Output voltage shows zero through voltage divider!

Hi everyone,
I managed to use an attiny84 5V output voltage to send it to a Bluetooth input pin which is 3.3V and that's why I used a voltage divider . 10 and 15k allows almost a half milli amp to go through, but, the problem is, the output voltage is almost zero ( about 23mV).
Can anyone please tell me what I missed?

Regards,
Tommy

Measure the resistors with a multimeter.

Does PA1 output 5V correctly when if everything (including connections to others and divider resistors) is removed?

I measured it with multimeter actually... about 23mV

It outputs 5V actually without removing any thing, because I have everything soldered on PCB

Can you make a photo of the resistors ?

No, if you have trouble, trying to desoldering it is one of the first steps.
This is because it is possible that the soldering is defective or the component is defective.
(The one component can't be true measured unless it is removed.)

Especially, If think it's not a code issue at only your think.
If not that, show all the code here...

I assume it measures OK before connecting to the Bluetooth board? A voltage divider will always work by itself if the voltage-input is correct and if the resistor values are correct. It's based on physics and laws of nature...

If so, the input on the Bluetooth board has too-low of an impedance and it's "loading" the voltage divider. (A lower resistance in parallel with R2 will lower the voltage.)

That could happen if it's not really an "input" pin, or maybe if the Bluetooth module isn't powered-up.

That is an input-pin, right? Not the 3.3V power input?

It's rather complicated
circuit.
R317 is 15K and which is connected to GND
R318 is 10K and which is connected to PA1
I checked the continue between the tiny84 output pin, the two resistors to the bluetooth module input pin. Everything is ok and it's not shorted to ground or something like that.

No, if you have trouble, trying to desoldering it is one of the first steps.
This is because it is possible that the soldering is defective or the component is defective.
(The one component can't be true measured unless it is removed.)

It's not easy, but, I'll try desolder the components one by one to discover the wrong part.

I assume it measures OK before connecting to the Bluetooth board?

Everything is soldered to the board. Please see the pic above.

A voltage divider will always work by itself if the voltage-input is correct and if the resistor values are correct. It's based on physics and laws of nature...

agree with u :+1:

If so, the input on the Bluetooth board has too-low of an impedance and it's "loading" the voltage divider. (A lower resistance in parallel with R2 will lower the voltage.)
That could happen if it's not really an "input" pin, or maybe if the Bluetooth module isn't powered-up.
That is an input-pin, right? Not the 3.3V power input?

It's input pin not powering pin with 3.3V tolerance.
It's the pin PIO4 on HM-10 Bluetooth module actually. By selecting the right mode, it turns this pin as an INPUT mode. I tested my circuit on that BL module before I soldered it, but I used an STM32 3.3V output pin to test the circuit instead of ATtiny84 with voltage divider.

It's a small confirmation, but the line you show in red is the path to the left of the resistor, right?
Probably a miswritten?

How is the voltage at this pin in image when outputting 5V?

It outputs 5V as it should do

With actual DMM measure?

Good but bad news....
The problem is with BT module. The voltage divider works after I desoldered the BT module.

Wow.
Thanks for trying the process I presented by #8.

Did it work BT module when prototyped with STM?
Did you solder a different module than when you prototyped it?

With actual DMM measure?

Yes!

Dontcha want 15/25ths of 5 volts, not 10/25ths?

a7

You're welcome

I didn't move to this part yet, but, I can confirm that th BT is working, I can connect my mobile to and its connected LED is on when it's connected

No, the module when I test the idea before designing this part.

What do u mean?
it's Vout= (Vin ×R2)/(R1+R2) which is in my case (5 ×15)/(10+15)= about 3V