Getting 3.3V from Nano 33 BLE I/O Pins

I want to provide power to the enable pin of a device using the output of a digital i/o pin on an arduino nano 33 ble. Since the i/o pins themselves cant provide enough power, im connecting the digital pin and 3v3 pin to a bc-547 transistor. However, my schematic feels lackluster.

Something in me is telling me this is wrong/dangerous, despite the fact that it works. Is this correct? should i add a resistor from the digital pin to ground? Here is my schematic, please help.

|500x214

probablyhuman: I want to provide power to the enable pin of a device using the output of a digital i/o pin on an arduino nano 33 ble. Since the i/o pins themselves cant provide enough power, im connecting the digital pin and 3v3 pin to a bc-547 transistor. However, my schematic feels lackluster.

Something in me is telling me this is wrong/dangerous, despite the fact that it works. Is this correct? should i add a resistor from the digital pin to ground? Here is my schematic, please help.

|500x214

Why does your device enable pin need "power". Normally an enable pin just needs a signal voltage. Paul

Because... it does?? i dont know what else to tell you; i tried with the current alone from the digital i/o pins and didn't work, then i saw it was meant for 3.3v, which is what im attempting right now. My schematic works, so my assumption was correct. I simply want to know if there's a more proper way to do this, say potentially with resistors.

You have an emitter follower configuration. It is not dangerous or wrong, but is not the best solution. Depending on the transistor you are using (assuming a general purpose NPN) you will have a voltage drop of about 0,7 volts from the base voltage to the output. Your 3.3V becomes 3.3V - 0.7V or approximately 2.6V. Your circuit will definitely give you more power. Assuming a gain of 100 and 10mA from D8 you will get 1 Amp at the emitter at about 2.6V - additional drops. If you replace it with a P-MOSFET (hard to find) that will enhance at about 2,5 volt you will get the 3.3 V out. The P-MOSFET behaves like a resistor which is the RDSon rating. A PNP transistor will give you about the same output voltage but it will need a base resistor. The less parts the better, the MOSFET is the best of both worlds.

"a P-MOSFET (hard to find)" Yes, that is certainly true for thru hole parts. If you can work with SMD, and perhaps a DIP or SIP adapter, there are plenty of surface mount part for switching current to loads such as DMP2045UQ-7 with low Vgs levels.

DMP2045UQ-7.JPG|1127x684

Because... it does?? i dont know what else to tell you;

With 46 posts, most forum members would expect you to have a [u]tiny[/u] bit of sense. Follow the forum rules and post a link to the device or better, the device data sheet. Which you should read, too.

The enable pin on some interfaces sources or sinks current to/from the LED of an optocoupler, and must handle a few mA of current.

Look, the "enable pin" which i refer to isn't exactly an enable pin per say, but acts the same. i simplified it to the explanation, as it has the same function as one, and from an electrical hookup standpoint isn't very different. I could add an unnecessary explanation as to what exactly it is, but i felt as if it would be easier and i would get more help if i had summarized it as such. These intricacies are not important to the efficiency of my circuit.