Data Sheet P12710EJ2V0DS00

8

TEST CIRCUIT

1 000 pF

1 000 pF

1 000 pF

L

4

6

1

2, 3, 5

50

Ω

50

Ω

OUT

IN

COMPONENTS OF TEST CIRCUIT

EXAMPLE OF ACTUAL APPLICATION COMPONENTS

FOR MEASURING ELECTRICAL

CHARACTERISTICS

Type

Value

Type

Value

Operating Frequency

Bias Tee

1 000 pF

Chip capacitor

1 000 pF

100 MHz or higher

Capacitor

1 000 pF

L

Chip inductor

100 nH

100 MHz or higher

L

Bias Tee

1 000 nH

10 nH

2.0 GHz or higher

INDUCTOR FOR THE OUTPUT PIN

The internal output transistor of this IC consumes 20 mA, to output medium power. To supply current for output

transistor, connect an inductor between the Vcc pin (pin 6) and output pin (pin 4). Select large value inductance, as

listed above.

The inductor has both DC and AC effects. In terms of DC, the inductor biases the output transistor with minimum

voltage drop to output enable high level. In terms of AC, the inductor make output-port-impedance higher to get

enough gain. In this case, large inductance and Q is suitable.

For above reason, select an inductance of 100

Ω or over impedance in the operating frequency. The gain is a

peak in the operating frequency band, and suppressed at lower frequencies.

The recommendable inductance can be chosen from example of actual application components list as shown

above.

Capacitors of 1 000 pF are recommendable as the bypass capacitor for the Vcc pin and the coupling capacitors

for the input and output pins.

The bypass capacitor connected to the Vcc pin is used to minimize ground impedance of Vcc pin. So, stable bias

can be supplied against Vcc fluctuation.

The coupling capacitors, connected to the input and output pins, are used to cut the DC and minimize RF serial

impedance. Their capacitance are therefore selected as lower impedance against a 50

Ω load. The capacitors thus

perform as high pass filters, suppressing low frequencies to DC.

To obtain a flat gain from 100 MHz upwards, 1 000 pF capacitors are used in the test circuit. In the case of under

10 MHz operation, increase the value of coupling capacitor such as 10 000 pF. Because the coupling capacitors are

determined by equation, C = 1/(2

πRfc).