TS4900(2002) 查看數據表（PDF） - STMicroelectronics

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TS4900
(Rev.:2002)

300mW at 3.3V SUPPLY AUDIO POWER AMPLIFIER WITH STANDBY MODE ACTIVE HIGH

STMicroelectronics 

TS4900

s BTL Configuration Principle

The TS4900 is a monolithic power amplifier with a

BTL (Bridge Tied Load) output configuration. BTL

means that each end of the load is connected to

two single ended output amplifiers. Thus, we have:

Single ended output 1 = Vout1 = Vout (V)

Single ended output 2 = Vout2 = -Vout (V)

And Vout1 - Vout2 = 2Vout (V)

The output power is :

Pout = (2 Vout RMS )2 (W)

RL

For the same power supply voltage, the output

power in BTL configuration is four times higher

than the output power in single ended

configuration.

s Gain In Typical Application Schematic

(cf. page 1)

In flat region (no effect of Cin), the output voltage

of the first stage is :

Vout1 = –Vin R-----fR--e---i-e-n---d---- (V)

For the second stage : Vout2 = -Vout1 (V)

The differential output voltage is

Vout2 – Vo ut1 = 2Vin R-----f-R-e---i-e-n---d---- (V)

The differential gain named gain (Gv) for more

convenient usage is :

G v = -V----o---u---t--2--V---–--i--nV-----o---u----t--1-- = 2 R-----fR--e---i-e-n---d----

Remark : Vout2 is in phase with Vin and Vout1 is

180 phased with Vin. It means that the positive

terminal of the loudspeaker should be connected

to Vout2 and the negative to Vout1.

s Low and high frequency response

In low frequency region, the effect of Cin starts.

Cin with Rin forms a high pass filter with a -3dB cut

off frequency

FCL = --2----π------R---1--i-n------C----i-n-- (Hz)

In high frequency region, you can limit the

bandwidth by adding a capacitor (Cfeed) in

parallel with Rfeed. Its form a low pass filter with a

-3dB cut off frequency

FCH = --2----π------R-----f--e---1-e---d------C----f--e---e----d- (Hz)

s Power dissipation and efficiency

Hypothesis :

• Voltage and current in the load are sinusoidal

(Vout and Iout)

• Supply voltage is a pure DC source (Vcc)

Regarding the load we have:

VO UT = VPEAK sin ω t (V)

and

IOUT

=

V-----O----U----T---

RL

(A)

and

PO UT = -V----P-2--E--R--A---LK----2--- (W)

Then, the average current delivered by the supply

voltage is:

ICC AVG = 2 V-----Pπ---E-R---A-L--K---- (A)

The power delivered by the supply voltage is

Psupply = Vcc IccAVG (W)

Then, the power dissipated by the amplifier is

Pdiss = Psupply - Pout (W)

Pdi ss = 2--------2----V-----c---c-- POUT – POUT (W)

π RL

and the maximum value is obtained when:

∂--∂--P--P--d--O--i--sU---s-T--- = 0

and its value is:

Pdiss max

=

2 Vcc2

π2RL

(W)

Remark : This maximum value is only depending

on power supply voltage and load values.

The efficiency is the ratio between the output

15/19

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