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MAX1584 Datasheet, PDF (20/29 Pages) Maxim Integrated Products – 5-Channel Slim DSC Power Supplies
5-Channel Slim DSC Power Supplies
IIND(PK) = 1.25 x IOUT / (1 - D)
Inductance values smaller than LIDEAL can be used to
reduce inductor size; however, if much smaller values are
used, inductor current rises and a larger output capaci-
tance might 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 cho-
sen to optimize control-loop stability. In some cases, it
helps to readjust the inductor or output capacitor value
to get optimum results. For typical designs, the compo-
nent 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 (fC) much less
than that of the right-half-plane zero.
The relevant characteristics for step-up channel com-
pensation are as follows:
• Transconductance (from FBSU to CCSU), gMEA
(135µS)
• Current-sense amplifier transresistance, RCS
(0.3V/A)
• Feedback regulation voltage, VFB (1.25V)
• Step-up output voltage, VSU, in V
• Output load equivalent resistance, RLOAD, in
Ω = VSUOUT / ILOAD
The key steps for step-up compensation are as follows:
1) Place fC sufficiently below the right-half-plane zero
(RHPZ) and calculate CC.
2) Select RC based on the allowed load-step transient.
RC sets a voltage delta on the CC pin that corre-
sponds to load-current step.
3) Calculate the output-filter capacitor (COUT) required
to allow the RC and CC selected.
4) Determine if CP is required (if calculated to be >10pF).
For continuous conduction, the right-half-plane zero fre-
quency (fRHPZ) is given by the following:
fRHPZ = VSUOUT (1 - D)2 / (2π x L x ILOAD)
where D = the duty cycle = 1 - (VIN / VOUT), L is the
inductor value, and ILOAD is the maximum output cur-
rent. Typically, target crossover (fC) for 1/6 of the
RHPZ. For example, if we assume fOSC = 500kHz, VIN
= 2.5V, VOUT = 5V, and IOUT = 0.5A, then RLOAD =
10Ω. If we select L = 4.7µH, then:
fRHPZ = 5 (2.5 / 5)2 / (2π x 4.7 x 10-6 x 0.5) = 84.65kHz
Choose fC = 14kHz. Calculate CC:
CC = (VFB / VOUT)(RLOAD / RCS)(gM / 2π x fC)(1 - D)
= (1.25 / 5)(10 / 0.3) x (135µS / (6.28 x 14kHz) (2/5)
= 6.4nF
Choose 6.8nF.
Now select RC 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 RC to provide transient gain. Since the
current-sense transresistance is 0.3V/A, the value of RC
that allows the required load step swing is as follows:
RC = 0.3 IIND(PK) / 6.75µA
In a step-up DC-DC converter, if LIDEAL is used, output
current relates to inductor current by:
IIND(PK) = 1.25 x IOUT / (1 - D) = 1.25 x IOUT x VOUT /
VIN
So for a 500mA output load step with VIN = 2.5V and
VOUT = 5V:
RC = [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 COUT
RLOAD pole cancels the RC CC zero:
COUT x RLOAD = RC x CC
For the example:
COUT = 68kΩ x 6.8nF / 10Ω = 46µF
Choose 47µF for COUT. If the available COUT is sub-
stantially different from the calculated value, insert the
available COUT value into the above equation and
recalculate RC. Higher substituted COUT values allow a
higher RC, which provides higher transient gain and
consequently less transient droop.
If the output filter capacitor has significant ESR, a zero
occurs at the following:
ZESR = 1 / (2π x COUT x RESR)
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