11/21/2023 0 Comments Emitter base voltage![]() Note: More technical information can be found in the BC550 datasheet, linked at the bottom of this page.īC547, BC548, BC549, 2n3904, 2SC5200,BC639, BC636, 2N222 TO-92, The load (ie 20k), since 200 times less current flows for "about" the same voltage drop as the emitter.Electrons Emitted from Emitter Collected by the CollectorĮlectrons emitted from the emitter into the first PN junction If the gain of the transistor is 200, then the base looks like 200 times Vary the base voltage and see the emitter voltage follow it almost linearly. Sorry if the questions are too silly.Ĭhange your circuit and add a 100 ohm resistor from emitter to ground. However I am unable to understand impedance at all!Īny help will be highly appreciated. Also, the primary application of this config is to change high input impedance to low output impedance. (I know that if there is load on the collector side the transistor will simply act as a switch.) But if I'm not wrong, when there is no load, the voltage between Emitter and Ground is 4.2V (5V - Voltage Drop), and not 12V (or 11.2V accounting for drop)! Why is this happening? When the transistor is saturated, shouldn't current flow from Collector to Emitter and voltage should be 12/11.2 V? The collector is at a fixed voltage, the emitter has a load, and the emitter voltage tracks the base voltage less about 0.6 to 0.7V. The base-emitter voltage can't rise much above 0.7V without large currents flowing from base to emitter.Īn emitter follower is also known as common-collector. ![]() The base-emitter junction has been massively overloaded and melted rapidly. Voltage at collector is +12V and Emitter is connected to common ground. I give a NPN transistor a base voltage of +5V. Can anybody explain it in easy to understand terms? Consider the following scenario: I know basic electronics but am totally unable to understand the Emitter-Follower configuration of transistor. The emitter resistor acts as a sort of negative feedback reducing the effective gain of the transistor to one. So the equilibrium point is reached when the emitter voltage matches the base voltage minus the voltage drop across a diode which is 0.7V for a silicon diode. You might think this equilibrium point is reached when the two voltages are equal, but in fact the circuit from base to emitter looks like a diode. This continues until an equilibrium is reached where there is just enough base current to sustain a voltage on the emitter and no more. This in turn leads to a smaller voltage difference between the base and the emitter and so reduces the base current. But as more current flows through the emitter resistor, there is more voltage dropped across the resistor. The size of this current is the base current times the gain of the transistor. The instant you apply a voltage to the base, current starts to flow from the base to the emitter and causes a bigger current to flow from collector to the emitter. Notice how a mere 123♚ at the base, controls a whopping 8.3mA at the Emitter!īTW: another name for the Emitter Follower is the Common Collector. So, if R 2 is 500Ω and Vin is 5V, and V BE is 700mV, and h FE = 70, then: In other words, using this transistor arrangement, you can mirror an AC voltage, at the output, with little current impact, on the input! This is what makes this configuration a "unity-gain buffer". ![]() The fact that the output is Vin - V BE, and that V BE hardly changes, is key, to understanding this circuit.Īn interesting feature of this is: because of the current gain of the transistor, signified by the parameter "h FE", the current at the Base is far lower than the current at the Emitter. In other words, the Emitter closely follows the voltage applied at the Base of the transistor. Notice how, in the Emitter Follower configuration, the Emitter voltage, is always one V BE less than the input voltage. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |