence of Lα will be a little short of R0 by an amount ΔR, but the Z-plane diagram above is highly exaggerated. In practice, the phase shift φ due to the capacitor network will be no more than a few degrees, and ΔR will be correspondingly tiny. Observe also that we have neglected the insertion impedance of the current transformer, and ΔR will go some way towards compensating for it. In fact, by judicious choice of Lα and Cα, it is possible to possible to offset the transformer insertion resistance exactly (but probably not worth the effort).
Ignoring the current transformer and Rv, the load seen by the generator will be:
Z = jX + ( R0 // jX )
This expands to:
Z = jX + [ jX R0 / ( R0 + jX ) ]
Multiplying numerator and denominator of the second term by the complex conjugate of the denominator gives:
Z = jX + [ jX R0 ( R0 - jX ) / ( R0² + X² ) ]
which rearranges to:
Z = jX + [ ( R0 X² + jX R0² ) / ( R0² + X² ) ]
In order to bring Z onto the resistance axis, we simply set the imaginary part of this expression to zero, i.e.:
X + [ ( X R0² ) / ( R0² + X² ) ] = 0
Now notice that if X² >> R0², this being realistic in practice, we may reasonably delete R0² from the denominator of the second term. Hence:
X + [ X R0² / X² ] ≈ 0
Thus:
X = -R0² / X
i.e.:
2πf Lα = 2πf Cα R0²
The correction is frequency independent provided that X² >> R0².
 Lα = Cα R0²    gives zero generator phase angle at balance when  XCα² >> R0²

It was mentioned earlier that the capacitive voltage sampling network should be chosen such that the generator phase angle never becomes more negative than -7°, i.e., if R0 is 50Ω, then -X should never be allowed to fall below 407Ω. Now, by the inclusion of Lα we may relax this criterion somewhat, thereby reducing the output impedance of the sampling network and making the bridge less susceptible to detector loading. Let us say, for example, that we decide to let -X fall to 300Ω at the highest frequency of operation. In the absence of a compensating inductance, this would give a generator phase angle of:
Arctan(50/-300) = -9.46°
which would be unacceptable. But the difference between (R0²+X²)=92500 and X²=90000 is only 2.7%. Hence a series inductance compensation for this phase shift will be nearly perfect, and the compensation will be even better at lower frequencies because the magnitude of X will increase. If the maximum frequency of operation is (say) 30MHz, then the capacitance that has a reactance of -300Ω at this frequency is:
Cα = -1/( 2πf X ) = 17.7pF
For this capacitance, the compensating inductance is:
Lα = Cα R0² = 44.2nH

>>> 3 or 4 turns of the coax inner conductor around a 4mm drill shank.

© D W Knight 2008.
David Knight asserts the right to be recognised as the author of this work.

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