datasheetbank_Logo
Integrated circuits, Transistor, Semiconductors Search and Datasheet PDF Download Site

TS419-4IQT View Datasheet(PDF) - STMicroelectronics

Part Name
Description
View to exact match
TS419-4IQT
ST-Microelectronics
STMicroelectronics ST-Microelectronics
TS419-4IQT Datasheet PDF : 32 Pages
First Prev 21 22 23 24 25 26 27 28 29 30 Next Last
TS419-TS421
APPLICATION INFORMATION
s BTL Configuration Principle
The TS419 & TS420 are monolithic power
amplifiers with a BTL output type. BTL (Bridge
Tied Load) 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 VoutRMS)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 3 of TS419-TS421 datasheet)
In the flat region (no CIN effect), the output voltage
of the first stage is:
Vout1= −Vin Rfeed (V)
Rin
For the second stage : Vout2 = -Vout1 (V)
The differential output voltage is
Vout2 Vout1 = 2 Vin Rfeed (V)
Rin
The differential gain named gain (Gv) for more
convenient usage is :
Gv = Vout2 Vout1 = 2 Rfeed
Vin
Rin
Remark : Vout2 is in phase with Vin and Vout1 is
phased 180° with Vin. This 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 the low frequency region, CIN starts to have an
effect. CIN forms with RIN a high-pass filter with a
-3dB cut off frequency .
FCL
=
1
2πRinCin
(Hz)
In the high frequency region, you can limit the
bandwidth by adding a capacitor (Cfeed) in
parallel with Rfeed. It forms a low-pass filter with a
-3dB cut off frequency .
FCH
=
2π
1
Rfeed
Cfeed
(Hz)
s Power dissipation and efficiency
Hypothesis:
• Load voltage and current are sinusoidal (Vout
and Iout)
• Supply voltage is a pure DC source (Vcc)
Regarding the load we have:
VOUT = VPEAK sinωt (V)
and
IOUT
=
VOUT
RL
(A)
and
POUT =
VPEAK 2
2RL
(W)
Then, the average current delivered by the supply
voltage is:
IccAVG
= 2 VPEAK
π RL
(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)
2 2 Vcc
Pdiss =
POUT POUT (W)
π RL
and the maximum value is obtained when:
Pdiss = 0
POUT
and its value is:
Pdiss max
=
2 Vcc2
π2RL
(W)
Remark : This maximum value is only dependent
upon power supply voltage and load values.
27/32
 

Share Link: 

datasheetbank.com [ Privacy Policy ] [ Request Datasheet ] [ Contact Us ]