transistor base emitter collector identification

#Transistor base emitter collector identification how to#

So in order for the transistor to provide load current R1 must provide enough current to supply both the zener and the transistor. For instance, the transistor has what is called current gain, the ratio of base current to collector current. This is all a gross simplification, of course, but at your stage of study take it as true. When the zener voltage equals 5.6, the load voltage equals 4.9, and everybody is happy. So what actually happens is that, for example, as the zener approaches its operating point, so does the corresponding load voltage, but always about 0.7 volts lower.

Then the base-emitter junction will be reverse-biased by about 1.7 volts, the transistor will be turned off, and the load voltage will be zero. But let's assume that it does, say, 5.9 volts. I hope you see that the load voltage cannot get to more than about 4.9 volts. The voltage across the load will increase with increasing current. Simulate this circuit – Schematic created using CircuitLab

You've left out a most important component - the load. Or, if you do try to control base voltage, use feedback to keep the voltage at exactly the right point.

#Transistor base emitter collector identification how to#

How to get around this? Don't try to control the base voltage, but rather the base current. If you try to control a transistor by explicitly controlling base voltage, as you have realized the effective resistance will change from very high to very low over a rather small span. Transistor stands for 'transfer resistor', and in principle that resistance (collector to emitter) can span a very large range, from microamps to amps. In other words you paint an impossible scenario. With 10 volts at the emitter (by some magical means) AND 5.6 volts at the base, the base-emitter region is completely 'off' and thus no collector current can flow. Voltage at the emitter be (5.6V - 0.7V = 4.9V) and not 10V? Zener voltage would still be something like 5.6V, why would the the emitter has to follow the base voltage minus the 0.7 volts (or so) that is required to forward bias the base emitter region.Īnd assuming the Vin (input voltage) is something like 10V, while the

This circuit is called an emitter follower i.e. If more current tries to pass from the collector to the emitter, the emitter voltage might start to rise and this would shut-off the base-emitter junction and reduces conduction. This places the base 0.7 volts higher than the emitter. Given that the base has 5.6 volts on it, the base-emitter becomes naturally forward biased when the emitter is at 4.9 volts. Think about the base and emitter - to get current through the collector to the emitter, the base-emitter region must be forward biased.