Author Topic: Amplifying MHz signals  (Read 6259 times)

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Offline gf

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Re: Amplifying MHz signals
« Reply #25 on: June 06, 2023, 05:15:41 am »
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.
 
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Offline armandine2

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Re: Amplifying MHz signals
« Reply #26 on: June 06, 2023, 10:21:50 am »
sometime ago I screenshot a common-emitter FRA example from the internet 

it looks like it may show the same errors

Funny, the things you have the hardest time parting with are the things you need the least - Bob Dylan
 
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Offline jasonRF

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Re: Amplifying MHz signals
« Reply #27 on: June 06, 2023, 01:57:34 pm »
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.
Somehow 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
 
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Offline RJSV

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Re: Amplifying MHz signals
« Reply #28 on: June 06, 2023, 08:41:55 pm »
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 the
   OP 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 closed
   loop gain (mostly) determines outcome.  That also means some degree of independence when some
   effect like ambient temperature is acting to change the OP AMP internal gain, as is natural.  So you get
   a 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
   
 
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Offline jasonRF

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Re: Amplifying MHz signals
« Reply #29 on: June 06, 2023, 10:06:39 pm »
-snip-
   Honestly, I don't see a need, though, for the transistor in the output, unless you need some extra drive ability.  IMO.
 
The 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
 

Offline InfravioletTopic starter

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Re: Amplifying MHz signals
« Reply #30 on: June 06, 2023, 10:21:08 pm »
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.
 

Offline TimFox

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Re: Amplifying MHz signals
« Reply #31 on: June 06, 2023, 10:40:10 pm »
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 high hfe, so I neglected any DC loading of the base current in your base voltage divider network.)
An elementary approximation is gm = (26 mV)/Ie, where Ie is 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 than Vbe DC value to reduce the dependence on the transistor.  With the minimum hfe value 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 expected Vbe DC 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.

« Last Edit: June 06, 2023, 10:53:42 pm by TimFox »
 
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Offline InfravioletTopic starter

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Re: Amplifying MHz signals
« Reply #32 on: June 07, 2023, 02:27:07 am »
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.
Thanks

EDIT: 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.
« Last Edit: June 07, 2023, 03:43:05 am by Infraviolet »
 

Offline TimFox

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Re: Amplifying MHz signals
« Reply #33 on: June 07, 2023, 02:47:06 am »
If the emitter resistor Re is unbypassed, the voltage gain is roughly Rc / Re , where Rc is the collector resistor.
This assumes that Re is much larger than (26 mV) / Ie , where Ie is the DC emitter current.
 

Offline jasonRF

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Re: Amplifying MHz signals
« Reply #34 on: June 07, 2023, 02:18:41 pm »

EDIT: 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.
Does 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
 

Offline TimFox

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Re: Amplifying MHz signals
« Reply #35 on: June 07, 2023, 07:24:00 pm »
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)/Ie
The 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.
 

Offline InfravioletTopic starter

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Re: Amplifying MHz signals
« Reply #36 on: June 08, 2023, 12:08:21 am »
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
 

Offline armandine2

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Re: Amplifying MHz signals
« Reply #37 on: June 08, 2023, 06:44:53 am »
https://www.amazon.co.uk/Electronics-Manchester-Physics-JM-CALVERT/dp/0471996394

https://www.amazon.co.uk/Intro-Electronic-Circuits-Martin-Hartley/dp/0521478790

https://www.amazon.co.uk/Electronic-Testing-Diagnosis-George-Loveday/dp/0582038650

I'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
Funny, the things you have the hardest time parting with are the things you need the least - Bob Dylan
 

Offline jasonRF

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Re: Amplifying MHz signals
« Reply #38 on: June 08, 2023, 11:33:06 am »
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

What 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 is

There 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 variables

4. 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, you
1. 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  Rc
3. pick Re to get the gain you want
4. 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
« Last Edit: June 08, 2023, 02:10:53 pm by jasonRF »
 

Offline gf

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Re: Amplifying MHz signals
« Reply #39 on: June 08, 2023, 02:08:55 pm »
Quote from: Infraviolet
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.

You 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.
 
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Offline TimFox

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Re: Amplifying MHz signals
« Reply #40 on: June 08, 2023, 03:40:33 pm »
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 (probably Vcc/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 than Vcc/2 .

The input and output capacitors, along with the resistances at the "right" side of each capacitor, will define the low-frequency response.
 
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Offline jasonRF

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Re: Amplifying MHz signals
« Reply #41 on: June 08, 2023, 05:38:35 pm »
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
 
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Offline bdunham7

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Re: Amplifying MHz signals
« Reply #42 on: June 08, 2023, 05:47:36 pm »
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. 

How did you physically construct your circuit?
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline TimFox

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Re: Amplifying MHz signals
« Reply #43 on: June 08, 2023, 05:47:50 pm »
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.
 

Offline jasonRF

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Re: Amplifying MHz signals
« Reply #44 on: June 08, 2023, 06:25:22 pm »
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. 

How did you physically construct your circuit?
Solderless 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



« Last Edit: June 08, 2023, 10:28:04 pm by jasonRF »
 

Offline InfravioletTopic starter

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Re: Amplifying MHz signals
« Reply #45 on: June 08, 2023, 11:33:31 pm »
Thanks all for the replies, a lot to think through.
armandine2, thanks for the links

jasonRF,
"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, thanks

I'll post more when I've made some furthe measurements based on all your advice.
« Last Edit: June 08, 2023, 11:35:45 pm by Infraviolet »
 

Offline gf

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Re: Amplifying MHz signals
« Reply #46 on: June 09, 2023, 10:48:47 am »
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,

You get less dependency if you increase negative feedback, i.e. if you reduce the closed loop gain. Currently you have approx. 5600/680 = 8.24

What 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.
 

Offline InfravioletTopic starter

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Re: Amplifying MHz signals
« Reply #47 on: June 10, 2023, 12:08:22 am »
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.
 

Offline gf

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Re: Amplifying MHz signals
« Reply #48 on: June 10, 2023, 07:22:23 am »
But with R_e=270 and R_c of 1K2 the practical gain is -4 gainst a theoretical of 4.4.

If 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.
 

Offline Picuino

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Re: Amplifying MHz signals
« Reply #49 on: June 10, 2023, 11:42:33 am »
You can look for video amplifier stages with negative feedback. They are very similar to what you are looking for.
« Last Edit: June 10, 2023, 11:47:13 am by Picuino »
 


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