EL5250 datasheet(13/18 Pages) RENESAS

13 / 18 page

EL5150, EL5151, EL5250, EL5251, EL5451

FN7384 Rev 7.00

Page 13 of 18

January 16, 2008

Power Supply Bypassing Printed Circuit Board

Layout

As with any high frequency device, a good printed circuit

board layout is necessary for optimum performance. Lead

lengths should be as short as possible. The power supply

pin must be well bypassed to reduce the risk of oscillation.

For normal single supply operation, where the VS- pin is

connected to the ground plane, a single 4.7µF tantalum

capacitor in parallel with a 0.1µF ceramic capacitor from VS+

to GND will suffice. This same capacitor combination should

be placed at each supply pin to ground if split supplies are to

be used. In this case, the VS- pin becomes the negative

supply rail.

Printed Circuit Board Layout

For good AC performance, parasitic capacitance should be

kept to a minimum. Use of wire wound resistors should be

avoided because of their additional series inductance. Use

of sockets should also be avoided if possible. Sockets add

parasitic inductance and capacitance that can result in

compromised performance. Minimizing parasitic capacitance

at the amplifier's inverting input pin is very important. The

feedback resistor should be placed very close to the

inverting input pin. Strip line design techniques are

recommended for the signal traces.

Application Circuits

Sallen Key Low Pass Filter

A common and easy to implement filter taking advantage of

the wide bandwidth, low offset and low power demands of

the EL5150. A derivation of the transfer function is provided

for convenience (see Figure 41).

Sallen Key High Pass Filter

Again, this useful filter benefits from the characteristics of the

EL5150. The transfer function is very similar to the low pass

so only the results are presented (see Figure 42).

FIGURE 41. SALLEN KEY LOW PASS FILTER

Holp

)

(

Holp

)

((

)

(

)

((

)

(















































Equations simplify if we let all

components be equal R = C

0.1µF

VOUT

V+

V-

U1A

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