MP3908 – HIGH EFFICIENCY BOOST CONTROLLER

MP3908 Rev.0.9

www.MonolithicPower.com

8

8/29/2008

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10V. The MP3908 has its gate driving resistance of

Based on the loss calculation above, the

conduction loss is around 0.629W. The

switching loss is around 0.171W, and the gate

drive loss is 0.015W.

Selecting the Output Diode

The output rectifier diode supplies current to the

inductor when the MOSFET is off. To reduce

losses due to diode forward voltage and

recovery time, use a Schottky diode. The diode

should be rated for a reverse voltage greater

than the output voltage used. Considering the

voltage spike during the commutation period,

the voltage rating of the diode should be set as

1.5 times the output voltage. For high output

voltages (150V or above), a Schottky diode

might not be practical. A high-speed ultra-fast

recovery silicon rectifier is recommended.

Observation of the boost converter circuit

shows that the average current through the

diode is the average load current, and the peak

current through the diode is the peak current

through the inductor. The average current rating

must be greater than 1.5 times of the maximum

load current, and the peak current rating must

be greater than the peak inductor current.

For the application in page 1, a Vishay SS16

Schottky diode or equivalent part is chosen.

Boost Converter: Compensation Design

The output of the transconductance error

amplifier (COMP) is used to compensate the

regulation control system. The system uses two

poles and one zero to stabilize the control loop.

the compensation capacitor (C3) and the

compensation resistor (R3). These parameters

are determined by the equations:

LOAD

1

P

R

C2

1

f

×

×

π

=

3

R

C3

2

1

Z

×

×

π

×

=

The DC mid-band loop gain is:

SENSE

2

OUT

ET

REF

LOAD

IN

EA

VDC

R

V

A

3

R

V

R

V

G

5

.

0

A

×

×

×

×

×

×

×

=

the error amplifier transconductance.

The ESR zero in this example locates at very

high frequency. Therefore, it is not taken into

design consideration.

exists in continuous conduction mode (inductor

current does not drop to zero on each cycle)

step-up converters. The frequency of the right

half plane zero is:

2

2

OUT

LOAD

IN

RHPZ

V

L

2

R

V

f

×

×

π

×

×

=

The right-half-plane zero increases the gain and

reduces the phase simultaneously, which

results in smaller phase margin and gain

margin. The worst case happens at the

condition of minimum input voltage and

maximum output power.

change the voltage gain. Make sure the

bandwidth is about 1/10 of the lower one of the

ESR zero and the right-half-plane zero.

R3

C3

2

1

R

C2

1

LOAD

×

×

π

×

=

×

×

π

ET

IN

REF

EA

SENSE

c

OUT

A

V

V

G

R

f

2

C

2

V

3

R

2

×

×

×

×

×

×

π

×

×

=

Based on these equations, R3 and C3 can be

solved.

ESR zero is much higher than the switching

3.18KHz and make the crossover frequency

8.5kHz, then R3=5kΩ and C3=10nF. Choose

5kΩ and 10nF.