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RestlessMind
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Does anybody happen to know what it is, or how to find out the voltage drop of a particular transistor?
Draven said:Does anybody happen to know what it is, or how to find out the voltage drop of a particular transistor?
Draven said:Here's the datasheet: http://www.fairchildsemi.com/ds/2N/2N3904.pdf
And it will be used to switch on an LED; the battery is 3.6V, 600mAh.
waht said:In this circuit the transistor is in saturation mode. If you turn it on at the base the collector voltage will be close to zero, and so you can treat the collector as a ground for most purposes. If off, the collector will remain open and no current will flow.
Draven said:Well, I'm a bit of a novice, but I know that there is a voltage drop when the current passes through the collector/emitter, so that if you send, for example, 10V through you might only get something like 9.5V of of the other side. So I want to know how to find out that voltage drop for a specific model of transistor, such as the 3904.
Are you saying that the 3904 requires a volt to the base to fully turn on?This is a clip from a 2N3904 data sheet.
Note the marked values for Vce and also the very low gain of the transistor at saturation.
Ic is only 10 times Ib.
Also in the next line, see the base voltage at saturation. Nearly a volt.
I... think I sort of get this. A beta of 100 means that emitter:collector is 1:100?You should think in terms of current when working with transistors. Get yourself familiar with the definition of beta. Beta is the ratio of emitter current to base current. With a typical beta of 100 we usually, for simplicities sake, assume the collector current and emitter current are the same. Suppose you have a load in series with the collector that draws 50 mA when 5 volts is across it (100 ohm resistor).
That's where you really lost me. Didn't you say that the load was 5V, and now its .2V? I'm confused. And not really sure how this applies to my problem.Suppose the transistor you have picked has a beta of 100. So in order to get the complete (or very close within .2 volts) supply voltage across the 100 ohm resistor you need to supply a current that is .05/100 (collector current/beta) into the base, or .5 mA. At this point, the transistor is said to be saturated. Any more current we put into the base will not cause any more current to flow in the collector circuit. However, if we decrease the 100 ohm resistor to 75 ohms, then it will take more base current to get the full 5 volts across the 75 ohm resistor. Work with currents when dealing with transistor circuits.
The 2N3904 transistor is a type of bipolar junction transistor (BJT) commonly used in electronic circuits to amplify or switch electrical signals. It is also used as a voltage regulator and in logic gates.
The "NPN" in "NPN 2N3904 Transistor" refers to the type of semiconductor material used in the transistor. NPN stands for "negative-positive-negative" and indicates that the transistor has an n-type semiconductor sandwiched between two p-type semiconductors.
The voltage drop in a 2N3904 transistor depends on the circuit it is used in and the amount of current flowing through it. Generally, the voltage drop across the base-emitter junction is around 0.7 volts, while the voltage drop across the collector-emitter junction can vary from 0.2 to 0.4 volts.
The maximum current that a 2N3904 transistor can handle depends on the manufacturer's specifications. Generally, it can handle a maximum collector current of 200 mA and a maximum power dissipation of 625 mW.
No, a 2N3904 transistor is not suitable for controlling high voltage circuits. It is designed to handle low to medium voltage applications and using it to control high voltage circuits can damage the transistor and the circuit it is connected to.