KH561

DATA SHEET

REV. 1A January 2004

7

part, relatively constant performance over supply voltage

is achieved. A current sense in the error current leg of

the 10X current mirror feeds back to the bias current

setup providing a current shutdown feature when the

output current approaches 250mA.

Figure 2: Simplified Circuit Diagram

Developing the Performance Equations

The KH561 is intended to provide both a controllable

voltage gain from input to output as well as a controllable

output impedance. It is best to treat these two operations

separately with no load in place. Then, with the no-load

gain and output impedance determined, the gain to the

load will simply be the no-load gain attenuated by the

voltage divider formed by the load and the equivalent

output impedance.

Figure 3 steps through the output impedance develop-

ment using an equivalent model of Figure 2. Offering an

equivalent, non-zero, output impedance into a matched

load allows the KH561 to operate at lower internal volt-

age swings for a given desired swing at the load. This

allows higher voltage swings to be delivered at the load

for a given power supply voltage at lower distortion levels

than an equivalent op amp needing to generate twice the

voltage swing actually desired at the matched load. This

improved distortion is specified and tested over a wide

range as shown in the specification listing.

using:

Figure 3: Output Impedance Derivation

go to infinity, the output impedance would go to 0. This

would be the normal op amp topology with a very high

setting the internal forward gain to be a low, well

controlled, value.

Developing the No-Load Gain Expression

Taking the output impedance expression as one con-

straint setting the external resistor values, we now need

to develop the no-load voltage gain expression from the

non-inverting input to the output as the other constraint.

Figure 4 shows the derivation of the no load gain.

19

10X Current Mirror

Current Limit

5pF

Q3

Q1

4

10X Current Mirror

Current Limit

5pF

Q4

Q2

21

23

8

X1

V-

+

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−

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=

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

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=+ =

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=

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R

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then

R

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I

RR 1

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note that R

R

G1

R0

i

g

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