AM6442...ALV datasheet(126/244 Pages) TI

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VDDS_OSC

MCU_OSC0_XO

Time

VSS

VDDS_OSC (min.)

VDD_CORE (min.)

VSS

VDD_CORE

AM65x_MCU_OSC_STARTUP_02

Figure 7-18. MCU_OSC0 Start-up Time

7.10.4.1.1.1 Load Capacitance

The crystal circuit must be designed such that it applies the appropriate capacitive load to the crystal, as defined

by the crystal manufacturer. The capacitive load, CL, of this circuit is a combination of discrete capacitors

CL1, CL2, and several parasitic contributions. PCB signal traces which connect crystal circuit components to

MCU_OSC0_XI and MCU_OSC0_XO have parasitic capacitance to ground, CPCBXI and CPCBXO, where the

PCB designer should be able to extract parasitic capacitance for each signal trace. The MCU_OSC0 circuits

and device package have combined parasitic capacitance to ground, CPCBXI and CPCBXO, where these parasitic

capacitance values are defined in Table 7-18.

AM65x_MCU_OSC_CC_05

PCBXI

PCBXO

PCB

Signal Traces

Crystal Circuit

Components

MCU_OSC0_XO

MCU_OSC0_XI

Device

Figure 7-19. Load Capacitance

Load capacitors, CL1 and CL2 in Figure 7-17, should be chosen such that the below equation is satisfied. CL in

the equation is the load specified by the crystal manufacturer.

CL = [(CL1 + CPCBXI + CXI) × (CL2 + CPCBXO + CXO)] / [(CL1 + CPCBXI + CXI) + (CL2 + CPCBXO + CXO)]

To determine the value of CL1 and CL2, multiply the capacitive load value CL by 2. Using this result, subtract the

combined values of CPCBXI + CXI to determine the value of CL1 and the combined values of CPCBXO + CXO to

determine the value of CL2. For example, if CL = 10 pF, CPCBXI = 2.9 pF, CXI = 0.5 pF, CPCBXO = 3.7 pF, CXO =

0.5 pF, the value of CL1 = [(2CL) - (CPCBXI + CXI)] = [(2 × 10 pF) - 2.9 pF - 0.5 pF)] = 16.6 pF and CL2 = [(2CL) -

(CPCBXO + CXO)] = [(2 × 10 pF) - 3.7 pF - 0.5 pF)] = 15.8 pF

AM6442, AM6441, AM6421, AM6412, AM6411

SPRSP56D – JANUARY 2021 – REVISED JULY 2022

www.ti.com

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