Help explaining exactly what a transistor does - Page 1
Forum section: electronics / beginners

author: tyrronic
author_rank: Newbie
timestamp: January 12, 2015, 12:20:39 am
content: So i'm working on a final project for school (also a newbie at electronics) and was going to include a tip31 transistor. From my understanding, a transistor is simply a switch using silicon as a semi-conductor, and when power is applied to the base the switch allows power to pass through it. Although, after experimenting with it, I found that if I applied power to the base with nothing connected to the collector, I could get an output from the emitter. Also, the amount of power i gave the base dictated the amount power coming out of the emitter when power was connected to the collector. I am now pretty confused, can someone explain what is happening here.

author: c4757p
author_rank: Super Contributor
timestamp: January 12, 2015, 12:54:04 am
content: I'll drop down to the internal physics a bit (though I'll keep it somewhat simple). Try not to TL;DR outCharge carriers: either an electron, or a "hole" (a space where an electron could fit).Free electrons: electrons that aren't locked to one atom. They can move around, forming electric current. Metal has lots of these, and insulators have almost none.Doped semiconductor:Hugely simplified - "intrinsic" (mostly pure) semiconductor material is "almost" conductive - that is, it has very few free electrons, but if you give some of the outermost electrons a bit of energy, they'll become free. Adding some impurities can add some free charge carriers without adding energy, because these atoms have either one too many or one too few with respect to the natural structure the electrons form. This way, doped semiconductorscanconduct current.Doped semiconductor with excess electrons is called N-type (negative), and with excess holes is called P-type (positive).PN junction:You can cram together an N-type block and a P-type block. When this happens, the excess electrons "fall into" the excess holes, canceling out the excess at the boundary. This gives a region that no longer has free carriers, so it can't conduct. It's called a "depletion region" because it's depleted of free carriers. You now have a device which will not conduct.However, if you apply a positive voltage to the P side and a negative voltage to the N side, you repel the holes and electrons inward from the ends. This shrinks the depletion region, and eventually, it's narrow enough that charge carriers can move past it. This is called "forward bias", and it causes the PN junction to become conductive.Applying a reverse voltage (positive to N, negative to P) widens the depletion region and makes it conduct even less.What you have is a device that conducts in only one direction - adiode.Bipolar junction transistor:Now, stick together three blocks:N P N. (You can do P N P as well, it just reverses all the polarities). Call them emitter, base, collector respectively.Assume, if you will, that there is a large positive voltage at the collector, with respect to the emitter. Why? Because it's useful, we put it there!Now, what happens if you put a small positive voltage on the base? The base is P, and the emitter is N, so you've forward biased the device from base to emitter. Electrons start to flow into the emitter, upwards towards the base and collector.The base is very narrow, though, by design. The electrons move into the base, but most of them overshoot toward the collector. They're negative, and the collector has a positive voltage on it, so they are attracted to it. The result is that if you apply a positive voltage to the base, you get a large current between thecollectorand the emitter - and a smaller one between the base and the emitter.Transfer function:Both PN junctions and bipolar transistors can be described by theEbers-Mollequation:I = ISexp(V / VT)I is the emitter current (or just the 'current' for a diode, it only has one). V is the voltage applied to the base (or just across the whole diode). Without getting to far into it, ISis the "reverse saturation current" which is a function of temperature and device construction, and VTis the "thermal voltage", purely a function of temperature.So a transistor can be said to convert VBE(base-emitter voltage) into an IC(collector current) proportional to the exponential of VBE. They're "exponential amplifiers".(For diodes, this same equation is often called the "Shockley diode equation", but it's the same bloody equation.)Another way of looking at it is:IC= beta IBSo the collector current is proportional to the base current, wherebetais the coefficient of proportion (anywhere from 20ish to 400ish depending on the transistor). This, of course, falls apart when it saturates, unless you allow for beta to decrease in saturation.Saturation:There's usually a limit to how much collector current you can draw. You put a positive voltage up there, but eventually, the source supplying it runs out. When this happens, the transistorsaturates. The positive voltage which attracts the collector current away from the base is gone, and all the current goes up the base instead.This is what happened when you applied power to the base with nothing from the collector. All the base current just went straight through.Switching:A lot of times, people start off learning these as switches. The real answer is that if you increase the base voltage, the collector current increases exponentially until it can't anymore - to switch, you just go between "no base voltage" (which gives no collector current) and "yup, that's enough base voltage" (which gives as much collector current as possible). Once the transistor saturates, all the excess current will come out the base instead of the collector, so you add a resistor in series so that the base voltage won't go any higher than itneedsto to saturate it.

author: dannyf
author_rank: Super Contributor
timestamp: January 12, 2015, 12:58:00 am
content: QuoteThink of a transistor as a trans-resistor: a resistor whose resistance transits / changes with a control signal.In the case of being a switch, this resistance goes from very high (so high that it practically cuts off the current flow) to so low that it is like a piece of wire.For amplification, this resistance varies anywhere from being very high to being very low.That's all it takes to understand a transistor.

author: IanB
author_rank: Super Contributor
timestamp: January 12, 2015, 12:58:43 am
content: A transistor is not simply a switch, it is an amplifier. A switch is one thing you can make out of it.There is no complete short answer to your question, however you may find some help if you try reading various topics on Hyperphysics:http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans.html#c1http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/amp.html#c2Follow the links and keep reading.Incidentally, "including" something in a project just because it seems cool is one of the biggest mistakes to make. Once you know something about what a transistor can do, you should try to think of some problem to solve with it. It doesn't have to be a real problem, necessarily. For example, here is someone using transistors to make a ghost detector:http://youtu.be/oA5cuLMHIsY

author: tyrronic
author_rank: Newbie
timestamp: January 12, 2015, 01:35:38 am
content: Thanks for the awesome responses and I actually understood all of it, first time using the forum and it is so far excellent.

author: Excavatoree
author_rank: Frequent Contributor
timestamp: January 12, 2015, 02:00:07 am
content: c4757p - great response. It should be a sticky in the beginner's section or a page in the Wiki.If anyone has insomnia, one can look at one of the "transistor operation argument" threads for pages and pages of riveting reading.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: January 13, 2015, 02:50:23 am
content: I actually did like that thread, to be honest. It's interesting to read something and think, "Ok. That makes sense." and then read a counterpoint that makes sense as well, and think, "Well, damn." and round and round we go. I learned a lot from that thread.

author: Zero999
author_rank: Super Contributor
timestamp: January 13, 2015, 08:14:13 pm
content: https://www.eevblog.com/forum/beginners/transitor-the-base-pin/

author: katzohki
author_rank: Frequent Contributor
timestamp: January 13, 2015, 09:36:00 pm
content: Quote from: tyrronic on January 12, 2015, 01:35:38 amWelcome! This is a great place to learn.

author: Paul Price
author_rank: Super Contributor
timestamp: January 14, 2015, 03:21:32 am
content: Just think of a transistor as an amplifier or a switch. It has two modes of operation.A transistor has three terminals, but most commonly the base is the input terminal, the emitter is a common terminal for both input and output, and the collector is the output terminal.A transistor works either as an amplifier or a switch. With small variations of the current flowing into the base to the emitter, large currents are caused to flow between collector to emitter. The maximum amplification is set by the "beta" of the transistor and can be adjusted down to a desired gain by choice of resistors at the collector and emitter.If a small voltage biases the base-emitter connection of bipolar NPN transistor to greater than approx .55V, the transistor starts to conduct current from collector to emitter. If a small change in base to emitter voltage (an input signal) is then applied to the base-emitter, this small change will be amplified by the transistor to produce a larger change in the collector to emitter voltage. A large current change appears a large voltage change across a "load" resistor on the collector resistor leading to the power supply, and depending on part values, a small to great amplification will be seen.If just even a little more positive voltage is applied to the base-emitter input (the base-emitter circuit), it will then cause a maximum large current to flow in the collector to emitter circuit. The current is limited by the value of the collector resistor. This amplified current will create a voltage drop across the collector resistor that will be seen to switch in value between almost zero and the power supply voltage. In this case the transistor is functioning as a switch.

author: c4757p
author_rank: Super Contributor
timestamp: January 14, 2015, 03:27:10 am
content: Quote from: Paul Price on January 14, 2015, 03:21:32 amNo. beta sets the maximum current gain when used as a switch. It's irrelevant for an amplifier, other than making sure it doesn't ruin your base biasing network. As an amplifier, the maximum signal gain is set by the transconductance, which is the partial derivative in VBEof that Ebers-Moll equation, or more simply, IC/VT.(At higher frequencies, the maximum gain may be dominated by the transition frequency fTor by the Miller capacitance, of course.)

author: Paul Price
author_rank: Super Contributor
timestamp: January 14, 2015, 03:44:36 am
content: Ok, but that explanation might be more suitable to help explain things to PHD physicists, not an electronics beginner.The beta of a transistor does relate directly to current gain and so limits and controls setting the current gain of a single common emitter amplifier. And since current gain translates into the effect of a voltage gain seen in the collector circuit, it sets the amplification in an amplifier that uses a simple resistive load to set the mid-supply voltage operating point of the collector. In this case the ratio of emitter resistance to collector does set the gain, but the maximum amplification of current, the current gain is explicitly determined and limited by the available beta.

author: c4757p
author_rank: Super Contributor
timestamp: January 14, 2015, 03:49:11 am
content: It has anywhere from zero to very mininal effect on voltage gain, regardless of how beginner-friendly you think your misconception might be. You're notsimplifying, you'relying(or you yourself have no idea how it works).Build yourself a voltage amplifier, and demonstrate it with a pathetic-beta transistor like a 2N3055, and then with a high-beta transistor like a 2N5089.If you're talking about voltage gain, or voltage-to-current gain (transconductance), that is gm= IC/VT, and that will hold for any BJT you place in that circuit.

author: Paul Price
author_rank: Super Contributor
timestamp: January 14, 2015, 03:55:41 am
content: Suppose i have small-signal transistor with a beta>400 operating in a common-emitter circuit and it shows a gain of 100. If I now switch the transistor in this same circuit to some H\v transistor switch having a low beta you would expect the same results. No.According to you, a large power transistor with a current gain of 3 will amplify a signal as much as single small signal transistor with a beta of >450 in the same circuit and have the same input and the same output impedance.Nonsense!

author: c4757p
author_rank: Super Contributor
timestamp: January 14, 2015, 04:03:28 am
content: Best I'll do tonight, as I'd rather sleep. The two transistors need different DC bias points due to the change in IS, but the differential voltage gain is nearly identical. I'd build one with real parts, but, you know,Now, if you're going to be a smart-ass with your example gain ofthree, do mind that whatever you set up to test it can supply the necessary base current. I already allowed for making sure that beta doesn't upset your bias points.

author: Paul Price
author_rank: Super Contributor
timestamp: January 14, 2015, 04:56:57 am
content: We can calculate the voltage and current gain, and the input and output impedance of acommon emitter amp using this model. Below is the equivalent circuit (without biasing network)for a CE amp attached to a voltage source (with resistance Rs) and load resistor (Rload).Current gain: The current gain (GI) is defined as: GI = Io/Iin.Using Kirchhoff's current law at the output side we have:hfeIin + Vohoe = IoUsing Kirchhoff's voltage rule at the output we have:Vo = -IoRloadPutting the two equations together we get:hfeIin = hoeIoRload + IoGI = Io / Iin = hfe / (1+ hoeRload)For typical CE amps, hoeRload << 1 and the gain reduces to familiar form:GI ª hfe = bVoltage gain:The voltage gain (Gv)is defined as: Gv = Vo/Vin.This gain can be derived in a similar fashion as the current gain. The result is:GV = Vo / Vin = -hfeRload / (DRload + hie)with D = hiehoe - hfehre ª10-2This reduces to a familiar form for most cases where DRload << hieGV = -hfe Rload/ hie = -Rload / rBESo, as you can see, Beta determines current and voltage gain of the most common of common-emitter single BJT amplifiers.

author: c4757p
author_rank: Super Contributor
timestamp: January 14, 2015, 08:53:27 pm
content: Dude... if you can't be bothered to use subscripts, can you at least put spaces between your variables?Anyway...Quote from: Paul Price on January 14, 2015, 04:56:57 amMathematically correct, but...QuoteNo.Current "gain", yeah, but you can hardly call that an amplifier, as the gain isn't exactly usefully controlled. As for voltage gain, which is the only gain that's actually used as signal gain with any regularity...GV = -hfeRload/ hieThe problem here is that (hfe/hie) isn't actually a function of hfe; it cancels out and leaves just the transconductance...hie= vbe/ ib,givenhfe= ic/ ib,given,thereforehfe/hie= (ic/ ib) / (vbe/ib)= ic/ vbe= gmOr to put it differently,hfe= ic/ ib,given,thereforeib= ic/ hfehie= vbe/ ib,given,thereforehie= hfevbe/ icgm= ic/ vbe,given,thereforehie= hfe/ gmGV = -hfeRload/ hie,given,thereforeGV = -hfeRload/ (hfe/ gm)hfetop and bottom? They cancel out."-hfeRload/ hie" is as much a function of hfeas it is a function of the capacitance of a banana because it can be written as:-hfeRloadCbanana/ (hieCbanana)(For Cbanana> 0, obviously)By the way - you're conflating hfeand hFE(and yes, I rolled with it).Edit: I suppose I should include something of a proof that it's not a function of hfe...GV = -hfeRload/ hie,givenhie= hfe/ gm,from aboveLet hfe2= hfe+ Delta hfehie2= hfe2/ gm= (hfe+ Delta hfe) / gmgm= hfe/hie,from abovegm2= hfe2/hie2= (hfe+ Delta hfe) / ((hfe+ Delta hfe) / gm)= gmgm= gm2GV = -RloadgmGV2= -Rloadgm2= -Rloadgm= GVTherefore, GV is not affected by replacing hfeby (hfe+ Delta hfe).QED.

author: dannyf
author_rank: Super Contributor
timestamp: January 14, 2015, 11:11:18 pm
content: QuoteThe maximum ***current*** amplification is set by the beta.The maximum ***voltage*** amplification depends primarily on Vbe vs. Ie and the external circuit and independent of beta. If you have an ideal transistor where its Vbe is invariant with regards to Ib, you can have voltage gains approaching infinity.

author: c4757p
author_rank: Super Contributor
timestamp: January 15, 2015, 12:33:38 am
content: Quote from: Paul Price on January 14, 2015, 03:55:41 am"and have the same input and the same output impedance" - I wonder if you added this later, or if I just didn't notice it. No, I'd never say that, that's just plain dumb.

author: free_electron
author_rank: Super Contributor
timestamp: January 15, 2015, 01:04:09 am
content: i have a theory that transistors are a conspiracy to set up engineers against each other in endless topics on whether they are current controlled or voltage controlled and what formula set to use. It keeps the engineers from inventing really usefull things like free energy, time travel and a cure for all that ails us. The governments are afraid to lose control . Furthermore ..... oops gotta go . black helicopter circling ...

author: c4757p
author_rank: Super Contributor
timestamp: January 15, 2015, 01:06:58 am
content: Quote from: free_electron on January 15, 2015, 01:04:09 amI didn't pick a side here. Like I said before, it's a dumb argument, the two go hand in hand.QuoteI've no problem with the hybrid model, either. The statement that voltage gain is a function of hfeiswrong, whatever formulas you want to use. And I couldn't just leave alone something that'swrong...

author: alterbaron
author_rank: Contributor
timestamp: January 15, 2015, 06:06:33 am
content: Quote from: tyrronic on January 12, 2015, 12:20:39 amHere's a simplified explanation for how bipolar junction transistors (BJTs) work.The base-emitter junction is like a diode. Current can flow one way, but not the other way.In an NPN transistor, when a positive voltage is applied to the base (relative to the emitter), a current flows between the base and the emitter. If you were to apply the higher voltage to the emitter and the lower voltage to the base, no appreciable current would flow.Now that's all well and good, but a transistor is more than just a diode. It has that special third terminal: the collector. Here's how it relates to the other two terminals.When a current flows between the base and emitter, electrons shoot out from the emitter and into the base (remember, conventional current is the flow of positive charge; electrons flow the other way!). Some of these electrons simply contribute to the current through the base terminal. BUT, some of these electrons can travel through the entire thickness of the base, and reach the collector. (The collector is so-named because it collects the electrons!)What do the electrons that reach the collector do? They cause a current to flow into the collector terminal!This current can be MUCH larger than the current that flows form the base to the emitter(typically between 10 and 100 times, depending on the transistor.)So a small current into the base can lead to a LARGE current into the collector. This is the basis for using a transistor as an amplifier.In summary, we can apply a voltage between the base and emitter of a BJT. This causes a small current to flow into the base terminal. This small base current gives rise to a larger collector current.There are other types of transistors too! One popular variety is the MOSFET, which is used in just about every electronic device that's made today. Read into them if you are interested!