common emitter biasing - Page 1
Forum section: electronics / beginners

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 09:49:19 am
content: With usual apologies, embarrassment etc...I'm playing around a common emitter setup: 9V battery, and a 2N3904. My signal is capacitively coupled. I'm using a 220k potentiometer to bias the base. I've got the collector resistor at 5.6k, and I've connected the emitter to ground. It doesn't matter to me if the output is clipped: I just want to see what happens.As I understand it, I have to bias the base according to certain criteria: make sure VBat least a diode-drop above VE; and allow enough room for my input signal to swing without going below ground (or below some other point?). Another point seems to be that I can use the base bias to set VEat a value that makes a nice quiescent current when it goes through REto ground.In my case REis zero. This seems to imply that VEis ground, and that VBmust be one diode drop above ground (e.g. 0.7V). If I deliberately bias the base with a higher voltage, what happens? Does the transistor have to drop more voltage between base and emitter? Or does it override the biasing? Or is there incredibly high current between emitter and ground?

author: mikerj
author_rank: Super Contributor
timestamp: June 10, 2015, 10:10:55 am
content: Re is never actually zero, since the emitter always has some inherent resistance. For a common emitter amplifier, you would normally be aiming to bias the collector voltage to half the supply voltage in order to maximise the available voltage swing.Too much voltage on the base will simply turn the transistor on harder, so your collector voltage will drop. Taken to extremes the transistor will saturate and the collector voltage will be close to ground, (typically 0.1-0.3v for a small signal transistor, dpending on collector current). Obviously you don't have an amplifier any more at this stage.

author: Zero999
author_rank: Super Contributor
timestamp: June 10, 2015, 10:18:14 am
content: Setting REto zero makes it almost impossible to stably bias the transistor because VBEis very dependant on the temperature and current.You could do with with a current mirror but both transistors need to be thermally coupled.EDIT: Schematic attached.The base resistors have been added to increase the impedance, which will still be very low. Without the base resistors, Tr1's collector will make the impedance far too low.

author: LvW
author_rank: Frequent Contributor
timestamp: June 10, 2015, 11:31:35 am
content: Quote from: mikerj on June 10, 2015, 10:10:55 amFor my opinion - a rather "dangerous" statement. Without an external emitter resistor there is no feedback and absolutely no stabilization of the DC operating point.I assume that mikerj refers to the symbol re=1/gm which - in some contributions - is called "inherent emitter resistance".However, this is a common misconception.It is NOT any kind of emitter resistance - it is a "transresistance" (invers of "transconductance" gm).And as such it relates theinputvoltage Vbe with theoutputcurrent Ic (gm=dIc/dVbe).This parameter has no stabilizing effect.

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 12:47:03 pm
content: Thank you for your speedy replies.If I were trying to make a good, temperature-stable amplifier for small signals, these comments would be very helpful indeed. As it happens, I'm not. I'm trying to get an idea of how the transistor deals with me doing the 'wrong' thing. (Wider agenda: I've been thinking about ways to make pulse-waves. There are other ways, of course, but I want to understand this one.)So, to pick up mikerj, if I don't have an amplifier, I suppose I have a switch? I can't remember seeing a transistor switch with explicit biasing. But supposing that's what I've done: divided my 9V supply and sent (e.g.) 3V to the base of the switch. That is saturating the transistor? So is VBstill 3V, and does the 'diode drop' relationship between VBand VEno longer pertain?

author: flynwill
author_rank: Regular Contributor
timestamp: June 10, 2015, 02:21:37 pm
content: If your goal is to get as much AC gain out of the transistor as you can the traditional way would be to bias the base with a resistor voltage divider, add a resistor in series with the emitter to set the DC collector current to a reasonable value and then bypass that emitter resistor with a large value capacitor so that for the frequencies of interest you get the full gain of the transistor.

author: Zero999
author_rank: Super Contributor
timestamp: June 10, 2015, 02:31:33 pm
content: The problem is, the gain is so high, it's nearly impossible to bias.Suppose the gain is 180 when VCE= 4.5V. This means to change VCEby 1V you need to alter VBEby just 5.56mV.Because VBEand the base impedance is very temperature dependant, it doesn't take much for the voltage to change by more than this as the device warms up.

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 02:47:52 pm
content: Thanks again.But again, I think you are making incorrect assumptions about what I am trying to achieve. I don't need improvements to the circuit: I'm just trying to understand, given this setup, what is actually going on in it. Precisely because it's a 'bad' circuit (as an amplifier) it's hard to get information about it. (It actually works well for the use I am putting it to).

author: LvW
author_rank: Frequent Contributor
timestamp: June 10, 2015, 02:51:46 pm
content: Quote from: socratidion on June 10, 2015, 02:47:52 pmSocratidio - may I ask: How does the circuit look like you are speaking on (circuit diagram) ? Otherwise, it is "problematic" to give explanations.

author: opty
author_rank: Regular Contributor
timestamp: June 10, 2015, 03:05:12 pm
content: Quote from: socratidion on June 10, 2015, 09:49:19 amIb= (Vb- 0.7) / Retherefore if you try to force Vbto higher than a diode drop then with near zero Recurrent will go crazy high...(it's really just a diode there - check transistor equivalent circuit)

author: LvW
author_rank: Frequent Contributor
timestamp: June 10, 2015, 03:53:09 pm
content: Quote from: opty on June 10, 2015, 03:05:12 pmOpty - can you explain your equation? (Are you sure it is correct?)

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 04:23:56 pm
content: Re the circuit diagram: there isn't any more to the circuit than I have described, except that I am inputting waves from a function generator, and reading the output on an oscilloscope. These are details I gave to make the question more concrete, but really it is the theoretical principle I'm asking about: input and output don't concern me, just what happens when there is an unusually large difference between VBand VE– or rather, whether it's even possible to have more than a diode drop.According to opty, it is possible. Excessive current is a possibility I've wondered about. I guess I should measure it...But I'm troubled: if I'm supposed to set up an amplifier for a quiescent DC, with a diode drop between base and emitter, then when I input an AC signal, that should push the voltage periodically a bit higher than the (roughly) 0.7 volts. Doesn't that (intermittently) create the same high-current conditions? Must my signal be limited to tiny amplitude (a few millivolts), in order to avoid blowing up the transistor?

author: Zero999
author_rank: Super Contributor
timestamp: June 10, 2015, 05:02:27 pm
content: The base-emitter junction is a diode and follows the diode equation which can be found using your favourite search engine.If the AC signal causes a significant current to flow into the base, then the transistor will saturate and the output will drop close to the negative rail.The problem with biasing a transistor in the manner is it's easy to unintentionally cause it to go into saturation.

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 05:34:51 pm
content: Chewing this over... I've had a look at the diode equation: it seems to suggest that current will get very high very quickly above a certain voltage.So, e.g. in the case of a transistor switch (which will probably involve voltages higher than 0.7 at the base), one must take precautions, like using a resistor at the base? I seem to see this often: I put resistors there myself without knowing why, exactly. Is this why? Does a base resistor drop the necessary voltage to make the base 0.7V? Or does it just limit current?Is it then current I have to guard against? In the given setup, the only DC current into the base is coming from the upper biasing resistor (probably more than 150k). I don't think the AC signal can deliver much. Am I in the clear?I'm still not clear whether VBis nailed to a diode drop away from VE, or whether it's 'stretchy' in extremities.

author: LvW
author_rank: Frequent Contributor
timestamp: June 10, 2015, 08:32:55 pm
content: Quote from: socratidion on June 10, 2015, 05:34:51 pmBoth at the same time. A B-E voltage which is rising above app. 0.75 volts would draw a current that will destroy the device. Thereforte, a properly designed series resistor causes a rising voltage drop that keeps the B-E voltage at a safe level.Example:*Desired current 10 mA,*B-E voltage 0.7 V*Supply: 10V*Series resistor: Rs=(10-0.7)V/10mA=930 Ohms

author: socratidion
author_rank: Contributor
timestamp: June 10, 2015, 10:02:30 pm
content: Thank you.Now that I've managed to get home to my equipment, I've done a simple test: no input (I even removed the capacitor): I disconnected the voltage-dividing 220k pot from the base, set it for a certain voltage, reconnected it to the transistor base, and took a reading of VBand IE. I tried voltages from 1V upwards in small increments. Results: VBstays at 0.66 volts for every biasing voltage up to 5V, where it is 0.67 volts. At 7V bias, the base voltage reads 0.68 volts. I stopped there (my 9V battery is more like 8V now). Current similarly stays quite flat around the 1.4 mA range, and gradually nudges up to 1.68 mA with the 7V bias. After that there's a point (I'm not sure where, exactly) where the current shoots of the scale, and I get I 'hot' smell from the circuit. Somewhere around where the voltage divider gets to Vcc, I guess.Well, it answers my question, more or less. Somehow the diode in the transistor forces the base voltage down. Not sure how it manages it, but at least I don't have to worry about excessive current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 10, 2015, 10:36:00 pm
content: Quote from: LvW on June 10, 2015, 11:31:35 amThis would be true if the input did not draw current at all (as in a MOSFET), however the base has the same exponential slope as the transfer curve, therefore it is not as unstable as you claim. The ratio between currents is known as hFE.The technique of biasing a transistor with a fixed base offset or current, and no emitter resistance, is commonly known assuicide biasing. While it won't cause destruction if the collector is loaded with a resistor, you can imagine the consequences if it had an inductive load instead (so that Vce ~= Vcc). The best case is to use shunt feedback (i.e., taking the base bias from the collector, so the variation in collector voltage stabilizes the base bias), or using an emitter resistor (which swamps the conductance and transconductance of the device, giving a stable bias point for any collector load that doesn't cause saturation).Tim

author: Zero999
author_rank: Super Contributor
timestamp: June 11, 2015, 08:01:34 am
content: Quote from: socratidion on June 10, 2015, 10:02:30 pmThe base lowers the voltage because your voltage source has a high input impedance. When the potentiometer is set to the middle (4V with an 8V supply) the impedance seen at the wiper will be half of the total value, 110k. When the wiper is at either extremes, the impedance will be near zero. When the voltage is set to near zero, hardly any current will flow despite the low impedance, but when the voltage is set to near the supply voltage, the current will be high, which is what caused your transistor to overheat.

author: socratidion
author_rank: Contributor
timestamp: June 11, 2015, 09:09:22 am
content: Thank you: if you mean high 'output impedance' from the point of view of the voltage source, that makes sense. Or if you meant input impedance from the point of the incoming signal. Either way, I see what you're saying, I think.Does this further imply that, say I biased the transistor with a much higher voltage, like 15V, made out of a divider coming from a 30V battery, maybe even keeping the transistor attached to the 8V supply, (as long as everything joined up at ground), then I would still see the base being held down at 0.66V... that is, that there's no particular reason why 7 or 8 volts should be the ceiling?Or, contrarily, is that the point of the 'maximum emitter-base voltage' value in the data sheet? For the 2N3904 that's 6V: do I take it then that the transistor in my setup can only properly deal with 6V bias? As in, it can only 'eat' 6 volts before it starts overheating?Is that indeed what diodes do? Eat voltage, make it disappear? I'm realising I've seen something like this phenomenon in 'diode clamps'.

author: LvW
author_rank: Frequent Contributor
timestamp: June 11, 2015, 09:27:18 am
content: Quote from: T3sl4co1l on June 10, 2015, 10:36:00 pmI must confess that I do not understand this part of your reply. What means "...not as unstable as you claim"? It was my only intention to make clear that the so-called "inherent emitter resistor" is (a) in fact, not a resistor and (b) can provide no stabilization effect at all. Something wrong?

author: Zero999
author_rank: Super Contributor
timestamp: June 11, 2015, 09:42:13 am
content: Quote from: socratidion on June 11, 2015, 09:09:22 amThe base-emitter junction acts as a diode clamp. Once the turn on voltage is reached, the voltage across a diode doesn't vary much, as the current is increased. If you add an emitter follower to your potentiometer, to give a low impedance variable power supply, you'll find the current increases rapidly beyond 0.6V and the transistor will overheat.The emitter-base voltage is the maximum reverse bias the base-emitter junction can withstand before it starts conducting and damages the transistor. Note how it's specified as emitter-base voltage, rather than base-emitter voltage. If it were specified as emitter-base voltage then it would be given as a negative number.

author: Ian.M
author_rank: Super Contributor
timestamp: June 11, 2015, 10:12:27 am
content: Although theGummel-Poon model, which is device physics based, and the simplified Ebers-Moll model treat the transistor as a voltage controlled device, to a first order approximation it can be treated as a current controlled device. Ic=HFE*Ib and due to the variation in the performance parameters of any specific transistor due to small process variations, the greater accuracy of the physics based models is rarely needed for circuit design.Your experiment was equivalent to applying a linearly varying bias voltage through a resistance varying from 110K when the pot was at 50% to near zero at the ends.Using the simple current controlled model, you can see that the only limit on base current was theThéveninequivalent resistance of the pot and the forward characteristic of the B-E diode junction. Ic was HFE*Ib while Ib was small until it was limited by the 5K6 collector resistor once HFE*Ib exceeded the current that could flow through the resistor at that supply voltage.N.B. HFE (large signal current gain) and Hfe (small signal current gain) vary considerably with collector current + from device to device which is why its vanishingly rare to see single base resistor biassing in any linear transistor circuit.Call me a dinosaur, but I learned how to bias a transistor to operate stably and with predictable gain from theMullard manual of transistor circuits. Its rather dated and oriented towards PNP germanium devices, but if you take any biassing circuit that can keep a Germanium device stable over a reasonable temperature range, flip the supply rails and any diodes, and allow for the higher Silicon Vbe, adjusting resistors accordingly, it will easily stabilise current NPN Silicon devices.

author: socratidion
author_rank: Contributor
timestamp: June 11, 2015, 10:43:14 am
content: Thank you both indeed: very helpful.To Hero999, re your remarks about the consequence of putting an emitter follower after the pot: in some versions of my circuit (as I say, I've been playing around), I put my input signal through a source-follower JFET, then a resistor, then a big cap, before it reaches the voltage divider. I suppose it might depend on the values (JFET is 2N5458, source resistor is 22k, output resistor is 10k, DC-blocking capacitor is 10µF), but would I be right in thinking this is a different scenario, and I don't run the same risks?To Ian.M, I shall have to ponder your analysis of my test. Certainly the RCwith an 8V supply would explain the emitter current sticking at around 1.4mA. I maybe have to do the test again and think explicitly about the parallel resistance of the two halves of my voltage divider, and compare it with base current, and see if I can convince myself that I understand you.Again, I should emphasise that the point here is not to make the transistor operate stably with predictable gain: it's just to see what's going on. If there's a wider agenda, I'm looking into the phenomenon of extreme clipping, as a way of producing pulse waves. At this point what I want to know is, am I going to blow anything up?

author: Ian.M
author_rank: Super Contributor
timestamp: June 11, 2015, 10:57:42 am
content: Quote from: socratidion on June 11, 2015, 10:43:14 amExamine datasheet and determine Ib_max, Ic_max, Vcebo_max and max power.Add a diode from emitter to base, cathode to base if the input can ever go negative so you don't have to worry about the reverse Vbe limit.Choose series base and collector resistors that give an adequate safety margin (e.g not more than 80% of max) for Ib and Ic. Make sure the collector voltage never goes over 80% of Vcebo_max - clamp or snub inductive loads and reduce supply voltage if you have to. Make sure that Vce*Ic never exceeds the max power. Don't forget to derate the power limit with temperature.For power devices its a little more complex as they can often handle significantly higher pulsed currents and power levels (for SHORT pulses) than their continuous rating.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 11, 2015, 11:10:23 am
content: Quote from: socratidion on June 11, 2015, 09:09:22 amMmm, "eat current" more likely. The voltage drop is low, so the diode causes the component in series with it to "eat the voltage". Whether it likes to or not. If that series component is a power supply's own internal resistance, the voltage might indeed be eaten by that resistance... but the current won't be very friendly. And maybe the diode will eat itself after a few milliseconds!A series diode could be said to "eat the voltage", because it has a high voltage drop when in reverse bias.Quote from: socratidion on June 11, 2015, 10:43:14 amHe was referring to this:Where the buffer is something like an ideal voltage source (i.e., a low output resistance, and capable of sourcing way more current than the base should ever see).An emitter follower is not a perfect buffer, but it's very close: about hFE times worse than the base should want to see.If you add a series resistor between buffer and base, you're back to the safe, current limited condition. (You also have a higher impedance, which means you can supply actual signals to it, say by coupling in with a DC blocking capacitor.)QuoteExtreme clipping as applied to what sort of input? If you start with sine waves, you can very easily apply enough gain and clamping to render a square wave, which might be considered "pulse waves". The common-emitter stage is generally less helpful for that goal, because the base draws current, which leads to clipping, yes, but also a shift in the DC offset of the stage (you will see a different voltage across the DC blocking capacitor depending on whether the source is 1mV, 1V or 10V).A good solution is to drive the emitter voltage, not with GND, but with another transistor, so that the B-E diodes worth together without upsetting your signal -- this is called a differential amplifier, for its useful properties.Diff amp stages can be cascaded to realize very high gain in the linear range, meaning it very quickly transitions from +saturation to -saturation, such as over a single mV at the input. These are called comparators.Tim

author: Ian.M
author_rank: Super Contributor
timestamp: June 11, 2015, 11:18:56 am
content: If you are interested in a pulse output you may find the trailing edge of the pulse is degraded due the the transistor's recovery time from saturation. If you need fast pulses, you may need aBaker clampto keep the transistor out of hard saturation..

author: socratidion
author_rank: Contributor
timestamp: June 11, 2015, 11:55:29 am
content: Aha, I am indeed finding the falling edge isn't as sharp as it could be. I shall look into your suggestion.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 12, 2015, 12:21:47 am
content: If your signal source is something sharper already (like a square wave, but maybe not very good risetime), you can skip all the biasing junk and use a voltage divider from input to base to GND. You design the voltage divider so that it has a base voltage of about 1.4V (assuming the base didn't act as a diode), and a Thevenin resistance of enough to deliver the required base current (usually, Ib = Ic / 20 or so). This way, when the source is 'on', the base gets +Ib, holding the transistor in saturation (a 2N3904 will turn on in ~30ns, depending on various conditions). When the source turns off, the base capacitance is discharged to GND, with about -Ib of current (a 2N3904 will probably turn off in 100ns under these conditions).As I mentioned earlier, you can cascade stages to increase the gain and reduce the risetime, at least to a point. A "switch" circuit like this is really just a linear amplifier, at some point of its swing -- it's just not intended to stay in that range very long, hence the lack of biasing! After a few stages, if the risetime is now limited by the circuit itself (and not the signal source), you need to change up the circuit. Usually, more -Ib, more Ic, and lower resistances is required to get there. Under ideal conditions, you can get ~10ns edges out of a 2N3904.Tim

author: socratidion
author_rank: Contributor
timestamp: June 12, 2015, 07:21:17 am
content: A lot of information in the last few posts – it may take me some time to digest and/or try out. I very much appreciate the advice and technical help.There are of course other ways of making pulse waves from other shapes (a comparator, yes; or a 40106, or indeed any logic gate IC wired the right way). I'm being methodical, trying to imagine every possibility, and seeing what comes out.The input signal is mostly lopsided triangle shapes at the moment. I've noticed that changing the base bias gives me control over the pulse width of the output. It strikes me that (with a few things added) this would be a way of making symmetrical waves out of unsymmetrical ones, e.g. squares out of pulses, triangles out of saws. (There are other ways of doing this too).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 12, 2015, 08:52:18 am
content: A single transistor is, of course, the simplest case of amplifier or comparator; all about how you use it. Ignoring Vbe, if you apply voltage to base with a current limited voltage source, and to the emitter with a constant voltage source, then the collector current is either on or off based on the input voltages.The best way to get a constant voltage source is to use an emitter follower, which gets you a differential amplifier topology (the conventional one if using the same polarity transistors, or a "complementary differential" if using opposite polarity -- which isn't very useful because you still have to mind base-to-base voltages, but comes up on rare occasion). Sometimes you can arrange the circuit so you have a low impedance already, and get away with just one transistor.Tim