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TS4962_10 Datasheet, PDF (31/44 Pages) STMicroelectronics – 2.8 W filter-free mono class D audio power amplifier
TS4962
Application information
4.3
Common-mode feedback loop limitations
As explained previously, the common-mode feedback loop allows the output DC bias
voltage to be averaged at VCC/2 for any DC common-mode bias input voltage.
However, due to a Vicm limitation in the input stage (see Table 3: Operating conditions on
page 4), the common-mode feedback loop can play its role only within a defined range. This
range depends upon the values of VCC and Rin (AVdiff). To have a good estimation of the
Vicm value, we can apply this formula (no tolerance on Rin):
Vicm
=
V-----C---C-----×------R-----i-n----+-----2-----×------V----I--C----×------1----5---0----k---Ω--
2 × (Rin + 150kΩ)
(V)
with
VIC
=
I--n----+----+-----I--n-----
2
(V)
And the result of the calculation must be in the range:
0.5V ≤ Vicm ≤ VCC – 0.8V
Due to the +/-9% tolerance on the 150 kΩ resistor, it is also important to check Vicm in these
conditions.
-V----C---C-----×------R-----i-n----+-----2-----×------V----I--C----×------1----3---6---.--5----k---Ω--
2 × (Rin + 136.5kΩ)
≤
Vi
cm
≤
V-----C---C-----×------R-----i-n----+-----2-----×------V----I--C----×------1----6---3---.--5----k---Ω--
2 × (Rin + 163.5kΩ)
If the result of the Vicm calculation is not in the previous range, input coupling capacitors
must be used. With VCC between 2.4 and 2.5 V, input coupling capacitors are mandatory.
For example:
With VCC = 3 V, Rin = 150 k and VIC = 2.5 V, we typically find Vicm = 2 V, which is lower than
3 V-0.8 V = 2.2 V. With 136.5 kΩ we find 1.97 V and with 163.5 kΩ we have 2.02 V.
Therefore, no input coupling capacitors are required.
4.4
Low frequency response
If a low frequency bandwidth limitation is requested, it is possible to use input coupling
capacitors.
In the low frequency region, Cin (input coupling capacitor) starts to have an effect. Cin forms,
with Rin, a first order high-pass filter with a -3 dB cut-off frequency.
FCL
=
------------------1-------------------
2π × Rin × Cin
(Hz)
So, for a desired cut-off frequency we can calculate Cin,
Cin
=
-------------------1--------------------
2π × Rin × FCL
(F)
with Rin in Ω and FCL in Hz.
Doc ID 10968 Rev 8
31/44