Basics of Common Emitter Amplifier Design - Page 1
Forum section: electronics / beginners

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 30, 2017, 02:00:13 am
content: Okay, so after posting a thread about an audio amplifier circuit, I was told to read a bit about transistor circuit design. It was excellent advice because I know nothing about the process, only going off of projects found online. I have Paul Scherz's and Simon Monk's Practical Electronics For Inventors, so I thought I would start there.After reading through the section on common-emitter design, I am a little confused as to the step-by-step process of designing, or more specifically, biasing the transistor.Let's say I have a +9VDC source to power the circuit, and my transistor choice is a 2N4401. If I want a gain of 80, according to the datasheet, the collector current Ic should be 10mA, and Vce should be 1.0V. Does this mean that to calculate the collector resistor Rc, the equation is Rc = (Vcc - Vce) / Ic? So, Rc would come out to be 800 ohms, or am I missing something here?Going from that point assuming I am on the right track, biasing the base voltage and current would be a matter of Ic/href = Ib, then Rb = (Vcc-Vb) / Ib, where Vb is my desired biased point, so for this example, 4.5 volts? So 10mA/80 = 125uA ==> (9 - 4.5) / 125uA = 36k, creating a voltage divider at the base of two 36k resistors (well 33k in real component) in series? Again, am I on the right track here, or am I way off?According to the book, a transistor has a transresistance that, usually, is not significant enough to worry about, and therefore Ic = Ie.Any help in understanding this foggy topic is greatly appreciated.

author: Cerebus
author_rank: Super Contributor
timestamp: September 30, 2017, 02:52:11 am
content: It's literally my bedtime, as I was having a last check before toddling off to the land of nob. If no one else has posted a cogent reply by tomorrow I'll come back and fill in the details.One quick comment, it sounds as if you're misreading things a bit judging byQuote from: sshoptaugh1991 on September 30, 2017, 02:00:13 amThat sounds as if you're looking at the current gain graph, finding a particular current gain and picking off the related conditions. No, you want tochoosethevoltagegain you need, thenchoosea collector current that's appropriate (which is going to be determined mostly by your desired output impedance and load). Don't expect too much voltage gain out of a simple common emitter amplifier, long story short and somewhat simplified the highest gain you'll get is 20 x Vcc.Really quick design run through: As we're talkingsimplecommon emitter amplifier, just pick your quiescent collector voltage to be 1/2 Vcc; your chosen collector current and ohms law tell you the value for the collector resistor, Rc = (0.5 * Vcc)/Icquiescent. Pick your emitter resistor to be Re = Rc/(chosen voltage gain). Work out what voltage that Re will drop at Icquiescent, bias the base to about 0.6-0.7 V higher than that voltage from Vee using a voltage divider between Vcc and Vee, pick resistors for that voltage divider than carry about 1/10 the current that Rc carries.I know that's rushed, but it's late. If you want to give it a go in your favourite simulator try these simple design parameters. Vcc = 10V, voltage gain = 10, collector current = 1 mA. Don't forget a DC blocker capacitor on the input, in a simulation it's OK to make it massive so pick 1 farad and we don't have to get into calculating you rlow frequency limit until another day.Damn, this version of BBcode doesn't support the  tag, so I can't calculate the resistor values and leave them hidden for you to check after you've had a go. I'll just put a lot of line feeds in. Scroll down for answer.Rc = 5k, Re = 500R, Top divider resistor = 88.5k, bottom = 11.5k.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: September 30, 2017, 07:16:29 am
content: The base voltage is set by potential divider at a somewhat arbitrary value of a volt or two. If you say 1.7v then that gives an emitter voltage of 1v which makes calculations easier.The emitter resistor determines the collector current, so 100R would give 10mA for example.The collector resistor needs to drop (Vcc-Ve)/2 or 4v, so at 10mA that would be a tenth of 4k. Call it 390R.The base potential divider is the slightly more difficult calculation. It needs to have a ratio of (9-1.7)/1.7 or 4.25 to 1.It needs to have an impedance low enough that it can supply the required base current without to much voltage droop., but must not be so low impedance that it loads down the signal. If Ic is 10ma and we assume a worst case gain of 80, then the base current requirement is 10/80 mA or 0.125mA.If we decide we can tolerate 0.25v low base voltage under worst case low gain conditions, then that means our potential divider cannot have an impedance much higher than 2k. Therefore we could use 2k7 lower, a 12k upper which will give 1.65v at the base 9*2.7/(2.7+12) which is near enough. Taking the upper resistor to 10k would give 1.9v which is a tad high.In most cases the gain will be well over 80 so the base current will not load this pair of resistors significantly. You need to allow for the odd bad sample with a low gain, though.So overall, 390R collector, 100R emitter, 2k7 and 12k base.-and of course you need an emitter resistor bypass cap. The value should ideally be enough to make the impedance less than the emitter resistor divided by the hFE of the transistor, at the lowest operating frequency. However, a fraction of an ohm at 20Hz would require a ridiculous size of cap. Something like 470uF will do, although this will cause some roll-off at bass frequencies, with an impedance of 17 ohms at 20Hz.BTW the voltage gain is not easy to calculate since what we are doing is to drive an AC signal voltage into a diode, the b-e junction, whose conductance will be nonlinear. There is no simple equation which will give us a definite value for this.If you are building a real project I'd suggest using a two-transistor 'opamp' stage instead. Vastly better performance for the sake of only two or three more components. The c-e amp mainly dates from the days when transistors were costly items.

author: Cerebus
author_rank: Super Contributor
timestamp: September 30, 2017, 01:44:15 pm
content: OK, back. Now I've time to tackle your questions point by point.Quote from: sshoptaugh1991 on September 30, 2017, 02:00:13 amAs I said above, it looks like you're picking out values from the data sheet for acurrentgain of 80. That's not usually what you want to design for in a common emitter amplifier (let's just sayCEAfrom now on as I know I'm gonna have to say it a lot). Usually your primary design parameter for aCEAis the voltage gain of the stage. The current gain of bipolar transistors is a very poor parameter to design around, it varies with temperature, VCE, VBEand varies wildly from one transistor to the next.So basically, if you want to keep your sanity, pickvoltage gainas your primary design parameter for aCEA. TheCEAtopology is almost always chosen in cases wherevoltage gainis what you're asking out of an amplifier at that point.I'll let you into a secret that no one who actually knew what they were doing told me until I found it out for myself. For simple amplifier circuits like theCEA, in small signal situations, in typical uses, in typical applications (usually audio frequency stuff) the choice of transistor to use just plain doesn't matter. A well designed small signal amplifier can be made relatively insensitive to transistor choice, and once you know a bit more and can design in some negative feedback, a small signal amplifier can be designed so that you can throw in almost any old transistor (within reason) and it will still work well. The point being that in the learning stage of things good design matters more than the parts chosen and to practice design of the basics you can use just about any old (small signal) transistors that you can lay your hands on.I mention the above because as a youngster, for years, I though there was some magic involved in picking parts and that a circuit design I'd picked up from someone else wouldn't work until I hadjustthe right part. That was bunkum, and a well designed transistor circuit will still work quite well even with parts that are distinctly the wrong choice.QuoteYou want topickthe quiescent collector voltage for yourself.To avoid assumptions about what terminology you understand,quiescentconditions are when the circuit is just sitting with DC bias voltages applied, but no signal applied to the circuit. So if we talk about the quiescent collector voltage (VCquiescent) we're just talking about a special case of the collector voltage (VC) when the circuit is in quiescent conditions. And VCis just the voltage on the collector, relative to ground, VCEis the voltage on the collector relative to the emitter, and so on. VCCis conventionally used to imply the positive rail voltage and VEEthe negative rail voltage, and most times it's assumed that VEEis also ground if no other contrary indication is given.The classic voltage to pick for VCquiescentis 0.5 * VCC. We pick that because it gives the collector the most room to move around (being halfway between the voltage rails), thus it means that we can get the biggest possible voltage swing going on the collector (with a signal input present) before it becomes limited by reaching the rail voltage.You can pick any arbitrary voltage, which you might do if you've got a DC coupled second stage and you want the first stage to provide bias voltage for the second stage. Whatever voltage you choose, you should always make sure that you take into consideration what voltage swing is available at that chosen voltage so that you don't get unintended signal clipping by running out of that range. If your stages arenotDC coupled then approximately 0.5*VCCis the obvious choice as it gives you the biggest potential voltage swing.You also need to chose ICquiescent. The choice is dictated by a number of things, desired output impedance, desired input impedance (the current gain of the actual transistor used comes into that), power budget, cooling budget and a bunch of other things. It this stage in the game, while you're feeling your way, I suggest you ignore practical drivers and make an arbitrary choice in the range 250uA - 5mA. After you've got a few paper or simulated designs under your belt you can come back and revisit this. Then you can design for some chosen impedances and pick ICquiescentto suit some practical aim, but until then the range I've given you won't hold any nasty surprises for VCC<= 15V.QuoteOK, yes, you're going a bit off-piste there because you've misunderstood the whole IC~ VBErelationship. I'm going to take a break here and we can come back to it later. Perhaps you can review what I've written, ask questions and I'll come back to it all in another message.We're halfway there. We've said:Pick a voltage gain GVPick a quiescent collector voltage VCquiescentPick a quiescent collector current ICquiescentNext we need to:Calculate a collector resistorCalculate an emitter resistorCalculate the bias voltage for the basePick a base biasing scheme and calculate the resistors for itBTW I assume by "Ic/href= Ib" you really mean "IC/HFE= IB"QuoteThe relationship between ICand IEis best summed up by IE= IC+ IB. Because IBis HFEtimes smaller than ICit can be ignored in a lot of calculations, but it's best to remember the relationship so that you don't forget that IBhas to go somewhere.

author: Cerebus
author_rank: Super Contributor
timestamp: September 30, 2017, 02:29:39 pm
content: Quote from: IanMacdonald on September 30, 2017, 07:16:29 amIan, when we're dealing with a novice, it's not particularly helpful to use a method that starts with a "somewhat arbitrary value". The novice's obvious next question is going to be "How do I pick this arbitrary value". Novices like definitiveness, certainty, for there to be a "right answer".Doing it the other way around, from an arbitrary VC, with a default choice of VC= 0.5 * VCCand a short explanation of why voltage swing makes that a good choice is much easier for a novice to follow.Try to remember that this fellow understands next to nothing, as you or I did when we were starting out. Try and put yourself in his boots. The "I" of now can understand why you put:Quote from: IanMacdonald on September 30, 2017, 07:16:29 ambut I'm pretty damn sure that when I was trying to understand the very basics of biasing a common emitter amplifier I would have had no clue why you picked "(Vcc-Ve)/2" there. You haven't explained your choice, and absent an explanation, an novice will grasp this as amagic formulato be learned and trotted out when required. You must remember when you were at this stage? I do, and I'm fricking ancient!QuoteErm, what's wrong with GV= RC/RE? Or if, as you have, you've over-complicated yourfirstexample for a novice by introducing an emitter bypass capacitor: GV= RC/(0.025/IC)?Oh, and "Nae Campbells!" by the way.

author: danadak
author_rank: Super Contributor
timestamp: September 30, 2017, 07:01:19 pm
content: Some ref material -http://www.radio-electronics.com/info/circuits/transistor/common-emitter-amplifier-design.phphttps://www.electronics-notes.com/articles/analogue_circuits/transistor/transistor-common-emitter-amplifier-circuit-design.phphttp://www.hunter.cuny.edu/physics/courses/physics222/repository/files/pdf/ElectronicsLab15.pdfhttp://ece.tamu.edu/~kentesar/ELEN326/lab1.pdfRegards, Dana.

author: Cerebus
author_rank: Super Contributor
timestamp: September 30, 2017, 07:11:46 pm
content: Sooner or later we're going to want a diagram for everybody to refer to, so here it is:

author: orolo
author_rank: Frequent Contributor
timestamp: September 30, 2017, 07:59:48 pm
content: As Cerebus said, you are confusing current gain with voltage gain. If you are beginning with transistor amplifiers, which is a really fascinating topic, the best you can do is thorougly read some good, organized material. BJTs are, at first sight, very wild components: collector current depends exponentially on base voltage. Their behavior is extremely dependent on bias point, temperature, working frequency and manufacturer variations. Simple linear models for small signals do not always extend well to stronger signals. They have three terminals you have take care of, neither two nor four, which would be more natural. And (many times overlooked by beginners) the signal going into (and out of) the transistor cannot be arbitrarily set: the transistor may alter (load) its input and be altered (loaded) by its output.If you are starting with transistors, take a look atthese very well written notes. Particularly, the transistor chapters1,2and3.I also find that very old (from the 1950's-60's) transistor manuals fromGeneralElectricandRCAare fascinating and charming. Discrete transistor amplifiers were a hot topic then, and it shows. The manuals are a bit germanium, PNP centric, but that only adds to the charm.

author: LvW
author_rank: Frequent Contributor
timestamp: October 02, 2017, 02:30:30 pm
content: Quote from: orolo on September 30, 2017, 07:59:48 pmYes - full agreement! That means: Physically spoken - the BJT is a voltage-controlled current source Ic=f(Vbe).(Not only "at first sight").However, it is important - in order to avoid confusion on the questioners side - to point to the fact that the document referenced by you describes the transistor as a current-controlled device, which is simply false! There is another fundamental error in the paper: It is mentioned that the temperature sensitivity of Ic would be caused by a change of beta, which also is false! Rather, it is the factor Io in Shockleys well known exponential expression Ic=Io*exp(Vbe/Vt), which is sensitive to the temperature.

author: csar
author_rank: Newbie
timestamp: November 07, 2017, 04:20:26 am
content: Hi guys,I am confused about the Common Emitter Amplifier. The resistor RE is added to stabilize the operational point of transistor. However, when we bypass this resistor to increase the gain, the gain becomes dependent of gm (BJT transconductance), which is temperature dependent. My question is: Does it make sense to use the resistor RE and the bypass capacitor at the same time? Does not one effect cancel the effect of the other?

author: Cerebus
author_rank: Super Contributor
timestamp: November 07, 2017, 01:07:07 pm
content: The resistor is setting DC bias and quiescent conditions, the capacitor is setting [small signal] AC gain. So, yes, it does make sense to use them together and they don't cancel each other out but rather they offer a frequency dependent transition from small signal AC to large signal DC conditions. Does that make sense of it for you?

author: csar
author_rank: Newbie
timestamp: November 07, 2017, 01:47:49 pm
content: Hi Cerebus,We use the resistor RE to stabilize the quiescent point so that the small signal parameters do not depend on temperature, beta etc. Right?With the capacitor, however, the gain of amplifier is dependent of transconductance that is dependent on temperature.So, in this case, is the resistor RE used just to stabilize the amplifier parameters (gain, input impedance, output impedance) in terms of beta?Thanks

author: Cerebus
author_rank: Super Contributor
timestamp: November 07, 2017, 02:40:22 pm
content: Quote from: csar on November 07, 2017, 01:47:49 pmYes, as far as it goes. Obviously it's not perfect, there will always be some variations.QuoteYes. The ultimate solution to stabilising everything (again, within limits) is overall negative feedback, but we're just talking about a single transistor stage here, so REis providing local negative feedback and generally does a pretty good job. It can be helpful to think of the parallel resistor and capacitor as a single effective frequency dependent impedance (ZE= (RE-1+ 2 pi f C)-1, if you will) rather than two separate components.QuotePersonally I prefer to think in terms of VBE, IEand the consequent V(RE) and ignore beta as a design parameter - you have to account for it, providing adequate base current and so on, but it's a big mistake to design around beta. Better to design so that a circuit is tolerant of a wide variation in beta.

author: LvW
author_rank: Frequent Contributor
timestamp: November 07, 2017, 04:17:24 pm
content: Quote from: csar on November 07, 2017, 01:47:49 pmIt is correct that the gain depends on the transconductance gm.However, which quantity determines the transconductance gm? It is theDC collector currentonly!Remember: gm is nothing else than the slope of the Ic=f(Vbe) characteristics gm=dIc/dVbe.And - fortunately - this DC current (and with it the OP point) ist stabilized by RE against temperature effects and device tolerances.This is the effect of negative voltage feedback.And note: To keep the transconductance gm (relatively) constant (and with it also the ac gain value) is the main reason for stabilizing the DC current Ic !!

author: csar
author_rank: Newbie
timestamp: November 07, 2017, 08:33:53 pm
content: Hi LvW and Cerebus,Thank you for your helpLvW,The transconductance depends on the collector current and temperature (gm = Ic/Vt, Vt = KT/q). The dependence of gm with Vt was what motivated my question.Cerebus,Is it correct to conclude that the commom emitter amplifier isolated is not a practical solution? (Since its gain is strongly dependent on temperature variations).

author: jmelson
author_rank: Super Contributor
timestamp: November 07, 2017, 09:06:59 pm
content: Quote from: csar on November 07, 2017, 08:33:53 pmI don't know where you get this. Without any emitter resistor, then it is not a stable or practical circuit. The AC gain is not the problem, it is the critical nature of the bias point.But, by putting in a modest emitter resistor, it solves this problem. Gain and the bias point will shift SLIGHTLY with temperature, but it is a completely practical circuit that is used many places, where absolute linearity and gain are not critical. It is unlikely you could ever see 5% gain variation over any temperature range you will experience, unless you work in cryogenics or down-hole well instruments.Jon

author: Cerebus
author_rank: Super Contributor
timestamp: November 07, 2017, 10:49:51 pm
content: Quote from: csar on November 07, 2017, 08:33:53 pmIt's notstronglydependent, the net effect of temperature (when you take the temperature dependence of saturation current into account) is a change of approx. -2.1mV/C in VBEto maintain the same collector current. Even a moderate REwill provide enough negative feedback to keep that under control in most circumstances. If you're working in an environment where you'll see significant temperature swings you can compensate the VBEbias source with a similar diode junction so that it moves at the same rate.It's not really that big an effect. Here's a simple common emitter amplifier:and here is its AC gain plot for 0, 25, 50 and 70 C:Not much difference, the 200R REworks!You've got to remember that the simple CEA (or indeed any other single transistor amplification stage) is a building block, you can't expect it to cope with all possible amplification demands. If you've got a design problem that is simple enough to solve with a single CEA stage, then it's probably good enough to do that job. If it's a bit more complex you might need some temperature compensation, or perhaps an active (current source) collector load. If you've got a bigger design problem then you'll have several building blocks connected together and in 9 out of 10 cases you'll be relying on overall negative feedback to tie all the competing circuit characteristics together and keep them under control.

author: Audioguru
author_rank: Super Contributor
timestamp: November 07, 2017, 11:31:10 pm
content: A common emitter transistor with its emitter resistor bypassed with a capacitor has very high voltage gain. This high gain causes very high distortion if the output voltage swing is high because the gm is reduced when the collector current drops on the positive parts of the signal. Here is an example of a transistor with its emitter resistor bypassed and not bypassed:

author: Cerebus
author_rank: Super Contributor
timestamp: November 08, 2017, 12:20:00 am
content: Quote from: Audioguru on November 07, 2017, 11:31:10 pmOr it could just be that you've got your quiescent collector voltage set to 6V, are feeding a 25mV peak signal into an amplifier with a gain of -160 (25mV * 160 = 4V) with a 10V rail and are therefore driving it way too hard? You're trying to drive the collector to 10V, the rail is 10V and there's a resistor in the way. If you grossly overdrive it of course you're going to get gross distortion.You make it sound as if there is something fundamentally wrong with that circuit topology. There is not, as long as it's being fed appropriate signal levels and given an appropriate load. Anyway, you can always reduce the gain by having a resistor in series with the emitter bypass cap - it's possible to tailor both the DC quiescent and AC small signal gain to suit your needs.

author: Audioguru
author_rank: Super Contributor
timestamp: November 08, 2017, 01:38:14 am
content: The transistor with high voltage gain and high distortion is far from being cutoff because its collector voltage reaches only 8.5V when at 10V it is cutoff. When the input signal causes its collector voltage to increase then its collector current decreases which decreases its gm so it needsmoregm or input signal for the output to swing higher.But when the input signal causes the collector voltage to decrease then its collector current increases which increases its gm.Then the output isverydistorted without negative feedback to clean it up.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: November 08, 2017, 03:56:57 pm
content: Isn't a question of clipping, rather that you are trying to drive a voltage into a conducting diode. Therefore the base current flow is bound to be nonlinear.Removing the emitter bypass cap reduces the distortion because the circuit then works partly as an emitter follower, with the emitter voltage nearly following the base voltage. Since that also kills the ability of the signal to drive current into the base, it also drastically reduces the gain. That could be seen as neg f/b. Result is the same anyway, less distortion but at the expense of a lot less amplification.A better solution is to add a PNP transistor and a couple more resistors to make a complementary amp. For negligible extra component count you have a much better all-round performance.Is the C-E amp a 'broken' arrangement? No, but it's a very sub-optimal arrangement. In the early days it was favoured because it got the most gain out of an expensive transistor. Nowadays transistors are cheap and quality electrolytics more costly, therefore if you can eliminate a few large caps you've saved money and board space.

author: Cerebus
author_rank: Super Contributor
timestamp: November 08, 2017, 04:59:25 pm
content: Quote from: IanMacdonald on November 08, 2017, 03:56:57 pmHereit's not a question of what makes a better arrangement,hereis where two beginners have come asking for help understanding aspects of the good ole common emitter amplifier. Heck, any op amp is probably a 'better arrangement' in the modern world but 'chuck an op amp at it' would not increase any student's understanding of the common emitter amplifier. This isn't intended to be having a go, just an attempt to keep on topic before we wonder off into "What's my favourite BJT amplifier topology?" (Baker's 'diamond' transistor, if you're asking).Quote from: IanMacdonald on November 08, 2017, 03:56:57 pmYet it lives on, the majority of the voltage gain in 99.5% of op amps is provided by a common emitter amplifier, with a constant current collector load and, ultimately, a boatload of feedback. It's a critical building block in a huge number of circuits and, crucially, it's usually the place where most people first begin to get to grips with "Howdothese transistor thingies work?" and encounter many of their own, and other people's, misunderstandings or simplifications of "What'sreallygoing on?".