ELEC6251 Assignment

Running head: ELEC6251 Assignment
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ELEC6251 Assignment
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ELEC6251 Assignment
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Forward converter
To get the desired output and ripple voltage to be 5V and 25mV respectively, the resistor had
to be increased 50k and the inductor 10mH, and the capacitor ratings of 10u. When the
MOSFET with the specified, the components changed to 2M R1, 1pF C1, and 50mH for L1.
The switching frequency is determined by closing the switch, and the choice is based on the
compromise that the quality that the power supply has with respect to the load and the
switching losses that is fed into the components (often has a direct proportionality to the
switching frequency).
The timing resistor gives a constant off-time of 6μs, and we have chosen around 80% duty
cycle, the on-time duty cycle gave 20μs, therefore the switching frequency came to 40kHz.
The change of value was discovered to be the supply voltage and the timing resistor, it was
changed. These parameters changed the duty cycle of the MOSFET we had in the circuit.
In calculations are shown below:

ELEC6251 Assignment
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Specification of input power
We must consider the line frequency (fL):
Parameters to be considered, estimated frequency (Eff) and max power output (Po)
Eff can be set to be a range of 0.7 to 0.75, when the voltage is low, and when the voltage is
high the range is 0.8 to 0.85. the MOSFET has a voltage stress that is twice the voltage input
P¿= Po
Eff
... ... ... ... ... ... .... (1)
Determine the reset method for the transformer and the max. duty ratio
One original limitation that the forward converter has is that there must be a reset of the
transformer when the MOSFET is at off state. This determination can be made by first,
getting the auxiliary winding reset, which the circuit diagram has, the schematic has some
advantages in terms of frequency, because the stored energy in the inductor goes to the input.
V ds
max=V DC (1+ N p
Nr )... ... ... ... ... ... ... ... (2 )
Dmax ≤ N p
N p + N r
... ... ... ... ... ... ... ... . ... ... (3)
Where Nr and Np are the turns number for the reset and primary winding respectively. In
equation (2) and (3), the maximum voltage shown on MOSFET is reduced by reducing the
value of Dmax. But when Dmax results are reduced where on the secondary side voltage stress is
increased, means that Nr and NP are equal, therefore we would set Dmax to 0.45. From that we
can determine the secondary and primary turns, reset turns, and turns ratio.
The Ripple factor of the inductor current is obtained by:
K RF= ∆l
2 lO
... ... ... ... ... ... ... ... ... ... ... ... ... ... (4)
Where, lO is the maximum current output, when we do simulation, we get IO to be 2.5μA, the
∆ l is the gradient of the current and voltage curve. Ripple factor shall be obtained and will
aid in the attainment of the peak current and the rms value of the MOSFET,
I ds
peak =I EDC ( 1+ K RF ) ... ... ... ... ... ... ... ... ..... ( 5 )
I ds
rms=IEDC √ (3+ K RF
2
) Dmax
3 ... ... ... ... ... ... ... ..(6)
Were;
I EDC= P¿
V DC
min ∙ Dmax
... ... ... ... ... ... ... ... ... ... .. ( 7 )

ELEC6251 Assignment
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Transformer windings
The first step to this is to obtain the turns ratio, to be the reference point.
n= N p
N sl
= V DC
min ∙ Dmax
V o 1+V F 1
... ... ... ... ... ... ... ... ... . ( 8 )
Where, Ns1 and NP are the turns number of the reference output and the primary side
respectively. V F 1 is the forward voltage drop of the diode, andV o 1 is the output voltage of the
point of reference.
After which, we obtain the integer numbers appropriate for Nsl for the value of N p achieved
be larger than N pmin which is gotten from (9) below:
N p
min= V DC
min ∙ Dmax
Ae ∙ f s ∙ ∆ B × 106 (turns)... ... ... ... ... ... ... ... ...(9)
The magnetizing inductance on the transformer’s primary side is obtained as follows:
Lm= AL × N p
2 × 10−9 ( H ) ... ... ... ... ... ... ... ..... ... .(10)
Where, AL is the value of AL without a gap measured in nH/tunrs2
The nth output turns are therefore obtained by
Ns ( n )= V o ( n ) +V F (n )
V o 1 +V F 1
∙ N s 1 ( turns ) ... ... ... ... ... ... ... ... . ( 11 )
Where, Vo(n) is the voltage output and VF(n) is the diode forward voltage drop that the nth
output has. These can be obtained from the simulation and the other previous calculation.