Hello, I would like to understand the working of the following circuit, principally why the input of the logic gate is connected both to vcc and the sensor's output

 

What is a "fan sensor"?
If the sensor output is a logic high then the output of the logic inverter is low.
If the output of the sensor is a logic low then the output of the logic inverter is high.

Maybe the resistor is to provide the inverter with a logic high input when the sensor is turned off?

 

OK, What it looks like is that the sensor has what is called an "Open Collector Output". What that means, usually, is that the sensor device output is the collector terminal of an NPN transistor, which has it's emitter tied to the power common negative circuit. THEN, the resistor connected from there to the +Vcc will pull the output up to Vcc when the transistor is not conducting, thus delivering a LOGIC HIGH to the logic inverter gate input. When the internal transistor is switched on, into conduction, the output is pulled low, sending a logic low into the inverter gate input. OPen collector outputs are frequently used, the benefit is that they can function with several different external voltages with no adaption required.

 

Hello, I would like to understand the working of the following circuit, principally why the input of the logic gate is connected both to vcc and the sensor's output

VCC is power. You probably know that. R18 holds pin 1 high when VOUT has no output. VOUT could be anything, we don't yet know what we're dealing with as far as a Fan Sensor goes. VOUT could be a High, Low or Open (floating). IF it's a High then there would be no change in U12 (? can't quite read it). Pin 1 would be high if either the sensor is high or VCC is high. IF VOUT is Low then the input to U12 pin 1 would be Low. That would result in a High on pin 2. IF VOUT is Open (floating) then pin 1 would be high due to the presence of VCC through R18.

 

Why is the DWG shown with signal flow backwards, i.e right to left and not conventional, vice-versa??
Also anyone's guess with no sensor details?

 

VCC is power. You probably know that. R18 holds pin 1 high when VOUT has no output. VOUT could be anything, we don't yet know what we're dealing with as far as a Fan Sensor goes. VOUT could be a High, Low or Open (floating). IF it's a High then there would be no change in U12 (? can't quite read it). Pin 1 would be high if either the sensor is high or VCC is high. IF VOUT is Low then the input to U12 pin 1 would be Low. That would result in a High on pin 2. IF VOUT is Open (floating) then pin 1 would be high due to the presence of VCC through R18.

thanks

 

Why is the DWG shown with signal flow backwards, i.e right to left and not conventional, vice-versa??
Also anyone's guess with no sensor details?

I do not know why. I found the diagram in a book.

 

It was quite obvious to those of us who have experienced open collector outputs .