Design Realisation - Continued
8. LV Side
In this section, we will derive the requirements for the low voltage side, starting secondary losses.
8.1 Secondary Losses
Two lots losses exist on the secondary side: synchronous rectifier section losses and magnetic losses in the transformer and LC section.
We start with the synchronous rectifier section.
8.1.1 Synchronous Rectifier Losses
We have said previously that a Synchronous Rectification scheme will be used. This is to reduce losses in the secondary side, maintain high efficiency and reduce heat generated. We will be using a bridge rectifier composed of 4 synchronous rectifiers.
In a bridge full-wave rectification setup, which we're using in this design, each rectifier will bear half the total output current I[SUB]o[/SUB].
Thus, MOSFET transistors need to be chosen to bear at least that much current, in addition to safety a margin of say 15 - 20%, taking account of turn-on, turn-off and conduction losses.
In our case, each synchronous rectifier MOSFET needs to be at least rated at an I[SUB]D[/SUB] of 300A.
For this work, the IRFB7434PBf from International Rectifier is chosen. Brief specs of this device:
I[sub]D[/sub]: 317A @ V[sub]DSS[/sub] = 40V and V[sub]GS[/sub] = 10V, measured at T[sub]c[/sub] = 25 ℃
R[sub]DS,on[/sub]: 1.25mΩ - 1.65mΩ
t[sub]r[/sub] = t[sub]f[/sub] = 68ns
Reverse Recovery Charge, Q[sub]rr[/sub] = 50 nC measured at reference current, voltage and temperature
To calculate the turn-on (P[SUB]on[/SUB]) and turn-off (P[SUB]off[/SUB]) losses, we will rely on the following formulae:
P[sub]on[/sub]
= (V[sub]d[/sub]I[sub]o[/sub]((t[sub]r[/sub] + t[sub]f[/sub])/2) + Q[sub]rr[/sub]V[sub]d[/sub] + ((Q[sub]rr[/sub]V[sub]d[/sub])/4))f[SUB]s[/SUB]
= ((12 . 300)((68 x 10[sup]-9[/sup] + 68 x 10[sup]-9[/sup]/2) + 50 x 10[sup]-9[/sup] + ((50 x 10[sup]-9[/sup]) . 12)/4)) . 100 x 10[sup]3[/sup]
= (0.0002448 + 50 x 10[sup]-9[/sup] + 15 x 10[sup]-10) x 100 x 10[sup]-10[/sup]
= 0.0002448515 x 100000
≈ 25W
P[sub]off[/sub] is simpler, expressed as:
= (V[sub]d[/sub]I[sub]o[/sub]((t[sub]r[/sub] + t[sub]f[/sub])/2) . f[sub]s[/sub]
= (12 . 300)((68 x 10[sup]-9[/sup] + 68 x 10[sup]-9[/sup]/2) . 100 x 10[sup]3[/sup]
= 0.0002448 x 100,000
≈ 25W
Conduction losses are derived as follows, ignoring Duty Cycle (D), since we are at the LV side:
We will assume R[sub]DS,on[/sub] to be assumed to be the maximum of 1.65mΩ and I[sub]DS[/sub] = I[sub]o[/sub] = 520A
Conduction losses, P[sub]c[/sub] can be approximated as:
P[sub]c[/sub] = R[sub]DS,on[/sub] . (I[sub]o[/sub]/2)[sup]2[/sup]
= 1.68 x 10[sup]-3[/sup] . 520
≈ 1W
∴ Total Synchronous Rectifier Losses = (P[sub]on[/sub] + P[sub]off[/sub] + P[sub]c[/sub]) x 4
= (25W + 25W + 1) x 4
= 104W
8.1.2 Magnetic Losses
As stated earlier, magnetic losses are composed of losses in the transformer core, added to losses in the inductor core of the LC section inductor.
8.1.2.1 Transformer Configuration
Primary side RMS current (I[sub]1,RMS[/sub]) is:
Assuming a 20:1 secondary to primary ratio:
I[sub]1,RMS[/sub] = (n[sub]2[/sub]/n[sub]1[/sub]) . I[sub]o,max . sqrt (D)
= 1/20 . 520 . sqrt (0.32)
= 14.7A
I[sub]1,RMS[/sub] ≈ 15A
Secondary side RMS current (I[sub]2,RMS[/sub]) is:
I[sub]2,RMS[/sub] = 0.5 . I[sub]o,max[/sub] . sqrt (1 + D)
= 0.5 . 520 . 1.5
I[sub]2,RMS[/sub] = 390A
Now to the Litz bundle. Recall we are aiming at a current density of J = 3A/mm[sup]2[/sup] and Litz wire with a cross section area A[sub]c,Litz[/sub] = 0.94mm[sup]2[/sup]
Primary Bundle:
X[sub]prim,INT bundled wires[/sub] = I[sub]1,RMS[/sub] / (J . A[sub]c,Litz[/sub])
= 15 / (3 . 0.94)
= 5.319
∴ X[sub]prim,INT bundled wires[/sub] ≈ 6
∴ New cross section area of primary bundle = 6 x 0.94
A[sub]c1,bundle[/sub]= 5.64mm[sup]2[/sup]
Secondary Bundle:
Similarly, for the secondary bundle, X[sub]sec,INT bundled wires[/sub]:
X[sub]sec,INT bundled wires[/sub] = 390 / (3 . 0.94)
= 138.297
∴ X[sub]sec,INT bundled wires[/sub] ≈ 139
∴ New cross section area of secondary bundle = 139 x 0.94
A[sub]c2,bundle[/sub]= 130.66mm[sup]2[/sup]
Now to calculate the window area, ∆A[sub]w[/sub], needed to satisfy the above bundle dimensions, recalling our n=20 secondary-to-primary ratio and a cooper filling factor k[sub]cu,1[/sub] = k[sub]cu,2[/sub] = 0.5:
∆A[sub]w[/sub] = (A[sub]c2,bundle[/sub] + nA[sub]c1,bundle[/sub]) / k[sub]cu,1
Thus, the area can be expressed as a function of the number of primary turns, N[sub]1[/sub] as follows:
A[sub]w[/sub] = N[sub]2[/sub]∆A[sub]w[/sub]
= (N1 x ∆A[sub]w[/sub]) / n
Research has indicated that core material 3C91 from Ferroxcube is feasible, having the following characteristics:
Material: 3C91
Saturation flux density (T): 0.47
Density (g/cm[sup]3[/sup]): 4.8
Curie Temp (℃): 220
(To be continued)