Amplifying MHz signals

[jasonRF]

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.

If 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 are
1. 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

[TimFox]

Quick addition to above:
The parameter r_e in jasonRF's analysis corresponds to (26 mV)/I_E in 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.

[Picuino]

Without feedback it is very difficult to maintain a constant amplification ratio.
A feedback circuit can correct these response curves with frequency.

[Infraviolet]

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.

[jasonRF]

Found a derivation of the bandwidth of the circuit including the emitter resistor
https://www.brown.edu/Departments/Engineering/Courses/En162/lecture_JackMiller12.pdf
It 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

[Picuino]

Simulated and working.

[TimFox]

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.

To first order, I like to think of a CE transistor amplifier as embedding an aetherial active device with infinite transconductance in series with that r_e in series with the external emitter circuit.
Similarly, if you use the CB connection, that gives you the input impedance into the emitter.

[Infraviolet]

Thank you Picuino, really like the look of that, which resistors control the gain?

[jasonRF]

Simulated and working.

I think Q2 needs to be flipped. 

[Picuino]

R2 and R3 controls the gain.

[Picuino]

Simulated and working.

I think Q2 needs to be flipped.

yes

[Picuino]

Working.
The gain is controlled by R2 and R3:
Gain = R2/R3 + 1

R4 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.

[jasonRF]

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

[Picuino]

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.

[Infraviolet]

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 splitter


Any thoughts of whether the principle ought to be able to work for such voltage ranges?
Thanks

[jasonRF]

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 splitter


Any thoughts of whether the principle ought to be able to work for such voltage ranges?
Thanks

If 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

[jasonRF]

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.html

jason

[Infraviolet]

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

[jasonRF]

Yes, Q2 and  Q3 are buffers with >100kOhm input impedance.   They might not be necessary. 

[Infraviolet]

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

[TimFox]

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.

[Terry Bites]

Do you know about cmos gate linear amplifiers?
The cheapest of them all:
CD4xxxUB series up to 15V Vdd
www.youtube.com/watch?app=desktop&v=e6q7graP-1Y

[mawyatt]

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,

[Infraviolet]

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.

[TimFox]

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.