5/17/2023 0 Comments Transistor biasedHence, a maximum ratio of Ic/Ib=10 is a good and safe INDICATION for saturation. And as a kind of "rule of thumbs" we require: The base current Ib should be at least one tenth of the collector current (which is 10 times larger than for linear opeeration, assuming a specified value B=100). Therefore, the current into the base must be much larger than under linear operation conditions (amplification). So - how we can be sure that this condition is realized?Īnswer: The current into the base node now must consist of two currents: The "normal" current to the emitter and - in addition - the current to the collector. The DEFINITION for saturation requires that the base-collector pn junction is forward biased. "Also in some tutorial it says "The amount of biasing current flowing through resistor R2 is generally set to 10 times the value of the required base current IB so that it is sufficiently high enough to have no effect on the voltage divider current or changes in Beta." What does that mean exactly?" so 100K or so would probably take care of that unless you're expecting a flood. You might also want to consider leakage across the switch etc. Even with R1 infinite it will typically leak only a few uA through the relay coil at the maximum junction temperature. To get an actual maximum value you'd have to calculate the leakage from Icbo and gain at the maximum operating temperature. Note: The value of R1 above is very conservative for most situations. So R1 = 0.7V/.2mA = 3.5K (maybe you use 3.6K as a standard value). R1 is used to ensure the transistor turns off fully so, you might want to make it such that the current through R1 is 1/20 of the base current, or 200uA. So, let's say we want Ib = 4mA, that means that R2 must be less than (12-0.7)/0.004 = 2.83K. DC biasing is used to establish fixed dc values for the transistor currents and. The current through R1 is subtracted from the base current, so R2 must provide even more current in order to feed both. A transistor must be properly biased in order to operate as an amplifier. When fully saturated, there should be a much lower drop. You want to drive the transistor deep into saturation to minimize the voltage drop across it, not provide the minimum current that (under nominal conditions) might drop a volt or two across the transistor. A forward-biased pn- junction is comparable to a low-resistance circuit element because it passes a high current for a given voltage. The emitter base is forward biased in the n-p-n transistor the collector base is reversed biased in the n-p-n transistor.They are saying that you should have Ic/Ib ~= hFE/10 or about 11 in this case. Biasing to the transistor ensures that it operates in the active region. Note: : In n-p-n emitter to base causes the flow of electron to n type. Emitter size is more than base but less than the collector. Section which collects the majority of charge carriers and supplied by the emitter is called collector. Emitter base injects a large amount of charge carrier to the base. In an n-p-n transistor is the movement of negative electrons, through the base region which constitute transistor action. Base is the most important factor in n-p-n transistors. In this transistor the flow of electrons is from emitter to collector, the base diode is slightly doped, the emitter diode is moderately doped and the collector diode is heavily doped. In base emitter transistors the current is flowing forward biased, whereas, in collector and emitter transistors the current is passing in reverse biased.Thus, the emitter-base junction is forward biased and collector-base junction is reverse biased.Īdditional information: In n-p-n transistor, there are three terminals namely base, emitter and collector. In the above figure, the base is connected to the p transistor, whereas collector and emitter are connected to the n transistor.
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