11

Application Circuit MGA-25203

1 RF In

2 GND

3 GND

4 BCTRL

GND 12

RF Out 11

GND 10

NC 9

BCTRL

RF In

RF Out

100pF

100pF

BSPLY BSW

Vdd1

100pF

Vdd2

100pF

100pF

0.1uF

100pF

0.1uF

10uF

47uF

10uF

1 RF In

2 GND

3 GND

4 BCTRL

GND 12

RF Out 11

GND 10

NC 9

Using 3.3V or 5V Supply and tying Vcc1, Vcc2, BSPLY and BCTRL

Vbat

Vcc1

Vcc2

BSPLY

BCTRL

3.3V Example :

40K



2.8V

=

*3.3V



= 7K



= 40K



Given :

= 3.3V

= 40K



= ?

5.0V Example :

20K



2.0V

=

*5.0V



= 30K



= 20K



Given :

= 5.0V

= 20K



= ?

Notes: BCTRL regulates the device current, thus R1 and R2 should have

good tolerance rating. If available, a voltage regulator is the preferred

method of bias.

In this example we set R2 at 40KOhm and solve for R1 with simple

voltage divider equation. Note this method will cause some leakage

current through R2.