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schematic

simulate this circuit – Schematic created using CircuitLab

Here's a little schematic I built for my 27MHz colpitts oscillator, which outputs approximately 120mVpp. Since the common base is less dependent on frequency, is less prone to distortion and has high gain I chose it over the CE amp. I used several different transistors in the place of 2n3904. Even some soviet ones with GBP pf 800-1000 MHz. I simulated the schematic in LTSpice and got this result.

enter image description here

The output is 2.4Vpp, so the gain is 20. And here's the first confusing part: So far I've learnt that the gain of CB amp is

A=Rc/re

Rc= 5000

re=25/Ie

Ie, in the DC state, according to the same simulation is about 1.44mA, so re should be about 17 ohms. The gain in that case should be about 300. I am not expecting this kind of gain in reality however it is a huge jump from 20 to 300. I do suspect that on high frequencies there is some effect that I am not aware of. The CB configuration should basically eliminate the Miller effect, and I don't know what else might cause the drop in gain.

Anyways it's not where the problem ends. After building the amp on the desk the gain was even smaller than expected 20. The output Vpp was around 300 mV, a bit more than 2 times larger than the input (every transistor behaved like that, even high GBP ones). I should also note that the same high GBP transistors performed much better when used in CE configuration (around 2Vpp), although with quite a bit of distortion, which was why I decided to use CB.

I disconnected the signal source to check if the DC biasing etc was done correctly, and the voltages on each node were around the same as in the simulation.

I decided to make a frequency domain analysis in spice and here's what I got: enter image description here This is with the 2N3904. I suppose that transistors with higher GBP will have different curve (although not sure about that). Anyways the gradual roll-off is very visible.

Since I used the transistors in CE configuration effectively I highly doubt that the transistor is the problem here. I suppose the problem lies within the schematic itself, although I fail to find it. The question is: why does the schematic behave like it does in real life? Also how can one fight the gradual roll-off if it is possible at all?

UPDATE:

After reading the suggestions, reduced the effect of low pass filters by reducing the resistor values, which also increased emitter current. Ended up using this circuit, which gave a gain of about 5. It's probably very suboptimal but, it worked better than the last one. The output was about 700 mVpp with input of 120 mVpp.

schematic

simulate this circuit

Anthropomorphous Dodecahedron
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    What kind of 10uF caps are you using? Low frequency, electrolytic caps may have a substantial inductive and resistive component and that might influence your circuit at 27MHz. In other words, low frequency caps could not be considered as pure capacitors at 27MHz! – LorenzoDonati4Ukraine-OnStrike Aug 27 '17 at 14:53
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    BTW, I'm not an high frequency guru, but a gain of 200 for a single transistor stage at high frequency seems a bit too much, unless you use transistors with an extremely high unity-gain frequency (ft on datasheets). – LorenzoDonati4Ukraine-OnStrike Aug 27 '17 at 14:56
  • They are indeed electrolytic and I suppose low frequency too. Would mica ones perform any better? Yes, I understand that 200 is a bit optimistic, or rather very much so, but a gain of 2? Seems likely that there's problem somewhere. – Anthropomorphous Dodecahedron Aug 27 '17 at 15:00
  • IIRC, as a rule of thumb, assuming a low-pass, single pole, behavior from the amplifier, multiply the cutoff frequency you need (say 50MHz, since you want to amplify a 27MHz signal) by the required gain. You get 50MHz x 200 = 1GHz. So you would need a bjt with at least an ft of 1GHz! (I hope my memory about HF circuits serves me well). – LorenzoDonati4Ukraine-OnStrike Aug 27 '17 at 15:00
  • Hm.. The datasheet does say they have ft of 800-1000 MHz. It may be not entirely accurate though, plus, does ft deteriorate with time? They are quite old, from the look of them. – Anthropomorphous Dodecahedron Aug 27 '17 at 15:03
  • If they are not mistreated (exposed to extreme temperatures or humidity, etc.), BJT's parameters should not change so much. – LorenzoDonati4Ukraine-OnStrike Aug 27 '17 at 15:09
  • Anyway, if the caps are low frequency electrolytic, you could try substitute them with 100nF high frequency ones (ceramic or polymeric): at 27MHz 100nF should be quite a short circuit, for the purpose of bypassing and signal coupling. – LorenzoDonati4Ukraine-OnStrike Aug 27 '17 at 15:12
  • Happened to have 100nF mica caps, the signal became more stable (it seems there was some self-oscillation before), however the gain did not change. Anyways, thanks for the suggestion. – Anthropomorphous Dodecahedron Aug 27 '17 at 15:20
  • You need to take into account your \$C_{BE}\$ and \$C_{BC}\$. Sure, the Miller effect doesn't apply here... it's CB. But those capacitances are significant with your collector resistor and frequency. You need to compute the \$\tau_{C_{CB}}\$ and \$\tau_{C_{CE}}\$ and then find \$\omega \approx \frac{1}{ \tau_{C_{CB}} + \tau_{C_{CE}}}\$. Using reasonable estimates, I don't get close to your selected frequency. – jonk Aug 27 '17 at 16:03
  • @jonk Could you please elaborate on what are tau and omega in the formula you mentioned? – Anthropomorphous Dodecahedron Aug 27 '17 at 16:23
  • @AnthropomorphousDodecahedron That you don't know, worries me. Even with what I wrote, it's not enough because at the frequencies you are talking about even better BJT models are really required. But using the overly simplified model, there is \$C_{BE}\$ and \$C_{BC}\$ treated as constants. \$\tau_{C_{CB}}=C_{BC}\cdot \left(r_e\mid\mid R_S\mid\mid R_E\right)\$ and \$\tau_{C_{BE}}=C_{BE}\cdot \left(R_C\mid\mid R_L\right)\$. The dominant one here was already talked about by G36. Although often modest, the voltage gain should incorporate the Early effect, too. Consider Cascode, to help gain. – jonk Aug 27 '17 at 16:35
  • Thanks, that explained it. Worries me too, but I started just recently, have a lot of catching up to do with the whole RF. – Anthropomorphous Dodecahedron Aug 27 '17 at 16:47
  • Collector Tau, at 5Kohm and 10pF on collector, is 50 nanoSeconds, or 20MegaRadians/second or 3MHz F3dB. – analogsystemsrf Aug 27 '17 at 17:21

4 Answers4

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At 1.44mA the transistor has a Re of 25/1.44 = about 17 ohms.

To this you must add the 50 ohm of R5, which you forgot ;)

So the gain would be 5000/(50+17) = 74.

Common base amplifiers have very low input impedance, and it depends on bias current, so the gain will depend on the output impedance of the source, which might also not be a perfect 50R resistor, but maybe the output impedance of a semiconductor...

If the RF source is connected by a long coax, then it gets a little bit more subtle. The RF source would have 50R output impedance, and we must terminate the cable on the transistor's side by adding a (50-17)=33R resistor. So the gain would be different, yet again.

Also we have not taken into account the imperfections of this transistor. Its datasheet fT is around 300 MHz... at 10mA, with a rather high Vce, and without a collector load.

At 1.4mA, and with the very low Vce used here, the figure will be different, the transistor will be slower. Its hFe will be quite low, and it will require a non-negligible base current.

Also, the collector-to-base capacitance is multiplied by Miller effect, which increases base current.

You have connected the base to a perfect simulated capacitor, which has negligible impedance at 27MHz. However, the transistor has an internal base resistance, which combined to base current, reduces your gain.

Also, the real capacitor you used in implementing the circuit will not be perfect. It will have ESL and ESR...

...and the output impedance of your amp is 5k which is very inconvenient at 27MHz, anything you connect there will have capacitance and suck all of the signal. This is why you need an emitter follower or buffer on the output.

Anyway,

  • If you want to convert your 120mVpp sine into a square wave google "clock conversion sine to square" and you will find schematics.
  • If you want to keep your sine but amplify it, I would suggest using a canned RF amplifier like these , there are many such chips available for low prices (less than 50 cents) and all the complicated design challenges have been handled by the manufacturer... all you have to do is not screw up the supply/biasing/bypassing.
  • You can also use a highspeed opamp, but they are finicky about layout.
  • There are also fixed-gain "video amplifiers" which will solve your problem, basically they're opamps with feedback resistors included in the package.
  • Or you can roll your own, but you will need more gain stages, you can't get a gain of 200 at 27 MHz with a single transistor mounted on a PCB.

Also here's a primer on capacitors at HF:

The inductance of the cap is the same as a piece of wire having the shape of the path taken by the current. Thus, a through-hole capacitor with e=5.08mm between the pins, with the internal foils sitting 2mm above the board, can be modeled as a 5mmx2mm square loop if we close the circuit with a trace on the board below. This loop has an inductance of 9nH.

2pi x 9nH x 27MHz = 1.5 ohms inductive

If the cap has low ESR then you get a LC tank circuit which may resonate, and since it drives the base of your BJT, you just built a Colpitts. It is very easy to unintentionally build an oscillator when combining devices with gain (ie, transistors) and layout parasitics... A transistor whose base is driven by a reactive impedance, without enough real (resistive) impedance in series may create some headaches...

This does not depend on capacitor type, chemistry, etc, only its mechanical construction (you will no doubt guess that axial caps are much worse than radials).

A leaded ceramic will have the same HF behavior as a modern good quality leaded electrolytic. Same inductance.

Basically, if you want a low-inductance capacitor, it needs to be low-profile, and lay flat against your ground plane to reduce the loop area. And therefore...

enter image description here

This means MLCC (ceramic) of X7R type for bypassing, or C0G type if you need an accurate value. Or SMD Film caps.

Then you add the inductance of the whole loop, traces, vias, ground plane etc.

enter image description here

So when you see audiophiles adding bypass caps...

enter image description here

Ickkkkkkk.

bobflux
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  • Thanks! I'm rolling on my own for now partly since we don't have these components (rf amps etc.) in the area and partly for education. In regards the problem at hand: I thought the common base configuration was largely independent of hFe. Where exactly does it come into play? – Anthropomorphous Dodecahedron Aug 27 '17 at 16:18
  • hFe falls with increasing frequency thus a higher base current is required, and since base impedance is not zero, this reduces the gain. Common Base is less sensitive to this because the base can be driven by a low impedance... but it is never zero... – bobflux Aug 27 '17 at 16:31
  • @peufeu this picture is hilarious – G36 Aug 27 '17 at 16:46
  • @G36 yeah especially since this is a single layer job without ground plane, the whole "supply bypassing" thingy is like, nope.... – bobflux Aug 27 '17 at 17:00
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The voltage gain of your circuit is around:

$$\frac{r_e}{r_e+R_5}*\frac{R_C}{r_e}= \frac{18Ω}{18Ω+50Ω}*\frac{5kΩ}{18Ω}=72$$

at the middle frequency.

Also from the data-sheet we can read that Cbc capacitance is around 4pF.

So Rc together with Cbc will form the low pass filter.

And the cut-off frequency is around:

$$Fc = \frac{0.16}{R_C *Cbc} = \frac{0.16}{5kΩ *4pF} = 8MHz$$

Hence the gain at the 28MHz is

$$72 * \frac{1}{ \sqrt{ 1+(\frac{28MHz}{8MHz})^2}} = 72*0.27 = 19.4 V/V $$

And this ballpark calculation\estimate do not included any additional parasitic capacitance present in the real world circuit except Cbc

Also do not forget about the input Cbe capacitance is:

$$Cbe = \frac{\frac{1}{r_e}}{2*\pi*F_T} = 35pF$$

And this capacitance also forms low pass filter.

G36
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  • Thanks, I completely neglected the low pass filter effect. However I don't quite follow with which resistors does Cbe form a low pass filter with. R2 and R3 in parallel? – Anthropomorphous Dodecahedron Aug 27 '17 at 16:14
  • Also, don't forget \$r_o\$ due to the Early effect! It applies here. (So will the load, which we don't understand from the OP yet.) – jonk Aug 27 '17 at 16:16
  • There is no load attached yet (other than oscilloscope probe), I plan to feed it into an emitter-follower or maybe another stage of CB. Depends on what I can achieve from this stage with corrected values. – Anthropomorphous Dodecahedron Aug 27 '17 at 16:20
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I've looked at your circuit. I'm not a formula guy. You should realize that almost all of the emitter current will have been coming from the collector. That's the idea of a transistor.

So when the input pulls your emitter down by some small voltage (we're doing small signal analysis here!) the current will move away from the steady state current by dI = dUin/Re. Now, apply that same current in the collector circuit and your output voltage swing will be dUout = Rc * dI = Rc * dUin/Re -> dUout/dUin = A = Rc / Re ! Neat! Same formula as you got!

Putting in the numbers gives me A = 5000/500 = 10. How you got 20 or 300 is beyond me. sorry.

user242579
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  • As far as I understand Vin = Ie*rc (notice the small r, or the base-emitter junction resistance Vt/Ie), since the Rc>>rc. Therefore the gain A=Rc/re, in my case re~17 ohm. At least that's how I learnt it, I may be wrong. – Anthropomorphous Dodecahedron Aug 27 '17 at 16:06
  • Hmm. Interesting.... Wouldn't the low "output" resistance of the emitter mean that now you're having a current-input amplifier? In that case, the input current would be about Vin/50 Ohms. – user242579 Aug 27 '17 at 16:22
  • I messed up the letters, sorry. It should be Vin=Ie*re(the base-emitter junction), because Re=500 > re=17. I can't comment on the input current with confidence although it seems intuitive that when small dVin occures, the input current that flows into the base would be dVin/re, although I might be entirely mistaken here. – Anthropomorphous Dodecahedron Aug 27 '17 at 16:32
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  1. the bypass caps and the base cap you can put higher caps like 1 uf; reason is that you want the transistor not affected (at lower frequencies; no need to cut off).
  2. Since the gain is Av= gmRc, since Rc is actually Rc in parrallel with the load this will affect the gain as well, so you should use a higher impedance at the load (mind that it wouldn't be that large at that frequency)
  3. input impedance is important as it can attenuate the incoming signal; you simulate 50 ohm source impedance (function gen or your oscillator) you should use a buffer stage (Common collector) before the common base stage.
  4. the voltage divider bias in not important in Ac analysis since its shorted to ground, however you should choose higher resistance values not to pass high current through resistors as well as transistor base, as a rule of thumb the R3 is under 1.7 volts (Ve = 1 volt + 0.7 Si forward voltage drop)
  5. use lower Ic value (since gm = Ic/26mv); you can use .5 mA
  6. for Rc subtract the Ve voltage from Vcc (10.4-1) and divide by 2; what you get then divide by Ic and find the lowest nearest standard resistor value.
mikeZ1
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