Pulse voltage drops trying to connect 2 devices

pulsevoltage

Hi I try to connect 2 devices. The first one sends a pulse to the second. The second has a 3.3 V input. And the output pin from the first device has a voltage of around 7.3V. VCC of the first is 12 V.
Information from the first device: the pin is active low. What does it mean?
I thought a voltage divider was a good idea to transform the 7.3V to 3.3V. (R1 =2k2 and R2=1k8)
But when I connect the output pin of the device 1 to R1 and his gnd to the gnd of the voltage divider, the 7.3 voltage drops and the voltate between R1 and R2 is 0.3V. Why does the voltage drop more than expected? is there a solution?

Best Answer

Active low means that the function of that pin is enabled when the signal on it drops to 0V of below the Vi(l) level specified in the data sheet. For instance, if a chip as a #EN or nEN line, this means that the device will be enabled when the signal on the #EN or nEN line drops low.

As for the voltage drop, you need to examine the data sheet for the pin impedance and whether it is classifies as Open collector/Drain. Without knowing which devices you are using, it is difficult to comment further. If the input side of your second device has some impedance, then it will appear in parrallel with R2. A better way to achieve level convertion is to use a level shifter or a simple low VGs fet arrangemententer image description here

Any FET may be used provided it satisfies the shift voltage and VGS(on) requirements.