CLC503 datasheet(4/6 Pages) NSC

4 / 6 page

CLC503 APPLICATIONS

APPLICATION CONSIDERATIONS

Theory of Operation

Figure 1 is a simplified schematic of the CLC503.

Figure 1: Simplified Block Diagram

The input voltage drives a unity gain buffer, B1, and

an inverting buffer, B2. These buffers drive emitter

followers, Q1 and Q2. Resistor, R3, is the gain set

resistor. The combination of B1, B2, Q1, Q2 and R3

form a transconductance stage.

The input voltage

across R3 is converted to an in-phase and out-of-

phase current through the collectors of Q1 and Q2. The

current through R3 is:

The common mode voltage across R4 is converted

to a current. Transistor Q5 has a collector current

equal to:

The common mode current is scaled and mirrored back

to Q1 and Q2. These currents, I, are converted back to

a voltage at the collector load resistors, R1 and R2.

This forms the common-mode output voltage.

Vocm = VCC – Vdiode – R1I

I = 16Icm

Vocm = Vcm

Figure 2 depicts the differential output voltage limits of

the CLC503.

Figure 2: Differential Output Voltage

Centered around Vcm, the outputs are derived from the

following equations.

The input to output relationship is shown in Figure 3.

Vin

Vodiff

±1V

±2V

2Vpp

4Vpp

±1.4V max

±2.8V max

Figure 3: Input vs. Output Relationship

Pulling the power down line (PDN) high decreases the

quiescent supply current. This

turns off the current

flowing in Q5, and therefore Q1 and Q2, allowing the

output voltages to drift high, to approximately 4.3V.

Since the signal is not significantly attenuated, PDN

does not effectively isolate the input from the output.

This part is not recommended for use as a multiplexer.

Refer to

Pin Descriptions - Power Down pin -

section. No damage occurs to the device when PDN is

high and the input is driven to the supply voltage.

Pin Descriptions

Figure 4: CLC503 Functional Pin Descriptions

-Vo

+Vin

VCC

-1

+Vo

Icm

VEE

PDN

Vcm

Current Mirror

R1 = R2

2 R1 = R3

R4 = 16R1

Input Voltage (V)

3.7

0.8

-1

2.25

3.25V

1.25V

Vcm

+Vo = Vcm +Vin

-Vo = Vcm -Vin

+Vo

Vin

PDN

VCC

-Vo

Vcm

VEE

VI R

I R

VI 2R

V2V

odiff

3 2

odiff

()

16R

diode

≈

−−

-V

V+V

-V

ocm

odiff

=−

–(

)

diode

=−

−

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