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USB Low-Full-High Speed Peripherals

mgre
New Contributor

Hello everyone,

 

i need a clarification on a detail regarding VCCIO=1V8 of CY7C65213.

As seen on the datasheet, the VCCD pin is the OUTPUT of the internal regulator for 1V8 and it cannot drive external loads.

Additionaly, on the datasheet  it is mentioned that when VCCIO is less than 2V, VCCD must be connected to VCCIO.

In this scenario, can the device be damaged during startup considering that the default configuration stored in its Flash doesn't disable the internal regulator (as shown on Table 2 in the datasheet)?

 

Lastly, always from Table 2 of the datasheet, it is hinted that CTS and DSR signals have an internal pull up. What about DCD and RI Signals? Do they have some internal pull up or pull down as well?

 

Thanks in advance.

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1 Solution
PranavaYN
Moderator
Moderator

Hi,

can the device be damaged during startup considering that the default configuration stored in its Flash doesn't disable the internal regulator (as shown on Table 2 in the datasheet)?

>>  Your understanding is correct. It will damage the chip. One need to pre-configure the chip before using it in different configuration.

Lastly, always from Table 2 of the datasheet, it is hinted that CTS and DSR signals have an internal pull up. What about DCD and RI Signals? Do they have some internal pull up or pull down as well?

>> All UART pins have internal pull ups.

Best regards,
Pranava

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3 Replies
PranavaYN
Moderator
Moderator

Hi,

can the device be damaged during startup considering that the default configuration stored in its Flash doesn't disable the internal regulator (as shown on Table 2 in the datasheet)?

>>  Your understanding is correct. It will damage the chip. One need to pre-configure the chip before using it in different configuration.

Lastly, always from Table 2 of the datasheet, it is hinted that CTS and DSR signals have an internal pull up. What about DCD and RI Signals? Do they have some internal pull up or pull down as well?

>> All UART pins have internal pull ups.

Best regards,
Pranava

View solution in original post

mgre
New Contributor

Hi PranavaYN,

Thanks for the reply.

Regarding the 1V8 operation:

this basically means that i need to start in the condition shown in the picture as CASE A, configure the device to be used in 1V8 and then switch to the CASE B circuit layout. Is this correct?

example.png

Is there perhaps another way to do this configuration without having to switch between the 2 circuit configurations?

Thanks in advance

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PranavaYN
Moderator
Moderator

Hi,

 

this basically means that i need to start in the condition shown in the picture as CASE A, configure the device to be used in 1V8 and then switch to the CASE B circuit layout. Is this correct?

>> Yes, the above setting is correct.

Is there perhaps another way to do this configuration without having to switch between the 2 circuit configurations?

>> This is the only way to configure and use it in other configuration. Since its one time configuration, you can get it configured in third party programming houses.

Best regards,

Pranava

Best regards,
Pranava