Channel slim dsc power supplies – Rainbow Electronics MAX1585 User Manual
Page 22
MAX1584/MAX1585
Significant MOSFET selection parameters are as fol-
lows:
• On-resistance (R
DS(ON)
)
• Maximum drain-to-source voltage (V
DS(MAX)
)
• Total gate charge (Q
G
)
• Reverse transfer capacitance (C
RSS
)
DL1 and DL3 swing between PVSU and GND. DL2
swings between INDL2 and GND. Use a MOSFET with
on-resistance specified at or below the DL_ drive volt-
age. The gate charge, Q
G
, includes all capacitance
associated with charging the gate and helps to predict
MOSFET transition time between on and off states.
MOSFET power dissipation is a combination of on-
resistance and transition losses. The on-resistance loss
is as follows:
P
RDSON
= D x I
L
2
x R
DS(ON)
where D is the duty cycle, I
L
is the average inductor
current, and R
DS(ON)
is the MOSFET on-resistance. The
transition loss is approximately:
P
TRANS
= (V
OUT
x I
L
x f
OSC
x t
T
) / 3
where V
OUT
is the output voltage, I
L
is the average
inductor current, f
OSC
is the switching frequency, and
t
T
is the transition time. The transition time is approxi-
mately Q
G
/ I
G
, where Q
G
is the total gate charge, and
I
G
is the gate-drive current (0.5A typ). The total power
dissipation in the MOSFET is as follows:
P
MOSFET
= P
RDSON
+ P
TRANS
Diode
For most AUX applications, a Schottky diode rectifies
the output voltage. Schottky low forward voltage and
fast recovery time provide the best performance in
most applications. Silicon signal diodes (such as
1N4148) are sometimes adequate in low-current
(<10mA), high-voltage (>10V) output circuits where the
output voltage is large compared to the diode forward
voltage.
AUX Compensation
The auxiliary controllers employ voltage-mode control
to regulate their output voltage. Optimum compensa-
tion depends on whether the design uses continuous or
discontinuous inductor current.
AUX Step-Up, Discontinuous Inductor Current
When the inductor current falls to zero on each switch-
ing cycle, it is described as discontinuous. The inductor
is not utilized as efficiently as with continuous current,
but in light-load applications, this often has little nega-
tive impact since the coil losses may already be low
compared to other losses. A benefit of discontinuous
inductor current is more flexible loop compensation, and
no maximum duty-cycle restriction on boost ratio.
To ensure discontinuous operation, the inductor must
have a sufficiently low inductance to fully discharge on
each cycle. This occurs when:
L < [V
IN
2
(V
OUT
- V
IN
) / V
OUT
3
] [R
LOAD
/ (2f
OSC
)]
A discontinuous current boost has a single pole at the
following:
F
P
= (2V
OUT
- V
IN
) / (2
π x R
LOAD
x C
OUT
x V
OUT
)
Choose the integrator cap so the unity-gain crossover,
f
C
, occurs at f
OSC
/ 10 or lower. For many AUX circuits,
such as those powering motors, LEDs, or other loads
that do not require fast transient response, it is often
acceptable to overcompensate by setting f
C
at f
OSC
/
20 or lower.
C
C
is then determined by the following:
C
C
= [2V
OUT
x V
IN
/ ((2V
OUT
- V
IN
) x V
RAMP
)] [V
OUT
/
(K(V
OUT
- V
IN
))]
1/2
[(V
FB
/ V
OUT
)(g
M
/ (2
π x f
C
))]
where:
K = 2L x f
OSC
/ R
LOAD
and V
RAMP
is the internal voltage ramp of 1.25V.
The C
C
R
C
zero is then used to cancel the f
P
pole, so:
R
C
= R
LOAD
x C
OUT
x V
OUT
/ [(2V
OUT
- V
IN
) x C
C
]
AUX Step-Up, Continuous Inductor Current
Continuous inductor current can sometimes improve
boost efficiency by lowering the ratio between peak
inductor current and output current. It does this at the
expense of a larger inductance value that requires larg-
er size for a given current rating. With continuous
inductor-current boost operation, there is a right-half-
plane zero, Z
RHP
, at the following:
Z
RHP
= (1 - D)
2
R
LOAD
/ (2
π x L)
where (1 - D) = V
IN
/ V
OUT
(in a boost converter)
There is a complex pole pair at the following:
f
0
= V
OUT
/ [2
π x V
IN
(L x C
OUT
)
1/2
]
If the zero due to the output capacitor capacitance and
ESR is less than 1/10 the right-half-plane zero:
Z
COUT
= 1 / (2
π x C
OUT
x R
ESR
) < Z
RHP
/ 10
Then choose C
C
so the crossover frequency f
C
occurs
at Z
COUT
. The ESR zero provides a phase boost at
crossover:
C
C
= (V
IN
/ V
RAMP
)(V
FB
/ V
OUT
)(g
M
/ (2
π x Z
COUT
))
Choose R
C
to place the integrator zero, 1 / (2
π x R
C
x
C
C
), at f
0
to cancel one of the pole pairs:
R
C
= V
IN
(L x C
OUT
)
1/2
/ (V
OUT
x C
C
)
5-Channel Slim DSC Power Supplies
22
______________________________________________________________________________________