5.11

Application n°4 : Differential inputs BTL power amplifier

In this configuration, we need to place these components : R1, R4, R5, R6, R7, C4, C5, C12.

We have also : R4 = R5, R1 = R6, C4 = C5.

The gain of the amplifier is:

GVDIFF=2R1R4

For Vcc = 5 V, a 20 Hz to 20 kHz bandwidth and 20 dB gain BTL power amplifier you could follow the bill of

material below.

Table 10. Components

Designator

Part type

R1

110 k / 0.125 W

R4

22 k / 0.125 W

R5

22 k / 0.125 W

R6

110 k / 0.125 W

R7

330 k / 0.125 W

R8*

(Vcc-Vf_led) / If_led

C4

470 nF

C5

470 nF

C6

100 µF

C7

100 nF

C9

Short-circuit

C10

Short-circuit

C12

1 µF

D1*

LED 3 mm

S1, S2, S6, S7

2 mm insulated plug 10.16 mm pitch

S8

3 pts connector 2.54 mm pitch

P1, P2

PCB phono Jack

U1

TS4871IS

5.12

Note on how to use the PSRR curves

We have finished a design and we have chosen the components values

•

Rin=Rfeed=22 kΩ

•

Cin=100 nF

•

Cb=1 μF

Now, on fig. 13, we can see the PSRR (input grounded) vs. frequency curves. At 217 Hz we have a PSRR value

of -36 dB.

In reality we want a value about -70 dB. So, we need a gain of 34 dB.

Now, on fig. 12 we can see the effect of Cb on the PSRR (input grounded) vs. frequency. With Cb=100 μF, we can

reach the -70 dB value.

The process to obtain the final curve (Cb=100 μF, Cin=100 nF, Rin=Rfeed=22 kΩ) is a simple transfer point by

point on each frequency of the curve on fig. 13 to the curve on fig. 12. The measurement result is shown on the

next figure.

TS4871

Application information

DS2547 - Rev 10

page 31/42