What is the V44? I've been looking for some time and I can't find an answer. LVC1G14 would fit the pins, that's all I found.

 

Maybe, but an LVC1G14 does not have two inputs. It only has 1 input, so I would keep searching. It is true that pin 1 is a no connect for that package so it is still possible, just unusual for the layout person to do that.

 

I found 2 matches for V44.

V44
IXD2110C441MR
IXYS
SOT-25
DC/DC voltage converter IC
PFM step-up, 100kHz, 4.4V±2.5%, 400mA, +CE

V44
KIC73A44M2
Korea Electronics
TSM
Voltage detector IC
4.4V±1%, -Reset PPO, Td=240ms

 

I have doubts about the drawing supplied.

If the output is generated from PB6 from an STM32F103 why would they need to buffer or invert the logic signal?
If the purpose is to drive the emitter in an opto-isolator, why the need for the 180Ω resistor to GND?
The emitter of NEC9587 forward voltage is 1.65V typical, 1.8V max. Hence it could be have been driven directly by the STM32F103.

The 390Ω resistor serves no purpose but to waste power.

There are many SN74LVC1Gxx gates that match that pinout, for example, SN74LVC1G00.
Vcc on pin-5 would need +V supply.

 

I have doubts about the drawing supplied.
View attachment 264821
If the output is generated from PB6 from an STM32F103 why would they need to buffer or invert the logic signal?
If the purpose is to drive the emitter in an opto-isolator, why the need for the 180Ω resistor to GND?
The emitter of NEC9587 forward voltage is 1.65V typical, 1.8V max. Hence it could be have been driven directly by the STM32F103.

The 390Ω resistor serves no purpose but to waste power.

There are many SN74LVC1Gxx gates that match that pinout, for example, SN74LVC1G00.
Vcc on pin-5 would need +V supply.

the scheme is correct

 

but 74AHC1G08Q

 

the scheme is correct

It is possibly correct as far as it goes but it is incomplete.

 

It is possibly correct as far as it goes but it is incomplete.

everything I found

 

everything I found

As has been pointed out, it is non-functional as you have drawn it, if pin 5 is not connected to a voltage source.
Vaya con Dios.

 

As has been pointed out, it is non-functional as you have drawn it, if pin 5 is not connected to a voltage source.
Vaya con Dios.

pin 5 is connected to + 5v

 

pin 5 is connected to + 5v

That is far from obvious from your diagram.

 

is good now ?

 

Go back and read post #4.

 

Go back and read post #4.

If the output is generated from PB6 from an STM32F103 why would they need to buffer or invert the logic signal?
If the purpose is to drive the emitter in an opto-isolator, why the need for the 180Ω resistor to GND?
The emitter of NEC9587 forward voltage is 1.65V typical, 1.8V max. Hence it could be have been driven directly by the STM32F103.

The 390Ω resistor serves no purpose but to waste power.


look at the picture

 

According to your photo your circuit is wrong.

 

According to your photo your circuit is wrong.

the second circuit v44 identical.
where is the mistake?

 

Your diagram shows 390 ohm resistor between Vcc and GND. That just wastes power.
The photo shows 390 ohm resistor between Vcc and NEC9587 pin-3.

 

View attachment 264934

Your diagram shows 390 ohm resistor between Vcc and GND. That just wastes power.
The photo shows 390 ohm resistor between Vcc and NEC9587 pin-3.

I changed......pb6 = +3.5 vcc ???

 

STM32F103 is powered at 3.3V.

The cathode of the opto-isolator emitter is pinned at 1.9V.
All of this is totally unnecessary as far as I can see.

 

Now the circuit is starting to look a little more "sensible".

The voltage divider on the cathode of the optocoupler allows you to switch off the LED a little faster with a negative voltage.