Question 4.2: Consider the problem of increasing the Q = 10 of an on-chip ...
Consider the problem of increasing the Q = 10 of an on-chip parallel resonance circuit to 20.
The resonance frequency, the inductance, and the capacitance are given as f_0 = 2 GHz, L = 10 nH and C = 0.633 pF, respectively.
The effective parallel conductance of the resonance circuit can be calculated as
G_{eff}=\frac{1}{L\omega_0 Q}=\frac{1}{10\times 10^{-9}\times 2\pi \times (2\times 10^9)}=7.96\times 10^{-4}which corresponds to R_{eff} = 1256.6 ohm.
It is obvious that to increase the Q factor to 20, the effective conductance must be decreased to G_{eff}^{'} = 3.98 \times 10^{-4} S and the necessary negative conductance to achieve this is G_n = -3.98\times 10^{-4} S.
From (3.17a)
g_i(\omega)=\frac{C_{gs}}{(C_{gs}+C)^2}(GC_{gs}-g_mC)\frac{1}{1+(\omega_p/\omega)^2} (3.17a)
we see that the negative input conductance of a source follower is maximum for g_mC\gg GC_{gs} and for \omega \gg \omega_p, and becomes equal to
g_i(\infty)=-\frac{CC_{gs}}{(C+C_{gs})^2}g_mand has its maximum value for C=C_{gs}. Under this condition the pole frequency given in (3.15) becomes
\omega_p=\frac{g_m+G}{2C_{gs}}\cong \frac{g_m}{2C_{gs}}\cong \frac{3}{2}\frac{\mu (V_{GS}-V_T)}{L^2}for g_m \gg G, which can be satisfied by using a current source instead of R, as shown in Fig 2.15.
From the last two expressions we obtain, for C=C_{gs}
g_i(\infty)\cong -\frac{g_m}{4}which offers the first design hint.
From Fig. 3.12, we see that w_0 must be chosen where the slope of the curve is small,
or in other words, close to the asymptote, in order to obtain a high negative conductance with a small sensitivity.
For \omega \gt 3\omega_p the slope of the curve (or the sensitivity
of the negative conductance against the frequency) is sufficiently small.
This means that for \omega_0 \gt 3\omega_p, the value of the negative conductance becomes approximately equal to g_i(\infty).
The aspect ratio of the transistor can be calculated in terms of G_n and the gate bias
voltage. From (3.17d) and (1.33)
which indicates a trade-off between gate overdrive and the aspect ratio.
(V_{GS}-V_T)= 1 V is a reasonable value for the AMS 0.35 micron technology, for which the maximum supply voltage is given as 3.3 V, and the aspect ratio can be calculated as
(W/L)=\frac{4\left|G_n\right|}{\mu_n C_{ox}(V_{GS}-V_T)}=\frac{4(3.98\times 10^{-4})}{374(4.54\times 10^{-7})1}=9.37For this aspect ratio and (V_{GS}-V_T)= 1 V, the drain current is
I_D\cong \frac{1}{2}\mu_n C_{ox}\frac{W}{L}(V_{GS}-V_T)=\frac{1}{2}374\times 9.3(4.54\times 10^{-7})=0.79 mANow L can be calculated from (3.15a). With \omega_0 = 4\omega_p as a value that satisfies
\omega_0 \gt 3\omega_p,
and
W=9.37\times 4/2\cong 39 \mu mThe circuit diagram together with the current source is given in Fig. 4.7.
The VDD supply is 3.3 V.
The gate of M1 is directly biased from V_{DD}.
To satisfy (V_{GS1}-V_T)= 1 V (or V_{GS1}) = 1.5 V, the DC voltage of the source of M1 must be 1.8 V, which is the drain–source voltage of the current source transistor, M2.
M2 must be biased such that its current is also 0.79 mA and is in the saturation region.
For V_{GS2} = 1 V, since I_{D1} = I_{D2}, the aspect ratio of M2 can be calculated as
(\frac{W}{L})_2=(\frac{W}{L})_1 \frac{(V_{GS1}-V_T)(1+\lambda_n . V_{DS1})}{(V_{GS2}-V_T)(1+\lambda_n . V_{DS2})}=9.37\frac{(1.5-0.5)^2(1+0.073\times 1.5)}{(1-0.5)^2 (1+0.073\times 1.8)}\cong 37and with L_2 = 0.35 \mu m, W_2 = 13 \mu m.
The final value to be calculated is C which is equal to ]C_{gs1}:
The designed Q-enhancement circuit is simulated with PSpice.
To adjust the current of M1 to 0.79 mA, it is necessary to adjust the gate voltage of the current source M2 to 1.22 V.
The frequency characteristics of the original resonance circuit and the Q-enhanced circuit are given in Fig. 4.7(b) as curve A and curve B, respectively.
The quality factor calculated from curve B is 17.8, which corresponds to 78% increase of
the quality factor, but not 100% as targeted.
One of the reasons is the loss associated with the body resistances of the devices that were not included in the analytical expressions.
The second reason is the value of gm, which is usually smaller than the calculated value, as discussed in Chapter 1.
To compensate it the channel width can be increased in order to reach the target value.
Another valuable possibility for this simple Q-enhancement circuit shown in Fig. 4.7(a) is to control the transconductance of M1 (and hence the negative conductance) with the gate bias voltage of the current source, M2.
In Fig. 4.7(c) the frequency characteristics of the circuit for different values of VG2 are shown.
Note that the value of C has been decreased to compensate the effect of the input capacitance of M1 and to bring the resonance frequency to the target value.
