参数资料
型号: MCP2050T-330E/SL
厂商: Microchip Technology
文件页数: 6/42页
文件大小: 0K
描述: IC TXRX LIN 3.3V LDO 14-SOIC
产品培训模块: Microchip MCP20xx LIN Transceiver Overview
标准包装: 2,600
系列: *
2012 Microchip Technology Inc.
DS22299B-page 14
MCP2050
EQUATION 1-1:
Assume VBATMIN = 8V. Equation 1-1 shows 10
EQUATION 1-2:
Assume ΔVRECCESSIVE = 1V and IREGMAX = 50 mA
EQUATION 1-3:
Assume
ΔSlope=15%,
VBATMIN=8V
and
IREGMAX=50 mA. Equation 1-2 shows 20
1.6.3
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 capacitor) may induce a significant
drop on the VBB pin. This drop is proportional to the
impedance of the VBAT connection (see Figure 1-14).
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 mini-
mum value the customer should use for CBAT:
EQUATION 1-4:
Equation 1-4 allows lower CBAT/CREG values than the
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, the result is 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.
For a highly inductive connection: RTOT = 0.1
and
L=1 mH: the CBAT/CREG ratio increases to 7!
For a highly resistive connection: RTOT = 10
and
L=0.1 mH: again the CBAT/CREG ratio increases to 7!
Figure 1-13 shows the minimum recommended CBAT/
CREG ratio as a function of the impedance of the VBAT
connection.
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
----------
++
--------------------------------
=
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