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Does anybody happen to know what it is, or how to find out the voltage drop of a particular transistor?
Depends on which voltage drop you're asking about, and under what conditions and which topology. Could you post a link to a 3904 datasheet, and talk some about the topology you are interested in?Does anybody happen to know what it is, or how to find out the voltage drop of a particular transistor?
Soooo, what voltage are you asking about? Where would you look on the datasheet to learn more...?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.
I'm not too sure about saturation; assuming the LED driving input is 3.6 V, the LED voltage is 2 V and V_ce(sat)=0.2 V ~ 0 V, I get i_b to be 66 uA and i_c to be 21 mA. The beta would then be 320 instead of a few dozen, or say 32.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.
I am not just being picky but it might help to get this a bit better sorted out. You don't "send Volts through" things. Volts are applied across things - they represent the energy used in getting a current through. The total supply volts are 'shared out' by the series components in the circuit. In electronics it is often important not to put the cart before the horse and to get your terms as precise as possible when you are trying to understand what is going on.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.