MAX1565 Datasheet, PDF (21/26 Pages) Maxim Integrated Products – Small, High-Efficiency, Five-Channel Digital Still Camera Power Supply

Small, High-Efficiency, Five-Channel
Digital Still Camera Power Supply
Discontinuous Inductor Current
When the inductor current falls to zero on each switching
cycle, it is described as discontinuous. The inductor is
not utilized as efficiently as with continuous current. This
often has little negative impact in light-load applications
since the coil losses may already be low compared to
other losses. A benefit of discontinuous inductor cur-
rent is more flexible loop compensation and no maxi-
mum 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_IN² (V_OUT - V_IN)/V_OUT3] [R_LOAD/(2 f_OSC)]
A discontinuous current boost has a single pole at:
f_P = (2V_OUT - V_IN)/(2π R_LOADC_OUTV_OUT)
Choose the integrator capacitor such that the unity-gain
crossover (f_C) occurs at f_OSC/10 or lower. Note that for
many auxiliary 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:
C_C = [2V_OUTV_IN /((2V_OUT - V_IN)V_RAMP)]
[V_OUT /(K(V_OUT - V_IN))]^1/2 [(V_FB/V_OUT)
(g_M /(2π f_C))]
where K = 2 L f_OSC/R_LOAD, and V_RAMP is the internal
slope compensation voltage ramp of 1.25V. The C_CR_C
zero is then used to cancel the f_P pole, so:
R_C = R_LOADC_OUTV_OUT/[(2V_OUT - V_IN) C_C]
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 larger
size for a given current rating. With continuous inductor
current boost operation, there is a right-plane zero at:
f_RHPZ = (1 - D)² R_LOAD /(2πL)
where (1 - D) = V_IN/V_OUT (in a boost converter). A com-
plex pole pair is located at:
f₀ = V_OUT/[2π V_IN (L C_OUT)^1/2]
If the zero due to the output capacitor capacitance and
ESR is less than 1/10 the right-plane zero:
Z_COUT = 1/(2π C_OUT R_ESR) < f_RHPZ/10
Choose C_C such that 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 /(2π Z_COUT))
Choose R_C to place the integrator zero, 1/(2π R_CC_C), at
f₀ to cancel one of the pole pairs:
R_C = V_IN (L C_OUT)^1/2/(V_OUT C_C)
If Z_COUT is not less than f_RHPZ/10 (as is typical with
ceramic output capacitors) and continuous conduction is
required, then cross the loop over before f_RHPZ and f₀:
f_C < f₀/10, and f_C < f_RHPZ/10
In that case:
C_C = (V_IN/V_RAMP)(V_FB/V_OUT)(g_M /(2π f_C))
Place 1/(2π R_CC_C) = 1/(2π R_LOADC_OUT), so that R_C =
R_LOAD C_OUT/C_C or reduce the inductor value for dis-
continuous operation.
Applications Information
LED, LCD, and Other Boost Applications
Any auxiliary channel can be used for a wide variety of
step-up applications. These include generating 5V or
some other voltage for motor or actuator drive, generating
15V or a similar voltage for LCD bias, or generating a
step-up current source to efficiently drive a series array
of white LEDs for display backlighting. Figures 5 and 6
show examples of these applications.
TO V_BATT
4.7µF
4.7µH
15V
100mA
22µF
1.1MΩ
OUTSU
DL_
FB_
AUX_
PWM
100kΩ
MAX1565
(PARTIAL)
Figure 5. Using an AUX_ Controller Channel to Generate LCD
Bias
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