ADP1851

Data Sheet

Rev. B | Page 18 of 24

There is also additional power loss during the time, known as

dead time, between the turn-off of the high-side switch and the

turn-on of the low-side switch, when the body diode of the low-

side MOSFET conducts the output current. The power loss in

the body diode is given by

O

SW

D

F

BODYDIODE

I

f

t

V

P

×

×

×

=

where:

of approximately 3 nF. The dead time is not fixed. Its effective

designs.

Therefore, the power loss in the low-side MOSFET is

BODYDIODE

CLS

LS

P

P

P

+

=

increasing temperature, with a typical temperature coefficient of

ambient temperature is

where:

LOOP COMPENSATION—VOLTAGE MODE

Set the controller to voltage mode operation by placing a

100 kΩ resistor between DL and PGND. Choose the largest

possible ramp amplitude for the voltage mode below 1.5 V.

The ramp voltage is programmed by a resistor placed between

RAMP

SW

IN

RAMP

V

f

V

R

×

×

−

=

pF

100

V

2

.

0

The voltage at the RAMP pin is fixed at 0.2 V, and the current

going into RAMP should be between 10 µA and 160 µA. Make

sure that the following condition is satisfied:

μA

160

V

2

.

0

μA

10

≤

−

≤

RAMP

IN

R

V

(3)

less than 73.8 kΩ.

Assuming that the LC filter design is complete, the feedback control

system can be compensated. In general, aluminum electrolytic

capacitors have high ESR; however, if several aluminum electrolytic

capacitors are connected in parallel and produce a low effective

ESR, then Type III compensation is needed. In addition, ceramic

capacitors have very low ESR (only a few milliohms), making

Type III compensation a better choice.

Type III Compensation

G

(dB)

PHASE

–90°

–270°

INTERNAL

EA

FB

COMP

–1

SL

OPE

–1

SL

OPE

+1

SL

OPE

Figure 26. Type III Compensation

If the output capacitor ESR zero frequency is greater than one-

half of the crossover frequency, use the Type III compensator as

shown in Figure 26.

Calculate the output LC filter resonant frequency as follows:

LC

π

2

1

=

(4)

Choose a crossover frequency that is 1/10 of the switching

frequency:

10

SW

CO

f

f =

(5)

Set the poles and zeros as follows:

SW

P2

P1

f

f

f

2

1

=

=

(6)

I

Z

SW

CO

Z2

Z1

C

R

f

f

f

f

π

2

1

40

4

=

=

=

=

(7)

or

I

Z

LC

Z2

Z1

C

R

f

f

f

π

2

1

2

=

=

=

(8)

Use the lower zero frequency from Equation 7 or Equation 8.

2

LC

IN

CO

Z1

RAMP

TOP

Z

f

V

f

f

V

R

R =

(9)

Z1

Z

I

f

R

C

π

=

2

1

(10)

Because of the finite output current drive of the error amplifier,