![]() ![]() That is 0.3V at I C=150mA according to the datasheet which is close enough to our target of 100mA to use. Next we look at the datasheet for a 2N2222 transistor and look up the at the V CE(sat) which means the voltage from the collector to the emitter when the transistor is in saturation (fully on). The voltage across the diode will be less than 1.7V according to the datasheet, so 5V - 1.7V = 3.3V. We could either use a 3.3V Vcc or a 5V Vcc. The datasheet says that the maximum continuous current is 100mA (your datasheet may specify some other number). The voltage and current for the IR diode, as mentioned above, we can get from the datasheet. In your case the load is the IR diode, and whatever current limiting resistor might be needed. Simulate this circuit Load resistor calculation The typical arrangement for driving this from a GPIO port (for any processor) looks like this: In fact, we seek to avoid it and operate the transistor solely in one of two regions: cut-off and saturation. However, if we're looking for a binary on/off operation, we are not interested in the linear range. That's useful if you're using a transistor as an analog signal amplifier. There's a linear region in which transistors act, well, linearly as the current amplifier described above. We want it "on" if the input voltage is high and "off" if the input voltage is low. What we often want from a transistor, as in your case, is for it to act like a switch. Everything's mostly the same for a PNP transistor except that the currents and voltages are reversed with respect to an NPN transistor. A 2N2222 is an NPN transistor, which is also the more common kind of bipolar transistor, so the rest of the discussion will assume NPN. Direction of the current depends on whether the transistor is PNP or NPN. That's split between the base (tiny) and collector (most of the current). The next thing to remember is I B + I C = I E all of the current flows through the emitter. The actual value of h FE varies, but a typical value is 100 or more. That is I B * h FE = I C (approximately). This a very simplistic way of thinking about it, but it's actually still useful in practice. ![]() The gain of a transistor, specified as h FE is roughly the ratio between the collector current I C and the base current I B. One useful way is to think of them as current amplifiers. There are many ways to think about transistors, though. If you're using a different IR diode, you'll have to find and look up its datasheet to get the specifics for your part. In this circuit, the forward voltage is around 1.6V (which is under the the maximum V F of 1.7V per the datasheet) and the current is around 85mA. Note that I've shown your four 10 ohm resistors as a single 40 ohm resistor for clarity. Simulate this circuit – Schematic created using CircuitLab The circuit you describe with the IR diode and no transistor looks like this: In your particular example, however, there is an IR diode as well. So if we had a 5V supply and a 100 ohm resistor across it, the current I would be V/R = 5V/100ohm = 0.05A = 50mA. Ohm's law expresses the relationship among resistance (R), current (I) and voltage (V): V = I * R. There's not quite enough information in your question to give a definitive answer, but let's go through the design steps so that not only can you figure out this one, but you might be better equipped to solve the next transistor question that occurs to you. ![]()
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