How to interface a Push Pull output to Arduino

Refer the sketch. This is a flow meter pickup with 24V dc supply. I want to interface the output which is said to be a push pull output to an Arduino digital pin.

Can i use a 1K potentiometer between the output and GND and set it for outputting pulses with 5V amplitude ?

Have always used open collector outputs and this is new to me. Any hints / traps to watch out welcome.

VHEF_Ckt.PNG

Safer to use a fixed voltage divider.
A 39k resistor from sensor to Arduino pin, and a 10k resistor from pin to Arduino ground will give 4.85volt on the Arduino pin. In a noisy environment you could lower these values to 22k and 4k7.
Don't forget to connect sensor ground to Arduino ground.
Leo..

Thanks Leo. So basically your response confirms that my idea to use a potential divider at the output of the signal conditioner is correct.

I fully agree with your point of using fixed resistors for the divider- much safer against accidental rotation of the potentiometer knob in the right direction.

But just one query, assuming I use the 22K + 4K7 combo. As per the formula given on the page, Voltage out from the Push Pull stage is (Ub-0.5) / (470 + Load in Ohm)

So in our case it will translate to 23.5 / (470 + 22000+4700) or in other words 0.86mV amplitude. Is this not too small ?

And to get a good amplitude suppose if I replace the 22K with 820 Ohm and 4K7 with 200 Ohm I will get a 4.5V output. Is my approach correct ??

So in our case it will translate to 23.5 / (470 + 22000+4700) or in other words 0.86mV amplitude. Is this not too small ?

No, maths fail! You misquote the current formula as the voltage formula. The current is 0.86mA, the voltage
is 23.5V - 470*0.00086 = 23.1V

I would use @Wawa’s suggestion (22k / 4.7k) but put an additional 4.7k R between the the tap point on the divider and the input pin, that way, if the ground connection failed, there would be 27k resistance between pin and 24V, limiting current through pin protection diode to < 1mA.

22k with accidently unconnected ground resistor will result in <1mA fault current into the pin.
Internal pin protection (to VCC) should be able to cope with that.
Leo…

MarkT:
No, maths fail! You misquote the current formula as the voltage formula. The current is 0.86mA, the voltage
is 23.5V - 470*0.00086 = 23.1V

Hmmm..looks like I belong to the 50% of the humanity who do not understand maths !!

I need to first find out the Iout using the second formula and plug that value into the first formula ?

Anyway now the 22K + 4K7 makes good sense and I shall try out with that when I get hold of the setup.

Thanks for the support.

PS : Incidentally once I see the waveform on the scope then the next job is to design a F-I convertor to work with a input frequency range of 4 to 525 Hz. and deliver 4-20mA. I know I can take a standard F-V convertor and then convert the V to I ....but looks long winded and there is no Arduino in that !! Any links to such circuits ?