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= = 0.062
2

2

2 = 0.062
2
= = 0.062

=

= = 0.062

= = 0.062
2
= = 0.062 35
2

2
1

2
EE|Times EUROPE — Power Electronics
= = = 16 Ω, use 20 Ω

0.062
2
2
Miniature Current Transformers Keep Pace with GaN

1

= 16 Ω, use 20 Ω

2 =
=

2
= = 0.0 2 2 1 = 16 Ω, use 20 Ω

162
= 16 Ω, use 20 Ω
=
=

0.062

0.062

of the pulse, exciting current is 0 and increases linearly with time Therefore, by combining, we can determine a burden resistance (R ),
b

(pulse duration). This increasing current multiplied by the winding

resistance creates an increasing voltage drop, reducing the output; 2 2 1 2 1 2
= 16 Ω, use 20 Ω

this is called droop. When using a current transformer to detect peak = = = 0.062 = 16 Ω, use 20 Ω
=

currents, the droop must be less than the rising peak current or the 0.062 1 = 0.050

=
peak will not be detected. the secondary current (I ), = 20

1
S
= = 1 = 0.050

= = 1 20 = 0.050
At higher operating frequencies, with the = = 20 = 0.050

20

unipolar pulses common in switching power

/N ).
supplies, it is important to reset the core and, finally, the turns ratio (N = S 1 = 0.050
P 1

= 0.050

=
=
=

before the next cycle begins. 20 20

5 ∙ 1

= = 0.050 = 100

= = 5 ∙ 1 = 100

5 ∙ 1

0.050
When calculating the Vµs rating, it needs to be less than the
= 100
5 ∙ 1

= 100

=
=
The difference caused by magnetizing inductance and droop can be transformer’s rating, which is based on the peak flux density (B), the
=
=

0.050
0.050

seen in Figure 5, which shows waveforms from identical setups but secondary turns (N ), and the core cross-sectional area (A ). Note that

e
s

uses two series of current transformers. The 749251050 series uses manufacturers use different values for B to calculate their volt-time

an EE5 core, whereas the 749252050 series uses a smaller, EE4.4 core. product. You should aim for a low value, ideally less than 0.2 T. The

5 ∙ 1
5 ∙ 1

= 100
= 100

The secondary inductance is 500 µH min. for the larger transformer series diode’s voltage drop (V ), which can vary widely, is added here as
=

=
=

=
f

0.050
0.050

and 205 µH min. for the smaller one, both using a 1N4934 diode and 1 V because it adds to the voltage across the coil, which will influence

0.45

= 1.80 <

= ( + )
= (1 + 1)
burden resistance of 20 Ω. However, the slightly lower output from the flux level. It does not affect the sense current or burden voltage.

500000
the smaller transformer can easily be compensated for by a slight 0.45 = 1.80 <

adjustment (increase) of the burden resistor value to match the = ( + ) = (1 + 1) 500000

0.45
0.45

= (1 + 1)
= ( + )
waveform exactly. = ( + ) = (1 + 1) 500000 = 1.80 <
= 1.80 <

The larger reset pulse results from the larger leakage inductance 500000

of the EE4.4 design. In this product series, the single-turn primary is Looking through the list of available current transformers in the

molded into the base of the coil former, placing the windings adjacent EE4.4 size series, the 749252100 with a rating of 7 A, a turns ratio of

0.45
0.45
= 1.80 <

0.45
( + )

)
= (1 + 1) 1)(
to each other rather than the more favorable concentric position. This 1:100, and a Vµs rating of 28.8 easily meets the requirements.
(1 +
= ( + )

= 1.80 <
= (1 + 1)
+ )
= (
500000
= 2.2

500000

=
500000
=

results in more energy stored for reset. To find the error from magnetizing current, first calculate the mag-
0.45

−6
)
( +
820 ⋅ 10
(1 + 1)(
)

At higher operating frequencies, with the unipolar pulses common netizing current using the maximum duty cycle (DC) and frequency

= 2.2
500000
0.45

0.45
=
( + )

), the series diode voltage drop (V ), and
in switching power supplies, it is important to reset the core before (f ), the burden’s voltage (V = (1 + 1)( 500000 ) ) f

( + )

−6
820
(1 + 1)( ⋅ 10

b
= 2.2
500000

=
the next cycle begins. This is caused by the series diode, which needs the secondary inductance (L = 820 ⋅ 10 −6 = 2.2
= ). Then divide by the ideal transferred
=

m
−6

to be fast (t < ~200 ns), ultra-fast (t < ~50 ns), or a Zener to control secondary current (I ), 820 ⋅ 10

rr
ST
rr

the reset voltage more precisely. Soft recovery is preferred.

0.45

Really fast signal diodes like the 1N4148 may be too fast because ( ( + ) (1 + 1)( 0.45 ) )
+ )
(1 + 1)(

500000

−6 = 2.2
they create larger voltages with harsher edges, which will contribute = = = 500000 = 2.2

=

820 ⋅ 10
to EMI noise. Schottky diodes may be limited by voltage rating. The 820 ⋅ 10 −6
2.2

response of several diodes is shown in Figure 6. It should be noted % = ∙ 100 = ∙ 100 = 4.4%

50
that the diode is necessary to prevent the negative reset pulse from 2.2 ∙ 100 = 4.4%

going to the controller, where it would damage it. % = ∙ 100 = 2.2
% = ∙ 100 = ∙ 100 = 4.4%
2.2
50

∙ 100 = 4.4%
% =
∙ 100 =

50
50

Three easy steps to current transformer selection are: where L = 820 µH minimum for the 749252100 transformer. Because the

m
• Determine the turns ratio inductance is specified as minimum, the actual error will be less.
• Calculate the flux density or Vµs rating If the accuracy is not adequate, an easy solution without resorting

2.2
2.2

∙ 100 = 4.4%
∙ 100 = 4.4%
• Calculate the error due to magnetization current to a larger transformer is to compensate by increasing the burden

% =
∙ 100 =
∙ 100 =
% =
50
resistance proportionally 50
— in this example, to 21 Ω. This will not sig-

The initial parameters will be determined by your application. nificantly affect the power dissipation or the Vµs required.
Let’s use: Miniature surface-mount current transformers such as the WE-CST
• Maximum current signal (voltage, V ) for controller: 1 V EE4.4 series are here today for your high-frequency GaN-enabled power
S
• Maximum primary current (I ) to measure: 5 A converters. They are at least as small as a sense resistor, dissipate less
P
• Operating frequency (f ) and maximum duty cycle (DC ): 500 kHz, power, operate at higher currents, provide galvanic isolation (which
0.45 means freedom of placement in the circuit), and provide a stronger,
• Power dissipation (P ) in burden resistor: 0.062 W (1/16 W) more noise-immune sense voltage without the need for any extra cir-
Rb
cuitry like an op amp. The outputs can be easily combined and can even
= = 0.062
2
Determine the turns ratio based on the given sense voltage and measure DC current. ■

power dissipation.
BIBLIOGRAPHY
Brander, T.; Gerfer, A.; Rall, B.; Zenkner, H. 2018. Trilogy of Magnetics.
2
= = 0.062 1 Waldenburg: Würth Elektronik eiSos GmbH & Co. KG.

= = 1 2 Designers” SEM1200 SLUP114:10-13, Texas Instruments.
Mammano, R. 1997. “Current Sensing Solutions for Power Supply

www.eetimes.eu | FEBRUARY 2021
2 2 1 2 1 2 = 16 Ω, use 20 Ω
= =
= = = 16 Ω, use 20 Ω
0.062
0.062

1
1 = 0.050
= 0.050
= =
=
= 20

20

5 ∙ 1
5 ∙ 1
= = = 100
= 100
= = 0.050

0.050

0.45

= ( + ) = (1 + 1) 0.45 = 1.80 <
= 1.80 <

= ( + ) = (1 + 1) 500000

500000

0.45

( + ) (1 + 1)( 0.45 )

( + ) (1 + 1)( 500000
)

= = 500000 = 2.2
−6 = 2.2

= = 820 ⋅ 10

820 ⋅ 10 −6

2.2
2.2 ∙ 100 = 4.4%
% = ∙ 100 =
∙ 100 = 4.4%
% = ∙ 100 = 50
50