12V external pull-up

[spittet]

Hi,

I have a MC050 with a MFIN input that I would like to add a +12V external pull-up resistor. The PinConfig is already set at "Activate w/ ground" "pull-up to internal 5V" because the input signal will be a ground when activated. Would there be any conflict between the external +12V pull-up resistor and the +5V internal pull-up resistor, hardware side?

I tried it on my MC050 and I was able to make it work correctly with a 4.7k external +12V pull-up resistor, so I know it can be done and it can work OK. My question is more about may I break something in the long term on this input pin doing it this way?

Have a good day!

[Marbek_Elektronik]

What is the voltage you get on the Pin with this 2 resistors to 12V and to 5V? 11V?

Why do you not use "Bias: 0 = DigIn activated with Sourcing Power"   ?
And then negate the signal?
And use an external pull up to 12V?

Range 1?
DigThresHigh: 4500mV   Low 3500mV?

[spittet]

What is the voltage you get on the Pin with this 2 resistors to 12V and to 5V? 11V?

Why do you not use "Bias: 0 = DigIn activated with Sourcing Power"   ?
And then negate the signal?
And use an external pull up to 12V?

Range 1?
DigThresHigh: 4500mV   Low 3500mV?

Yeah, I got about 10.3V for a 12.2V source.

I was looking for a solution where only the harness would need to be touched and not the program. That's the reason why I wanted to keed the program side the same/untouched.

Still looking for an answer about if this could cause long-term problems.

[Marbek_Elektronik]

Ok, I understand.
So, you have (12,2V -10,3V) / 4700 Ohm = 0,4 mA to Controller-pin, if your switch is open, no connection to ground.
In case of your pin configuration, the controller switches a resistor Rx to 5V and a resistor Rx to 0V and you get 2,5V at the pin, if it is open.

So, the 0,4mA flows from 10,3V to 5V and to 0V. Theoretically you have 1/3 from 0,4mA to 5V power supply = 0,13mA.

I don't think, this damage anything. Resistor from 5V to output is about (10,3V-2,5V)/0,4mA*2=39000 Ohm.

Only danfoss devoloper can tell you if it is OK to do so. But, I think, it is OK.

[Tor]

Hi,

I don't think this should be a problem since the MF input can handle 36V.
But to be sure I have sent the question to a Hardware developer. I hope they don't have to much to do, and have the time to answer.

I'll get back to you.

[Tor]

Hi,

I got an answer from the hardware team and it's more or less what Marbek_Elektronik mentioned.

"There should only be one pullup in the system. So, when the external pullup is present, the pin config enabling the internal one should not be used.
If the internal pullup is enabled, the system software would simply invert the pin’s logic in the .DigIn element, that is something the application can do as well. It should use an inverter to generate a TRUE from the digital input when it is pulled low."

About long term issues:
"If there is only the external pullup, I don’t think this will have any negative long-term effects. Important is either the internal or an external pullup shall be enabled, but not both at the same time."

I hope that answers the question, if not, please let us know.

[Marbek_Elektronik]

Thanks.
But what about the long term issue, if ther are 2 pullups? (We shouldn't do this, I know.)

He doesn't want to change software and he added an external resistor.

Does it damage the 5V regulator? I think no. But what say the developers?

[Tor]

Hi again,

Yes, forgot about the small thing about NOT changing the software. I'm sorry about that, but I'm used to be able to change the software.

Anyway, here's the answer from the hardware developer:
Q: "The customer wants to connect an external pullup to 12V to an input of a MC050-010 where the internal pullup to 5V is enabled. Seems like the customer does not want to change the software to disable the internal pullup. Would that have any negative effect on the hardware?"

A: "There will be no negative effects for this.  The pin is rated for 36V continuously.  It really makes no difference if the internal pull up is enabled or not since there are internal protection diodes to 5V on the side of the 15k pullup resistor not tied to the pin.
There will be up to 2mA ((36V-5V)/15k) going into the 5V supply with 36V applied, but this was accounted for in the design."

I hope this clarify most of the questions. If not, get back to us.

Have a nice day.

[spittet]

Hi Tor, that is the exact answer I was looking for.

Thank you!

Sam