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| 型号： | MCP2021AT-330E/MD |
| 厂商： | Microchip Technology |
| 文件页数： | 3/48页 |
| 文件大小： | 0K |
| 描述： | IC TXRX LIN 3.3V LDO 8-DFN |
| 产品培训模块： | Microchip MCP20xx LIN Transceiver Overview |
| 标准包装： | 3,300 |
| 系列： | * |
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2012 Microchip Technology Inc.
DS22298A-page 11
MCP2021A/2A
EQUATION 1-1:
EQUATION 1-2:
Assume ΔVRECCESSIVE = 1V and IREGMAX = 50 mA
Equation 1-2 shows 20
.
EQUATION 1-3:
Assume ΔSlope = 15%, VBATMIN = 8V and IREGMAX =
50 mA. Equation 1-2 shows 20
.
1.5.3
CBAT CAP
Selecting CBAT = 10* CREG is recommended, however,
this leads to a high value cap. Lower values for CBAT
cap can be used with respect to some rules. In any
case, the voltage at the VBB pin should remain above
VOFF when the device is turned on.
The current peak at start-up (due to the fast charge of
the CREG and CBAT capacitors) may induce a
significant drop on the VBB pin. This drop is
proportional to the impedance of the VBAT connection
(see Figure 1-9).
Assume that the VBAT connection is mainly inductive
and resistive, and that the customer knows the resistive
and inductive values of the connection.
The following formula gives an indication of the
minimum value the customer should use for CBAT:
EQUATION 1-4:
10* ratio we recommend.
Let’s assume that we have a good quality connection
with RTOT = 0.1
and L = 0.1 mH.
Solving the equation gives CBAT/CREG = 1.
If we increase RTOT up to 1
the result becomes CBAT/
CREG = 1.4.
But if the connection is highly resistive or highly induc-
tive (poor connection), the CBAT/CREG ratio greatly
increases.
Highly inductive connection: Let’s have RTOT = 0.1
and L = 1 mH: the CBAT/CREG ratio increases to 7!
Highly resistive connection: Let’s have RTOT = 10
and
L = 0.1 mH: again the CBAT/CREG ratio increases to 7!
CREG ratio as a function of the impedance of the VBAT
connection.
FIGURE 1-8:
Minimum Recommended
CBAT/CREG Ratio
250 mA is the peak current at power-on when VBB
=5.5V
RTP <= ΔVRECCESSIVE / IREGMAX.
ΔVRECCESSIVE is the maximum variation tolerated on
the recessive level
ΔSlope is the maximum variation tolerated on the
slope level and IREGMAX is the maximum current the
regulator will provide to the load.
VBATmin>VOFF + 1.0V.
RTP
VBATmin 5.5V
–
250mA
--------------------------------------
5.5VVOFF 1.0V
+
=
RTP
Slope
VBATmin 1V
–
Iregmax
---------------------------------------------------------------
where L is in mH and Rtot in .
Rtot = Rline + RTP.
CBAT
CREG
-------------
100L
2
Rtot
2
+
1 L
2
Rtot
2
100
----------
++
--------------------------------
=
10
C
BA
T
/C
RE
G
RBAT=0.1
RBAT=0.3
RBAT=1
1
10
0.1
1
C
BA
T
/C
RE
G
VBAT Line Inductance [mH]
RBAT=0.1
RBAT=0.3
BAT
R=1
RBAT=2
RBAT=4
RBAT=10
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