[Design View / Design Solution]
The Right Calculations Add Up To The Ideal Power Current Transformer
Unlike the voltage transformer, a current transformer measures current from one circuit to another, which requires different calculations to create the optimal design.
Current transformers measure current or transfer energy from one circuit to another, so their design requires calculations different from their voltage transformer cousins. The reason for the difference is that current transformer magnetizing current is the load current itself, unlike voltage transformers, where magnetizing current is "separated" from the load current and has a value that's a small fraction of the total current at full load.
We want to supply the load with current IL from a current source, producing primary winding current IP(Fig. 1). We're dealing with a current source, so the load voltage is set up by the load itself if it consumes current (like an incandescent lamp, voltage regulator, or zener diode).
To design the current transformer, we need to know its core shape, size, and material, as well as the number of turns to use. Select a toroidal core because it delivers minimum losses and provides appropriate flux coupling between the primary and secondary windings if the number of turns is high enough to cover most of the core surface.
According to Ampere's Law, current IP flowing through a winding with NP turns (Fig. 2) produces a magnetic field depending on the magnetic line length lM, which is described by the differential equation: dIp × Np = H × dlM(1)
Integrating over the whole magnetic line length and assuming usage of a toroidal core, obtain: Ip × Np = H × lM(2)
where lM = median length of a magnetic line, which, for a toroidal core is: lM = π × DMED = π × (DOUT + DINN)/2(3)
where DMED = median diameter; DOUT = the toroidal core's outer diameter; and DINN = the toroidal core's inner diameter.
Magnetic field H produces magnetic flux in the core, which has a density of B. This flux density depends on the core material relative permeability μR: where B = μR × μ0 × H, and μ0 = 4π 10–7 (H/meter = permeability of vacuum)
Therefore, Equation 2 may be rewritten as: Ip × Np = B/(μR × μO) × lM(4)
To transfer energy with minimal losses, the magnetic core should not impose excessive losses while the energy transfer lasts. In other words, it should not saturate. Hence, the core flux density, B, should not exceed the saturation limit value of BSAT. In addition, the primary side current IP, which produces this BSAT, should always be below some maximum value IPMAX. Therefore, Equation 4 can be rewritten as: IpMAX × Np = BSAT/(μR × μO) × lM(5)
In our consideration, we're dealing with a constant current, which may be assumed as IPMAX. Therefore, we should either assign a value to NP (to cover as much of the core surface as possible) and calculate lM, and then the core dimension DMED, or select a core and derive a proper NP. The BSAT value is a core datasheet item, as is the μR.
We don't care about the core cross-sectional area. Therefore, we can use a core of any thickness — only the core diameter is of interest.
There's a paradox associated with the relationships depicting current transformer operation: Ip × Np = IL × NS(6)
where NS = number of turns on the secondary winding. IP is fixed, so the secondary side load current is also fixed and: IL = Ip × (Np/NS)(7)
What would my total power drawn be from a 120V line ideal transformer. With only a RP and no RL.
Anonymous -February 01, 2008
An interesting article. However the calculations assume that only the primary is carrying current. In fact the secondary current acts to reduce the flux in the core. Hence many more turns can be used then would be suggested by these calculations. The minimum number of turns needed is related to the core cross-section, the maximum allowable flux density, the frequency, and the maximum voltage needed. See any article on transformer design.Too many turns give more copper loss.
Alex Hiley -May 04, 2007 (Article Rating: )
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