 TX to Ae Ch 6 Contents << 6.2 Part 2 6.3 Part 1 >> 6.2 Current Transformers: Part 3.

 6.2-xx. Multiple secondary windings: >>> work in progress . For some of the bridge designs to be discussed later, we will also have need for current transformers with two secondary windings. In this case we simply observe that the primary MMF is equal to the total secondary MMF, and derive our equations accordingly. A current transformer with two identical outputs is shown below:
 To make the analysis consistent with that for the single-output transformer, the turns, load resistances, and output voltages for each secondary have been defined as half the total, but the definition of the secondary current remains the same. Using the ampere-turns rule, we may then write: Ip = ( Is N/2) + ( Is N/2 ) = N Is and we may obtain the total secondary voltage thus: Vs = 2( Is Ri/2 ) = Is Ri hence:
 Vs = Ip Ri / N
Which, by virtue of the definitions used, is the same as for the single-output current transformer. Notice also that if the start of one secondary winding is connected to the end of the other, a centre-tapped transformer results. We may then evoke the rather unsurprising rule; that if the transformer secondary has a centre tap, and the load resistance has a centre tap, joining the two centre taps together has no effect upon the total output voltage. Note incidentally, that the current transformer voltage output can be split simply by tapping into the load resistor; but by tapping into the transformer winding as well, an imbalance current can be drawn from the transformer tap, making the voltage ratio less-susceptible to the effects of unequal loading and component tolerances. One small disadvantage of tapping (i.e., taking a wire out and then back in to the transformer) is that it increases the leakage inductance slightly, but this is not usually problematic provided that the wires are kept short.

 Bifilar winding (Don't do it): Because the transformer of a TRAB is bifilar wound to give the best possible symmetry, it is often assumed that tapped or double-secondary current-transformers should be bifilar wound. This is a serious misunderstanding. In a tapped current-transformer, bifilar winding is equivalent to winding the coil in two layers. It serves no useful purpose, increases the secondary capacitance, and degrades the high-frequency phase performance (and twisted bifilar winding is even worse). In a double-secondary transformer, one winding being used (say) for a forward-power bridge, and the other for a reverse power bridge, bifilar winding gives rise to a spurious capacitive coupling between the two bridges. >>>>>

 6.2-xx. The Tapped Current-Transformer Paradox. In the previous section, on the subject of current transformers with multiple secondary windings, we derived the secondary voltages using the ampere-turns rule. In the case of a transformer with two identical secondaries, with identical load resistors, it was shown that by choosing to define the number of turns on each winding as N/2, the load across each winding as Ri/2, and the output from each winding as Vi/2, we end up with exactly the same expression for output voltage as was the case for a current transformer with a single secondary, i.e., Vi = I Ri / N This expression, of course, relates only to the mid-band analysis, and at some point we must introduce the transformer secondary inductance in order to carry out the low-frequency analysis. Some vigilance is required at this point because there is a certain subtlety in the behaviour of current transformers.      The issue to be raised here might appear to be a paradox, but like all paradoxes it is actually a fallacy. Recall that the inductance of a coil is given by the expression: L = AL N² Hence, if N is taken to be the total number of turns on the transformer secondary, then the total secondary inductance, Li, obtained when both windings are connected in series (start to end), is also ALN². If the two half-windings are treated separately however, then the inductance of a half-winding (of N/2 turns) is given by AL(N/2)² = ALN²/4. The half-winding has only one quarter of the inductance of the whole winding. So how should we analyse this circuit? If the two secondary windings are connected in series as shown above, then the circuit is equivalent to a transformer with a centre-tapped secondary and a centre tapped load, with the two centre taps joined together. Since the two taps are at the same potential, it makes no difference whether they are connected or not. Hence we can derive the output voltage in the usual way: Vi = V (Ri // jXLi ) / ( N ZA ) and the output from one of the secondary windings is: Vi / 2 = V (Ri // jXLi ) / ( 2 N ZA )     . . . . . (x.1) . . . If on the other hand, we analyse the two windings separately, we get: Vi / 2 = Ii [ (Ri/2) // (jXLi /4) ] where Ii, is given by the ampere-turns rule: Ii = I / N = V / ( N ZA ) Hence, apparently: Vi / 2 = V [ (Ri/2) // (jXLi /4) ] / ( N ZA ) which rearranges to: Vi / 2 = V [ Ri // (jXLi /2) ] / ( 2 N ZA )     . . . . . . (x.2) . . . × which is not the same as 6-20a.1. So which analysis is correct? One way to find out is to perform a test of a somewhat preposterous prediction of equation 6-20a.2, which is that the output voltages will change if the connection between the transformer tap and the load tap is broken.      A current transformer was constructed on an Amidon FT50-61 toroid (AL=68.8nH/turn² nominal), the Faraday-shielded primary being a stub of RG108 (50Ω Ag-PTFE) coax, and the two secondary windings being each 6 turns of 0.9mm diameter enamelled copper wire. With the secondary windings connected in series (start to end), the total inductance was measured to be 9.7μH ±5% at 1.5915MHz. When the two 6-turn windings were measured individually (with the other open-circuit), the inductance was found to be 2.7μH ±10% in both cases. Hence, within the measurement error, the inductance of a 12-turn winding was found to be 4 times the inductance of a 6-turn winding. Here we will not get bogged down in error analysis, it not being necessary for the point to be demonstrated; we will simply take the total secondary inductance Li to be 9.7μH (this being the better measurement), and the inductance of the separate windings to be 9.7/4 = 2.425μH in each case.      The current transformer was mounted in a test jig and connected into the test circuit shown below. Phase-difference measurements were then made using the method described in section 11.  Here the Oscilloscope Y1 amplifier is presented with a voltage that is in-phase with the primary current, and this is also used to trigger the timebase. Note that the cable leading to the Y2 amplifier is properly terminated, and so the input to the cable is a good approximation to a pure 50Ω resistance. Hence, with the switch open, a 12 turn secondary winding of 9.7μH inductance is loaded with 100Ω. The Y2 voltage measurement is effectively made at a tap half-way down the load resistance, but since this network is a resistive potential divider, the relative phase recorded is the same as the phase of the voltage across the transformer secondary. With the generator (a radio transmitter) operating at 1.6MHz, the oscilloscope timebase, gain, shift and trigger-level controls were manipulated so that one cycle of the waveform was 10cm long. The phase difference between the primary current and the secondary voltage was then found to be 13±0.5mm on this scale, with the secondary voltage leading (inductive). Hence the phase difference in degrees was:
(13±0.5/100)×360° = +46.8±1.8°
The calculated reactance of the full 12 turn secondary winding at 1.6MHz is:
2π × 1.6×106 × 9.7×10-6 = 97.5Ω
and with a load of 100Ω, the calculated phase angle is:
Arctan(100/97.5) = 45.7°
This is perfectly in agreement with the measured value (which is why we don't need to bother taking the error in the inductance measurement into account).
Finally, operating the switch to join the centre taps made no observable difference to the phase measurement or the output voltage. Hence equation (x.1) is correct.

We have at this stage, resolved the paradox, but we have yet to explain it. With that in mind, a further experiment was conducted with the test jig rewired as per the circuit below: Now, one half of the secondary winding is 'completely isolated' from the other (or not, as the case may be). With the switch open, the phase difference between the primary current and the secondary voltage at 1.6MHz was found to be:
(18±0.5/100)×360° = +64.8±1.8°
The calculated reactance of the 6 turn (2.425μH) winding at 1.6MHz is:
2π × 1.6×106 × 2.425×10-6 = 24.4Ω
and with a load of 50Ω, the calculated phase angle is:
Arctan(50/24.4) = 64°
Now the interesting part: when the switch was closed, the phase angle changed back to 46.8±1.8° exactly as in the previous experiment. The phase angle for 9.7μH in parallel with 100Ω is exactly the same as the phase angle for 4.85μH in parallel with 50Ω. Hence, Closing the switch has the effect of doubling the inductance of the secondary winding connected to the oscilloscope.
An explanation is called for, and it is this: reactance is indicative of energy storage. Apart from parasitics, which only have an effect at higher frequencies, when a secondary winding is disconnected, its inductance ceases to exist. When no current flows in a winding, no energy can be stored in the inductance of that winding. When the switch is closed however, a current flows in the floating winding and an energy storage mechanism is activated. The two secondary windings moreover, regardless of any DC path between them, are intimately connected by the closed magnetic circuit of the toroidal core. Because a current transformer controls its own input voltage according to how it is loaded, closing the switch alters the flux density in the core. In effect, by transfer of energy between the two secondary windings; when the switch is closed, an extra energy storage mechanism becomes available to the winding connected to the oscilloscope.
If we call the separate inductances of the two secondary windings La and Lb, the total inductance of the windings is:
Li = La + Lb + Mab + Mba
where the Mab and Mba are the mutual inductances. If there is no leakage inductance (as is true to a 98% good approximation for high-permeability toroidal cores) then:
La = Lb = Mab = Mba
which is why the inductance of the two windings in series is 4 times larger than that of a single winding (when the windings are connected start to end), and is also why the inductance of a coil is proportional to N². Hence, when both windings are loaded, the effective inductance of a single winding is not Li/4, but La+Mba = Li/2, where Mba represents the auxiliary inductance provided by the other winding. Hence, the correct model for the split-secondary current transformer, regardless of whether there is a DC connection between the secondaries, is as shown below: Vi/2 = V [ (Ri/2) // (jXLi/2) ] / ( N ZA )  TX to Ae Ch 6 Contents << 6.2 Part 2 6.3 Part 1 >>