PCF8574A Source through digital lines

[ElectoNewbie]

Hi,

Currently I am using PCF8574A (https://www.nxp.com/docs/en/data-sheet/PCF8574_PCF8574A.pdf).

I am able to use the pins P0 to p7 as a current sink and light LEDs ( i.e. the LEDs are connected to voltage source and then to the pins)
But I am unable to ( or rather do not know how to) do the opposite. i.e. use P0 to P7 as a voltage source to power the LEDs

The datasheet says the the pins are quasi-bidirectional.
Also Page 7 shows the different configuration.

My questions:

1. Does the default configuration (current sink) mean 'Output LOW' ?
2. To convert the PIN to a source I will need it to be in 'Input HIGH' config ?
3. Is my assumption that I have to attach a pull-up resistor at the pins to +5V to convert the pin to a source correct ?
4.If the above 3. is correct, then how do I calculate how much resistance to add? in the pull-up?

 

[Pommie]

You can't source current with that type of I/O - it's known as open collector and can only pull the pin low. Actually, it can pull it high but with a very tiny current capability (100μA).
1. Yes.
2. Yes but it can only supply 100μA.
3. This is built into the chip. Edit, this is what supplies the 100μA.

Mike.

 

[ElectoNewbie]

Thanks a lot for the reply.

It is not that there is something particular I have accomplish with this circuit but just demonstrate conceptually that it can be done even if the current is 100μA.

The thing is not being a electrical or electronic engineer sometimes what can be considered common sense in this domain is not exactly present for me. ( Example: the setup is called an open collector.)

I had a followup question:

2. Shouldn't it be OUTPUT HIGH (rather than INPUT HIGH) for me to obtain the 100μA.

Also, is the voltage at the pin with 100μA is 5v ? that is the supply voltage ?

 

[Pommie]

Think of it as there only being a switch from the output to ground and a very high resistance(10k) from the output to 5V. When the switch is turned off the pin acts as an input AND a very high impedance output. If you connected an LED from the pin to ground then the pin will be pulled down by the LED - probably to 2V and so your current will drop to 50μA - the power illuminating the LED will be 100μW. I doubt you will be able to see any light from the LED. I can't find any data on LEDs at such microscopic power levels.

Note, input high turns the switch off and so the output will go high via the resistor.

Mike.