NPN 2N3904 Transistor Voltage Drop

RestlessMind

Does anybody happen to know what it is, or how to find out the voltage drop of a particular transistor?

berkeman

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?

(BTW, 3904s are our friend...)

RestlessMind

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.

berkeman

Soooo, what voltage are you asking about? Where would you look on the datasheet to learn more...?

RestlessMind

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.

waht

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.

MATLABdude

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.

Maybe this would work better with a 4k7 instead of a 47k? I think, as is, the LED would still turn on, but probably just not as brightly.

vk6kro

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.

sophiecentaur

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.
A lot of the following repeats previous posts but I am putting it in a different way so that, between us all, you may get what we're all on about! I was half way through writing this when the previous post arrived with a clip from the data sheet but I don't want to waste what I've written.
In the case of the LED driver circuit, you already have 330 Ohms in series with the LED (to limit the current) and the majority of the voltage drop from 9V needs to be across that resistor; If you have a red LED, the LED voltage drop will be about 2V, and the transistor (from a 2N3902 datasheet), assuming it is driven into saturation (being used as a switch), will drop about 0.2V so there would be about 6.7V across the 330 Ohm resistor - giving about 20mA. You would need to have about 2mA driven into the base for this (according to the sheet`) so that implies a base input resistor of only about 1k. You won't blow anything up at this base current and the transistor will be driven hard on. The advantage of driving the transistor hard on is that the current through the LED is set by the top resistor and is more or less independent of the characteristics of the transistor - also, in higher power circuits, the power dissipated by the transistor is minimal because there is such a small PD across it. With 47k in the base, you wouldn't be driving the transistor anything like into saturation and you may only get a few mA through the LED.

RestlessMind

Lemme make sure I've got this straight... when a transistor is saturated, it means it is turned on fully, and "saturation voltage" is how much voltage is needed to do so/dropped?

Are you saying that the 3904 requires a volt to the base to fully turn on?

sophiecentaur

It seems counter intuitive but the Vce sat can be very low, indeed - less than Vbe (which is weird).
Vbe can also be very high but there's no point in running it that way if all you need is 20mA! Saturated 'enough' (!?!) is all you need.

People mostly use transistors as black boxes. As a switch, when it's ON it's on (ignore the resistance) when it's OFF, its resistance is high enough to be considered as infinite.

Most times you use a transistor as an amplifier, you rely on the fact that the current gain of a modern transistor is very high and that you can use plenty of feedback to get over any inherent non-linearity and still have some gain.

RestlessMind

Ah, I see... though I'm not exactly sure what Vce and Vbe mean.

Maybe I can make this easier... on a little hobbyist site, there is a circuit using MPS2222A transistors, and it states that these have a ".7 - .8 voltage drop across them" and that has to be taken into account upon circuit design. Below is a simple circuit showing what LEDs and battery I have and puts the transistor to work:



But I'm afraid that the LED won't light up fully because of this ".7 - .8 voltage drop"; and instead of getting 3.6V, getting ~2.8V or something.

Luckily, I have a bunch of spare transistors, including the 2N3904, and maybe if this MPS2222A won't work right, maybe one of those will have a lower voltage drop and I won't have to go buy MOAR parts. o_0

sophiecentaur

Use a 1k resistor in the base instead. That should give more base current which should increase the Collector current and give less Vce drop. The LED should get a bit brighter.

Vbe is the voltage measured between b and e. Vce is the c - e voltage etc..

Alternatively, use a lower voltage (different colour, probably) LED.

RestlessMind

Hm. Why would using a resistor a little less than half than the one specified decrease the voltage drop of the transistor? Just curious. Lucky for me I have some 1k ohm resistors! =D

If that doesn't work there's always the option of using an LED that requires less voltage like you said; from my calculations an LED between 2.8V and 3.25V would work (probably is a yellow one).

Averagesupernova

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). 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.

RestlessMind

I... think I sort of get this. A beta of 100 means that emitter:collector is 1:100?

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.

sophiecentaur

Why do you guys have to do it this way?
Just spend some time with Wikipedia on the subject and try to figure it out for yourself. You'll never get this if you have to ask a question at every step. There must be thousands of sources for learning elementary electronics. The Internet has really not helped at all in this respect. People now seem to think they can learn something by just asking a stream of questions.
Start with Ohm's Law then sort out Kirchoff 1 and 2 and use them to solve elementary resistor problems. Only then can you hope to figure out what a transistor is doing. There is no quick way to understanding.