Determine the output voltage

Question

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\$.

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.

Answer 1

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

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:

by differentiation you can find Re like this

so

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

Or using equation 3 into 4 to get the expression

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

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.

Answer 2

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.