Fundamentals of Analog VLSI Design
Exercise 7 - Solution
Gain and Bandwidth Enhancement
1 Problem 1: Gain Enhancement
The amplifier shown in Figure 1 allows to enhance the DC gain by reducing the load conductance seen by the differential pair M1a-M1b [1] [2].
1.1 Small-signal analysis
The cross-coupled transistors M3a-M3b which are redrawn in Figure 2, allow to implement a negative conductance. When combined with transistors M2a-M2b, they allow to reduce the conductance seen by the differential pair M1a-M1b.
The small-signal circuit corresponding to Figure 2 is shown in Figure 3.
With \(\Delta I_{in} \triangleq \Delta I_1-\Delta I_2\) and \(V_{in} \triangleq V_1-V_2\) is then straightforward to show that \[\begin{equation} G \triangleq \frac{\Delta I_{in}}{\Delta V_{in}} = -G_{m3}. \end{equation}\] The circuit of Figure 2 therefore implements a negative conductance equal to the transconductance of M3a-M3b.
Note
The small-signal circuit of Figure 1 is shown in Figure 4, where \(G_o\) is the total conductance at the output node given by \[\begin{equation} G_o = G_{m2} + G_{ds2} + G_{ds3} \cong G_{m2}. \end{equation}\]
In differential operation we have \[\begin{align} \Delta V_{i1} &= \frac{\Delta V_{id}}{2},\\ \Delta V_{i2} &= -\frac{\Delta V_{id}}{2}, \end{align}\] where \(\Delta V_{id}\) is the small-signal differential input voltage. If we assume a perfect matching between the transistors in left and right branches, then \(\Delta V_S = 0\) and the small-signal circuit of Figure 4 simplifies to the one shown in Figure 5. We can then use the half-circuit shown in Figure 6 to calculate the small-signal differential transfer function. It is easy to show that the differential transfer function is given by \[\begin{equation} A_d(s) \triangleq \frac{\Delta V_{od}}{\Delta V_{id}} = \frac{A_{dc}}{1 + \frac{s}{\omega_p}}, \end{equation}\] where \[\begin{align} A_{dc} &= -\frac{G_{m1}}{G_{od}-G_{m3}} = -\frac{G_{m1}}{G_{m2}+G_{ds1}+G_{ds2}+G_{ds3}-G_{m3}},\\ \omega_p &= \frac{G_{od}-G_{m3}}{C_L} = \frac{G_{m2}+G_{ds1}+G_{ds2}+G_{ds3}-G_{m3}}{C_L}, \end{align}\] with \[\begin{equation} G_{od} = G_o + G_{ds1} = G_{m2}+G_{ds1}+G_{ds2}+G_{ds3}. \end{equation}\] The gain-bandwidth product or unity gain frequency \(\omega_u\) is then simply given by \[\begin{equation} \omega_u \triangleq |A_{dc}| \cdot \omega_p = \frac{G_{m1}}{C_L}. \end{equation}\] Contrary to the DC gain, the gain-bandwidth product is not increased and is the same as the amplifier without M3a-M3b.
From the expression of the pole \(\omega_p\), we see that for it to be located in the left half-plane, we need \(G_{m3} < G_{od} = G_{m2} + G_{ds1} + G_{ds1} + G_{ds3}\).
We can rewrite the DC gain as \[\begin{equation} A_{dc} = -\frac{G_{m1}}{G_{m2}} \cdot K, \end{equation}\] where \(K\) is the DC gain enhancement factor compared to the circuit without the cross-coupled pair M3a-M3b given by \[\begin{equation} K = \frac{1}{\frac{G_{od}}{G_{m2}}-\frac{G_{m3}}{G_{m2}}} = \frac{1}{1+\frac{G_{ds1}}{G_{m2}}+\frac{G_{ds2}}{G_{m2}}+\frac{G_{ds3}}{G_{m2}}-\frac{G_{m3}}{G_{m2}}}. \end{equation}\] If we set \(G_{m3} = G_{m2}\), the gain-enhancement factor becomes \[\begin{equation} K = \frac{G_{m2}}{G_{ds1} + G_{ds1} + G_{ds3}}, \end{equation}\] which seems large. But looking at the DC gain, it reduces to \[\begin{equation} A_{dc} = -\frac{G_{m1}}{G_{ds1}+G_{ds2}+G_{ds3}}. \end{equation}\] We see that it is basically the gain of a common-source stage loaded by an active load (current source). So we have indeed increased the gain compared to \(G_{m1}/G_{m2}\), but we don’t really improve the DC gain compared to a simple CS stage. We could have achieved the same gain with replacing the diode-connected transistors M2a-M2b with current sources.
To really boost the gain we need to compensate for more than \(G_{m2}\) but making sure that \(G_{m3} < G_{od}\) to avoid any instability. Because of variability and mismatch, it is however not easy to control \(G_{m3}\) such that it always remains smaller than \(G_{od}\) in presence of mismatch and instability.
1.2 Noise analysis
The small-signal half schematic of the amplifier of Figure 1 including all the noise sources is shown in Figure 7. The output noise voltage is then simply given by \[\begin{equation} V_{nout1} = \frac{I_{n1}+I_{n2}-I_{n3}}{G_{od}-G_{m3}}. \end{equation}\] The corresponding output noise resistance is then given by \[\begin{equation}\label{eqn:gain:rnout} R_{nout} = 2\,\frac{G_{n1}+G_{n2}+G_{n3}}{(G_{od}-G_{m3})^2}, \end{equation}\] where \[\begin{equation} G_{ni} = \gamma_{ni}\,G_{mi} + G_{mi}^2\,\frac{\rho_i}{f\,W_i\,L_i}\quad\textsf{for $i=1,2,3$}. \end{equation}\]
The noise can be referred to the input by dividing \(\eqref{eqn:gain:rnout}\) by \(A_{dc}^2\) resulting in \[\begin{equation} R_{nin} = \frac{R_{nout}}{A_d^2} = 2\,\frac{G_{n1}+G_{n2}+G_{n3}}{G_{m1}^2} = 2\frac{G_{n1}}{G_{m1}^2} \; (1+\eta) \end{equation}\] where \[\begin{equation} \eta = \frac{G_{n2}}{G_{n1}} + \frac{G_{n3}}{G_{n1}} \end{equation}\] corresponds to the noise contributions of M2a-M2b and M3a-M3b relative to that of the differential pair M1a-M1b. If we have no gain enhancement (i.e. remove transistors M3a-M3b), the noise is increased by the contribution of M2a-M2b only. The cross-coupled pair M3a-M3b does not cancel the noise of M2a-M2b but simply adds some noise resulting in the additional term in \(G_{n3}/G_{n1}\) in \(\eta\) further degrading the input-referred noise.
We can now have a closer look at the input-referred thermal noise resistance.
1.2.1 Input-referred thermal noise resistance
The input-referred thermal noise resistance \(R_{nt}\) is given by \[\begin{equation} R_{nt} = 2\,\frac{\gamma_{n1}}{G_{m1}} \, (1+\eta_{th}), \end{equation}\] where \[\begin{equation} \eta_{th} = \frac{\gamma_{n2}}{\gamma_{n1}} \, \frac{G_{m2}}{G_{m1}} + \frac{\gamma_{n3}}{\gamma_{n1}} \, \frac{G_{m3}}{G_{m1}}, \end{equation}\] represents the contributions of M2a-M2b and M3a-M3b to the input-referred thermal noise relative to that of M1a-M1b. If we set \(G_{m3} = G_{m2}\), the contribution of M3a-M3b is then equal to that of M2a-M2b. The value of \(\eta_{th}\) is then twice that of of the circuit without M3a-M3b. Note that \(G_{m1}/G_{m2}\) represents the voltage gain without M3a-M3b, the contributions of M2a-M2b and M3a-M3b should remain reasonable.
1.2.2 Input-referred flicker noise resistance
The input-referred flicker noise resistance \(R_{nf}\) is given by \[\begin{equation} R_{nf} = 2\,\frac{\rho_n}{f\,W_1\,L_1} \, (1+\eta_{fl}), \end{equation}\] where \[\begin{equation} \eta_{fl} = \frac{\rho_p}{\rho_n}\,\left(\frac{G_{m2}}{G_{m1}}\right)^2\,\frac{W_1\,L_1}{W_2\,L_2} + \frac{\rho_p}{\rho_n}\,\left(\frac{G_{m3}}{G_{m1}}\right)^2\,\frac{W_1\,L_1}{W_3\,L_3}, \end{equation}\] represents the contributions of M2a-M2b and M3a-M3b to the input-referred flicker noise relative to that of M1a-M1b. For the same reasons as discussed above, the contributions of M2a-M2b and M3a-M3b should remain small thanks to the square of the \(G_m\) ratios. They can be further reduced by making the gate area of M2a-M2b and M3a-M3b larger than that of M1a-M1b.
1.3 Conclusion
Adding the cross-coupled pair M3a-M3b seems very attractive to boost the DC gain. However, if we set \(G_{m3} = G_{m2}\), we have seen that the gain is indeed increased but only to that of a simple CS stage. The same gain would be easier to reach with replacing the diode-connected transistors M2a-M2b by simple current source. In addition, the cross-coupled pair M3a-M3b, although canceling part of the output conductance \(G_{od}\), adds more noise than what we get from a simple CS stage.
2 Problem 2: Bandwidth enhancement
The fully differential OTA shown in Figure 8 shows a larger gain-bandwidth product compared to the case without the cross-coupled transistors M2a-M2b [1] [6].
2.1 Small-signal analysis
The OTA of Figure 8 uses the differential transconductor shown in Figure 9 which allows to increase the equivalent transconductance compared to the case of a simple differential pair.
In small-signal differential operation (i.e. \(V_{i1}=\Delta V_{id}/2\) and \(V_{i2}=-\Delta V_{id}/2\)), assuming that transistors M1a-M1b and M2a-M2b are perfectly matched, the common source voltage of M2a-M2b does not change \(\Delta V_{S2} = 0\) and can be considered as an AC ground. By symmetry we also can state that \(\Delta V_{G2a} = \Delta V_{D2b} = -\Delta V_{D2a}\). The VCCS modeling M2a is therefore controlled by its drain voltage \(\Delta V_{D2a}\) which is equal to \(-\Delta V_{G2a}\). This means that the VCCS can be replaced by a negative conductance \(-G_{m2}\). The small-signal circuit then simplifies to the half-circuit shown in Figure 10
From the circuit shown in Figure 10, it is easy to derive the equivalent transconductance \(G_{meq}\) as [6] [7] \[\begin{equation} G_{meq} \triangleq \frac{\Delta I_{od}}{\Delta V_{id}} = \frac{G_{m1}\,(G_{ds2}-G_{m2})}{G_{ms1}+G_{ds1}+G_{ds2}-G_{m2}}. \end{equation}\] We see that setting \(G_{m2}=G_{ms1}+G_{ds1}+G_{ds2} \cong G_{ms1}\) gives an infinite transconductance \(G_{meq}\). We can now distinguish two cases \(G_{m2} < G_{ms1}\) or \(G_{m2} > G_{ms1}\). To investigate whether both choices are possible, we need to have a closer look at the small-signal circuit including the parasitic capacitance \(C_p\) at the sources of M1a-M1b. The latter is equal to the parasitic capacitances of M2a-M2b \[\begin{equation} C_p \cong C_{GS2} + 4\,C_{GD2}. \end{equation}\] The factor 4 comes from splitting the \(2\,C_{GD2}\) capacitances in two capacitances of value \(4\,C_{GD2}\) which are connected in series with the middle point connected to ground. They therefore come in parallel to \(C_{GS2}\) on each side.
To derive the transadmittance we can use the half-circuit shown in Figure 11.
The resulting transadmittance is then given by \[\begin{equation} Y_{meq} = G_{meq} \, \frac{1-s/\omega_z}{1+s/\omega_p}, \end{equation}\] where \[\begin{align} G_{meq} &= \frac{G_{m1}\,(G_{ds2}-G_{m2})}{G_{ms1}+G_{ds1}+G_{ds2}-G_{m2}} \cong -\frac{G_{m1}\,G_{m2}}{G_{ms1}-G_{m2}},\\ \omega_z &= \frac{G_{m2}-G_{ds2}}{C_p} \cong \frac{G_{m2}}{C_p},\\ \omega_p &= \frac{G_{ms1}+G_{ds1}+G_{ds2}-G_{m2}}{C_p} \cong \frac{G_{ms1}-G_{m2}}{C_p},\\ C_p &= C_{GS2} + 4\,C_{GD2}. \end{align}\] From the expression of the pole \(\omega_p\), we can deduce that for the pole to be in the left half-plane requires \(G_{ms1} > G_{m2}\). We can then rewrite \[\begin{align} G_{meq} &\cong -\frac{G_{m1}}{G_{ms1}/G_{m2}-1} = -\frac{G_{m1}}{\alpha-1},\\ \omega_z &\cong \frac{G_{m2}}{C_p},\\ \omega_p &\cong \frac{G_{m2}}{C_p}\,\left(1-\frac{G_{ms1}}{G_{m2}}\right) = \frac{G_{m2}}{C_p}\,(\alpha-1), \end{align}\] where \[\begin{equation} \alpha \triangleq \frac{G_{ms1}}{G_{m2}} > 1 \end{equation}\] Notice that the equivalent transconductance \(G_{meq}\) becomes negative for \(\alpha > 1\).
The equivalent transconductance of a differential pair is therefore increased by a factor \(K\) given by \[\begin{equation} K = \frac{1}{\alpha-1}. \end{equation}\] To achieve a transconductance increase of \(K\), the transconductance ratio \(\alpha = G_{ms1}/G_{m2}\) needs to be equal to \(1+1/K\).
If transistors M1a-M1b and M2a-M2b are all biased in strong inversion then \[\begin{align} G_{ms1} &= \sqrt{2\,n_1\,\beta_1\,I_{D1}},\\ G_{m2} &= \sqrt{\frac{2\,\beta_2\,I_{D2}}{n_2}}. \end{align}\] Since M1a-M1b and M2a-M2b share the same current \(I_{D1}=I_{D2}=I_b\) and the transconductance ratio is then given by \[\begin{equation} \alpha = \frac{G_{ms1}}{G_{m2}} = \sqrt{\frac{n_1\,n_2\,\beta_1}{\beta_2}}, \end{equation}\] which only depends on the \(\beta_1/\beta_2\) ratio.
Choosing \(\beta_1 = \beta_2\) leads to \(\alpha =\) 1.30 and a transconductance increase \(K =\) 3.333 for \(n_1 = n_2 =\) 1.30.
To minimize the \(V_{GS}\) voltages of M1a-M1b and M2a-M2b and better control the transconductance ratio \(\alpha = G_{ms1}/G_{m2}\) by a current ratio, we should bias them in weak inversion [6]. In this case the transconductances are given by \[\begin{align} G_{ms1} &= \frac{I_{D1}}{U_T},\\ G_{m2} &= \frac{I_{D2}}{n_2\,UT}. \end{align}\] The transconductance ratio is then proportional to the current ratio \[\begin{equation} \alpha = \frac{G_{ms1}}{G_{m2}} = n_2\,\frac{I_{D1}}{I_{D2}} \end{equation}\] The equivalent transconductance then writes \[\begin{equation} G_{meq} = \frac{G_{m1}}{n_2\,\frac{I_{D1}}{I_{D2}}-1} = \frac{I_{D1}}{n_1\,U_T}\,\frac{1}{n_2\,\frac{I_{D1}}{I_{D2}}-1}. \end{equation}\] Note that \(I_{D1}\) is actually half the total bias current.
If M1a-M1b and M2a-M2b share the same bias current \(I_b\) as it is the case in the circuit of Figure 9 then \(I_{D1}=I_{D2}=I_b\) and \(\alpha = n_2 > 1\) which is larger than 1. For \(n_2 =\) 1.30, we get \(K =\) 3.333.
We can compare the equivalent transconductance to the transconductance achieved in weak inversion by M1a-M1b \[\begin{equation} G_{mwi} = \frac{I_{D1}}{n_1\,U_T}, \end{equation}\] by defining the ratio \[\begin{equation}\label{eqn:alpha} \frac{G_{meq}}{G_{mwi}} = K = \frac{1}{n_2\,\frac{I_{D1}}{I_{D2}}-1}. \end{equation}\] The \(K\)-factor is plotted versus the current ratio \(I_{D1}/I_{D2}\) in Figure 12.
From \(\eqref{eqn:alpha}\) and Figure 12, we see that we only gain if the drain current ratio \(I_{D1}/I_{D2}\) is smaller than \(2/n_2 =\) 1.538. On the other hand, the condition \(\alpha > 1\) requires \(I_{D1}/I_{D2}\) to be larger than \(1/n_2 =\) 0.769. The drain current ratio \(I_{D1}/I_{D2}\) needs therefore to satisfy the following inequalities \[\begin{equation} \frac{1}{n_2} < \frac{I_{D1}}{I_{D2}} < \frac{2}{n_2}, \end{equation}\] which for \(n_2 =\) 1.300 gives 0.769 \(< I_{D1}/I_{D2} <\) 1.538. This actually leaves little margin. Choosing \(I_{D1} = I_{D2}\) is a good choice because of its simple implementation (it does not require to add any bias current to increase the current in M1a-M1b).
The transfer function of the \(G_m\)-enhanced OTA of Figure 8 is given by \[\begin{equation} A_d \triangleq \frac{V_{od}}{V_{id}} \cong \frac{G_{meq}}{G_{ds3}+s\,C_L} = \frac{A_{dc}}{1 + \frac{s}{\omega_p}}. \end{equation}\] with \[\begin{align} A_{dc} &\cong = K \cdot \frac{G_{m1}}{G_{ds3}}.\\ \end{align}\] The gain-bandwidth product is then given by \[\begin{equation} \omega_u =|A_{dc}|\;\omega_c \cong K \cdot \frac{G_{m1}}{C_L}. \end{equation}\] We see that adding the cross-coupled pair M2a-M2b increases both the DC gain and the gain-bandwidth product by the factor \(K\) compared to the same circuit of Figure 8 but without the cross-coupled pair M2a-M2b.
We can rewrite the DC gain as \[\begin{equation}\label{eqn:gm:adc} A_{dc} \cong -\frac{G_{m1}}{G_{ds3}}\,\frac{1}{1+\frac{G_{ds1}}{G_{ds3}}-\frac{G_{ms1}}{G_{m2}}} \end{equation}\] From \(\eqref{eqn:gm:adc}\), we see that in order to also increase the DC gain, we need to make \(G_{ms1}/G_{m2}\) sligthly larger than 1 but still smaller than \(1+G_{ds1}/G_{ds3}\) to avoid instability. Indeed if we set \(G_{ms1}/G_{m2}=1\) the DC gain becomes \(-G_{m1}/G_{ds1}\) which corresponds to the gain of a simple common-source stage. We have compensated the output conductance of M3a-M3b but not that of M1a-M1b. To boost the DC gain we need \[\begin{equation} 1 < \frac{G_{ms1}}{G_{m2}} < 1 + \frac{G_{ds1}}{G_{ds3}}. \end{equation}\] It is however dangerous to set \(G_{ms1}/G_{m2}\) sligthly larger than 1, because we don’t control the \(G_{ds1}/G_{ds3}\) very weel due to variability and mismatch.
To really take advantage of the \(G_m\)-enhancement also for the DC gain, the latter should be developped at another node than the drain of M1a-M1b. An example of such circuit is shown in Figure 14 which is close to the one proposed in [6].
2.2 Noise analysis
The small-signal schematic of Figure 9 including all the noise sources is shown in Figure 13. The output noise current \(I_{nout1}\) is given by \[\begin{equation} I_{nout1} = H_{n1} \cdot I_{n1} + H_{n2} \cdot I_{n2}, \end{equation}\] with \[\begin{align} H_{n1} &= \frac{G_{m2}}{G_{m2}-G_{ms1}} = -\frac{1}{\alpha-1},\\ H_{n2} &= \frac{G_{ms1}}{G_{ms1}-G_{m2}} = \frac{\alpha}{\alpha-1}, \end{align}\] where \(\alpha = G_{ms1}/G_{m2}\). The differential output noise current is simply \(I_{nout} = 2\,I_{nout1}\) and the output noise conductance then writes \[\begin{equation} G_{nout} = 2\,\left(|H_{n1}|^2\,G_{n1}+|H_{n2}|^2\,G_{n2}\right), \end{equation}\] with \[\begin{equation} G_{ni} = \gamma_{ni}\,G_{mi} + G_{mi}^2\,\frac{\rho_i}{f\,W_i\,L_i}\quad\textsf{for $i=1,2$}. \end{equation}\]
The input-referred noise resistance is then given by \[\begin{equation} R_{nin} = \frac{G_{nout}}{G_{meq}^2} = 2\frac{G_{n1}}{G_{m1}^2}\,(1+\eta) \end{equation}\] where \[\begin{equation} \eta = \left(\frac{G_{ms1}}{G_{m2}}\right)^2\,\frac{G_{n2}}{G_{n1}} = \alpha^2\,\frac{G_{n2}}{G_{n1}} \end{equation}\] corresponds to the noise contributions of M2a-M2b relative to that of the differential pair M1a-M1b. We see that the input-referred noise is degraded by M2a-M2b.
2.2.1 Input-referred thermal noise resistance
The input-referred thermal noise resistance \(R_{nt}\) is given by \[\begin{equation} R_{nt} = 2\,\frac{\gamma_{n1}}{G_{m1}} \, (1+\eta_{th}), \end{equation}\] where \[\begin{equation} \eta_{th} = \frac{\gamma_{n2}}{\gamma_{n1}} \, \frac{n_1\,G_{ms1}}{G_{m2}} = \frac{\gamma_{n2}}{\gamma_{n1}} \, n_1 \, \alpha, \end{equation}\] represents the contributions of M2a-M2b to the input-referred thermal noise relative to that of M1a-M1b. If both transistors M1a-M1b and M2a-M2b are biased in weak inversion then \[\begin{equation} \alpha = \,n_2\,\frac{I_{D1}}{I_{D2}} \end{equation}\] and \[\begin{equation} \eta_{th} = \frac{\gamma_{n2}}{\gamma_{n1}}\,n_1\,n_2\,\frac{I_{D1}}{I_{D2}} \cong n_1\,n_2\,\frac{I_{D1}}{I_{D2}}, \end{equation}\] In the case \(I_{D1}=I_{d2}=I_b\) then \[\begin{equation} \eta_{th} \cong n_1\,n_2. \end{equation}\] For \(n_1=n_2=\) 1.30, we get \(\eta_{th} =\) 1.690, which means that M2a-M2b contribute 1.690-times more than M1a-M1b to the input-referred thermal noise.
2.2.2 Input-referred flicker noise resistance
The input-referred flicker noise resistance \(R_{nf}\) is given by \[\begin{equation} R_{nf} = 2\,\frac{\rho_n}{f\,W_1\,L_1} \, (1+\eta_{fl}), \end{equation}\] where \[\begin{equation} \eta_{fl} = n_1^2\,\frac{W_1\,L_1}{W_2\,L_2}, \end{equation}\] represents the contributions of M2a-M2b to the input-referred flicker noise relative to that of M1a-M1b. We see that if M1a-M1b and M2a-M2b have the same gate area, their contribution to the input-referred flicker noise is about the same. For \(n_1=\) 1.30 and \(W_1\,L_1 = W_2\,L_2\), we get \(\eta_{fl} =\) 1.690, which means that M2a-M2b contributes 1.690 more than M1a-M1b. We can of course reduce this contribution by making \(W_2\,L_2\) larger than \(W_1\,L_1\).
The above analysis has shown that the \(G_m\)-enhanced OTA of Figure 8 can increase the gain-bandwidth product and eventually the DC gain compared to the same OTA without the cross-coupled pair M2a-M2b, but this comes at the cost of a higher noise.
2.3 Example
2.3.1 Design
We want to design the single-ended \(G_m\)-enhanced OTA shown in Figure 14 [6] for the specifications given in Table 1. The circuit of Figure 14 corresponds to a symmetrical OTA to which we have added the cross-coupled pair M2a-M2b. It is close to the original OTA proposed by Castello [6]. As explained above, this circuit takes advantage of the \(G_m\)-enhancement provided by the cross-coupled pair M2a-M2b to increase both the gain-bandwidth product and the DC gain. The reason is that the DC gain is realized at the output node instead of the drain of M1a-M1b like it is the case for the OTA of Figure 8.
We need to find the bias current and size the transistors to achieve the specs Table 1. We will design the amplifier for a generic 180nm bulk CMOS process. The physical parameters are given in Table 2, the global process parameters in Table 3 and finally the MOSFET parameters in Table 4.
| Specification | Symbol | Value | Unit |
|---|---|---|---|
| Input common mode voltage | \(V_{{ic}}\) | 1.2 | \(V\) |
| DC gain | \(A_{{dc}}\) | 80 | \(dB\) |
| DC gain | \(A_{{dc}}\) | 10000 | - |
| Gain-bandwidth product | \(GBW\) | 10 | \(MHz\) |
| Load capacitance | \(C_L\) | 1 | \(pF\) |
| Parameter | Value | Unit |
|---|---|---|
| \(T\) | 300 | \(K\) |
| \(U_T\) | 25.875 | \(mV\) |
| Parameter | Value | Unit |
|---|---|---|
| \(V_{DD}\) | 1.8 | \(V\) |
| \(C_{ox}\) | 8.443 | \(\frac{fF}{\mu m^2}\) |
| \(W_{min}\) | 200 | \(nm\) |
| \(L_{min}\) | 180 | \(nm\) |
| Parameter | NMOS | PMOS | Unit |
|---|---|---|---|
| sEKV parameters | |||
| \(n\) | 1.27 | 1.31 | - |
| \(I_{{spec\Box}}\) | 715 | 173 | \(nA\) |
| \(V_{{T0}}\) | 0.455 | 0.445 | \(V\) |
| \(L_{{sat}}\) | 26 | 36 | \(nm\) |
| \(\lambda\) | 20 | 20 | \(\frac{{V}}{{\mu m}}\) |
| Overlap capacitances parameters | |||
| \(C_{{GDo}}\) | 0.366 | 0.329 | \(\frac{{fF}}{{\mu m}}\) |
| \(C_{{GSo}}\) | 0.366 | 0.329 | \(\frac{{fF}}{{\mu m}}\) |
| \(C_{{GBo}}\) | 0 | 0 | \(\frac{{fF}}{{\mu m}}\) |
| Junction capacitances parameters | |||
| \(C_J\) | 1 | 1.121 | \(\frac{{fF}}{{\mu m^2}}\) |
| \(C_{{JSW}}\) | 0.2 | 0.248 | \(\frac{{fF}}{{\mu m}}\) |
| Flicker noise parameters | |||
| \(K_F\) | 8.1e-24 | 8.1e-24 | \(J\) |
| \(AF\) | 1 | 1 | - |
| \(\rho\) | 0.05794 | 0.4828 | \(\frac{{V \cdot m^2}}{{A \cdot s}}\) |
| Matching parameters | |||
| \(A_{{VT}}\) | 5 | 5 | \(mV \cdot \mu m\) |
| \(A_{{\beta}}\) | 1 | 1 | \(\% \cdot \mu m\) |
| Source and drain sheet resistance parameter | |||
| \(R_{{sh}}\) | 600 | 2386 | \(\frac{{\Omega}}{{\mu m}}\) |
| Width and length parameters | |||
| \(\Delta W\) | 39 | 54 | \(\,nm\) |
| \(\Delta L\) | -76 | -72 | \(\,nm\) |
The DC gain also benefits from the current gain \(B\) in the pMOS current mirrors. The equivalent transconductance is then given by \[\begin{equation} G_{meq} = B \cdot K \cdot G_{m1}. \end{equation}\] where \[\begin{equation} K = \frac{1}{\alpha-1} \end{equation}\] with \[\begin{equation} \alpha = \frac{G_{ms1}}{G_{m2}} = n_2. \end{equation}\] For the chosen technology this gives \(\alpha =\) 1.271 and \(K =\) 3.685.
The small-signal differential DC voltage gain is approximately given by \[\begin{equation} A_{dc} \cong - \frac{G_{meq}}{G_o} \end{equation}\] with \[\begin{equation} G_o = G_{ds3d} + G_{ds4b}. \end{equation}\]
We additionnally set the current gain of the pMOS current mirrors to \(B =\) 3. With \(K =\) 3.685 we have a total \(G_m\)-enhancement factor \(B \cdot K =\) 11.055. From the specified gain-bandwidth product and the load capacitance we can deduce the transconductance of M1a-M1b \(G_{m1} =\) 5.848 \(\mu A/V\). Setting the inversion coefficient of M1a-M1b to \(IC_1 =\) 0.1, we get the bias current \(I_b =\) 204 \(nA\), which we slightly increase to \(I_b =\) 210 \(nA\) in order to have some margin. The differential pair bias current \(2I_b =\) 420 \(nA\).
Note
Note that if we want to achieve the same gain-bandwidth product \(GBW =\) 10 \(MHz\) without the cross-coupled pair and current gain, keeping the same inversion coefficient for M1a-M1b, we would need a bias current \(I_b =\) 2.256 \(\mu A\), which is 11.055 times larger than the bias current with the cross-coupled pair and the current gain.
We can now deduce the \(I_{spec}\) and \(W/L\) for M1a-M1b as \(I_{spec1} =\) 2 \(\mu A\) and \(W_1/L_1 =\) 2.937. Choosing \(L_1 = L_{min} =\) 180 \(nm\) we get and \(W_1 =\) 0.71 \(\mu m\).
The transconductance of M2a-M2b is \(G_{m2} = G_{ms1}/\alpha =\) 5.848 \(\mu A/V\). For a current \(I_{D2} = I_b =\) 210 \(nA\), we get \(G_{m2}\,n_2\,U_T/I_{D2} =\) 0.916 leading to \(IC_2 =\) 0.100. The specific current \(I_{spec}\) and \(W/L\) for M2a-M2b are given by \(I_{spec2} =\) 2.100 \(\mu A\) and \(W_2/L_2 =\) 2.937. Choosing \(L_2 = L_{min} =\) 180 \(nm\) we get and \(W_2 =\) 0.34 \(\mu m\).
For M3a to M3d, choosing \(IC_3 =\) 10, we get \(I_{spec3} =\) 21 \(nA\) and \(W_3/L_3 =\) 0.121. If we split \(G_o\) equally between M3d and M4b \(G_{ds3d}=G_{ds4b}\), then \(G_{ds3d} =\) 3.142 \(nA/V\). This leads to \(L_3 =\) 10.10 \(\mu m\) and \(W_3 =\) 1.16 \(\mu m\).
Similarly for M4a-M4b, choosing \(IC_4 =\) 10, we get \(I_{spec4} =\) 63 \(nA\) and \(W_4/L_4 =\) 0.088. With \(G_{ds4b} =\) 3.142 \(nA/V\). This leads to \(L_4 =\) 10.10 \(\mu m\) and \(W_4 =\) 0.83 \(\mu m\).
This gives very big transistors for M3a to M3d and M4a-M4b that will add large parasitic capacitances at the mirror nodes and at the output penalizing the gain-bandwidth product and the phase margin. This is even more the case when accounting for the current gain \(B =\) 3 which makes the width of M3c-M3d 3-times larger than the width of M3a-M3b. In order to minimize the parasitic capacitances at the mirror nodes and at the output, we choose for M3a-M3b \(W_3 = W_{min} =\) 200 \(nm\), which leads to \(L_3 =\) 2.17 \(\mu m\). Note that the width of M3c-M3d is 3-times that of M3a-M3b.
Similarly for M4a-M4b, we choose \(W_4 = W_{min} =\) 200 \(nm\), which leads to \(L_4 =\) 2.79 \(\mu m\). This will obviously reduce the DC gain below the target value. The estimated DC gain is now only \(A_{dc} \cong\) 67.45 \(dB\). Simulations will show that this value is pessimistic and that we loose less DC gain, but still do not achieve the target value. Of course we could cascode the M3d and M4b to save the DC gain without penalty on the current consumption (except the additional bias circuit for biasing the cascode transistors).
Finally, choosing \(IC_5 =\) 10 for M5a-M5b, we get \(I_{spec5} =\) 42 \(nA\) and \(W_5/L_5 =\) 0.059. Choosing \(W_5 = W_{min} =\) 200 \(nm\), we get \(L_5 =\) 4.14 \(\mu m\).
2.3.2 Transistor information
The transistor sizes and large-signal variables are summarized in Table 5, whereas Table 6 gives the small-signal and thermal noise parameters. An Excel table is generated with more information (e.g. all the parasitic capacitances).
| Transistor | \(W\;[\mu m]\) | \(L\;[\mu m]\) | \(I_D\;[nA]\) | \(I_{{spec}}\;[nA]\) | \(IC\) | \(V_G-V_{{T0}}\;[mV]\) | \(V_{{DSsat}}\;[mV]\) |
|---|---|---|---|---|---|---|---|
| M1a | 0.71 | 0.18 | 210 | 5169 | 0.041 | -64 | 104 |
| M1b | 0.71 | 0.18 | 210 | 5169 | 0.041 | -64 | 104 |
| M2a | 0.34 | 0.18 | 210 | 2636 | 0.080 | -50 | 105 |
| M2b | 0.34 | 0.18 | 210 | 2636 | 0.080 | -50 | 105 |
| M3a | 0.20 | 2.17 | 210 | 21 | 10.000 | 127 | 194 |
| M3b | 0.20 | 2.17 | 210 | 21 | 10.000 | 127 | 194 |
| M3c | 0.60 | 2.17 | 630 | 54 | 11.651 | 138 | 205 |
| M3d | 0.60 | 2.17 | 630 | 54 | 11.651 | 138 | 205 |
| M4a | 0.20 | 2.79 | 630 | 63 | 10.000 | 130 | 194 |
| M4b | 0.20 | 2.79 | 630 | 63 | 10.000 | 130 | 194 |
| M5a | 0.20 | 4.14 | 420 | 42 | 10.000 | 130 | 194 |
| M5b | 0.20 | 4.14 | 420 | 42 | 10.000 | 130 | 194 |
| Transistor | \(G_{{spec}}\;[\mu A/V]\) | \(G_{{ms}}\;[\mu A/V]\) | \(G_m\;[\mu A/V]\) | \(G_{{ds}}\;[nA/V]\) | \(\gamma_n\) |
|---|---|---|---|---|---|
| M1a | 199.777 | 7.811 | 6.143 | 58.333 | 0.644 |
| M1b | 199.777 | 7.811 | 6.143 | 58.333 | 0.644 |
| M2a | 101.884 | 7.556 | 5.943 | 58.333 | 0.650 |
| M2b | 101.884 | 7.556 | 5.943 | 58.333 | 0.650 |
| M3a | 0.812 | 2.193 | 1.679 | 4.847 | 0.812 |
| M3b | 0.812 | 2.193 | 1.679 | 4.847 | 0.812 |
| M3c | 2.090 | 6.164 | 4.720 | 14.542 | 0.816 |
| M3d | 2.090 | 6.164 | 4.720 | 14.542 | 0.816 |
| M4a | 2.435 | 6.578 | 5.174 | 11.296 | 0.790 |
| M4b | 2.435 | 6.578 | 5.174 | 11.296 | 0.790 |
| M5a | 1.623 | 4.385 | 3.449 | 5.067 | 0.790 |
| M5b | 1.623 | 4.385 | 3.449 | 5.067 | 0.790 |
2.3.3 Simulation
Operating point
We first write the parameter file for this specific design for running the ngspice simulations. Before running the AC and NOISE simulations, we first need to check the quiescent voltages and currents and the operating points of all transistors by running an .OP simulation. The node voltages are extracted from the .ic file and presented in Table 7.
| Node | Voltage |
|---|---|
| vdd | 1.8 |
| inp | 1.2 |
| inn | 1.2 |
| id | 0 |
| ic | 1.2 |
| out | 0.662637 |
| 1 | 0.230944 |
| 2 | 0.683804 |
| 3 | 0.683804 |
| 4 | 1.1074 |
| 5 | 1.1074 |
| 6 | 0.662637 |
| 7 | 0.678442 |
From Table 7, we see that the ouput voltage is equal to \(V_{outq} =\) 663 \(mV\) for an input common mode voltage \(V_{ic} =\) 1.2 \(V\). We need to check wether this quiescent output voltage is in the high gain region for running the AC simulation. This is done below by looking at the large-signal transfer characteristic.
The operating point information for all transistors coming from the EKV2.6 compact model are extracted from the ngspice .op.dat file. The data is split into the large-signal operating informations presented in Table 8, the small-signal operating point informations shown in Table 9 and the noise operating point informations in Table 10.
| Transistor | \(I_D\;[nA]\) | \(I_{spec}\;[nA]\) | \(IC\) [-] | \(n\) [-] | \(V_{DSsat}\;[mV]\) |
|---|---|---|---|---|---|
| M1a | 206.9 | 5021.8 | 0.041 | 1.27 | 114 |
| M1b | 206.9 | 5021.8 | 0.041 | 1.27 | 114 |
| M2a | 206.9 | 2811.6 | 0.074 | 1.27 | 118 |
| M2b | 206.9 | 2811.6 | 0.074 | 1.27 | 118 |
| M3a | 206.9 | 18.7 | 11.110 | 1.31 | 276 |
| M3b | 206.9 | 18.7 | 11.110 | 1.31 | 276 |
| M3c | 535.4 | 48.3 | 11.120 | 1.31 | 276 |
| M3d | 535.4 | 48.3 | 11.120 | 1.31 | 276 |
| M4a | 535.4 | 57.5 | 9.347 | 1.27 | 262 |
| M4b | 535.4 | 57.5 | 9.347 | 1.27 | 262 |
| M5a | 420.0 | 38.2 | 11.024 | 1.27 | 275 |
| M5b | 413.8 | 38.0 | 11.013 | 1.27 | 275 |
| Transistor | \(G_m\;[\mu A/V]\) | \(G_{ms}\;[\mu A/V]\) | \(G_{ds}\;[nA/V]\) |
|---|---|---|---|
| M1a | 6.366 | 7.581 | 130.964 |
| M1b | 6.366 | 7.581 | 130.964 |
| M2a | 5.960 | 7.340 | 120.751 |
| M2b | 5.960 | 7.340 | 120.751 |
| M3a | 1.526 | 2.053 | 2.180 |
| M3b | 1.526 | 2.053 | 2.180 |
| M3c | 3.944 | 5.306 | 4.691 |
| M3d | 3.944 | 5.306 | 4.691 |
| M4a | 4.384 | 5.643 | 6.153 |
| M4b | 4.384 | 5.643 | 6.153 |
| M5a | 3.207 | 4.129 | 3.157 |
| M5b | 3.076 | 4.112 | 152.354 |
| Transistor | \(R_n\;[k\Omega]\) | \(\sqrt{S_{ID,th}}\;[nA/\sqrt{Hz}\) | \(\gamma_n\) [-] | \(\sqrt{S_{ID,fl}}\) at 1Hz \([nA/\sqrt{Hz}]\) |
|---|---|---|---|---|
| M1a | 96.861 | 40.070 | 0.617 | 110866 |
| M1b | 96.861 | 40.070 | 0.617 | 110866 |
| M2a | 108.924 | 42.492 | 0.649 | 155855 |
| M2b | 108.924 | 42.492 | 0.649 | 155855 |
| M3a | 561.542 | 96.479 | 0.857 | 122362 |
| M3b | 561.542 | 96.479 | 0.857 | 122362 |
| M3c | 217.183 | 60.000 | 0.857 | 76256.2 |
| M3d | 217.183 | 60.000 | 0.857 | 76256.2 |
| M4a | 187.923 | 55.812 | 0.824 | 38419.7 |
| M4b | 187.923 | 55.812 | 0.824 | 38419.7 |
| M5a | 257.913 | 65.385 | 0.827 | 31396.5 |
| M5b | 281.811 | 68.347 | 0.867 | 31396.5 |
Large-signal differential transfer characteristic
We can now simulate the large-signal DC input-output transfer characteristic. The simulation result is presented in Figure 15.
We see from Figure 15 that the quiescent output voltage is in the linear region. We nevertheless can impose the extracted offset voltage \(V_{os} =\) 32.2 \(\mu V\) to center the output voltage at \(V_{outq} = V_{DD}/2 =\) 0.9 \(V\). We also see that the OTA offers a rather large output voltage swing \(V_{swing} =\) 1.0 \(V\).
Open-loop gain
After having checked the operating point information and making sure that the OTA output is not saturated and in the high gain region, we can now perform the AC simulation. The simulation results are compared to the theoretical estimations in Figure 16.
From Figure 16, we see that the simulated gain-bandwidth product is right on target and larger than the theoretical estimation. The simulated DC gain is slightly smaller than the specs, but larger than the theoretical estimation.
2.4 Conclusion
The OTA of Figure 14 shows that the DC gain and gain-bandwidth product can both be increased without spending more current thanks to the cross-coupled pair M2a-M2b [6]. However, this achievement comes at the cost of a larger voltage required to bias the input transconductor M1a-M1b and M2a-M2b. Indeed, the minimum supply voltage is now increased by the \(V_{GS}\) voltage of M2a-M2b which is typically \(V_{T0n} =\) 455 \(mV\). In addition, the OTA of Figure 14 shows more noise because the cros-coupled pair contributes \(n_1\,n_2\)-times more than the differential pair. For \(n_1 = n_2=\) 1.27 the cross-coupled pair contributes 1.62-times more than the differential pair, which is significant. We also have to remember that the \(G_m\)-enhancement is directly related to transconductance ratio \(\alpha = G_{ms1}/G_{m2}\) which depend on the current ratio \(I_{D1}/I_{D2}\). This current ratio might not be well controlled and subject to variability and mismatch. Note that taking \(I_{D1}=I_{D2}\) would result in \(\alpha \cong 1\) which could lead to instability.
3 Conclusion
In the exercise we have investigated two techniques to boost the gain and bandwidth of an OTA. They are both based on the use of a crsoss-coupled pair either at the drain for gain-enhancement or at the source for \(G_m\)-enhancement. Although these techniques look attarctive at first glance they don’t really provide more performance compared to other techniques like simple cascode. Indeed cascode or its improved versions regulated cascode and gain-boosting turn out to be the simplest and most efficient gain-enhancement techniques [8] [9] [10].
4 References
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