Nonlinear circuit and RLC

I'm trying to solve the following question from "Communication Circuits: Analysis and Design" by Donald T. Hess:

For the circuit shown in Fig. $$\1.P1\$$, determine an expression for $$\v_o(t)\$$(Q1 and Q2 are identical).

My try: Q2 and Q3 form a current mirror. We can determine the current $$\I_{C2}\$$ which is equal to $$\I_{E1}\$$ since $$\1\mu F\$$ capacitor is an ac short circuit(So here are we assuming that ac and dc analysis are separable? But I think this is true only for linear circuits). Since Q2 and Q3 are identical,
$$\frac{0 - (-10+V_{BE})}{2} = \alpha I_E + \frac{2I_E}{\beta + 1}$$ $$\alpha = \frac{\beta}{\beta + 1}$$
Which leads to $$I_E = \frac{99\times 4.65}{100} \implies I_C = \frac{99}{100}\times \frac{99\times 4.65}{100} \approx 4.557mA$$ At this point we should use $$i_E = I_{ES} e^{\frac{v_{BE}}{x}} \\ x = \frac{kT}{q} \approx 26mV$$ I think $$\v_{BE} = -V_{dc} -v(t)\$$ but according to the solution because of the capacitor there is no dc voltage. Is this really correct? If we accept my expression for $$\v_{BE}(t)\$$, we have $$i_E = I_{ES}e^{\frac{-V_{dc}}{x}}e^{\frac{-v(t)}{x}} = I_pe^{-10\cos\omega t} \\ I_p = I_{ES}e^{\frac{-V_{dc}}{x}}$$ Using the Fourier series we have $$e^{-10\cos \omega t} = I_0(10) + 2\sum_{n = 1}^{\infty}(-1)^n I_n(10) \cos (n\omega t)$$ Where $$\I_n(x)\$$ is the modified Bessel functions. The biasing circuitry demands $$I_pI_0(10) = 4.557 \times 10^{-3} \implies I_p = \frac{4.557 \times 10^{-3}}{I_0(10)} \approx 1.6 \times 10^{-6}$$ Since $$\i_C = \alpha i_E\$$ the output voltage should be $$v_o(t) = 10 - \alpha I_p (I_0(10)Z_L(j0\omega) + 2\sum_{n = 1}^{\infty}(-1)^n I_n(10)Z_L(jn\omega) \cos (n\omega t))$$Where $$\Z_L(jn\omega)\$$ is the impedance of the parallel RLC at harmonics. The impedance of the RLC circuit to dc is zero and the resonance frequency is $$\omega_0 = \frac{1}{\sqrt{LC}} = 3\times 10^7$$ Assuming $$\\alpha \approx 1\$$ the answer should be $$v_o(t) \approx 10 - 3.2\times 10^{-6}(-I_1(10)Z_L(j\omega)\cos(\omega t) + I_2(10)Z_L(j2\omega)\cos(2\omega t)-I_3(10)\times 10^3\cos(3\omega t))$$From the table we know that $$\I_3(10) = 1758.4\$$ which implies the third term is $$-5.62\cos(3\times 10^7 t)$$and we can determine other terms. One solution manual gives $$v_o(t) \approx -0.155\cos (\omega t) + 0.424\cos(2\omega t) - 5.45\cos(3\omega t)$$And Hess's book gives $$v_o(t) \approx -0.28\sin(10^7 t) + 7.3\cos(2\times 10^7 t)+0.33\sin(3\times 10^7 t)$$The first answer seems very close to my answer but there is no $$\10\$$ dc voltage in the final answer. The second answer differs considerably from my answer. So what's the correct answer?

Here is the summary of the problems:

1. Are we assuming that ac and dc analysis are separable? If yes, why is this possible? Obviously the circuit is nonlinear and superposition doesn't hold.
2. Is there a dc voltage across BE junction?
3. Is my final answer for the $$\v_o(t)\$$ correct?
• The Hess answer is certainly wrong with the high Q near the 3rd harmonic and similarity, I would expect the 2nd harmonic more than the fundamental at this level input of distortion. At 4.6mA Vbe will be around 630 mV with the input at 520 mVpp is a massive amount of distortion Mar 1 at 0:15
• @TonyStewartSunnyskyguyEE75 Thanks. So there is a dc voltage across BE junction, right? Mar 1 at 0:29
• Remember/verify that Vbe =600 mV @ Ic=1mA and this uses a current mirror to Ic and Voltage source AC is 0 Ohms driving a low impedance common base so Vbe is modulated which modulates Ic with asymmetric exponential current. Mar 1 at 1:41
• The common base has an output impedance of rE/hFE since there is no base resistance which makes it , I guess <<1 Ohm so a true voltage source is used and 1uF is low enough above1MHz. This forces Vbe changes directly and that forces the exponential current swings with large modulation of the input voltage. the collector load is tuned above the 3rd harmonic so there is a high slope of gain on the harmonics up to the 3rd then attenuated as the 4th is on the down slope of the resonant filter and so on, You did a decent job with your method, but there is more to it. Such as temperature effects. Mar 1 at 3:08
• Bulk resistance will also factor into this with different chip sizes so I did not do the math, but your approach is good enough unless you want to simulate with different devices. I have never used this circuit as I don’t see the benefit of such non-linearity with an off-tuned filter. They let you assume Vbe=0.7 For simplicity but I always found 1mA is always closer to 600mV and this uses 4.6mA yet >10mA would be all over the map with a wide range of bulk resistance, on different devices. So I estimated 630 mV from experience. I’m not a teacher, but I would give full marks. Mar 1 at 3:19