LT8580 Ver la hoja de datos (PDF) - Linear Technology

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componentes Descripción

Fabricante

LT8580

Boost/SEPIC/Inverting DC/DC Converter with 1A, 65V Switch, Soft-Start and Synchronization

Linear Technology 

LT8580

Applications Information

As with any feedback loop, identifying the gain and phase

contribution of the various elements in the loop is critical.

Figure 5 shows the key equivalent elements of a boost con-

verter. Because of the fast current control loop, the power

stage of the IC, inductor and diode have been replaced by

a combination of the equivalent transconductance ampli-

fier gmp and the current controlled current source which

converts IVIN to (ηVIN/VOUT) • IVIN. gmp acts as a current

source where the peak input current, IVIN, is proportional

to the VC voltage. η is the efficiency of the switching

regulator, and is typically about 85%.

Note that the maximum output currents of gmp and gma are

finite. The limits for gmp are in the Electrical Characteristics

section (switch current limit), and gma is nominally limited

to about +15µA and –17µA.

–

gmp

+

VC

CF RC RO

CC

IVIN

gma

η • VIN

VOUT

• IVIN

CPL

1.204V

REFERENCE

R2

FBX

R2

VOUT

RESR

RL

COUT

R1

8580 F05

CC: COMPENSATION CAPACITOR

COUT: OUTPUT CAPACITOR

CPL: PHASE LEAD CAPACITOR

CF: HIGH FREQUENCY FILTER CAPACITOR

gma: TRANSCONDUCTANCE AMPLIFIER INSIDE IC

gmp: POWER STAGE TRANSCONDUCTANCE AMPLIFIER

RC: COMPENSATION RESISTOR

RL: OUTPUT RESISTANCE DEFINED AS VOUT DIVIDED BY ILOAD(MAX)

RO: OUTPUT RESISTANCE OF gma

R1, R2: FEEDBACK RESISTOR DIVIDER NETWORK

RESR: OUTPUT CAPACITOR ESR

η: CONVERTER EFFICIENCY (~85% AT HIGHER CURRENTS)

Figure 5. Boost Converter Equivalent Model

From Figure 5, the DC gain, poles and zeros can be cal-

culated as follows:

Output Pole: P1=

2

2 • π • RL • COUT

Error Amp Pole: P2 =

1

2 • π • [RO +RC ] • CC

Error Amp Zero: Z1=

1

2 • π • RC • CC

DC Gain:

(Breaking Loop at FBX Pin)

ADC

= AOL(0) =

∂VC

∂VFBX

•

∂IVIN

∂VC

•

∂VOUT

∂IVIN

•

∂VFBX =

∂VOUT

(gma

•

R0

)

•

gmp

•

⎛

⎜h

⎝

•

VIN

VOUT

• RL

2

⎞

⎟•

⎠

0.5R2

R1+ 0.5R2

ESR Zero: Z2 =

1

2 • π • RESR • COUT

RHP Zero:

Z3

=

4

•

VI2N

π•

• RL

VO2UT

•

L

High Frequency Pole: P3 > fS

3

Phase Lead Zero: Z4 =

1

2 • π • R1• CPL

Phase Lead Pole: P4 =

1

R1• R2

2

•

π

•

R1+

2

R2

•

CPL

2

Error Amp Filter Pole:

P5

=

2•π

•

1

RC • RO

RC + RO

, CF

• CF

< CC

10

The current mode zero (Z3) is a right-half plane zero

which can be an issue in feedback control design, but is

manageable with proper external component selection.

8580f

14

For more information www.linear.com/LT8580

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