increased without limit. Refer to the Reference Designs sec-
tion for examples of input capacitors.
Output Capacitor
The output capacitor is required for loop stability (compensa-
tion) as well as transient response. During sudden changes
in load current demand, the output capacitor must source or
sink current during the time it takes the control loop of the
LP2975 to adjust the gate drive to the pass FET. As a general
rule, a larger output capacitor will improve both transient re-
sponse and phase margin (stability). The value of COUT may
be increased without limit.
OUTPUT CAPACITOR AND COMPENSATION: Loop com-
pensation for the LP2975 is derived from COUT and, in some
cases, the feed-forward capacitor CF (see next section).
COUT forms a pole (referred to as fp) in conjuction with the load
resistance which causes the loop gain to roll off (decrease) at
an additional −20 dB/decade. The frequency of the pole is:
fp = 0.16 / [ (RL + ESR) × COUT]
Where:
RL is the load resistance.
COUT is the value of the output capacitor.
ESR is the equivalent series resistance of COUT.
As a general guideline, the frequency of fp should be ≤ 200
Hz. It should be noted that higher load currents correspond to
lower values of RL, which requires that COUT be increased to
keep fp at a given frequency.
DESIGN EXAMPLE: Select the minimum required output ca-
pacitance for a design whose output specifications are 5V @
1A:
fp = 0.16 / [ (RL + ESR) × COUT]
Re-written:
COUT = 0.16 / [fp × (RL + ESR) ]
Values used for the calculation:
fp = 200 Hz, RL = 5Ω, ESR = 0.1Ω (assumed).
Solving for COUT, we get
157 μ
F (nearest standard size would
be 180 μF).
The ESR of the output capacitor is very important for stability,
as it creates a zero (fz) which cancels much of the phase shift
resulting from one of the poles present in the loop. The fre-
quency of the zero is calculated from:
fz = 0.16 / (ESR × COUT)
For best results in most designs, the frequency of fz should
fall between 5 kHz and 50 kHz. It must be noted that the val-
ues of COUT and ESR usually vary with temperature (severely
in the case of aluminum electrolytics), and this must be taken
into consideration.
For the design example (VOUT = 5V @ 1A), select a capacitor
which meets the fz requirements. Solving the equation for
ESR yields:
ESR = 0.16 / (fz × COUT)
Assuming fz = 5 kHz and 50 kHz, the limiting values of ESR
for the 180 μF capacitor are found to be:
18 mΩ ≤ ESR ≤ 0.18Ω
A good-quality, low-ESR capacitor type such as the Pana-
sonic HFQ is a good choice. However, the 10V/180 µF ca-
pacitor (#ECA-1AFQ181) has an ESR of 0.3Ω which is not in
the desired range.
To assure a stable design, some of the options are:
1) Use a different type capacitor which has a lower ESR such
as an organic-electrolyte OSCON.
2) Use a higher voltage capacitor. Since ESR is inversely
proportional to the physical size of the capacitor, a higher
voltage capacitor with the same C value will typically have a
lower ESR (because of the larger case size). In this example,
a Panasonic ECA-1EFQ181 (which is a 180 µF/25V part) has
an ESR of 0.17Ω and would meet the desired ESR range.
3) Use a feed-forward capacitor (see next section).
Feed-Forward Capacitor
Although not required in every application, the use of a feed-
forward capacitor (CF) can yield improvements in both phase
margin and transient response in most designs.
The added phase margin provided by CF can prevent oscil-
lations in cases where the required value of COUT and ESR
can not be easily obtained (see previous section).
CF can also reduce the phase shift due to the pole resulting
from the Gate capacitance, stabilizing applications where this
pole occurs at a low frequency (before cross-over) which
would cause oscillations if left uncompensated (see later sec-
tion GATE CAPACITANCE POLE FREQUENCY).
Even in a stable design, adding CF will typically provide more
optimal loop response (faster settling time). For these rea-
sons, the use of a feed-forward capacitor is always rec-
ommended.
CF is connected across the top resistor in the divider used to
set the output voltage (see Typical Application Circuit). This
forms a zero in the loop response (defined as fzf), whose fre-
quency is:
fzf = 6.6 × 10−6 / [CF × (VOUT / 1.24 − 1) ]
When solved for CF, the fzf equation is:
CF = 6.6 × 10−6 / [fzf × (VOUT / 1.24 − 1) ]
For most applications, fzf should be set between 5 kHz and
50 kHz.
ADJUSTING THE OUTPUT VOLTAGE
If an output voltage is required which is not available as a
standard voltage, the LP2975 can be configured as an ad-
justable regulator (see Typical Application Circuits). The ex-
ternal resistors R1 and R2 (along with the internal 24 kΩ
resistor) set the output voltage.
The use of any external resistors to alter the LP2975 pre-set
output voltage is outside the guaranteed operating conditions.
Output voltage accuracy with external resistors will be inferior
when compared to the tolerances of the LP2975 pre-set volt-
ages options, and some external trim mechanism may be
needed to achieve an acceptable initial accuracy of the cus-
tom output voltage. Users of this methodology are strongly
encouraged to confirm that their custom circuit meets all of
their performance requirements.
It is important to note that the external R2 is connected in
parallel with the internal 24 kΩ resistor (typical). If we define
REQ as the total resistance between the COMP pin and
ground, then its value will be the parallel combination of R2
and 24 kΩ:
REQ = (R2 × 24k) / (R2 + 24k)
It follows that the output voltage will be:
VOUT = 1.24 [ (R1 / REQ) + 1]
Some important considerations for an adjustable LP2975 de-
sign:
13 www.national.com
LP2975