AND8112 Datasheet, PDF (3/12 Pages) ON Semiconductor – A Quasi-Resonant SPICE Model Eases Feedback Loop Designs

AND8112/D
Averaging Input / Output Voltages
From the inductor volt−second balance approximation, we know that the average voltage across an inductor operated in a
steady−state converter is null. By looking at the V(L_P) sketch, we obtain the following equation:
t V(LP) u+ d(t)
t Vg(t) u * ^dȀ(t)
t
N
V(t)
u
+
d(t)
t Vg(t) u * ^{(1 * d(t))} _N ^{t V(t) u} + 0
(eq. 8)
which lets us extract the classical output / input voltage ratio
t V(t) u
t Vg(t) u
+
N
d(t)
(1 * d(t))
(eq. 9)
and as a result, the duty−cycle expression:
d(t)
+
t
V(t)
t V(t)
u)N
u
t
Vg(t)
u
(eq. 10)
Now, by plugging equation 10 in equation 6, we obtain the average voltage across the primary switch terminal: <V_1(t)> =
ǒ Ǔ t V(t) u ) N t Vg(t) u
N
(1
*
d(t))
+
t
V(t)
u
)
N
N
t Vg(t) u
1
*
t
V(t)
t V(t)
u)N
u
t
Vg(t)
u
(eq. 11)
+t Vg(t) u
which agrees with the inductor volt−second balance approximation (from Figure 1 since, by definition, <V(L_P)> = 0, then V_g
appears across the switch terminals).
To reveal <V_2(t)>, let us plug equation 10 into 7: <V_2(t)> =
[t V(t) u ) N t Vg(t) u]
d(t) + [t V(t) u ) N t Vg(t) u]
t
V(t)
t V(t)
u)N
u
t
Vg(t)
u
+t
V(t)
u
(eq. 12)
which again could be deduced from Figure 3 since the average voltage across the secondary inductance is zero…
Averaging Input / Output Currents
The peak inductor current depends on the time during
which V_g is applied over L_P. If we recall that this time
(actually t_on) is d x T_S, then:
IP
+
Vg
LP
d
TS
(eq. 13)
From Figure 4, the average current <I_1(t)> can be obtained
by evaluating the triangular area (charge in Coul_omb) and
dividing by the switching period. This is expressed by
equation 4. Now plugging equation 13 in 4, we obtain:
t
I1(t)
u+
1
2
t Vg(t) u
LP
t Vg(t) u
d(t) TS d(t) +
d(t)2 TS
(eq. 14)
2 LP
however, from equation 11, we know that <V_1(t)> = <V_g(t)>
thus equation 14 turns into:
t
I1(t)
u+
t
V1(t)
u
2
d(t)2
LP
TS
(eq. 15)
Applying the same technique to the secondary current
I_2(t), leads to:
ŕ^TS
t
I2(t)
u+
1
TS
I2(t)
@
dt
+
1
2
IP
N
dȀ_(t)
d.TS
(eq. 16)
plugging equation 13 in 16 leads to:
t I2(t) u+ ^{t Vg(t) u}
d(t)
2N
(1 * d(t))
LP
TS ₊
(eq. 17)
t V1(t) u d(t) (1 * d(t)) TS
2 N LP
A 100% Efficiency Power Transfer…
Assuming that 100% of the primary stored energy is
released to the secondary side, then we can use equations 11
and 12 to write:
<P_(t)> = <V_1(t)> X <I_1(t)> = <V_2(t)> X <I_2(t)> (eq. 18)
From equation 15, we can see that a current is generated
by a voltage multiplied by a term. This term is obviously
homogenous to the inverse of an impedance. By
re−arranging equation 15, we obtain:
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