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I'm trying to solve the following question:

Determine the output-tuned circuit voltage in Fig. 4.P-7. The tuned circuit is resonant at \$\omega_0\$ and has \$Q_T = 20\$.enter image description here

My try: I don't know how to apply KVL and KCL correctly. Solving for DC: $$I_E = \frac{12 - V_{EB}}{12} = \frac{12 - 0.7}{12} \approx 0.941 \ \text{mA}$$Is the following equation true for AC analysis? $$\frac{v_1 - v_{eb}}{50} = i_e + \frac{v_{eb}}{12\times10^3} \\ v_1 = 0.3\cos(\omega_0 t)$$ Also we have $$i_E = I_{ES}e^{\frac{v_{EB}}{x}} = I_{ES}e^{\frac{v_{eb} + V_{EB}}{x}}= I_{ES}e^{\frac{V_{EB}}{{x}}}e^{\frac{v_{eb}}{{x}}} = I_Ee^{\frac{v_{eb}}{{x}}} \\ x = \frac{kT}{q} \approx 26 \ \text{mV}$$I don't know how to proceed further. Obviously these equations hold: $$v_1 - 50i_1 - v_C - v_{EB} = 0 \\ 12 - 12\times10^3i_2 - v_{EB} = 0 \\ i_1 + i_2 = i_E \\ i_E = I_{ES}e^{\frac{v_{EB}}{x}} \\ i_1 = C\frac{dv_C}{dt} \\ v_1 = 0.3\cos(\omega_0 t) \\ x = \frac{kT}{q}$$ but it's really difficult to solve them directly and some simplifying assumptions should be made.

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    \$\begingroup\$ +1 for showing effort to solve. \$\endgroup\$ Mar 7 at 1:08
  • \$\begingroup\$ @MathKeepsMeBusy Thanks. I hope there will be an answer to this problem. \$\endgroup\$
    – S.H.W
    Mar 7 at 1:15
  • \$\begingroup\$ Nobody can answer yet? \$\endgroup\$ Mar 7 at 2:52
  • \$\begingroup\$ Nobody can answer because it depends on unstated assumptions for the transistor bulk resistance which affects Rbe and effects of Vbc saturation. yet you can predict a range of voltage gains for some assumptions. Basically it is a nonlinear equation of base Rbe modulated gain, which may be beyond the scope of this question and perhaps my ability to articulate. \$\endgroup\$ Mar 8 at 13:53
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Before throwing too much effort in solving the complicated equations, there are a few basic considerations which are easy to do and as seen, it is worthwhile the effort.

The common base has a current gain of close to one because the base current is very small, so what goes in to the emitter comes out of the collector - gain close to 1. The input current into the emitter is

enter image description here

where Re is the input impedance of the emitter. Rg is the 50 ohm. (note 1) Re can be found from the diode equation which you have listed:

enter image description here

by differentiation you can find Re like this

enter image description here

so

enter image description here

By combining 1 and 3 we have the iE current. The collector current ic is about the same as mentioned above.

Because we know the frequency is the same as the resonant frequency of the LC circuit we know that the impedance of the LC part in common is very high so we will ignore the L and C. In the output remains only the resistor which I call Ro. So the relation between vo and v1 will be

enter image description here

Or using equation 3 into 4 to get the expression

enter image description here

So the small signal voltage gain from input to output at the resonant frequency of the LC tank is:

enter image description here

With an input peak amplitude of 300mV the output will be 22V peak!

Now here is the tricky part: the amplitude cannot exceed about 12V because the collector base diode will be conducting hard and clamp the voltage to about one diode drop above gnd.

Secondly, it is given that the Q is 20. This tells us that the tank is filtering the signal pretty well and thus the output is mostly sinusoidal at the fundamental frequency.

So the answer is Vo = 12.6V peak.

Notice how a lot of details are left out such as the emitter bias resistor - it only sets the dc current to about 1mA needed to estimate the emitter resistance. Also note that the collector output is assumed to have high impedance relative to Ro. This value is however not so easy to estimate compared the Re. But as you see from the result - its not that critical because the signal is limited by other factors in stead. Likewise the actual Q value is not important - just that it is fairly high for an LC, is important to know.

You seem to indicate this was some sort of theoretical exercise. If you can, please give my "Chapeau" to the one who made up the question. It is a real practical engineering challenge and not only a textbook exercise - well done. It illustrates that without appropriate approximations you end up with equations you cannot solve within a reasonable time. So you better make some educated shortcuts and at least have some result to work from.

Note 1: Always label relevant components so the equations can be generic.

All images above created by me.

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  • \$\begingroup\$ This is a trick question, because the emitter AC current is greater than the DC current on -300mV thus raising Vbe and Ic alot while Rbe is around 27 Ohms at DC it is modulated a lot because +300mV almost turns off Vbe as the input current drops to <~1mA due to the -(10V+0.6)/10k. So the Q is 20 for small signal but the Rce drops reducing Q for negative input loading the tank, then Rce rises significantly for Vin+. I see up to 19Vpp happening but possible a bit less. in any case gain is nonlinear at this input as it pumps Vbe with 2~3mA AC only on the Vin- peak. \$\endgroup\$ Mar 7 at 16:08
  • \$\begingroup\$ The amplitude is centred around Vcc and makes no difference in the Vpp out, nor does hFE as this becomes a current pump with gm(Rbe+Re). So the output swing can go to 20Vpp with a bit more input from 0.2 to 19.8 or so and Q is reduced by overdrive the input, making the assumption of Q being 20 constant non-linear and wrong. \$\endgroup\$ Mar 7 at 16:11
  • \$\begingroup\$ Vin is 600mVpp after input cap , but at the emitter is not the same. It drops to -400mV as Rbe attenuates the input and Vbe rises 50mV or so at the emitter as Vbe rises from 600mV to 650mV @2.5mA AC peak. This design is a good instruction to RF pumped common base designs. \$\endgroup\$ Mar 7 at 16:19
  • \$\begingroup\$ This answer is close , so I agree Av might be 70 for a 1mVpp input but at 600mVpp input or 400mVpp for Vbe AC the Av drops significantly perhaps Av=30 x 600mVpp= 18Vpp output. And this will not change much with Vcc. Even if it went to 20Vdc. Proof is a little more complex, Rbe(min)=26/2.6mA= 10 Ohms \$\endgroup\$ Mar 7 at 16:31
  • \$\begingroup\$ +1 for good answer but not perfect. \$\endgroup\$ Mar 7 at 16:38
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My Assumptions

\$\dfrac{R_c}{X_L} = \dfrac{R_c}{X_C}= Q\$
\$I_C=I_E, ~h_{FE} > 100\$ ...< 1% error
if \$I_E=1mA, V_{EB}= 600~mV_{dc}\$... 1% error
\$ T= \text{room temp or}~ 25°C, ~ r_{BE}= \dfrac{26 }{I_C} \$

The input AC current \$I_{{E}{ac}}=\dfrac{v_{in}}{R_{s}}, R_s= 50+r_{BE}\$

Calculations

We expect Ie near 1mA DC but not for AC .

\$I_{E_{dc}}=\dfrac{12~V-0.60~V}{12~k\Omega}= 0.95 mA_{dc}~,~\$
\$ r_{BE}=27 \Omega\$

\$I_{E{dc}} Rs = 50+27 =77 Ω\$ but with -300mV peak applied to PNP, Rs, if AC peak was Ie=300mV/77 = 3.9mApk AC yet the minimum Rbe would then be 26/Ic=7.1 Ohms but the maximum Rbe will be much higher as the Vbe almost turns off so the average Rbe depends on the duty cycle of 7.1 Ohms at this current so this is incorrect. The base-emitter conducts and acts as a clamp to the input capacitor so its Voltage drop matched the average Vbe and thus changes the DC voltage across it according to 0 DC current in a Cap. But the input current waveform will be distorted at this voltage level with an asymmetrical AC current from the asymmetrical Rbe. Thus Rbe actually modulates from 7.1 to > 1k Ohms because of the Vbe modulation and might end up averaging more like 35 Ohms which reduces the AC input current somewhat.

Nevertheless an idle Tank circuit with a Q of 20 won't stay at this Q with hard driving of the Vbe but you might consider the voltage gain of Rc/Rs= 5600/(50+35)= 66 and thus the output might be 600mVpp*66= 44Vpp !!! This cannot happen due to CB diode forward conduction.

These results will vary greatly with specs for transistor Rce saturation , Rbe saturation and Vee bias current, but not Vcc or hFE.

Conclusion

This is a trick question. There is no solution except the gain will be less than < Rc/Rs ~66 because it has a non-linear voltage gain when you drive a common base with a shorted Base and 600mV AC input with 50 Ohms with a source impedance nearly as low as the Rbe minimum yet << Rbe maximum. It will tend towards half the gain as Rbe max is so large so the AC current has injected only a fraction of the non-sinusoidal time.

So there is no universal Voltage gain result but with will be a large swing < 20Vpp and depends on the transistor. You could get 100Vpp and drive the Vbe harder with a 5V supply but your initial Q degrades rapidly when Rbe changes over a wide dynamic range with a large input swing.

For those not afraid of TMI too much information.

Final comment

here is the circuit in question with 600mVpp input and ~15Vpp output. Now prove why the gain is only 26.67 with 600mVpp and changes with input from Vbe 2t% sinus duty cycle of conduction.

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  • \$\begingroup\$ Stewart: "You could get 100Vpp and drive the Vbe harder with a 5V". No this is not possible - the collector base diode will clamp the voltage to gnd. As you also write yourself previously so why do you write this? \$\endgroup\$ Mar 8 at 12:46
  • \$\begingroup\$ With a gain of 66 or 70 and the emitter current , raising Vcc prevents saturation as the Collector Vdc is the same as Vcc. Then by injecting more current ( with more voltage/(50 Ohms +Rbe) The output swing increases to almost Vce =2x Vcc and Vce=0 Thus for Vcc=50Vdc you can synthesize a 100V swing, but for practical reasons the power dissipation must be lower than rating. I can prove this and verified already. Have you ? \$\endgroup\$ Mar 8 at 13:51
  • \$\begingroup\$ but Vcc is specified at 12V so it is not relevant what happens at 50V \$\endgroup\$ Mar 8 at 20:36
  • \$\begingroup\$ But it is possible. In order to understand nonlinear gain it is useful to make it more linear then prove it before going to non linear. \$\endgroup\$ Mar 8 at 22:56
  • \$\begingroup\$ The important thing to note is Vcc only affects amplitude slightly if Vce becomes partially saturated dropping hFE significantly otherwise not. A darlington works too. \$\endgroup\$ Mar 8 at 23:36

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