Simple BJT Amp Stage - Page 1
Forum section: electronics / beginners

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 12:56:09 am
content: I put this schematic together hoping to look at the THD side of things regarding this configuration, but only to discover vout is lopsided, the positive peak is around 0.8v, and the negative around -1.2v.I don't have an IV curve for 2n2222, and don't know what a good operating point is to set it up with, but here are the DC voltages and currents as they are:Vc=2.53VIc=3.44mAVb=0.806VIb=34.4uAand the output wave is lopsided.The THD is about 9.6%. Is this high number caused by the lopsided output?Did I do something wrong in the schematic, or with the LTspice simulation setup? I couldn't figure out why this is, all helps appreciated!

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2014, 01:29:26 am
content: Here's a nice tutorial:http://thesignalpath.com/blogs/2012/09/23/tutorial-on-the-theory-design-and-characterization-of-a-single-transistor-bipolar-amplifier/To my non-expert eye, it doesn't look like you have any feedback in the amplifier. Feedback would be used to stabilize the gain and make the circuit more linear.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 02, 2014, 02:13:29 am
content: Hi,I repeated the LTspice simulation and I got the same result.The distortion is from the non-linearity in the transistor and is to be expected.If you modify the circuit to this:The THD is reduced to 1.6%Regards,Jay_Diddy_B

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 05:03:38 am
content: Quote from: IanB on August 02, 2014, 01:29:26 amthank you for the pointer!i was hoping to solve the lopsided mystery first, and then come back looking at the THD stuff, but i'll read the tutorial, thank you for helping!

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 05:14:39 am
content: Quote from: Jay_Diddy_B on August 02, 2014, 02:13:29 amthank you very much for helping, it's nice to have the simulation result verified by someone else!i'll give this configuration a try. R5 and C1 are there to set the gain, yeah? are there any other ways of controlling gains for this configuration?the output waveform of my circuit is lopsided, the positive peaks are having a smaller swing than the negative ones, |0.8|v vs |1.2|v, that's terrible distortion of the input signal, yeah? any ideas about why that is? what's wrong with it and how can i get it corrected?

author: LvW
author_rank: Frequent Contributor
timestamp: August 02, 2014, 08:13:32 am
content: Questron - if you like to stay with your principle of dc feedback (resistor between collector and base) you can add signal feedback to your circuit simply by adding a additional resistor between signal source and coupling capacitor. This, of course, reduces the gain - but this is the normal result of signal feedback which cannot be avoided.

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 02:53:32 pm
content: Quote from: LvW on August 02, 2014, 08:13:32 amthank you! i'll give that a try. could you please explain how a resistor in series with C1 can provide signal feedback?is this signal feedback the thing that made the output lopsided, unequal positive/negative output swings, how so?Quotereduced gain is no issue for now, i'd like to understand what makes the positive/negative output swings unequal first, get that right, get the circuit working right, before worrying about the gain.

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 03:17:04 pm
content: Quote from: LvW on August 02, 2014, 08:13:32 amAdded a 100ohm resistor, R1, at the input like this,TDH did get reduced to 5.3% by just doing that. but the output waveform is still unequal, about 0.7v/0.85v, vs about 0.8v/1.2v earlier. but this reduced gain is no biggie now, i'd like to understand why the output waveform is not having the same swing amplitude, and have that corrected first.I'm not trying to build anything for real, just playing with simulations, in order to understand what's going on in the circuit.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 02, 2014, 03:21:18 pm
content: Quote from: questron on August 02, 2014, 02:53:32 pmHave you ever looked at op-amps? If so, think about the inverting amplifier, where the inverting input is a virtual earth. It's a pretty similar idea here.QuoteIt's all tied up together...A transistor is a voltage-operated device (despite what many people/text books say) - voltage in, current out.The relationship between Vin (base-emitter voltage) and Iout is exponential, not a linear straight-line curve. Which is not what we want from an amplifier. But this simple fact explains why the output is unsymmetrical. If that isn't completely clear immediately, draw a graph with an exponential curve, and "wiggle" the X axis (Vbe) equally either side of an arbitrary point (0.6V might be a useful one!), and see how the Y axis (Ic) varies.When you apply negative feedback, an amplifier becomes more linear. The greater the difference between the gain with no feedback and the gain with feedback - i.e. the more feedback you apply, the more linear the amplifier becomes. Negative feedback theory is relatively straightforward to understand with a bit of study and practice - but that can wait a day or twoWith a simple common-emitter amplifier, you can apply negative feedback by adding resistance to the emitter (that was the first suggestion). As the collector current varies, a voltage is developed across this emitter resistance, and this voltage subtracts from the applied input voltage (remembering that the transistor responds to Vbe).The suggestion to add a resistor in series with the base achieves a similar result, but in a different way. In conjunction with the resistance from base to collector, you now have a shunt feedback circuit that is similar to the inverting op-amp configuration, as mentioned above. The transistor wants to see 0.6V at its base, and will wiggle the collector to achieve that. If you build the circuit, you'll see that the voltage at the base will be quite small, and distorted (of course, that's because of the exponential relationship between Vbe and Ic - the the output at the collector is reasonably undistorted, then the base has to be because of that graph I described above).There is a simpler way to think of this. The relationship between base current and collector currentisa straight-line, because of transistor Hfe. We all know that Hfe is a bit of a shifty parameter that varies for fun, and can't be trusted, but put that to one side for now. Consider that the input resistor is converting the input voltage to a current, and then you should see why the output appears to be more linear...

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 04:58:58 pm
content: Quote from: Mark Hennessy on August 02, 2014, 03:21:18 pmThanks a lot for the help! i got you, you are saying the asymmetrical output is caused by the nonlinearity of the BJT.With 0.8v vs 1.2v, the difference is 0.4v, comparing to 0.8v that's 50%. Is that normal when a BJT is set at a reasonable operating point? Could it be that this rather large distortion is caused by the operating point being set on the shoulder or toe of the IV-curve? I don't have the IV-curve for 2N2222, so i just guessed maybe Ic about 3mA and Vc about 3v is ok, and i even reduced the signal level down to 1.0mV, but still asymmetrical output to the same degree.Sticking to this circuit arrangement is not my purpose here at all, it's just that this arrangement is now showing this particular characteristics and I just want to understand it. Of course i don't know much about the way BJT's behave, but if a BJT is set at or close to a normal operating point that satisfies the normal small signal condition, a 50% output distortion is rather suspicious of something else at work, instead of merely being a result of the BJT's intrinsic nonlinearity. Does this sound reasonable? or I'm missing something here? or the BJT nonlinearity is really this bad without negative feedback? the book I read says the shunt bias resistor, R2, does provide some feedback, so this form of feedback is by nature inferior to the voltage divider arrangement as Jay suggested, by several orders, to the degree of producing a 50% distortion? or something else?

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2014, 05:13:20 pm
content: Instead of feeding a sine wave into the circuit, why not vary the input voltage over a small range around the DC operating point and plot the output voltage against the input voltage? See what shape that graph has. (Obviously without AC coupling capacitors or load resistor.)

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 05:41:14 pm
content: Quote from: IanB on August 02, 2014, 05:13:20 pmThank you Ian again!I did that earlier, and thought the BJT should be working within the region between the two thick white lines I put in, although that's a surprisingly narrow range, the output waveform in my first post is with Vb sitting at about 806mV, and the Vc and Ic crossing point is around 3v and 3mA, so i didn't think much about this set of curves. is anything there suspicious?I also attempted to plot out an IV-curve for 2N2222, so that operating point could be eliminated as a nonissue, but failed, I don't know how to achieve that with LTspice. is there a way?

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2014, 06:01:22 pm
content: Well I'm learning as much as anyone here. But looking at the range of Vc between the mid-point of the two white lines it appears to rise two divisions (1.2 V) to the left, but to fall four divisions (2.4 V) to the right. That non-linearity would suggest a reason for the amplified sine wave to extend further downwards than upwards if it follows the same curve.

author: LvW
author_rank: Frequent Contributor
timestamp: August 02, 2014, 08:07:04 pm
content: Quote from: Mark Hennessy on August 02, 2014, 03:21:18 pmThank you Mark for this clear statement.It is really funny - but I don´t know why some people and - yes - even some textbooks believe that the BJT would be a current-controlled device. It´s simply false!Just because the relationship Ic=B*Ib ? Didn´t they never hear about Shockley´s famous equation ?As an example: It is impossible to explain some basic circuits and effects based on the current-control approach (diff. amplifier, current mirror, emitter resitance feedback,...).

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 02, 2014, 09:24:49 pm
content: Quote from: LvW on August 02, 2014, 08:07:04 pmAbsolutely - the Ebers-Moll equation holds true with startling accuracy of many decades of collector current.The key point is that the base current is an inconvenientconsequenceof putting a voltage across a PN junction - it's not the fundamental mode of operation that determines the collector current. If the transistor was really controlled by its base current, then Hfe (Beta) wouldn't be so variable.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 02, 2014, 09:48:03 pm
content: If it hasn't been mentioned already... the "lopsided waveform" IS a distorted waveform. You don't just hook up a spectrum analyzer or FFT to an otherwise unassuming waveform and say "oh, that's distorted", no: the distortion is the result of the voltage (or current, as the case may be) not following a sinusoidal profile. If you can see it's distorted (lop-sided, kinked, flattened, etc.), you know it has a lot of distortion!It's difficult to even resolve voltage differences in the 1% range. For instance, a graphical display 256 pixels tall -- whether it's a non-magnified graph in LTSpice, or the waveform display on an 8 bit digital oscilloscope -- only resolves 0.4%, and whether you can decide by eye if it's right or not, another few percent usually. That's what the distortion analyzer is for -- it resolves the measurement differently, so you can find smaller errors with higher precision.Anyway, this circuit illustrates one of the fundamental limitations of transistors, and nonideal amplifiers in general. You can get high gain, and high distortion, or low gain and low distortion. It's a straightforward tradeoff. Most amps (op-amps as they are called) use many transistors inside, resulting in very high gain, and dubious distortion (some are very good even at high gain, but they use even more transistors, so you get the idea), but very low distortion under most use cases.Tim

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 02, 2014, 10:08:38 pm
content: Quote from: questron on August 02, 2014, 04:58:58 pmYes, absolutely.This non-linearity is fundamental to how it operates. You'd get similar problems with any other active device (e.g. FET, MOST-FET, thermionic valve), as none of them have a linear relationship between voltage in, and current out.However, that's why we love negative feedbackQuoteWhen setting an operating point for a BJT, we think about things differently to this approach...Firstly, we don't look up "IV curves" for specific BJTs, as they aren't really a device-specific thing.It's a bit of a long story, this one, so I'll try to be concise...Firstly, it's conventional to bias the BJT so the collector is in the centre of the available voltage swing. We do that on the assumption that signals are symmetrical, which is usually a fair assumption. YMMV. So for a 6V PSU, let's make Vc equal 3V...Next, we need to decide on the collector current. For a small-signal amplifier, 1mA is a good starting point (more on this in a moment). So, we can make the collector resistor equal to 3k (Vsupply minus Vc).Why 1mA? Basically to make the numbers easier!Next,If you explore the Ebers-Moll equation I mentioned in my previous post, it allows us to derive a simplified equation for gm.What's gm? Simply, gm is a value of gain, but it's a little bit different to (voltage out) over (voltage in). Rather, it's (current out) over (voltage in). Remember, the BJT, just like FETs, MOS-FETs and valves/tubes are voltage-in, current-out devices... The "posh" word for this is "transconductance", FWIW... We use gm to express that. Unlike regular voltage gain, which is just a ratio, gm has a unit - it's Amps per Volt. Note that this is the reciprical of resistance (V/I), and the unit is Siemens.In transistor circuits, it's convenient to work in milliamps rather than amps...From Ebers-Moll:For a BJT working at room temperature, gm=40.IcFor a BJT working at a higher temperature, gm=35.IcI tend to use the latter, as gear often runs warm. But some textbooks will use the 40Ic version. Not a huge deal either way...So, if we set the collector current to 1mA, we have a gm of 35mA per volt.The next part of the story:Your collector resistor is converting the current from the BJT into a voltage (simply Ohm's law). So the voltage at the collector is Ic times Rl. Now, Ic has already been set to 1mA to establish the correct DC conditions, and the signal itself is just a variation either side of 1mA. So,...if gm is 35 times Ic, Ic is Vbe times gmAndVoutis Ic times Rlgain = gm times RlFor a typical circuit where Ic is 1mA, and Rl is 3k, then the overall gain should be 105.In practice, it will be a little lower - there are always secondary effects (that can be considered later). But you'd probably get around 90 or so...Note, BTW, that to calculate a value of gain, we didn't need to consider Hfe/Beta. That's lucky, as Hfe/Beta varies enormously from device to device. In a typical circuit, the biggest effect Hfe has is on input impedance, not the raw gain.So what we see here is that getting large gain is very easy. OK, there is distortion that is readily visible on a 'scope, but that's where negative feedback comes in. Imagine if we had two of these stages - you'd have a gain of 90 times 90, or 8000 (simplifying a hugely). Far more gain than you'd actually need, but all that "excess gain" will, via the magic of negative feedback, be "thrown away" to produce a very linear amplifier indeed.Many op-amps have just two gain stages. OK, they pull some "tricks" to get lots of open-loop gain, but essentially, all amplifiers start with the principles outlined above.Personally, I feel that any time invested at this level is time well spent. The more you understand the very basics, the easier everything else becomes. But that much you already knowI do have some standard advice for people at this stage: spend some time learning how to separatesignalsfromDC conditions. What we do here is establish the correct DC conditions, and then wesuperimposethe AC signal on top of the DC conditions. Often, superposition is taught with no obvious "buy-in", but this is it - any signal can be regarded as a mixture of AC+DC, and in a simple transistor amplifier, that is very much the case - the DC conditions are an inconvenient pre-requisite of a practical amplifier. We're lucky today - we have thousands of great op-amps to choose from, and in most op-amp circuits, the DC components are usually 0 volts, so we're not even aware that we're dealing with them. Very different with discrete transistor circuits.Hope this helps, and good luck,Mark

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2014, 10:19:11 pm
content: I took a screen capture of Shahriar's example transistor amplifier output signal and made a rough measurement of its symmetry. I simply reflected a positive half wave about the measured half way point to see if it had the same amplitude as the negative half. To within the visual limits of this experiment, it does. Therefore Shahriar has managed to construct a reasonably linear amplifier with a gain of about -60 around a 2N2222 NPN transistor. This shows that the results are all about the details of the design, which Shahriar outlines in his video (with a few details skipped over, I believe).

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 11:04:31 pm
content: Thank you very much Mark for the detailed explanation, great design practice advice as well!I'm glad you encourage understanding on this level, I'm a beginner and very much do need understandings on this level, otherwise I'm unable to go on understanding more complicated circuits. opamps still remain way above me now,I understand BJT's are non-linear, I just never imagined they are this bad, and need quantitative confirmations about it, that's all.I seem to have seen this basic shunt bias configuration in many places, cheap radio RF amp stages for example. is it normal practice where 9.6% THD per stage is accepted for lower costs?how can I control the gain of this basic shunt bias amp circuit, schematic as per the first post? I'd like to reduce the gain of this exact circuit and see how much THD improvement can be obtained, therefore have a comparison and see whether this configuration is useful or useless as an amp. Thank you!

author: questron
author_rank: Regular Contributor
timestamp: August 02, 2014, 11:11:53 pm
content: Quote from: IanB on August 02, 2014, 10:19:11 pmThank you Ian for the help!Actually you have nailed it in your last posting, the non-linearity is really that bad.the question now for me is how could the THD be lowered to a reasonable level, but still maintain this basic shunt bias configuration, just for the sake of exploring.oh, also, is Shahriar's circuit also a shunt bias one, could you post a schematic of it if so? Thanks a lot!

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 02, 2014, 11:36:02 pm
content: Quote from: questron on August 02, 2014, 11:04:31 pmIt's partly a factor of the output swing. In many applications, the output is much smaller.Also, an RF or IF stage in a radio will have a tuned collector load, so the harmonics will be filtered away to a reasonable extent.QuoteYou've already had the answer to this one - you need a base resistor.Try using 50k. The gain should be very close to unity, but the results should be pretty OK. Try 10k; the gain should be close to 5.Oh, and you might try increasing the 3k9 load resistor. This circuit has a moderately high output impedance.And it would be worth investigating the fully stabilised circuit, as posted by Jay_Diddy_B earlier. Here, the gain is the ratio of Rl and Re (which at signal frequencies is 1k in parallel with 100 Ohms), provided the answer to that is about 10 or less. For higher "closed-loop" gains, you need to use some proper negative feedback theory (not difficult, but not tonight)

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 02, 2014, 11:39:31 pm
content: Quote from: IanB on August 02, 2014, 10:19:11 pmHaving just watched this, I'd question the wisdom of the overall design. OK, the point of the video was to demonstrate transistor design - I get that - but ultimately, a single op-amp would be more linear. I doubt it would be any noisier also - in fact, probably the opposite. I do a lot of audio design, and having a single transistor ahead of an op-amp, outside of any global feedback, is a technique that died in the early 1980s...Remember, just because something looks linear on a 'scope, it doesn't mean it's good enough for quality audio. Typically, you can see about 1% THD on a 'scope, if lucky.He could have got the required gain by simply raising the supply voltage. If the point was to teach transistors first and foremost, that's what he should have done when designing the lesson. The op-amp is an unnecessary diversion.Personally, when teaching I prefer to build up slowly in "layers". This was a full-on "brain-dump" - I understood it all, but I'm not sure how well a beginner would follow it. Still, I admire anyone who has the guts to make a video, so I don't intend to sound as critical as I probably do.

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2014, 11:47:34 pm
content: Quote from: questron on August 02, 2014, 11:11:53 pmIt's best to watch Shahriar's video tutorial. He walks through all the design steps from a statement of requirements through to circuit build and testing. It's much better to watch that than just to look at the schematic:http://youtu.be/Y2ELwLrZrEMYou will see that he doesn't use a circuit simulator, he does the design calculations on paper and then builds the circuit to see if it matches the calculated predictions.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 02, 2014, 11:55:01 pm
content: Hi group,The simulation that was performed by questron can be expanded to show another property:I have introduced the .step command to change the input amplitude. I have stepped the input amplitude 0.1mV, 0.316mV, 1mV, 3.16mV 10mVThe transient response shows:The circuit appears to have better linearity, lower distortion, for small signals.I have snipped the THD measurements from the LTspice error log and I see:If I plot THD versus signal size I get:So THD is proportional to the signal amplitude.Regards,Jay_Diddy_B

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 12:05:25 am
content: thank you very much Mark, it helped a whole lot!i tried 50k, 10k, and 2.2k base resistors. with 2.2k and Rload changed to 100k, THD=0.17%, and the gain is about 13, that's much better. with base resistor being 500ohms, THD=1.1%, and the asymmetrical output waveform is obvious in LTspice, but i get the idea now.Quotesay in radio RF stages, what's the normal interstage output voltage roughly speaking? just want to have a general idea. why do designer not fully utilize the gain capability of BJT's, is it because of negative feedbacks?I have not investigated Jay's circuit yet, becasue I know the emitter resistor is going to reduce the gain, then I won't be able to tell whether any THD reduction is due to reduced gain, or circuit configuration, or the BJT itself. I'd like to nail down the BJT first, the gain of a particular circuit configuration the next, and then different circuit configuration the next.

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 12:08:14 am
content: Quote from: IanB on August 02, 2014, 11:47:34 pmThank you very much Ian! i'll watch it.I can't do it that way though, I don't have the necessary equipment, although that's surely the most fun way of doing it.

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 12:23:27 am
content: Thank you Jay for the great work!I learned about how to step a sine wave signal as well!is 10mv, as in my circuit, a reasonable assumption for real applications, radio RF stages for example? 10mv seems to give too high a THD number as your study shows.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 03, 2014, 01:47:43 am
content: Hi,I decided to switch from LTspice to hardware to see if I could see the same property.First I built the circuit on a solderless breadboard. I used a 2N3904 transistor.If I look the output waveform with a scope, I see a reasonable waveform, this is actually 3% THD.I then used my HP 35665A Dynamic Signal Analyzer. Here is a screen shot showing the spectrum of the output. You can see the fundamental and the second harmonic.I then measured the THD versus signal amplitude.With 1mV peak I measured around 1%:With 3mV peak I measured around 2%:With 5mV peak I measured around 3.2%:You can see that the THD is increasing with the signal amplitude.This makes sense, because for very small signals the non-linearity is less important.Regards,Jay_Diddy_B

author: LvW
author_rank: Frequent Contributor
timestamp: August 03, 2014, 06:56:29 am
content: Hi questron,for your information, I like to add some general remarks to the two feedback schemes discussed above.1.) Resistor between collector and base plus input series resistor: This principle is called "voltage controlled current feedback" and causes a DECREASE in input as well as output impedance of the amplifier stage.2.) Using a signal relevant emitter resistance (resp. a corresponding Re-Ce combination) is called "current controlled voltage feedback". This feedback principle has the advantage of an increased input impedance. However, the output impedance (from the collector into the transistor) is also increased - however, this is not so relevant because you have to consider Rc in parallel. That means: The total output resistance primarily is determined by Rc (some kohms).(Remark: This feedback principle exploits the fact that the BJT is a voltage-controlled device).------------In many cases, these considerations are important for selection of a suitable feedback scheme.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: August 03, 2014, 11:10:38 am
content: Quote from: questron on August 03, 2014, 12:05:25 amThank you those who contributed to that point while I was snoozing last nightQuoteIn a radio, the situation is complicated by the interstage coupling transformers - these improve the impedance matching between stages, which reduces losses. They are also tuned to allow only 10.7MHz through, so distortion is less of an issue - these will attenuate 21.4Mhz and 32.1MHz, etc, by a large amount.Also, don't forget that FM IF amplifiers (at least the last one) are run into saturation to reduce the sensitivity to AM.At the input to the discriminator coil, it's not uncommon in my experience to see several volts of 10.7MHz.Rest assured; the designers did all they could to maximise sensitivity, and they generally were very successful. When I restore old transistor radios, I'm always very impressed at the performance achieved from just 5 or 6 BJTs. The challenges were noise and stability (the risk of amplifiers turning into oscillators), not signal level. To understand this, imagine 4 transistor amplifiers connected in series, all with a relatively low gain of 10 to keep the numbers simple, and with an overall output voltage of 5 volts. The input to the last amplifier will be 0.5V. The input to the penultimate amplifier is 50mV. And the input to the antepenultimate amplifier will be 5mV. So as you can see, signal "swing" is only an issue for the very last amplifier in a multistage circuit.QuoteQuite simply, the distortion comes from the exponential transfer characteristic of the BJT, and NFB - however applied - reduces this distortion.To further lower the distortion, you need more open-loop gain. But this is limited by the gm of the transistor (set by the collector current) and the load resistance. To increase gm, you could double the collector current. But if you've kept the supply voltage the same, this will require a lower value of Rl, so you're back where you started!Don't forget that the impedance of the following circuit (or 3k9 in your first schematic) is in parallel with Rl, reducing the gain. Everything is against youTo take the performance to the next level, wouldn't it be nice if we had something that passed a current, but had a very high impedance? That, in a nutshell, is why current sources or current mirrors are often used as a collector load for transistor amplifiers. Or, back in the days when BJTs were expensive, we did "bootstrapping" - an extremely elegant technique.

author: onlooker
author_rank: Frequent Contributor
timestamp: August 03, 2014, 11:50:19 am
content: QuoteIf I did not miss something, my comments would be:The video did not talk about the signal input and output impedances in relation to the voltage gain. If voltage gain were the only goal, a transformer can do it simpler.The 1st important effect of the input impedance is the loading down of the signal source. The input impedance is depending on the Bf of the transistor, and espetially so if the design were to have max voltage gain (no negative signal feedback). This Bf dependency is to be avoided in any real design.Lastly, a minor point, the vedio demoed a 2 stage amp, but titled as a single stage amp.

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 08:09:34 pm
content: thank you Jay for your great work again, it's really nice to see the real thing at work plus a real frequency spectrum! too bad i don't have fancy equipment like that.as per Mark's radio example, 10mv input for the last stage of a radio circuit is not out of the ordinary, so probably some sort of formal NFB is a must for a reasonable low THD figure suitable for audio applications, i'm guessing. so let's wait for Mark to get a chance when he could give a lesson on NFB, that'd be great at least for me

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 08:32:12 pm
content: Quote from: LvW on August 03, 2014, 06:56:29 amthank you LvW, alright we are finally in NFB land now!one question, if BJT's are voltage controlled devices, how does a whatever-controlled-CURRENT-feedback work? doesn't any controlling thing have to use voltage to control a voltage-controlled device?suppose the given is that a BJT is on already, then we pump currents into its base, that will cause a corresponding change in Ic, could we say that this additional pumped-in current caused the base-emitter P-N junction resistane to drop, as the BJT is behaving as a trans-resistor, which in turn caused the Ic to rise? so at least for this part, with the issue of how a BJT is turned on ignored, the working mechanism is based on current, could we say that?

author: questron
author_rank: Regular Contributor
timestamp: August 03, 2014, 08:56:54 pm
content: thank you very much Mark, a lot of useful information again!Quotedidn't know that one before, one more pending topic to be explored. i'm mainly interested in audio related topics such as radios and audio power amps, etc., so i guess i'd be interested in learning about how to have control over THD and noise, on top of all necessary basic circuit topics. when you get a chance, please do give a lesson on formal NFB, many thanks in advance!Quotei see, thank you! no wonder people talk about high supply rails with discretes, they say high supply voltages is the whole point of using discretes. so increasing the supply voltage is certainly one of the legitimate ways of lowering THD, yeah?Quotewow, bootstrapping, current-mirror, these are certainly the topics i'd like to learn about. i've just finished my reading on current-mirror, now i can apply that to lower THD, just great!all great topics, very interesting to me. could we talk about each one by one? i'd like to hear formal NFB first, but it's all up to you, whichever topic, whenever you feel like it. thank you very much!

author: LvW
author_rank: Frequent Contributor
timestamp: August 04, 2014, 07:20:16 am
content: Quote from: questron on August 03, 2014, 08:32:12 pmNo - we never "pump" a current into the base. Base current is always a current than cannot be avoided, but in any case - as mentioned by you - it is the base-emitter voltage that controls Ic. However - how is this voltage developped?Remember the classic inverting opamp circuit: Two currents meet at a common node (inverting terminal) - the current driven by the input voltage and the current driven by the output voltage (to be computed using the superposition principle). This combined current produces a voltage at the inverting terminal which causes a correponding output voltage.And exactly the same applies in our case to the signal currents (only signals because of the capacitor): The collector voltage drives a current through both resistors and the signal source which adds to the signal input current (in the reverse direction). Both currents generate a signal voltage at the base which is amplified.(In this description, I intentionally have neglected the base current which also contributes up to a certain degree to the base signal voltage).The "working mechanism" of a BJT is always based on physical laws and cannot depend on any external circuitry.These considerations do even apply for the DC current into the base. It is common to say that such a single resitor Rb between collector and base "injects" a DC current Ib into the base. But this is not true! It is the collector VOLTAGE which is connected to a voltage divider consisting of Rb and the base-emitter region of the BJT where Vbe is developped.