Question 8.5: Discuss the frequency components (DC, fundamental, harmonics......

In time domain, the following voltage and current Kirchoff equations hold:

\begin{aligned} v_L(t)+V_B & =v_{D S}(t) \\ i_L(t)+i_D(t)+i_r(t) & =I_{D D} ,\end{aligned}

where V_B is the voltage across the DC block (ideally DC only), I_{D D} is the current from the bias (ideally DC only because of the bias T RF blocking inductor), i_r is the resonator current. Balancing the harmonics we have at DC:

\begin{aligned} V_{D S}(D C) & =V_B=V_{D D} \\ V_L(D C) & =0 \\ I_L(D C) & =I_r(D C)=0 \\ I_D(D C) & =I_{D D}. \end{aligned}

In fact, at DC the load is shorted by the resonator, so that the load DC current is certainly zero; but since the sum of the load and resonator DC currents is also zero owing to the DC block, also the resonator DC current vanishes. The DC drain current coincides with the bias current. Concerning the fundamental f_0 we have:

\begin{aligned} V_{D S}\left(f_0\right) & =V_L\left(f_0\right) \\ I_L\left(f_0\right) & =-I_D\left(f_0\right), \end{aligned}

i.e., the fundamental component of the load and output voltages coincide, while the load current is the drain current with a minus sign. At the harmonics we finally have:

\begin{aligned} V_{D S}\left(n f_0\right) & =V_L\left(n f_0\right)=0 \\ I_L\left(n f_0\right) & =0 \\ I_r\left(n f_0\right) & =-I_D\left(n f_0\right); \end{aligned}

in fact, the device output is shorted at nf_0 by the DC block and by the resonator, and so is the load; as a consequence, the load current is zero and the harmonics of the drain current flow in the resonator. Taking into account that:

V_L\left(f_0\right)=V_{D S}\left(f_0\right)=R_L I_L\left(f_0\right)=-R_L I_D\left(f_0\right),

we can write the time-domain v_{D S} as:

v_{D S}(t)=V_{D D}-R_L I_D\left(f_0\right) \cos \omega_0 t

with components at DC and at the fundamental, while the harmonic distortion is in the drain current:

i_D(t)=I_{D D}+I_D\left(f_0\right) \cos \omega_0 t+I_D\left(2 f_0\right) \cos 2 \omega_0 t+\ldots

The load voltage and current are of course purely sinusoidal at the fundamental. Notice that if the drain current harmonics are negligible we can write:

\begin{aligned} v_{D S}(t)-V_{D D} & =-R_L I_D\left(f_0\right) \cos \omega_0 t \\ i_D(t)-I_{D D} & =I_D\left(f_0\right) \cos \omega_0 t ,\end{aligned}

i.e., combining:

\left(i_D-I_{D D}\right)=-\frac{\left(v_{D S}-V_{D D}\right)}{R_L} . \hspace{30 pt} \text{(8.5)}

Eq. (8.5) describes the device load line; the instantaneous working point is the intersection of the load line with the device characteristic:

i_D(t)=I_D\left(v_{D S}(t), v_{G S}(t)\right) .