Example: calculating a transformerFlyback transformers are designed like coils. They must store energy in the primary coil during the first part of the cycle, to be returned through the secondary in the second part of the cycle. They are actually coils with two separate windings.
For this reason flyback transformers need a relatively large air gap, which is the area of the magnetic circuit that stores the most energy.
All other transformers operate by making a direct transfer of energy, so they do not need to store it. The air gap in these cases will be very small or not exist at all.
This is the type of transformer that we are going to calculate in this post.
Core:E 30/15/7gaps: 0mm in one core and 0mm in other core (ungapped)
Material:
N87Parameters:A = 60 mm
2 = 60 · 10
-6 m
2A
min = 49 mm
2 = 49 · 10
-6 m
2AL = 1900 nH = 1.9 · 10
-6 H
B
max = 200 mT (below saturation)
Copper window (between former and core): 17mm x 5mm = 85mm
2Current density of copper: 5 A/mm
2 rms
Frecuency of operation: 20 kHz
Calculations and results:We are going to try to fill with copper wire all the window between the former and the core.
Half of the window with the primary coil and the other half with the secondary coil.
We will start with a round wire of 0.5mm diameter, which will fill the hole of the window as if it were a square section wire.
d
1 = 0.5 mm (diameter of primary wire)
N
1 = 0.5 · Window / d
12 = 0.5 · 85mm
2 / 0.5
2 = 170 Turns
d
2 = 0.5 mm (diameter of secundary wire)
N
2 = 0.5 · Window / d
22 = 0.5 · 85mm
2 / 0.5
2 = 170 Turns
Lenght of one turn of wire (median):
L
Turn = 4 · (9mm + 19mm) / 2 = 56mm
Lenght of primary wire:
L
N1 = N1 · L
turn = 170 · 0.056m = 9.52m
Cross section of primary wire:
S
N1 = 0.5
2 · π / 4 = 0.1963 mm
2Estimated resistance of primary wire:
R
N1 = ρCu · L
N1 / S
N1 = 0.020 · 9.52 / 0.1963 = 0.97 Ω
Maximum current allowed by copper:
I
rms_max_N1 = 5A/mm
2 · S
N1 = 5 · 0.1963 = 0.982 A
rmsMaximum power losses in copper:
P
N1 = I
rms_max_N12 · R
N1 = 0.982
2 · 0.97 = 0.935 Watt
P
N2 = P
N1 = 0.935 Watt
P
Cu_N1+N2 = 1.87 W
Transformer excitation inductance:
L
excitation = N
12 · AL = 170
2 · 1.9 uH = 54.91 mH
Maximum excitation current allowed by magnetic core:
I
max = B
max · A
min / (N · AL) = 0.2 · 49 · 10
-6 / (170 · 1.9 · 10
-6) = 30.34 mA
Maximum primary supply voltage:
V
rms_max = I
max · sqrt(2) · π · f · L = 0.03034 · 4.44 · 20000 · 0.05491 = 148 Volts
Power constant:Now, if I change the number of turns of the primary by half and the wire section by double (for example, by using double parallel wires to make the winding) what is achieved is to double the maximum admissible current and halve the maximum admissible voltage.
Here again we can see a constant which is the power that the transformer can handle with this core at calculated frequency:
P = I · V = 148 · 0.982 = 145 Watt (Constant for this core @ f = 20kHz)