![]() ![]() For example, if the calculated power dissipation is 100 mW, choose a resistor whose power rating is ≥ 200 mW. Therefore, you need to use the transistor's saturation beta \$\beta_)\$. You do not want to operate the transistor in its forward active mode (small signal amplification). When using a BJT as an ON|OFF switch, you want to drive the transistor into hard saturation for the ON state. This handy YouTube video also helped me understand some of the logic or what I think is the logic. Arduino driven LED stays on, touching resistor makes LED come on, and 3.3V line comes on as expected. I also noticed simply touching the resistor R b also triggers the LED, albeit quite dim.Įdit: I should add, I did stumble across this post, and even copied the resistor values from that, and came up with the same issues. ![]() When it's set to HIGH, I see ~3.3V as expected. I go back to the Arduino and check, when the pin is set to LOW, my multi-meter sees no voltage. I have a breadboard power supply that outputs 5V and 3.3V, so I take a lead and touch 3.3V to the R b and LED comes on. Remove the Arduino and make the circuit really simple. Right, so plug in the Arduino into where 3.3V is, and set the pin to HIGH, and LED is on. Using Ohm's law again, and factoring in the 0.7v drop from V b to V e we have 12k Ohms. My understanding is that I c is I b x h fe. Using Ohm's law, that gives us a resistor of 150 ohms, correct so far?įor the sake of easy math, we'll say h fe is 100. Using a standard, run of the mill, red LED, we say the voltage drop is 2V, and requires 20mA. Simulate this circuit – Schematic created using CircuitLab To test out the circuit, I figured I'd substitute an LED for the time being. Obviously the Arduino I'm using doesn't have the voltage to drive the 5V relay, so that's where the 2N2222 comes in. I'm looking to drive a relay on a 5V rail from an Arduino 3.3V pin.
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