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Collector emitter breakdown voltage12/19/2023 ![]() ![]() ![]() An electric field causes a large current to flow, so metals have low resistivity, making them good conductors. For example, in metals one or more of the negatively charged electrons in each atom, called conduction electrons, are free to move about the crystal lattice. The force of the field causes the charge carriers within the material to move, creating an electric current from the positive contact to the negative contact. An electric field is created across a piece of the material by applying a voltage difference between electrical contacts on different sides of the material. A conductor is a substance which contains many mobile charged particles called charge carriers which are free to move about inside the material. (point being there is never going to 5V or 12V on the base because it is going to be 0.Materials are often classified as conductors or insulators based on their resistivity. You know is not going to happen because if there a resistor the voltage is going to be 12V-0.7V = 11.3V and the base voltage is still going to be 0.7V) ? (hence the 'reverse' part),īut why would the OP be asking this question if his intention is to use 12V on the base (which, as Reverse Breakdown Voltage means the base is 5V less than the emitter. I'm working on a small 12volt logic gate with switching and I'm aiming for more plug and play instead of using regulators and a lot of extra things that may not me necessary.Īnd what does this have to do with applying 12V to the base ? Imagine that the BE junction is a ~5.6volt zener diode.Ī ~0.65volt diode one way and a ~5.6volt zener diode the other way.īut in some other research I read that its more current based. It's the reverse voltage you can put between BE, without resistor, and without anything happening. I base Sat = 10 times the small signal current, so depending on the collector current My guess is that you want to use the transistor as a driver and therefore want to saturate it Or if your calculating the resistor value, Now if instead of 5V, you wanted to drive it with 12V, then you would simplyįollow the above but use 12V for Vin, yielding : ![]() The base resistor R base = (Vin - 0.7V)/I base Look at the base=emitter saturation voltage, which for this transistor is specified for a collectorĬurrent of 500 mA and a base current of 50mA, in which case the base resistor would be: If you want to use it as a driver instead of amplifier you would want to saturated it and would H fe is the transistor small signal gain, which is generally a spec used for small signalĪpplications where you are not using saturating the transistor as you would driving a relay. Was 500mA (which is really pushing it for this transistor) then the base current should be:Īnd the base resistor R base = (Vin - 0.7V)/I base The 2N2222 collector current max is 600 mA and h fe 300 so if for example your load So what should you do if, say, you wanted to drive the transistor with 5V ? So if the forward voltage is 0.7V (typ) and as you already know, it is CURRENT based, and theĥV you referred to is actually the Breakdown Voltage (BV EBO) (the voltage at which the transistor junction is destroyed), and there is NO resistor if for example, you apply 5V to it without a resistor, the the base junction will conduct as much current as your 5V supply can deliver (800+ mA in theĬase of an arduino) and the base junction will fry. So if it is current based, couldn't I apply 12v to the base as long as I use the correct resistor?īecause the base-emitter forward voltage drop is a diode voltage drop (0.5 - 0.8V) becauseĪ transistor consists of diode like junctions. All of these transistors say 5 volt max on the base via the datasheets. ![]()
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