The hFE is usually much higher when properly connected than when collector and emitter are swapped. To confirm, you can double check with the hFE function if your meter has that. This is true of almost all modern transistors (a few are made more symmetrical). The slightly lower of the two voltages will correspond to the collector-base junction, and the other will be the emitter-base junction. Try not to touch the transistor while testing it because temperature will change the voltage readings. To distinguish collector from emitter, you look at the voltage shown in the diode function with the one successful test that found the base. Symb ol V CBO V CEO V EBO IC P tot Parameter Collector-Base Voltage (IE 0). Now you have identified the base pin and whether it is PNP or NPN. 2n2222 - Free download as PDF File (.pdf), Text File (.txt) or read online. ![]() If the red test lead is on the base, then it is NPN, if the black is on the base, then it is PNP. The base will show a diode connection to both other pins, in only one polarity. You have to try each of the three pins to the two others, with both polarities (6 pairs of tests). You can search on this site to find out more about reverse active mode.You can determine the pinout and NPN/PNP of an unknown BJT with the diode function of a digital multimeter in seconds. There is also reverse active mode, but I won't go into it. So Vce will be larger than in the saturation case. But in active mode, the base collector junction is reverse biased. As long as you drive the base hard, you don't need to worry about accidentally letting Vce get too low. Once the collector drops below the base voltage, you are in saturation, and base current will start to increase (if you hold the base voltage constant).Īll of this just happens. You drive the base hard into forward bias, and the transistor will pull the collector down as low as it can, allowing maximum current to flow through the load. It is not like you need to do some calculations to avoid accidentally letting the voltage get too low. So the point is that if you maintain the bias conditions and current flow, you will get your few tenths of a volt. In saturation Vbc will be a bit lower than Vbe because of the geometry and doping concentrations in the various regions and there will be net current flow from collector to emitter (in NPN). We are trying to get Vce (NPN) as low as possible so that the transistor approximates a mechanical switch. ![]() Saturation is the mode we use when we are trying to just turn on some load. This is often given as the true definition of saturation. In order for an npn transistor to be in its saturated mode, both the base emitter junction and the base collector junction must be forward biased. But there is another way to look at this. You are hung up on quantifying the "few tenths". It can generally be considered a constant for most purposes, however just be aware that it will be lower than the datasheet value if you operate at much lower collector currents than the test value used in the datasheet, and higher if you operate at much higher currents. The transistor's datasheet will have a quantity listed called the saturation voltage, \$V_\$ is actually not a constant, and varies logarithmically with collector current. ![]() ![]() If Vc is too close to Ve, the transistor is saturated, not in forward (or reverse) active mode.
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