R34DS0034EU0101 Rev.1.01

Page 22

Mar 13, 2024

ISL73006SLH Datasheet

RCOMP value is set by transient response requirement. We need to know Equation 6 and the transient step value

ΔIOUT.

We also need error amp transconductance (gm = 0.923mA/V) and modulator transconductance (GM = 4A/V,

which means 250mV voltage step at COMP node causes 1A output current step). Calculate RCOMP using

Equation 6.

Internal compensation is set in such a way as to ensure ±2% VOUT transient response for ±0.5A load current step.

CCOMP defines compensator zero frequency:

Set fz to ft/10 to maximize phase margin. However, this slows down transient response recovery time. You can

reduce this time by increasing fz (at the expense of the phase margin). In general, zero frequency should not

exceed ft/3 (12.7deg loss of phase margin).

When RCOMP is determined, use Equation 8 to calculate the output capacitance, where gm = 0.923mS,

GM = 4A/V, VREF = 0.6V, and unity gain frequency ft is typically fSW/10.

Equation 8 does not guarantee that transient response is met in all cases. The main reason is the nonlinear nature

of the switching regulator. To derive equations, approximate the modulator with a simple (and linear) GM stage,

which means any fast dV/dt at the input of GM produces equally fast dI/dt at the output. Because the output

inductor (L) limits dI/dt (dI/dt = V/L), in some cases (typically extremely low D or extremely large D), the current

slew rate dI/dt = V/L might get limited by V/L in which case transient response is going to be larger than expected.

In those cases, you must reduce L to increase dI/dt or increase COUT to slow down dV/dt at the GM input.

In the case of internal compensation (set for ±2% VOUT transient response with ±0.5A load current step),

calculate COUT_MIN using Equation 9:

Equations are derived for ideal COUT. Treat MLCCs as ideal capacitors because of small parasitic components

(ESR and ESL). In cases where you cannot use them, carefully consider the ESR value. In the case of extremely

fast transients (1A/ns for microprocessors), voltage drop (ESR x dI) appears extremely quickly, and the regulation

loop cannot react that fast. In those cases, you need to increase COUT. Transient response effectively has two

components (ESR and COUT). The solution is to reduce COUT transient by the ESR x dI product value. For

example, if 2% transient is required and ESR x dI causes 0.5% transient response, 1.5% transient should be used

to determine RCOMP.

Regarding loop stability, ESR zero must be canceled by a pole created with CPOLE such that:

(EQ. 5)

(EQ. 6)

(EQ. 7)

(EQ. 8)

(EQ. 9)

(EQ. 10)

k

V

OUT

VOUT

--------------------

=

RCOMP

I

OUT

kVREFgmEAGM

---------------------------------------------

=

fz

1

2

R

COMPCCOMP

---------------------------------------------------

=

COUT_MIN

VREFgmGMRCOMP

2

ftVOUT

--------------------------------------------------------

=

COUT_MIN F



40000

2

f

t kHz

V

OUT V



----------------------------------------------------

=

ESR COUT



RCOMPCPOLE

=