NCP1010, NCP1011, NCP1012, NCP1013, NCP1014
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18
The Flyback transfer formula dictates that:
Pout
1
(eq. 19)
which, by extracting
Ip and plugging into Equation 19, leads to:
Tsw
Lp
2 · Pout
1
Vin
1
N · (Vout
Vf)
(eq. 20)
Extracting Lp from Equation 20 gives:
Lpcritical
(Vin · Vr)2·
2 · Fsw · [Pout · (Vr2
2 · Vr · Vin
Vin2)]
(eq. 21)
, with Vr = N . (Vout + Vf) and the efficiency.
If Lp critical gives the inductance value above which
DCM operation is lost, there is another expression we can
write to connect Lp, the primary peak current bounded by
the NCP101X and the maximum dutycycle that needs to
stay below 50%:
DCmax · Vinmin · Tsw
Ipmax
corresponds to the lowest rectified bulk voltage, hence the
longest ton duration or largest dutycycle. Ip max is the
available peak current from the considered part, e.g. 350 mA
typical for the NCP1013 (however, the minimum value of
this parameter shall be considered for reliable evaluation).
Combining Equations 21 and 22 gives the maximum
theoretical power you can pass respecting the peak current
capability of the NCP101X, the maximum dutycycle and
the discontinuous mode operation:
Tsw2· Vinmin2· Vr2·
Fsw
(2 · Lpmax · Vr2
4 · Lpmax · Vr · Vinmin
2 · Lpmax · Vinmin2)
From Equation 22 we obtain the operating dutycycle
Ip · Lp
Vin · Tsw
current circulating in the MOSFET:
d
3
obtain the average dissipation in the MOSFET:
1
losses shall be added.
If we stick to Equation 23, compute Lp and follow the
above calculations, we will discover that a power supply
built with the NCP101X and operating from a 100 Vac line
minimum will not be able to deliver more than 7.0 W
continuous, regardless of the selected switching frequency
(however the transformer core size will go down as
Fswitching is increased). This number increases
significantly when operated from a single European mains
(18 W). Application note AND8125/D, “Evaluating the
Power Capability of the NCP101X Members” details how
to assess the available power budget from all the NCP101X
series.
Lpmax
(eq. 22)
where Vinmin
Pmax :
·
(eq. 23)
d
(eq. 24)
which lets us calculate the RMS
IdRMS
Ip ·
(eq. 25)
. From this equation, we
Pavg
(eq. 26)
to which switching
Example 1. A 12 V 7.0 W SMPS operating on a large
mains with NCP101X:
Vin = 100 Vac to 250 Vac or 140 Vdc to 350 Vdc once
rectified, assuming a low bulk ripple
Efficiency = 80%
Vout = 12 V, Iout = 580 mA
Fswitching = 65 kHz
Ip max = 350 mA – 10% = 315 mA
Applying the above equations leads to:
Selected maximum reflected voltage = 120 V
with Vout = 12 V, secondary drop = 0.5 V
→
Np:Ns = 1:0.1
Lp critical = 3.2 mH
Ip = 292 mA
Dutycycle worse case = 50%
Idrain RMS = 119 mA
P
MOSFET
= 354 mW at R
DSon
= 24 (T
J
> 100
°
C)
P
DSS
= 1.1 mA x 350 V = 385 mW, if DSS is used
Secondary diode voltage stress = (350 x 0.1) + 12 = 47 V
(e.g. a MBRS360T3, 3.0 A/60 V would fit)
Example 2. A 12 V 16 W SMPS operating on narrow
European mains with NCP101X:
Vin = 230 Vac
15%, 276 Vdc for Vin min to 370 Vdc
once rectified
Efficiency = 80%
Vout = 12 V, Iout = 1.25 A
Fswitching = 65 kHz
Ip max = 350 mA – 10% = 315 mA
Applying the equations leads to:
Selected maximum reflected voltage = 250 V
with Vout = 12 V, secondary drop = 0.5 V
→
Np:Ns = 1:0.05
Lp = 6.6 mH
Ip = 0.305 mA
Dutycycle worse case = 0.47
Idrain RMS = 121 mA
P
MOSFET
= 368 mW at R
DSon
= 24 (T
J
> 100
°
C)
P
DSS
= 1.1 mA x 370 V = 407 mW, if DSS is used below an
ambient of 50
°
C.
Secondary diode voltage stress = (370 x 0.05) + 12 = 30.5 V
(e.g. a MBRS340T3, 3.0 A/40 V)
Please note that these calculations assume a flat DC rail
whereas a 10 ms ripple naturally affects the final voltage
available on the transformer end. Once the Bulk capacitor has
been selected, one should check that the resulting ripple (min
Vbulk) is still compatible with the above calculations. As an
example, to benefit from the largest operating range, a 7.0 W
board was built with a 47 F bulk capacitor which ensured
discontinuous operation even in the ripple minimum waves.