7
UCC2585
UCC3585
ESR
V
I
OUT
OUT
===
∆
∆
Ω
0 018
05
0 026
.
.
.
A 220µF, 6.3V Sprague 594D capacitor has an ESR of
75mΩ. Three of these in parallel will result in an overall
ESR of 25mΩ. (C9, C10, and C11 in Fig. 1). Since the
output ripple current is so low, the capacitor’s ripple cur
-
rent rating of 1.45A is not a concern.
To check the assumption that the capacitor’s impedance
at the switching frequency is dominated by the ESR and
not the capacitor’s capacitance value, calculate the im
-
pedance and compare it to the ESR.
Z
FC k
mC
S
=
••=••
=
1
2
1
2 350 220
2
ππ µ
Ω
The ESR of the capacitor is 37 times that of the imped
ance of the capacitor at the switching frequency, so the
earlier assumption was valid.
4) Before selecting the switching MOSFETs, the current
that will be flowing through them must first be determined.
II
I
A
DOUT
OUT
PK
==+
∆
2
38.
The RMS of this current in Q1 is
IQ I A
DRMS D
PK
128==δ .
And in Q2
IQ I A
DRMSD
PK
2125=−=δ .
5) Since this regulator must be able to operate from a
3.3V source, the MOSFETs used must have a gate
threshold level of no more than 2V.
For Q1, an IRF7404 is selected. It has an R
DS(on)
of
0.04Ω, a total gate charge (Q
G
1) of 50nC, and a turn
OFF (t
OFF
1) time of 65ns. The conduction loss in Q1 will
be:
PQ IQ R Q W
DONDRMS DS
ON
1 1 1 0 593
2
=•=.
The gate drive losses will be
PQ Q V F mW
DGATEG IN S
158
1
=••=
And finally the turn OFF losses are estimated
PQ V IQ T F W
D OFF IN D PK OFF S
1
1
2
1014
1
=• • • • =.
The total power loss for Q1 is the sum of these three:
PQ W
DTOTAL
105= .
6) Q2 has been selected to be an IRF7401, which has an
R
DS(ON)
of 0.03Ω, and a total gate charge (QG2) of 48nC
and a body diode turn OFF switching time (t
OFF
2) of
59ns. In this topology, the N Channel MOSFET, Q2, is
turned OFF prior to the turn ON of Q1, so when Q2 is
turned OFF, current is being re-routed from the channel
of the device into the intrinsic body diode. Therefore Q2’s
intrinsic body diode incurs switching loss during the turn
OFF interval.
The conduction loss in Q2 is:
PQ IQ R Q W
DOND RMS DS
ON
22 202
2
=•=.
The gate drive losses will be
PQ Q V F mW
DGATE G INS
255
2
=••=
And the body diode turn OFF loss:
PQ V I T F W
D D OFF IN D OFF S
PK
2
1
2
0132_.=• • • • =
The total power loss for Q2 is the sum of these three:
PQ W
DTOTAL
204= .
7) Thus far the power loss in the two MOSFETs and the
output inductor total 1.0W. The average input current is:
I
VIP
V
A
IN
OUT OUT LOSS
IN
AVG
=
•+
=22.
The peak to peak ripple in the input capacitors is the
peak current less the average input current during Q1’s
ON time, and equal to the average input current during
Q1’s OFF time.The RMS value of this current is then:
IIII A
IN CAP D IN IN
RMS PK AVG AVG
_
()()().=− •+ •−=
22
119δδ
8) After the input capacitor’s input ripple current is
known, select the input capacitors. Again, Sprague 594D
Solid Tantalum capacitors are chosen. A single 150µF,
10V capacitor has a ripple current rating of 1.35A RMS.
Two in parallel (C1 and C2) will have a combined capabil
ity of 2.7A, and a total ESR of 40mΩ. The losses in the
capacitors are:
P I ESR W
DIN CAP IN CAP
RMS
__
.=•=
2
014
Adding the capacitor loss to that previously found, the to
tal losses are now 2.1W.
9) The overall efficiency of the power train is then
E
VI
VI
FF
OUT OUT
OUT OUT
=
•
•+
=21084..
The losses are dominated by the MOSFETs Q1 and Q2.
One way to improve the efficiency would be to reduce the
conduction loss in Q1, either by choosing a device with a
APPLICATION INFORMATION (cont.)