LPC2368 Input Resistance

[Kvel]

I'm connecting a 3.3v signal to the GPIOs on Port 2 of the LPC2368.

Apart from the usual pullup/pulldown resistors, I suspect there needs to be an additional resistance between the voltage pickup point and the input pin. I could not find details of this in the LPC2368 manuals, any guidance on resistance values would be greatly appreciated.

The board I'm using an ETT CP-JR Arm7 LPC2368 board, and in the board manual it states that they use a 470ohm resistor for the ADC input, and a 1kohm resistor for GPIOs when in output mode.

 

[alexan_e]

Why do you think that you need an input resistor to connect your signal?
If it is in the correct range (below 5v) then you don't have to put any resistance, you can if you want but it doesn't matter, the input resistance is very high anyway, the datasheet says that the high/low level input current is 3uA.

Alex

 

[Kvel]

Was looking in the manual, but somehow skipped the datasheet. Apologies for the trivial question, I'm a beginner of sorts.

Reason I was thinking that was because I had no idea of the current draw/ internal input resistance and didn't want to risk frying any pins.


Thanks alot for your input.

 

[alexan_e]

As long as the pin is in input mode (the default at startup ) you have no problem , if you think that there is a case where you may set the pin as an output (by mistake) while a signal is connected to that pin then you can use the resistor, 1K or 10K should be fine.

The resistor may be useful for your output pins, depends on the circuit you drive from each pin.

Alex

 

[Kvel]

With the output pins I'll be driving two SSRs with a current draw of 16mA, which should be no problem to connect directly if my research is correct.

 

[alexan_e]

Page 34 of the datasheet says 4mA for 0.4 voltage drop in the output, 16mA is high and you will get a big voltage drop
https://www.keil.com/dd/docs/datashts/philips/lpc236x_ds.pdf

I can't find the current vs output voltage for your mcu but the 17xx series have the same behavior, you will get a very low voltage, take a look below.



Alex

 

[Kvel]

I see that voltage drop will be a problem, will need to see how to work around that. I think a simple mosfet should take care of it. Thanks again for the heads up.