(usually about 100 times R2 which assumes a minimum gain of 100 for the transistor). Typical values for R1 would be in the thousands to tens of thousands of Ohms. The value of the resistor depends upon the curent needed - more current, smaller value. The emitter terminal is connected to the left side n-type layer. The npn transistor has three terminals: emitter, base and collector. The p-type semiconductor layer is sandwiched between two n-type semiconductor layers. Typical values for R2 would be in the range of a few hundred Ohms to a couple of thousand Ohms depending upon supply voltage. The npn transistor is made up of three semiconductor layers: one p-type semiconductor layer and two n-type semiconductor layers. The device can be used as a switch and also as an amplifier to change the values or control the passing of an electrical signal. The voltage at the emitter follows the voltage at the base (but about 0.6V less. A transistor is a semiconductor device with three terminals that can be connected to an external circuit. The circuit you have used is called an emitter follower it is not using the transistor as a switch. One other point to remember about Bipolar NPN Transistors. Calculate the base current Ibrequired to switch a resistive load of 4mA. This is bad design.Īdding series resistance will prevent damage due to excessive current as the voltages are increased. A bipolar NPN transistor has a DC current gain, (Beta) value of 200. You haven't destroyed the transistor and/or LED because your voltages are too low to do any damage (i.e. The voltage at the collector needs to be higher than the base, when fully turned ON this is at least another 0.2 volts so the minimum in this circuit should be 2.6V (V1) The minimum voltage (V2) that needs to supply the base is 0.6 + 1.8 = 2.4V There are so many things you just got wrong.įor an NPN (silicon) transistor to turn ON the base needs to be at least 0.6V more positive than the emitter.įor a (red) LED to turn ON you need at least 1V8 across it.
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