![]() So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. The BJT's base-emitter junction current is an exponential relationship. The base resistor's voltage drop is a simple linear relationship to the current passing through it. This is why a resistor is often applied to the base circuit. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.) Of course it was getting hot! It was dissipating serious power. So the BJT was being literally flooded with base current. So far, so good.īut the idea of directly tying \$+5\:\text\$!! This is way, way, way above what you should have been using. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.) Hope all is correct, the last time I did these calcs was over 40 years ago. The 200 Ohm resistor will limit the LED-C-E current to 14.5 mA. This is less than the 600 mA Ic allowed by the data sheet. Use a standard 1.5k (1.47k) Ohm which will allow 2.67 mA Base current which means the Ic current will be ~ 534 mA (with Hfe =200). A value of 430 Ohms would be ~ 10 mA Base current, but. That is why you MUST PUT a current limiting RESISTOR in series with the BASE. This circuit has no reverse voltage, BUT with 5 V on the Base and ~ 0.7 V drop across the B-E junction, you will have 4.3 V into a short circuit (~ 0.05 Ohms for wiring) which is 56 Amps / 369 Watts! Neither the transistor or the power supply will last more than a microsecond. The Base Emitter junction is a small diode with a reverse breakdown voltage specified at 6 V. This means that if you have a current flow of 1 mA in the base-emitter, there will be a current flow of 200 mA through the collector - emitter. On the ON Semi Datasheet, see hfe on page 2 for the gain, which is between 50 and 375 for this transistor. A small current from the Base to Emitter causes a large current from the Collector to Emitter. However, I had the original circuit built improperly there, and I'm now wondering if the 2N2222 can take 5 V on its base pin (according to the datasheet).īi-Polar transistors multiply current. * Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source?. I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5 V - but that may have been due to being wired improperly, I'm not sure. Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V on it? From reading the datasheet (below) would you expect that to be too much?Īlso, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40 V(?). Simulate this circuit – Schematic created using CircuitLabĪccording to the datasheet the emitter-base maximum voltage is 6.0 V. I've been struggling with understanding datasheets for transistors. I have created the following circuit to better understand how to use a transistor as a switch*.
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