Amplifying MHz signals - Page 1
Forum section: electronics / beginners

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 01:41:52 am
content: I need to amplify some 3MHz sine wave signals by a factor of around 5, with an amplification method which keeps signal magnitudes proportional (so a 0.1V peak to peak would come out as 0.5Vpp, a 0.3Vpp to 1.5V and so on).The trouble is the rail to rail op-amps I've got to hand (MCP6024 and MCP6294, GBWP 10MHz) can't keep up with it, their maximum slew rate is around 6V/us, and the signal varies about this fast even when unamplified.I'm in a situation where I've only got 5V and ground power rails, no negative rails or higher voltages available.I can drive a few mA of current from the unamplified signal if necessary, so a method for amplifying does not need supr high impedance inputs.The key things I need to preserve in the amplified signal are the peak voltage of the sine wave (accurately), and whether it is in phase or in anti-phase with another similar sine signal. I would be ok with getting something as distorted as a square wave out so long as peak voltage and phase were preserved.My assumption is at these frequencies one starts using methods more similar to those for RF amplification rather tha using op amps, unfortunately I know very little about RF, so any suggestions of circuit types to search for would be helpful.Can anyone suggest how I might do this?Thank you

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 03:42:42 am
content: You can use coupling capacitors at MHz frequencies, and keep the DC level midway between your rails.Fast op amps are ubiquitous.

author: Vovk_Z
author_rank: Super Contributor
timestamp: June 05, 2023, 04:58:47 am
content: You may try low-voltage fast (almost jelly-bean except price) opamp AD8066.

author: RJSV
author_rank: Super Contributor
timestamp: June 05, 2023, 05:10:06 am
content: To create a virtual ground you simply make a divider between your zero and +5 volts, that could be a 10 k resistor up, to +5, and another 10 k down, to zero. Put a fairly large cap, like 5 ufd or larger, and you have an ac reference point.To obtain more details perhaps try to search on some similar analog application notes, for another type of IC, like maybe LM358.I've used circuits with the (quad pack) LM324, although that one can't deliver output all the way up to your + rail... Still lots of info in various application notes...maybe too much at first.

author: tszaboo
author_rank: Super Contributor
timestamp: June 05, 2023, 11:30:27 am
content: Quote from: Infraviolet on June 05, 2023, 01:41:52 amNah, you don't need RF stuff for that, regular, but fast opamp will do. Read the datasheet, there is someting called GBW, gain bandwidth, you need minimum of 15 MHz of that, since you have a 3MHz signal and your gain is 5. And then you calculated the slew rate, which is ballpark 1.42 x f x u = 21V/us. I would try the OPA2350, which would be my goto opamp for such a task, but there are a few that you can try. You also want all the resistors in the 1 KOhm range and not significantly higher so your stray capacitances are not turning it into a filter (or oscillator).

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 02:13:32 pm
content: TimFox, RJHayward, can you give some links to webpages about those capacitor based amplification methods.ThanksEDIT: I might have misinterpreted your responses. I already have a virtual ground rail at 2.5V, that is what the signal is oscillating about, but I need to amplify its peak to peak, not just shift the DC reference level which it oscillates about.P.S. LM324 and LM358 already have far worse slew rates than the MCP6024 I'm already using, do you mean there are app notes about how to make them work above their slew rate which might be applicable to be trying togo above the slew rate of the chips I've got?

author: DavidAlfa
author_rank: Super Contributor
timestamp: June 05, 2023, 02:47:10 pm
content: The specified bandwidth is at -3dB gain (Gain 0.5), so you need something way faster, at least 50-60MHz I'd say.

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 03:39:50 pm
content: Quote from: Infraviolet on June 05, 2023, 02:13:32 pmCan you re-phrase that? What do you mean by "amplify its peak to peak"?

author: bdunham7
author_rank: Super Contributor
timestamp: June 05, 2023, 04:07:49 pm
content: Quote from: Infraviolet on June 05, 2023, 01:41:52 amIf you want to preserve phase and amplitude (scaled) you simply need a faster op-amp and they are available at operating voltages of +/-2.5V or less. You'll need to read the datasheets to find a suitable one, in addition to supply voltage and GBWP there will be a minimum stable gain with faster op-amps that are decompensated. This means that your design needs to be set up with at least that much gain, and since that might be 10 or more, you might need to attenuate the input signal to make all that work.Here is one that meets all your criteria and has a minimum gain of 6--and would probably work OK at 5 but it wouldn't be guaranteed. So you'd need to attenuate your input 6:5 with a divider, or perhaps you can make a gain of 6 work for your application.https://www.ti.com/lit/ds/symlink/opa607.pdf?HQS=dis-mous-null-mousermode-dsf-pf-null-wwe&ts=1685981084108&ref_url=https%253A%252F%252Fwww.mouser.com%252F

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 04:11:08 pm
content: TimFox, I've got a signal of say peak-to-peak of 0.4V centred around 2.5V (min 2.3V, max 2.7V), I want to amplify it, staying centred on the 2.5V rail so it will be peak-to-peak 2V (min 1.5V, max 3.5V). If the input were 2.4V to 2.6V I'd want to be outputting 2V to 3V... The signal is a sine wave, roughly, I need to give it a larger min-max voltage range and must preserve a linear relationship between V_peak_in and V_peak_out such that, within limits, if the amplitude of the sine wave entering doubles so does the amplitude of the output.Might a single transistor common base amplifier with series coupling caps on the input and output be worth looking in to for these frequencies?

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 04:16:30 pm
content: To clarify my question above:If all you need is the AC waveform of a signal in the 3 MHz range, the normal approach is "AC coupling", or capacitor coupling.Take the input through a suitable capacitor to a suitable resistor from your Vcc/2 "virtual ground rail" bias voltage, connecting that to the amplifier input.Connect the output through similar capacitor to a similar resistor to ground.You will lose any DC component on your input this way, but will have an output with zero DC value (mean voltage).The requirement on both C-R networks isRxC>> 1/(2pi xfLF)wherefLFis the lowest frequency of interest ("bass response" in an audio amplifier).If you want the output centered on +2.5 V, then the resistor on the output should go to your bias voltage rail.Alternatively, if the DC offset of your amplifier circuit is reasonable, and the input is biased to +2.5 V by the C-R input network above, then you do not need the output C, since the amplifier output will still be centered at about +2.5 V.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 04:30:43 pm
content: TimFox, have I badly misunderstood what you've been saying, maybe we've been trying to describe different things. Can that coupling method which eliminates DC do any amplification itself, it doesn't sound like it, or is that just pre-conditioning to do to a signal before feeding to to an op amp?

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 04:50:47 pm
content: A C-R circuit is merely a coupling mechanism that preserves the AC component while rejecting any DC component: amplification is required to increase the AC component.The loss of amplitude at low frequencies is governed by theRCproduct ("time constant").Fundamental concept: there is no DC current through a capacitor; therefore, the mean (DC) voltage after a C-R coupling network is determined by the resistor load (voltage at other end plus any voltage from DC current flowing in the resistor from other places such as amplifier input bias current).An op amp circuit is convenient, but if you use C-R coupling at both ends of the amplifier, its DC offset is irrelevant and a single active device (traditional AC-coupled amplifier) will work.If you need the DC component of the input signal (down to zero frequency), you probably want to use split power supplies instead of single-rail, or you need to look at the output with respect to your bias rail.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 05:09:02 pm
content: Thank you, yes I'd misunderstood an earlier post and thought you implied coupling could somehow serve as an amplifier. I was trying to understand what sort of system you were imagining. Now I understand you just meant it as a way of eliminating DC offsets, yes I'm know of them and can add those where needed. Feeding it to an ac coupled common-emitter class-A type single NPN transistor amplifier with an emitter degenration resistor now looks like a possibility for me, that ought to be fesible in the 3MHz or so frequency range so long ad my transistor can handle this frequency? Ought I try to make the R_e and R_c resistors as small as I can (up to the largest current the transistor can handle without heating up) to give this amplifier the best output slew rate?

author: TimFox
author_rank: Super Contributor
timestamp: June 05, 2023, 05:45:37 pm
content: In a simple amplifier, the slew rate is determined by the available device current and the capacitance into which it flows.You must also consider the "Miller effect", determined by the capacitance from output to input of an inverting amplifier and the source impedance driving it.One reason to use an op amp, is that the relevant parameters are often easier to find in the full data sheet, while if you build an amplifier from discrete parts you need to calculate them yourself.There is a large literature from days of yore about "video amplifiers" using discrete components in capacitively-coupled stages (transistor or tube), with interesting output networks (including inductances) to optimize bandwidth.Example from the vacuum-tube days (1959):https://worldradiohistory.com/BOOKSHELF-ARH/Technology/Rider-Books/Rider-Video-Amplifiers-Alexander-1951-Schure.pdf

author: DavidAlfa
author_rank: Super Contributor
timestamp: June 05, 2023, 07:00:55 pm
content: If you want it centered at 2.5V, simply remove C2 in my example.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 05, 2023, 07:28:36 pm
content: Quote from: Infraviolet on June 05, 2023, 05:09:02 pmA simple textbook common-emitter might work fine for this. When i put together a simple function generator, I used one to get a voltage gain of about 12 with something like 2.5 MHz -3 dB point, and I did no work to try and maximize bandwidth. I wanted some roll-off anyway to mitigate dds artifacts, and even had a cap in parallel with the collector resistor. I did it that way because it was easy to keep stable on a solderless breadboard and on perfboard, which don’t tend to be friendly towards high-speed opamps.If you don’t get enough bandwidth from the common-emitter you might try a cascode.Jason

author: RJSV
author_rank: Super Contributor
timestamp: June 05, 2023, 08:01:43 pm
content: Yes it's a virtual ground I was mentioning. The two op amps I had used...all the way back 45 years ! (Whew). You would simply search on that: 'Op Amp voltage follower, ' op amp simple audio gain'Ideally to learn, say with gain of X5 using negative feedback, one circuit does positive gain, the other with (AC) gain but as inverter.The DC gain perhaps X1 (unity gain), so DC input at 2.5 volts makes a DC output at 2.5 volts..The AC rides on top of that.Generally restrict your AC gain to less than 20, or you start having phase problems. So, for 2X gain the negative feedback is going to be 1/2, for 5X gain the feedback is 1/5 etc.Unity gain is often called a follower, and you would see feedback of one....The idea is closely analogous to a servo: In the case of 5X circuit your output is 'controlling' to get the two differential inputs match.So, if the 1/5 divided down output drifts upwards, very slightly, the OP amp magnifies that and feeds back to cause the 1/5 X feedback term to be controlled to go back down.Same for case when that 1/5 term drifts low: the output very vigorously responds the other way, to cause the input to move up, thus you have an error canceling effect, very much like a simple servo.As long as not too high frequency, because at higher frequencies the phase relations get messy.That open loop gain is very high...ideally infinite but can be 100,000. Ditto for that little error, between the differential inputs...ideally at 'zero'.but in practice could be a few micro volts.Confusing, but try doing search terms like Inverting AC amplifier, unity gain amplifier etc.- - Rick

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 08:04:24 pm
content: jasonRF, That textbook common-emitter design is the same as a "class A" layout?I'm trying to breadboard a test of such an amplifier at lower frequencies, yet can't get it to amplify at all, even copying examples found online with all the same component values.I've tried this topology:The only output I get is identical to the input on the base, until the base signal's amplitude gets past a certain point, at which point huge distortion occurs and still nothing like an amplification of the proper signal. I should be getting an amplified and inverted copy at the collector, but don't. Can't understand why.

author: Slh
author_rank: Regular Contributor
timestamp: June 05, 2023, 08:11:39 pm
content: The bias is wrong on that circuit. Increase R1 to 11k and it will probably work ok.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 05, 2023, 11:39:32 pm
content: That was an example of the topology, not the only set of values I tried, but I seemed to have the same problem whatever I tried. I was calculating that the bias was supposed to be sitting at around 0.5*Vcc (i'm using 5V not 12V) with some alterations to account for the typical 0.7V drop, and my calculations (following the method in Horowitz and Hill (Sec 2.2.5 A) ) all suggested having R1 similar to R2.Given your note I've now tried a larger ratio though and seem to be able to get amplification of the output voltage, never realised the biasing has to be such as to make V_base very low without an incoming AC signal.Now to try at the full frequency I need.

author: gf
author_rank: Super Contributor
timestamp: June 05, 2023, 11:55:43 pm
content: To get a DC collector voltage of about 6V, the base voltage needs to be about 1.2...1.3V (0.6V voltage drop across RE + VBE).

author: jasonRF
author_rank: Regular Contributor
timestamp: June 06, 2023, 12:31:18 am
content: Quote from: Infraviolet on June 05, 2023, 11:39:32 pmYou can tell that your posted values had problems by assuming it is biased properly then working out the voltages across each element. Which means that (neglecting the base current) the voltage at the base of the BJT should be about 5.7 volts, which means that there is about 5 Volts across Re from the bias current. But that same bias current flows through Rc, which is 10x Re so there would be 50 Volts across Rc. Given your 12 volt supply this makes no sense.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 06, 2023, 01:49:56 am
content: I've got a solution to this, it seems.With a 10K and 47K biasing potential divider and a 470 ohm and 4.7K R_emitter and R_collector (found I ended up needing 10 factor gain as the circuit draws a bit of current from the input which lowers its voltage) I'm able to amplify my 3MHz signal. I also added a 10nF from the emitter to ground in parallel with the 470 ohms to allow the output to swing further.I'll upload a schematic diagram soon.Can I just check, is this type of class A type transistor amplifier one of those devices which is highly suspectible to individual peculiarties in components? With an op amp type circuit you know that if one of them works then so long as as all the components used in other are within tolerances any copy will work, is this true for single transistor amplifiers? I understand this type is not supposed to depend on things like the transistor's beta property, which varies wildly between two transistors of the same model and batch, but is it dependent on anything else highly variable? I have found that using a BC337 for this transistor gives a rather weaker amplification than if a 2n3904 is used, and the actual amplification is somewhat below the expected -R_c/R_e ratio , does this mean this circuit is probably not something I can reliably reproduce later? or should it be about as reproducible as op amp based circuits are?Thanks

author: jasonRF
author_rank: Regular Contributor
timestamp: June 06, 2023, 04:43:52 am
content: Quote from: Infraviolet on June 06, 2023, 01:49:56 amGlad you found a solution!When I simulate your choice of resistor values, with the bc337-40 LTSpice predicts the -3 dB point is somewhere near 4.3 MHz, so you are already down ~1.7 dB at 3 MHz; the 2n3904 gives a -3 dB point of about 12 MHz so the response has not rolled off at 3 MHz. Perhaps I will breadboard it tomorrow to see if measurements agree. The issue is that your bias current is quite low; if you increase the bias you will increase the bandwidth with both of these devices and they probably will be more similar at 3 MHz.I breadboarded a 5V, x5 gain version with a bc337-40 biased at 17 mA that was only down 0.2 dB at 20 MHz (the max I could measure with the setup I was using), so you can get a lot of bandwidth if you want it. When I swapped different transistors (tried 2n2222, 2n3904, bc549c) I don't think the gain at 3 MHz changed by more than perhaps 0.2 dB, but I didn't record any numbers as I was just goofing around.I think you forgot about re (intrinsic resistance) in your gain formula. At your bias current of about 0.5 mA, re = 26/0.5 = 52 Ohms, so gain is -Rc/(Re+re) = -9. Is that close to what you are getting?Regarding dependence on the device - yes, the circuit will have some dependence on device parameters. You can calculate the dependence (I would think most "Microelectronics" type of books will show how) and verify your circuit design is robust enough for your needs. Biasing is the key, and for starters you really want R1||R2 << beta*Re.jason

author: gf
author_rank: Super Contributor
timestamp: June 06, 2023, 05:15:41 am
content: Output impedance is of course high, so already a scope probe with say 10pF attached to the output leads to a cut-off at about 3MHz.If you bypass R_e with a capacitor, then you give up negative feedback and the amp operates at open loop gain, which certainly makes it very dependent on transistor tolerances.

author: armandine2
author_rank: Frequent Contributor
timestamp: June 06, 2023, 10:21:50 am
content: sometime ago I screenshot a common-emitter FRA example from the internetit looks like it may show the same errors

author: jasonRF
author_rank: Regular Contributor
timestamp: June 06, 2023, 01:57:34 pm
content: Quote from: gf on June 06, 2023, 05:15:41 amSomehow I missed the bypass cap! Also, good catch on the issues with probing the high output impedance.I made a 5V x5 gain version with R1=4.3k, R2=1.5k, Re=33, and Rc=180, driving a 10k load. It has a low input impedance so the OP would need an emitter follower before a higher bias version like this. In any case, at 3 MHz it is reasonably insensitive to choice of device. I have attached some Bode plots, limited to 20 MHz from the gear I was using. The particular transistor I was using is in the filename. Bandwidths are all greater than 20 MHz. I did not fiddle with the values after I did the original design on paper, so I am sure it could be improved.Jason

author: RJSV
author_rank: Super Contributor
timestamp: June 06, 2023, 08:41:55 pm
content: My response is only a bit of isolated fragment, but here my thoughts about recent comments:1.) The speculation about OP AMP circuit being able to work satisfactory, if one works then all of theOP AMP ICs should work; That comes from the 'infinite gain' aspect.You see, a gain of 100,000 X open loop vs something like '110,000 X' is not significant, as your closedloop gain (mostly) determines outcome. That also means some degree of independence when someeffect like ambient temperature is acting to change the OP AMP internal gain, as is natural. So you geta good degree of independence.2.). Putting a discrete transistor on the output, with gain controlled by resistor, gets you out of that independence, so now you have to consider individual parts, measure as you go, with some variations of outcome.3.). Saying that the overall circuit has some 'loading effect' on input is...troubling but only to a minor extent. That really isn't necessary, IMO.4.). When viewing a common emitter connected transistor stage any emitter resistor will cause some negative feedback, and reduce gain but also will give some measure of independence from the variation parts to parts, in individual transistors (please see #2 above). That works because your output current will elevate the transistor emitter (voltage) and thus directly raise the base voltage, ultimately reducing the base to emitter voltage, and subsequent output swing. (that being a good, stabilizing to thing).======================================================================Honestly, I don't see a need, though, for the transistor in the output, unless you need some extra drive ability. IMO.I would rate my analog skills at 'moderate', meaning general comprehension but subject to some 'imperfection', there.Check out RON WIDLAR for good background info.He developed a substitute for very large resistances, in IC internal circuits, that idea being to use a current source instead....sounds very non-intuitive, lol

author: jasonRF
author_rank: Regular Contributor
timestamp: June 06, 2023, 10:06:39 pm
content: Quote from: RJHayward on June 06, 2023, 08:41:55 pmThe OP needed gain at 3 MHz and had no suitable opamps that would work from a 5V supply. A simple common-emitter allowed him to get a solution with parts on-hand.jason

author: Infraviolet
author_rank: Super Contributor
timestamp: June 06, 2023, 10:21:08 pm
content: Here is the diagram of my transistor amplifier setup. You're suggesting that keeping the R7 to R8 ratio the same but shrinking both of them could let me get amplification closer to the gain that the R9 to R10 ratio would imply, and let me remove that output cap which risks making things dvice charactertistics dependent?With the warning about needing the biasing resistors in parallel to be <<beta*R_emitter I can see why 10K and 47K aren't too good.I'll give this a go shortly, thanks.

author: TimFox
author_rank: Super Contributor
timestamp: June 06, 2023, 10:40:10 pm
content: You need to simulate or calculate that circuit carefully:My rough calculation indicates that your desire for 0.4 V pk-pk input will drive that circuit (with bypassed emitter resistor) into heavy saturation with that value of collector resistor, since a rough value for small-signal voltage gain is 75 V/V.With an unbypassed emitter resistor ("emitter degeneration", a nice term), that voltage gain falls to maybe 7.5 V/V, giving 3 V pk-pk at the output for 0.4 V pk-pk input, since the transconductance falls by roughly 10x.With a DC voltage across the collector resistor of roughly 2.2 V, that output would be possible.(I assumed a very highhfe, so I neglected any DC loading of the base current in your base voltage divider network.)An elementary approximation isgm= (26 mV)/Ie, whereIeis the DC emitter current.This applies to the case where the emitter resistor is fully bypassed. The effect of an unbypassed series resistor is left as an exercise for the reader, but simple algebra will yield it.You have maybe 0.25 V across that emitter bias resistor, giving roughly 0.4 mA.Hint: start with your desired emitter current. Choose a voltage across the emitter resistor to be several times higher thanVbeDC value to reduce the dependence on the transistor. With the minimumhfevalue for your transistor, calculate the maximum base current. Choose a DC value for the current through the base voltage divider to be several times higher than that, and to give a base-ground voltage equal to the emitter-ground voltage plus the expectedVbeDC value (0.6 to 0.7 V, typically). Go to the textbooks to get better values and approximations, but this should be a reasonable start.Then choose the DC voltage across the collector resistor to be half of the supply voltage minus the DC voltage from emitter to ground; that gives you the collector resistor for your specified DC emitter current.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 07, 2023, 02:27:07 am
content: I tested with 1K and 4.7K for base biasing. With the emitter cap still in place the signal was the same as with 10K/47K, once that cap was removed the signal became rather weaker again. But the lower resistances on the base and the removal of that cap make this weak output consistent between different models of transistor.I am strongly suspecting the weak output from the transistor amplifer might be due to the signal source having enough resistance associated with it that the input current drawn by the circuit is reducing the input waveform peak-to-peak voltage quite a lot. I'm going to try putting an NPN emitter follower between the signal source and the input to this amplifier as a current bossting buffer, then see if I get better results.Thanks for the further explanation of the equations, I'm pretty new to using discrete transistors as anything other than digital switches, I think I must have interpreted the equations I found online incorrectly and thought that the R_e to R_c ratio was the only thing controlling the amplifier gain. I'll work through them again and see what values I come out with for each resistor.ThanksEDIT: putting an emitter follower voltage buffer between the signal source and the amplifier's input cap has really helped. An input signal peaking at 300mV and -300mV about its centre is now being output from +960mV to -960mV. With my Rc/Re ratio I'd be expecting almost 2.5V either way, but I guess the 5V supply and the voltage drop of the transistor are limiting this from amplifying any further? And it seems to stay consistent when swapping transistor type in either place. I haven't yet changed the resistors around the voltage amplifier to improve the gain further yet (still 1K and 4K7 biasing, 680 and 5K6 emitter and collector). Presently I'm getting, at DC and all relative to ground, 0.3V on the emitter, 0.9V on the base, 2.85V on the collector.

author: TimFox
author_rank: Super Contributor
timestamp: June 07, 2023, 02:47:06 am
content: If the emitter resistorReis unbypassed, the voltage gain is roughlyRc/Re, whereRcis the collector resistor.This assumes thatReis much larger than (26 mV) /Ie, whereIeis the DC emitter current.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 07, 2023, 02:18:41 pm
content: Quote from: Infraviolet on June 07, 2023, 02:27:07 amDoes your output still look like a sinusoid, or is it distorted? As TimFox notes in his excellent post, at some point the bjt will be in saturation. If this is happening then you will certainly distort the signal.And if you now have 600 mV p-p, then you need less gain, yes?Jason

author: TimFox
author_rank: Super Contributor
timestamp: June 07, 2023, 07:24:00 pm
content: An addition to my post above:When using a bypass capacitor across the emitter resistor (or analogous cathode resistor for a tube or source resistor for a FET), the reactance at the lowest frequency of interest must be compared to the inverse of the transconductance, not the actual emitter resistor.I.e.,XC< 1/gm= (26 mV)/IeThe resulting capacitor value will be substantially higher than if calculated to short out the external emitter resistor, which can easily be 10x greater than given here.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 08, 2023, 12:08:21 am
content: To be clear, since I put the voltage buffering emitter follower transistor after my signal source and before the caleading in to this amplifier I have REMOVED the cap which was parallel to the emitter resistor. The behaviour with the buffer and no emiter resistor bypass cap is what i was discussing in reply #32.The output sine wave doesn't look distorted, certainly no huge distortions, perhaps a slight element of the rising/falling slopes being slightly more linear (hard to be sure but maybe slightly more constant-ish gradient when crossing the centre) than a perfect sine, but peaks still curve like a sine wave. No clipping, no huge distortion, no added spikes or ripples.As far as gain goes, I'm ideally looking to gain this signal up as far as possible* without clipping from my 5V and ground rails, what I'm getting now is adequate, but more (up until the limitations of my voltage rails and any effects of the voltage drop in the transistor) would be desirable. I'd like to gain as much as I can, but only up to a point before I'd end up sacrificing accuracy to clipping or other phenomena. But it seems with the present situation I still can't seem to increase the gain above what it presently is by further adjusting resistors**, can anyone point me to a good webpage guide to not only calculating these but also to getting a proper feel for what is going on, my intuitions keep just considering transistors as digital switches despite all the other wonders I know they can accomplish. I have got Horowitz and Hill's book which has a BJT transistors chapter, but it perhaps discusses too many aspects of them at once.*originally I had intended five fold because of what the input signal magnitude was originally, but as other sections of the circuit outside this thread's discussion have changed the signal input magnitude altered somewhat, it is now +300mV and -300mV about a centre and will stay such with any further changes to parts of circuits not discussed in this thread.**might I have run in to limitations of the voltage drop already, if so I can tolerate it but if not I'd like to get more gain

author: armandine2
author_rank: Frequent Contributor
timestamp: June 08, 2023, 06:44:53 am
content: https://www.amazon.co.uk/Electronics-Manchester-Physics-JM-CALVERT/dp/0471996394https://www.amazon.co.uk/Intro-Electronic-Circuits-Martin-Hartley/dp/0521478790https://www.amazon.co.uk/Electronic-Testing-Diagnosis-George-Loveday/dp/0582038650I'm currently dabbling with the common emitter, mostly using above books. The last two are pretty good introductions IMO -I'm also using Hayes and Horowitz 2nd edition Lab Book

author: jasonRF
author_rank: Regular Contributor
timestamp: June 08, 2023, 11:33:06 am
content: Quote from: Infraviolet on June 08, 2023, 12:08:21 amWhat do you mean by "limitations of the voltage drop"?Just to make sure I understand the complete circuit that for some reason you haven't shown us, you now have an emitter-follower feeding the common-emitter, and where they are connected you are measuring a 600 mV p-p signal going into the common-emitter. Is that correct?I have attached three images from an LTSpice simulation of your circuit as I understand it. One is a frequency-response plot, one shows the input and output voltages, and one shows the voltages at the three legs of the BJT.The maximum gain that will work depends on the input signal, so that knowing that really matters. Right now I calculate a mid-band gain of about 7.4, which agrees with the LTSpice frequency-response plot. But the plot shows that at 3 MHz your circuit response is already down -2 dB, and of course the Spice model will not be perfect for your particular bc337 so you may be down more (or less!). The fact that the input and output signals in the time-domain aren't exactly 180 degrees out of phase indicate the extra phase shift you expect to see in a low-pass filter when the response starts to fall.What phase difference do you see between the input and output of the common-emitter? Could you post a screen shot?By the way, a gain of 7.4 is unrealistic for a 5V rail and 300 mV input signal. You will need to reduce the nominal mid-band gain to optimize this circuit. I would think that a gain much more than about 6 or so would be tough.One thing I would try immediately is to drop the collector and emitter resistors (by a factor of 10 perhaps) to increase the bias. This will increase the bandwidth of the common-emitter and also make probing easier and let you know if you have been trying to use your circuit above its operating band.Another issue with the circuit is the bias point set by your voltage divider. The emitter bias point is at about 0.3 Volts, so if you input a 0.3 Volt sinusoid you may be driving the BJT into near cutoff when the input sine is at its most negative point. You may want to move the bias point to higher voltage. This will also make the circuit operation less dependent on the details of the particular transistor you pick. The plot I attached that shows the three signals at the BJT legs looks like it may be approaching cutoff, and at some point the base-collector junction is forward biased so it is entering saturation.EDIT: In general I am unable to recommend web sites since that would require me to search for and read them to decide which ones were helpful. One video on the basics of the emitter-follower that shows AC and DC analysis of a BJT circuit isThere must be a good description of a common-emitter somewhere online... I learned this stuff from Microelectronics by Millman and Grabel, but the text by Sedra and Smith looks better. They do require that you already know basic circuit theory, though, and go through lots of math. If you want to maximize the gain from this circuit then, as pointed out by TimFox, you really need to learn and do the math or try trial and error with a simulation program. Note that this circuit will have limitations since you do not have a lot of control. I have attached a schematic that includes an extra bypassed resistor. The idea is that Rbp gives you control over the DC bias in a way that does not effect the gain. The design proces is similar to what TimFox indicated:1. Fix the bias point - ideally would want the emitter to be 1.3 V or so (to make much less dependent on the transistor details) but to maximize voltage swing you might make the circuit less robust and lower it to 0.5V. You also need to pick a bias current (should probably be at least a few mA - so 10x what you have now) and a nominal gain.2. Setup the voltage divider bias. Probably want the current through it to be at least 1/20 of the current through the transistor.3. Do some algebra to determine the quiescent (DC) operating point with no signal input, leaving the resistor values as variables4. Write a condition to avoid saturation. This is when Vcollector-Vemitter > Vsat where Vsat is the saturation voltage of the device (often 0.2V or so). This analysis will require an analysis of the AC circuit to add to the DC quiescent point. More algebra will yield a condition on the maximum allowed collector resistor, Rc.5. Pick a collector resistor from a standard value that satisfies the condition from (4), then compute Re to get the gain you want. Once you know that you can then compute Rbp. Pick Cbp to be large enough so that at signal frequencies it is effectively a short.If you don't have the time/inclination to learn the math, you can do this trial-and-error in a simulation program. In this case, though, you1. setup the voltage divider to get something like 0.5V at the emitter (again, 1 V or more would make it less sensitive to devices). You need to pick a nominal bias current through the transistor in order to pick the sizes of the resistors for this voltage divider. Finally, you need to pick the gain.2. pick Rc3. pick Re to get the gain you want4. pick Rbp to set the bias current.5. run the simulation and see if you like the output. If you don't you can try a) reduce the gain and re-pick Re and Rbp. or b) return to 2 and pick a smaller Rc. Note that if you do b) then you need to keep the bias current and gain the same as you iterate. If you change them, then the maximum allowed value for Rc changes and you will do trial and error forever.jason

author: gf
author_rank: Super Contributor
timestamp: June 08, 2023, 02:08:55 pm
content: Quote from: InfravioletYou can't get rail to rail output with this circuit. I would not expect more than say 3.5Vpp w/o driving the transistor into saturation.

author: TimFox
author_rank: Super Contributor
timestamp: June 08, 2023, 03:40:33 pm
content: Reviewing my comments on biasing with the other peoples' comments on simulation of transistor circuits with other practical considerations such as load capacitance, we see that designing a single-transistor circuit for 3 MHz amplification is not rocket science, but not a trivial matter of just drawing wiring diagrams.The designers of operational amplifiers have already done the necessary design work using good design methods, and a decent data sheet contains all the information you need to see if a given device will work for you.Suggested procedure:1. Define your requirements: specifically bandwidth (conventionally at -3 dB), required output swing (V pk-pk), and voltage gain.2. Use an input capacitor to isolate your input DC voltage (probablyVcc/2) from the DC output resistance of your generator (maybe 50\$\Omega\$), and an input bias network to set the input DC level well within the common-mode input range of the device.3. The required output swing will require a supply voltage greater than the pk-pk swing.4. The gain and bandwidth together define the GBW product (or unity-gain frequency) required to obtain that gain at high frequency.5. The output swing and frequency determine the required slew rate (dVout/dt), another data-sheet parameter.6. Use a feedback network (output to inverting input) with a capacitor to ground to obtain unity gain at DC, but the required gain at your frequency. The total resistance of this network will draw current from the output, so you need to use high enough resistors not to overload the output, but otherwise not too high to cause high-frequency roll-off with stray capacitance across the feedback resistor. Capacitances around this network can be tweaked to optimize the response.7. An output capacitor is a good idea to remove any DC error, or to have the output swing around ground rather thanVcc/2 .The input and output capacitors, along with the resistances at the "right" side of each capacitor, will define the low-frequency response.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 08, 2023, 05:38:35 pm
content: 100% agree with TimFox, who clearly understands this stuff better than I do. Op amps are much easier to use and provide superior performance. And (as Dave has a nice video about) if you don't have ones with enough bandwidth you can often cascade a few lower-gain stages to get the needed gain at the frequency of interest.In any case I went ahead and did a pencil-and-paper design for a single-transistor amp that simulates and measures as expected. Was a fun exercise, at least for a hobbyist who happens to find discrete more 'fun' than opamps despite the fact that I am not an expert and they usually aren't the best technical solution.jason

author: bdunham7
author_rank: Super Contributor
timestamp: June 08, 2023, 05:47:36 pm
content: Quote from: jasonRF on June 08, 2023, 05:38:35 pmHow did you physically construct your circuit?

author: TimFox
author_rank: Super Contributor
timestamp: June 08, 2023, 05:47:50 pm
content: In JasonRF's simulations, note that the .tran simulation is required to see actual waveforms to look for clipping, etc., but the .AC simulation is used to measure the (small-signal) frequency response.The partially-bypassed emitter resistor allows some independence in setting the bias (quiescent) voltages and current, with more gain than with an unbypassed emitter resistor.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 08, 2023, 06:25:22 pm
content: Quote from: bdunham7 on June 08, 2023, 05:47:36 pmSolderless breadboard - picture attached.Also have a better measurement showing the signals at all 3 terminals of the BJT, color-coded to agree with the simulations. Since I only have a 2-channel scope the base is a 'reference waveform' I saved a minute before the emitter collection.EDIT: and for a simple explanation of the circuit. The collector is biased at about 3 Volts and the emitter at about 0.5 Volts. With 0.3V input sinusoid, the emitter will swing from 0.2V to 0.8 V and the gain is about 6 so the collector swings from about 1.2V to 4.8V. So Vce should be at least 0.4V which is high enough for many transistors to stay out of saturation, and the signals have about 0.2V 'margin' at the top and bottom to avoid cutoff and clipping.This type of thinking could be used to more easily design this than the math-heavy approach that I used: if want at least Vce>=0.4V to avoid saturation, and 0.2V 'margins' on top and bottom, and and we have a 0.6V peak-to-peak input signal, then for 5V supply the maximum peak-to-peak output voltage would be 5V - 0.4V - 0.2V - 0.2V - 0.6V = 3.6V so a gain of 6. The collector bias point would be 5V-0.2V-1.8V = 3V, and the emitter bias point would be 0.2V + 0.3V = 0.5V. Now pick a bias current, then pick Rc to give the 3V bias point, then select Re to give a gain of 6, and then pick Rbp to cause the emitter to be biased at 0.5V. Finally, make the voltage-divider biasing set the base at about 0.5V + Vbe or about 1.2V. Viola! You may need to iterate a little if the emitter bias voltage isn't quite what you want and it changes the current, etc., but these should be minor adjustments.The emitter voltage at only 0.5 V does of course make the operation more dependent on the BJT characteristics than if it were higher. But the approach is still easy to use for designing a more robust circuit. If the emitter is at 1.4V, the maximum peak-to-peak output voltage is 5V - 0.2V - 0.4V - 1.4V - 0.3V = 2.7V. So the collector should be biased to 5 - 0.2 - 1.35 = 3.45 Volts. Gain should be 4.5.For the gain calculation I did include the effect of the load and the intrinsic resistance of the BJT, by the way, so on paper the gain should be 5.95. My circuit actually has the emitter at 0.45V, which is why the 220 Ohm resistor sets the collector at 3V.jason

author: Infraviolet
author_rank: Super Contributor
timestamp: June 08, 2023, 11:33:31 pm
content: Thanks all for the replies, a lot to think through.armandine2, thanks for the linksjasonRF,"complete circuit that for some reason you haven't shown us"Sorry, it always takes so much longer to prepare an image than just post words, here's the current schematic. Note that the incoming signal is already centred at 2.5V, roughly, so the emiter follower voltage buffer in theory shouldn't need any furtherbase biasing.I will post details of the phase differences tomorrow. I'll give a go to reducing the R10 and R9 while preserving ratio, thanks."I would think that a gain much more than about 6 or so would be tough."thanks for clarifying, guess I'll be acepting the current gain levels then. But I'd like to try making the emitter voltage higher if that makes it less dependent on transistor properties, I'll give that a go and post what I find.I'll work through the maths of the various values again, I might have been confused by letting intuitions about transistors as switches overtake the reality of this application, and might have mistaken some of the terminology too, in the references some methods made to which resistor one was calculating at a give point in the process.gf, thanks for confirming,seems I am getting about as good as possible then.TimFox, thanksI'll post more when I've made some furthe measurements based on all your advice.

author: gf
author_rank: Super Contributor
timestamp: June 09, 2023, 10:48:47 am
content: Quote from: Infraviolet on June 08, 2023, 11:33:31 pmYou get less dependency if you increase negative feedback, i.e. if youreducethe closed loop gain. Currently you have approx. 5600/680 = 8.24What is actually the intended load impeance/capacitance? If the load impedance is not significantly higher than Rc, then output voltage will drop (and a capacitive load will let it drop at high frequencies). Note that a lowpass with 5600 Ohm and only (say) 10pF cuts off at 2.84 MHz.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 10, 2023, 12:08:22 am
content: Ok, so I've found that, as suggested, reducing the sizes of both the emitter resistor and collector resistor whilst maintaining their ratio increases the gain I get in practice.With the base biasing pair set to 3.9K and 1K, if one uses R_e=820 ohms and R_c=5.6K then whilst the gain in theory is -8.2, the reality is more like -2.7. But with R_e=270 and R_c of 1K2 the practical gain is -4 gainst a theoretical of 4.4. Increasing the current in the collector and emitter resistors is making the real gain come closer to the theory, but I'm still confused as to why. Isn't the gain just -R_c/R_e, is there some phenomena where if the current in R_c and R_e is low enough then full predicted gains can't be achieved?In either case I'm getting V_emitter at about 0.4V in the quiescent state, when the largest magnitude input wave is present it still stays just above ground on its low swings. Trying to bring the base any higher, as a way to bring the emitter up too, results in clipping of the lowest points of the output waveform when the input is near maximum.P.S. regarding the phases the output lags the input by about 200 degrees, or leads by 160 depending how it is considered. Close to the 180 expected.

author: gf
author_rank: Super Contributor
timestamp: June 10, 2023, 07:22:23 am
content: Quote from: Infraviolet on June 10, 2023, 12:08:22 amIf you don't neglect the internal Re of the transistors, then the theoretical gain is rather -4 either. Better simulate with LTspice (or any other circuit simulator of your choice). They use more accurate models. R_c/R_e ist just a first order approximation for the gain -- rule of thumb.

author: Picuino
author_rank: Super Contributor
timestamp: June 10, 2023, 11:42:33 am
content: You can look for video amplifier stages with negative feedback. They are very similar to what you are looking for.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 10, 2023, 02:01:03 pm
content: Quote from: Infraviolet on June 10, 2023, 12:08:22 amIf the emitter resistor is 0.4V, then in the first case the current is 0.49 mA so (assuming your load impedance is much much larger than 5.6k) the intrinsic emitter resistance is 26mV/0.49mA = 53 Ohms. Since the ideal gain for very large beta is -(Rc||Rload)/(Re + re), if Rload is infinite then I get an ideal gain of about 6.4. I will let you work out the ideal gain for the second case.But the larger effects are1. The gain-bandwidth product is dependent on the bias current through the transistor. I would expect the bandwidth at 5 mA to be much larger than the bandwidth at 0.5 mA. Some BJT datasheets show plots of this.EDIT: should have noted that analysis of the standard hybrid-pi model shows the unity-gain-bandwidth (in Hz) is approximately $$f_T \approx \frac{1}{2 \pi \, r_e \, C_{internal}}$$, where \$C_{internal}\$ is an internal capacitance of the transistor. Since \$r_e\$ is proportional to \$1/I_{bias}\$, increasing the bias increases the bandwidth.2. As others have noted, with 5.6k output impedance, when you probe it the capacitance of your scope probe will create a lowpass filter so the signal amplitude your scope shows may be lower than the output signal amplitude when it is not being probed.I did a couple of simulations of a circuit similar to yours, one with Rc=5.6k and Re=820, and the other with Rc=560 and Re=82. I hung an emitter follower off of the output to buffer it. The simulation plots show that the bandwidth of the 560 / 82 case is much greater than the bandwidth of the 5.6k/820 case. Also, they show that the frequency response of the output of hte emitter follower is the same as for the common-emitter, as expected in simulation.I then built these circuits and measured their frequency responses, both at the common-emitter output and the emitter-follower output. I have attached the plots, where the filenames tell you the resistances used, and the _EF in the filename indicates the measurements after the emitter-follower. You again see that the bandwidth of the 560/82 version is much higher than for the 5.6k/820 version. In both cases, the measured bandwidth after the emitter-follower was larger than before, most likely due to the effects of the probing.I was using very sloppy probing on a solderless breadboard, by the way. I had wires plugged in at the measurement points then clipped the probes to the other ends of the wires. So the effects of probing could probably be reduced.jason

author: TimFox
author_rank: Super Contributor
timestamp: June 10, 2023, 03:42:34 pm
content: Quick addition to above:The parameterrein jasonRF's analysis corresponds to (26 mV)/IEin my discussion above, and comes from the derivative of the I-V curve in the Schockley equation for PN diode operation.As he points out, one should compare it to external series resistance in the emitter circuit.

author: Picuino
author_rank: Super Contributor
timestamp: June 10, 2023, 05:33:00 pm
content: Without feedback it is very difficult to maintain a constant amplification ratio.A feedback circuit can correct these response curves with frequency.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 10, 2023, 05:39:51 pm
content: TimFox, thanks for stating that like that, I kept just seeing 26mV and thinking "oh that's not much against 5V", now I can see to see it as an effective resistance 26mv/I_emitter, more of a problem the smaller the emitter current is.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 10, 2023, 06:33:12 pm
content: Found a derivation of the bandwidth of the circuit including the emitter resistorhttps://www.brown.edu/Departments/Engineering/Courses/En162/lecture_JackMiller12.pdfIt indicates the cutoff frequency should instead be at$$ f \approx \frac{1}{2\pi \, (r_e \parallel R_E)\, C_\pi}$$where \$C_\pi\$ is an internal capacitance of the transistor. Usually \$R_E >> r_e\$ so \$R_E \parallel r_e \approx r_e\$, but I thought I should correct it. The prior result I posted is one I looked up and it did not include the emitter resistor... oops!cheers!jason

author: Picuino
author_rank: Super Contributor
timestamp: June 10, 2023, 06:38:53 pm
content: Simulated and working.

author: TimFox
author_rank: Super Contributor
timestamp: June 10, 2023, 06:45:21 pm
content: Quote from: Infraviolet on June 10, 2023, 05:39:51 pmTo first order, I like to think of a CE transistor amplifier as embedding an aetherial active device with infinite transconductance in series with thatrein series with the external emitter circuit.Similarly, if you use the CB connection, that gives you the input impedance into the emitter.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 10, 2023, 08:34:48 pm
content: Thank you Picuino, really like the look of that, which resistors control the gain?

author: jasonRF
author_rank: Regular Contributor
timestamp: June 10, 2023, 09:04:12 pm
content: Quote from: Picuino on June 10, 2023, 06:38:53 pmI think Q2 needs to be flipped.

author: Picuino
author_rank: Super Contributor
timestamp: June 11, 2023, 07:43:01 am
content: R2 and R3 controls the gain.

author: Picuino
author_rank: Super Contributor
timestamp: June 11, 2023, 07:43:54 am
content: Quote from: jasonRF on June 10, 2023, 09:04:12 pmyes

author: Picuino
author_rank: Super Contributor
timestamp: June 11, 2023, 07:49:03 am
content: Working.The gain is controlled by R2 and R3:Gain = R2/R3 + 1R4 and R5 control the bias point.R1 controls the Ic of Q1 (more current, more speed).R3 controls the Ic of Q2 (more current, more speed) and output impedance.

author: jasonRF
author_rank: Regular Contributor
timestamp: June 11, 2023, 05:09:22 pm
content: Hi Picuino,Have you noticed any stability issues with the video amp? When I have used that topology (which I really like, by the way) I typically needed a capacitor on Q2 from base to collector to keep it stable. In one case the instability was obvious oscillations, and in another case it was most noticeable during square-wave testing where there was high-frequency 'fuzz' after the falling edge of a square wave.jason

author: Picuino
author_rank: Super Contributor
timestamp: June 11, 2023, 06:06:59 pm
content: I have not tested this topology many times nor have I studied it thoroughly in stability. Therefore I do not know how to answer reliably. The times I have used this topology it has worked without problems.The stability will also depend on the transistors selected and the PCB layout.Edit: Also, as in other feedback circuits, at low gains you will have more instability than at high gains.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 14, 2023, 02:19:13 am
content: Related to these amplifiers I've also been having a go at making an analog phase splitter, set up as an amplifier with unity gain (Re=Rc).I've been having really serious clipping though, whatever I do with it. Are phase splitters of this kind impractical for supply voltages of 5V with input waveforms as large as 2.7Vpp? Like so many other discrete transistor circuits, the online examples are very often using 12 or 20 volt supplies. Can't seem to find any way to get the quiescent voltage at the collector to be as low as feasible while having the quiescent voltage of the emitter as high as possible, both of which would be necessary to get output waveforms of a substantial fraction of the supply voltage.This shows the overall topology of the phase splitterAny thoughts of whether the principle ought to be able to work for such voltage ranges?Thanks

author: jasonRF
author_rank: Regular Contributor
timestamp: June 14, 2023, 03:14:32 am
content: Quote from: Infraviolet on June 14, 2023, 02:19:13 amIf you had 2.7vpp input then you will have 2.7 x 2 = 5.4v of signal output if the circuit is operating as you desire. With only a 5v supply that is -0.4 volts across the transistor, even if the signals are literally touching ground and the 5v rail. The transistor will be in saturation well before then.The maximum output will be possible when the emitter is at about 5/3v, and the collector is at about 10/3v. The maximum output should be when the emitter is at about 1/4 of the supply voltage and the collector at 3/4 of the supply voltage. Not sure what I was thinking! I would expect 2Vpp input or so to be the max you could hope for. And to keep the gains the same on the two legs, they should have the same (or very high) load impedance. The Art of Electronics discusses this circuit as well. One way to do that is to send them both to emitter-followers or other buffers.Attached is an example circuit that might be worth starting with, although I did no optimization really.Edit: I recommend reading ch2 of The Art of Electronics. They list a few conditions for the bjt to operate as an amplifier. The first is that (for npn) the collector needs to be at a higher voltage than the emitter. This would be violated with a 2.7v input with a 5v supply, regardless of the biasing.jason

author: jasonRF
author_rank: Regular Contributor
timestamp: June 14, 2023, 12:08:35 pm
content: Thought it might be helpful to show the emitter and collector voltages for 2vpp input and 2.7Vpp input. The circuit is happy with the 2Vpp, but you see problems with 2.7Vpp.I am also attaching the input file for LTSpice. It is a free circuit analysis program that I recommend you download since it makes it very quick and easy to try out different circuit ideas, and also makes it easy to visualize all of the voltages and currents in a way that would be difficult with a physical setup.https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.htmljason

author: Infraviolet
author_rank: Super Contributor
timestamp: June 14, 2023, 09:00:37 pm
content: The type of clipping is the same shape as I observed when trying it with just the Q1 amplifier of the circuit, although I got that effect starting from inputs rather below 2Vpp, my biasing might have been different at the time though, will rebuild again and see how it comes out this time.To clarify, Q2 and Q3 both act purely as signal buffers, if you were using a high enough resistance load (to ground) on both channels you could just have that in their places?Thanks

author: jasonRF
author_rank: Regular Contributor
timestamp: June 14, 2023, 10:12:49 pm
content: Yes, Q2 and Q3 are buffers with >100kOhm input impedance. They might not be necessary.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 17, 2023, 03:45:40 pm
content: To check, for both the simple common emitter amplifier (not with a bypass cap across R_e), and the video amplifier circuit design, what equations govern how the gain and the centre-level of the output signal will vary with changes in V_be.Might a two transistor differential amplifier be a better idea, with a fixed voltage as the reference rail. I understand those can achieve pretty much any gain one could wish for, and that they have some amount of inherent negative feedback which cancels out V_be variations, but I don't know whether those would find themselves unable to give outputs where the amplitude of the wave would bring it close to the power and ground rails, for differential amplifiers is this capability mroe limited than for common emitters and/or video amplifier designs.My biggest concern really is clipping, if I end up clipping the outgoing signal at either the positive or negative ends of the signal then the system cannot work for its required purpose, so if V_be variation could easily take me in to clipping situations then that would be very bad. In my tests so far I've had no clipping for situations where I am able to get a maximum peak-to-peak output voltage of about 2.8V (peak-to-centre 1.4V), but I want to ensure the finished system is resilient against entering clipping if V_be changes within plausible limits, and I don't want to sacrifice gain to do so (getting substantially less than 2.8Vpp out for the maximum input amplitude is not desirable). Looks like I might be in a fine balance between trying to get a decent amplitude and avoiding the risk of clipping?Thanks

author: TimFox
author_rank: Super Contributor
timestamp: June 17, 2023, 04:07:04 pm
content: For the single transistor and unbypassed emitter resistor, see my posts above.If you cannot tolerate clipping on a 2.8 V pk-pk signal, you should probably use more than 5 V supply voltage to be conservative.If you live on the edge, then note that for the single transistor, you should have at least 1.5 V across the collector resistor and 1.5 V across the transistor (marginal for +/- 1.4 V swing on each), which leaves 2 V for the voltage across the emitter resistor (again, marginal but it might work).This is only 0.1 V of margin at the output.Increasing the supply voltage to 9 V will allow 3 V across the collector resistor, 3 V across the transistor, and 3 V across the emitter resistor.Bypassing or partially bypassing the emitter resistor can increase the gain, but will not affect the clipping levels at the collector.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 17, 2023, 05:12:06 pm
content: Do you know about cmos gate linear amplifiers?The cheapest of them all:CD4xxxUB series up to 15V Vddwww.youtube.com/watch?app=desktop&v=e6q7graP-1Y

author: mawyatt
author_rank: Super Contributor
timestamp: June 17, 2023, 06:23:32 pm
content: Here's a simple single transistor wideband topology called Series Shunt Feedback that we've used with Si, SiGe and InP bipolar transistors over the years.Edit: Adding a Peaking Inductor and/or Capacitor can extend the response as shown.Best,

author: Infraviolet
author_rank: Super Contributor
timestamp: June 17, 2023, 07:20:29 pm
content: TimFox, maybe I'm misunderstanding again, but 2V across the emitter resistor for a 5V supply would make it impossible to have a large enough ratio of R_c to R_e resistances to get a gain of even 2? I am very much fixed on using a 5V supply, the end use for this is a building block within other projects and I've reached a sort of "standard" for how they get supplied, there isn't >5V available.Terry Bites, sounds a VERY clever trick, using a digital inverter IC as an analog amplifier. As he seems to be calibrting gain simply from resistor ratios I'd guess he'd get consistent performance from such use, without much dependency on device manufacturing variation or current or temperature? Looks ideal, assuming I could then feed the output to a buffer transistor so that the eventual load the signal goes to wouldn't disrupt the carefully selected load. I'll give this a go, and see if it can work at 5V supply. I have plenty of NOT gates to hand.mawyatt, the idea with the Series Shunt feedback being that the biasing of the base is altered by the level of the output? That kills off any V_be dependencies? I'll try to find the equations for adjusting the resistors, thanks.

author: TimFox
author_rank: Super Contributor
timestamp: June 17, 2023, 07:53:34 pm
content: If you have 2 V DC across the emitter resistor, and that gives insufficient gain at high frequencies, you can connect a series circuit of a smaller resistor and appropriate capacitor in parallel with the emitter resistor.Or, as discussed above, you can split the emitter resistor into two series resistors and put a bypass capacitor across one of them, to get a smaller resistance at high frequencies.It is normal to use highish voltages and resistances to assure the DC quiescent point, and then partially bypass appropriate resistors with a capacitor for high-frequency gain.jasonRF's reply (#38) shows the two series resistors circuit, but if you want to play with DC and AC separately, putting the series RC in parallel allows you to adjust them independently.

author: Infraviolet
author_rank: Super Contributor
timestamp: June 17, 2023, 08:07:09 pm
content: That separated handling of DC and AC will ensure that V_be and other temperature/current/manufacturing variabilities in the transistor are mostly made insignificant?

author: TimFox
author_rank: Super Contributor
timestamp: June 17, 2023, 08:39:05 pm
content: The DC part reduces the effect of any change or difference inVBEon the DC (quiescent) operating point.Note that if you partially bypass the emitter resistor (either circuit), the required capacitor is governed by the (smaller) AC resistance needed, not the (larger) DC resistance.Specifically, if the AC resistor isRACand the transistor's internal resistance isrE= (26 mV)/IE, then the capacitor's reactanceXCmust be substantially less than (RAC+rE) to short out the DC resistorRDCat the lowest frequency of interest.If you really need that much peak-to-peak output with that small a supply voltage, you are probably better off with a fast op-amp circuit, where the DC operating point can be set precisely by the usual circuitry, including capacitors where needed to obtain a lower DC gain (for stability) than the AC gain (for required amplification).

author: Infraviolet
author_rank: Super Contributor
timestamp: June 17, 2023, 09:12:05 pm
content: I can get my quiescent V_e emitter voltage up to 1.2V this way without compromising gain when I pick appropriate values, at these levels am I in a situation where V_be variations and other variables properties of the transistor will no longer have much effect? Which equation specifically governs V_be's effect on the gain and the quiescent DC levels of the various transistor pins? Am I along the right lines thinking of:V_b_qui is set by the biasing resistor ratio,V_e_qui is always V_be less than V_b_qui by V_be,R_c is controlling the current I_c,and R_e*I_c (as I_c is virtually I_e) is controlling how high V_c_qui isV_be changes will then alter V_e_qui, which would serve to shift V_c_qui around, but not affect gain unless V_c_qui 's shifting brought it to a value where clipping became possible?Every 0.1V increase in V_be raises V_c_qui by 0.1V? 0.1V of drop to V_be lowers V_c_qui by 0.1V?EDIT: might have just found what I neededsetting R_e>(2.5mV)*expected_temp_range/(proportion_error_tolerable*current) is supposed to ensure temperature variation won't interfere with a common emitter amplifier by more than the specified proportion.http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/npnce3.html#c2

author: mawyatt
author_rank: Super Contributor
timestamp: June 17, 2023, 10:03:16 pm
content: Quote from: Terry Bites on June 17, 2023, 05:12:06 pmThese CMOS inverters make wonderful amplifiers and limiters, we've used them in these modes since the early renditions of the RCA CMOS CD4000 series.Best,

author: Infraviolet
author_rank: Super Contributor
timestamp: June 18, 2023, 08:28:33 pm
content: I've got some 74HCU04 chips, the U indicates unbuffered so in theory a good choice. Will test soon.

author: Infraviolet
author_rank: Super Contributor
timestamp: July 11, 2023, 11:59:48 pm
content: Just thought I'd add:I'm ordering a PCB with the transistor version of this amplifier setup on it, all worked on breadboard. But even with SMD parts, 0603 passives and sot-23 and soic, the transistor one turns out to need a remarkably large area. I'm going to check all the practical results of how the system works with the transistor version, then later return to further testing of 74HCU04 CMOS inverters as amplifiers*, because it would be nice to get a future version of the PCB down to a more compact size.* I had some odd behaviours and lots of noise on output signals where the transistor versions had looked to give clean signals for the same situations, may have been due to poor probing or breadboard layout at the time though, I'll try again after fully verifying the function of the design using the transistorised PCBs