LM388 datasheet(4/8 Pages) NSC

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Application Hints (Continued)

is open If pins 2 and 6 are bypassed then R as low as 2 kX

can be used This restriction is because the amplifier is only

compensated for closed-loop gains greater than 9 VV

INPUT BIASING

The schematic shows that both inputs are biased to ground

witha50kX resistor The base current of the input transis-

tors is about 250 nA so the inputs are at about 125 mV

when left open If the dc source resistance driving the

LM388 is higher than 250 kX it will contribute very little

additional offset (about 25 mV at the input 50 mV at the

output) If the dc source resistance is less than 10 kX then

shorting the unused input to ground will keep the offset low

(about 25 mV at the input 50 mV at the output) For dc

source resistances between these values we can eliminate

excess offset by putting a resistor from the unused input to

ground equal in value to the dc source resistance Of

course all offset problems are eliminated if the input is ca-

pacitively coupled

When using the LM388 with higher gains (bypassing the

135 kX resistor between pins 2 and 6) it is necessary to

bypass the unused input preventing degradation of gain

and possible instabilities This is done with a 01 mF capaci-

tor or a short to ground depending on the dc source resist-

ance on the driven input

BOOTSTRAPPING

The base of the output transistor of the LM388 is brought

out to pin 9 for Bootstrapping The output stage of the am-

plifier during positive swing is shown in

Figure 3 with its

external circuitry

R1 a R2 set the amount of base current available to the

output transistor The maximum output current divided by

beta is the value required for the current in R1 and R2

(R1 a R2) e bO

(VS 2) b VBE

IO MAX

Good design values are VBE e 07V and bO e 100

Example 1 watt into 8X load with VS e 12V

IO MAX e

02PO

500 mA

(R1 a R2) e 100

(122)b07

J e1060X

To keep the current in R2 constant during positive swing

capacitor CB is added As the output swings positive CB lifts

R1 and R2 above the supply maintaining a constant voltage

across R2 To minimize the value of CB R1 e R2 The pole

due to CB and R1 and R2 is usually set equal to the pole

due to the output coupling capacitor and the load This

gives

CB j

4Cc

Example for 100 Hz pole and RL e 8X Cc e 200 mF and

CB e 8 mF if R1 is made a diode and R2 increased to give

the same current CB can be decreased by about a factor of

4 as in

Figure 4

For reduced component count the load can replace R1 The

value of (R1 a R2) is the same so R2 is increased Now CB

is both the coupling and the bootstrapping capacitor (see

Figure 2 )

Typical Applications

TLH7846 – 3

FIGURE 1 Load Returned to Ground

(Amplifier with Gain e 20)

TLH7846 – 4

FIGURE 2 Load Returned to VS

(Amplifier with Gain e 20)

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