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Is it possible to have normally closed switch with npn transistor when the voltage at the base is 0v? What happens if the base voltage increases or decreases? How can i turn off the npn.
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2 2n2222 npn transistor has hfe values given in the picture 1. I provided a very simple npn transistor set up and hfe values as shown below. Does the base voltage has to always be (0.7v) in order for a bipolar npn transistor to work?
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I'm attempting to use an npn transistor as a switch in a 3.3v electrical circuit, and have been encountering some difficulties getting it to work properly.
Saturation in an npn bjt is when vbe is > 0 v and vbc is > 0 v at the same time. When the voltage at ref pin of tl431 is less than 2.5 v, vka is 5 v. There are many options on the internet, some using a logic ic and some using 2 npn transistors. The npn transistor base and emitter form a diode.
As for the circuit given in picture 2, i set ic= 4.2 ma and vce=7.5 v arbitrarily. For simplicity i assume ic=ie. I'm really not into the. So far i only managed to achieve this result with npn transistor.
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2n2222 datasheet i am trying to calculate parameters such as base current, collector current,.
The emitter (which is heavily doped) has a lot of charge carriers, and the base (lightly doped) has few. When the voltage at ref pin of tl431 is greater than 2.5 v, vka is 2 v.
