Rainbow Electronics MAX1567 User Manual
Page 28
MAX1566/MAX1567
Six-Channel, High-Efficiency, Digital
Camera Power Supplies
28
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Schottky diode from the battery to PVSU. See the
Minimum Startup Voltage vs. Load Current graph in the
Typical Operating Characteristics.
Step-Up Inductor
In most step-up designs, a reasonable inductor value
(L
IDEAL
) can be derived from the following equation,
which sets continuous peak-to-peak inductor current at
1/2 the DC inductor current:
L
IDEAL
= [2V
IN(MAX)
x D(1 - D)] / (I
OUT
x f
OSC
)
where D is the duty factor given by:
D = 1 - (V
IN
/ V
OUT
)
Given L
IDEAL
, the consistent peak-to-peak inductor cur-
rent is 0.5 I
OUT
/ (1 - D). The peak inductor current,
I
IND(PK)
= 1.25 I
OUT
/ (1 - D).
Inductance values smaller than L
IDEAL
can be used to
reduce inductor size; however, if much smaller values
are used, inductor current rises and a larger output
capacitance may be required to suppress output ripple.
Step-Up Compensation
The inductor and output capacitor are usually chosen
first in consideration of performance, size, and cost. The
compensation resistor and capacitor are then chosen to
optimize control-loop stability. In some cases, it may
help to readjust the inductor or output-capacitor value to
get optimum results. For typical designs, the component
values in the circuit of Figure 1 yield good results.
The step-up converter employs current-mode control,
thereby simplifying the control-loop compensation.
When the converter operates with continuous inductor
current (typically the case), a right-half-plane zero
appears in the loop-gain frequency response. To
ensure stability, the control-loop gain should cross over
(drop below unity gain) at a frequency (f
C
) much less
than that of the right-half-plane zero.
The relevant characteristics for step-up channel com-
pensation are as follows:
• Transconductance (from FB to CC), gm
EA
(135µS)
• Current-sense amplifier transresistance, R
CS
(0.3V/A)
• Feedback regulation voltage, V
FB
(1.25V)
• Step-up output voltage, V
SU
, in V
• Output load equivalent resistance, R
LOAD
, in
Ω =
V
OUT
/ I
LOAD
The key steps for step-up compensation are as follows:
1) Place f
C
sufficiently below the right-half-plane zero
(RHPZ) and calculate C
C
.
2) Select R
C
based on the allowed load-step transient.
R
C
sets a voltage delta on the C
C
pin that corre-
sponds to load-current step.
3) Calculate the output-filter capacitor (C
OUT
) required
to allow the R
C
and C
C
selected.
4) Determine if C
P
is required (if calculated to be
>10pF).
For continuous conduction, the right-half-plane zero fre-
quency (f
RHPZ
) is given by the following:
f
RHPZ
= V
OUT
(1 - D)
2
/ (2
π x L x I
LOAD
)
where D = the duty cycle = 1 - (V
IN
/ V
OUT
), L is the
inductor value, and I
LOAD
is the maximum output cur-
rent. Typically target crossover (f
C
) for 1/6 of the RHPZ.
For example, if we assume f
OSC
= 500kHz, V
IN
= 2.5V,
V
OUT
= 5V, and I
OUT
= 0.5A, then R
LOAD
= 10
Ω. If we
select L = 4.7µH, then:
f
RHPZ
= 5 (2.5 / 5)
2
/ (2
π x 4.7 x 10
-6
x 0.5) = 84.65kHz
Choose f
C
= 14kHz. Calculate C
C
:
C
C
= (V
FB
/ V
OUT
)(R
LOAD
/ R
CS
)(gm / 2
π x f
C
)(1 - D)
= (1.25 / 5)(10 / 0.3) x [135µS / (6.28 x 14kHz)] (2/5)
= 6.4nF
Choose 6.8nF.
Now select R
C
so transient-droop requirements are
met. As an example, if 4% transient droop is allowed,
the input to the error amplifier moves 0.04 x 1.25V, or
50mV. The error-amp output drives 50mV x 135µS, or
6.75µA, across R
C
to provide transient gain. Since the
current-sense transresistance is 0.3V/A, the value of R
C
that allows the required load-step swing is as follows:
R
C
= 0.3 I
IND(PK)
/ 6.75µA
In a step-up DC-to-DC converter, if L
IDEAL
is used, out-
put current relates to inductor current by:
I
IND(PK)
= 1.25 I
OUT
/ (1 - D) = 1.25 I
OUT
x V
OUT
/ V
IN
So, for a 500mA output load step with V
IN
= 2.5V and
V
OUT
= 5V:
R
C
= [1.25(0.3 x 0.5 x 5) / 2)] / 6.75µA = 69.4k
Ω
Note that the inductor does not limit the response in this
case since it can ramp at 2.5V / 4.7µH, or 530mA/µs.
The output filter capacitor is then chosen so the C
OUT
R
LOAD
pole cancels the R
C
C
C
zero:
C
OUT
x R
LOAD
= R
C
x C
C
For the example:
C
OUT
= 68k
Ω x 6.8nF / 10Ω = 46µF
Choose 47µF for C
OUT
. If the available C
OUT
is sub-
stantially different from the calculated value, insert the
available C
OUT
value into the above equation and