16F690 MCLR as Input

[jackberg1]

Hello everyone,

a very basic question,

I have a 16F690 with the RA3 MCLR configured as input, and the RA3 pin

is connected to a 4k7 resistor pulldown and a switch tied to the VDD and it's working fine when the switch

is open I detect a low and when the switch is close I detect a high.

with the MCLRE bit #5 to = 0 the MCLR is tied to VDD

(0 = RA3/MCLR/VPP pin function is digital input, MCLR internally tied to VDD)

if the MCLR is tied to VDD, how come my pulldown + switch are working as it is.

does the MCLR has already an internal pullup resistor

if I remove the pulldown resistor with the switch open, the MCLR act as if it is high.

it need the pulldown resistor to be a low.

any suggestion ?




Also the internal EEPROM of the 16F690 is divided into 256 bytes location with address from 0 to 255
if I write 0, word B, C, does it mean the address 0 and 1 will contain the word variable B and at address 2 will contain variable C (byte)

Thanks for all comments.

[HenrikOlsson]

if the MCLR is tied to VDD, how come my pulldown + switch are working as it is.

Because when you configure RA3 to act as a normal input the MCLR signal is disconnected from the pin and instead connected to Vdd internally. In other words, it's not the PIN that's connected to Vdd but the MCLR-signal. Does that make sense?

if I write 0, word B, C, does it mean the address 0 and 1 will contain the word variable B and at address 2 will contain variable C (byte)

Yes.

/Henrik.

[jackberg1]

many thanks Henrik to clarify this post.

that make more sense to me now with further test, when the pin RA3
has nothing connected to it, the input is floating thus
when adding a pulldown it's state is stable.


I found from the datasheet a diagram of the MCLR input circuit, my guess is the red line indicate the path of the input configuration.