MAX1566-MAX1567 Datasheet, PDF (28/35 Pages) Maxim Integrated Products – Six-Channel, High-Efficiency, Digital Camera Power Supplies

Six-Channel, High-Efficiency, Digital
Camera Power Supplies
The inductor is typically selected to operate with contin-
uous current for best efficiency. An exception might be
if the step-up ratio, (V_OUT / V_IN), is greater than 1 / (1 -
D_MAX), where D_MAX is the maximum PWM duty factor
of 80%.
When using the step-up channel to boost from a low
input voltage, loaded startup is aided by connecting a
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 / (2π 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 = 10Ω. If we
select L = 4.7µH, then:
f_RHPZ = 5 (2.5 / 5)² / (2π 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 / 2π 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 = 69.4kΩ
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
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