Why SC in IC; resistor simulation; summer and integrators

Why resistors are replaced by switched capacitors in IC technology

• Area: MΩ resistors take huge silicon area; an equivalent switched capacitor is about 1 pF.
• Accuracy: absolute R and C on chip vary about ±20 % independently, so RC products are poor. SC time constants $C_F/(f_cC_R)$ depend on capacitor ratios (about 0.1 % accurate) and a crystal clock.
• Temperature/voltage tracking: ratios of matched capacitors stay constant; resistors drift.
• Tunable by the clock frequency, and fully compatible with MOS digital circuits on the same chip.

Simulation of a resistor

Consider a capacitor $C_R$ with two MOS switches driven by non-overlapping clocks $\phi_1$ and $\phi_2$ of frequency $f_c = 1/T_c$:

       phi1          phi2
 V1 ----o/o----+----o/o---- V2
               |
              ===  CR
               |
              GND

1. Phase $\phi_1$ (left switch closed): $C_R$ charges to $V_1$; stored charge $q_1 = C_RV_1$.
2. Phase $\phi_2$ (right switch closed): $C_R$ charges/discharges to $V_2$; stored charge $q_2 = C_RV_2$.
3. Charge moved from node 1 to node 2 in one clock period:

$$\Delta q = q_1 - q_2 = C_R(V_1 - V_2)$$

4. If $f_c$ is much higher than the signal frequency, $V_1$ and $V_2$ are nearly constant over a period, so the average current is

$$\begin{aligned} I_{avg} &= \frac{\Delta q}{T_c} = C_Rf_c(V_1 - V_2) \\ R_{eq} &= \frac{V_1 - V_2}{I_{avg}} = \frac{1}{f_cC_R} = \frac{T_c}{C_R} \end{aligned}$$

Example: $C_R = 1$ pF at $f_c = 100$ kHz simulates 10 MΩ.

SC equivalent circuits

Inverting summer. The active-RC summer gives $V_o = -\left(\frac{R_F}{R_1}V_1 + \frac{R_F}{R_2}V_2\right)$. Replacing each resistor by a switched capacitor ($R = 1/(f_cC)$) gives $R_F/R_i = C_i/C_F$, so

$$V_o = -\left(\frac{C_1}{C_F}V_1 + \frac{C_2}{C_F}V_2\right)$$

              +--[CF-SC]---+
              |            |
 V1 -[C1-SC]--+   |\       |
              +---|-\      |
 V2 -[C2-SC]--+   |  >-----+---> Vo
               +--|+/
               |  |/
              GND

The gains depend only on capacitor ratios.

Inverting lossy (damped) integrator. Active-RC form: input resistor $R_1$, feedback $R_2 \parallel C_F$:

$$T(s) = -\frac{1/(R_1C_F)}{s + 1/(R_2C_F)}$$

Replacing $R_1$ and $R_2$ by switched capacitors $C_1$ and $C_2$ ($1/R = f_cC$):

$$T(s) = -\frac{f_cC_1/C_F}{s + f_cC_2/C_F}$$

                    CF
              +-----||-----+
              |            |
              +--[C2-SC]---+
              |            |
              |   |\       |
 Vi -[C1-SC]--+---|-\      |
                  |  >-----+---> Vo
               +--|+/
               |  |/
              GND

DC gain $= -C_1/C_2$ and pole $\omega_p = f_cC_2/C_F$, both set by capacitor ratios. It is the first-order low-pass section used in SC ladders and biquads.

Non-inverting integrator. Here the switched capacitor is connected so that its charge is reversed before it reaches the op-amp: during $\phi_1$, $C_1$ is charged to $V_i$ (top plate to $V_i$, bottom plate to ground); during $\phi_2$, the top plate is grounded and the bottom plate is connected to the virtual ground. Charge $-C_1V_i$ is pushed into $C_F$, so the switched capacitor acts as a negative resistor $R = -1/(f_cC_1)$:

$$T(s) = \frac{V_o}{V_i} = +\frac{f_cC_1}{sC_F}$$

       phi1   C1   phi2           CF
 Vi ---o/o--+-||-+--o/o--+    +---||---+
            |    |       |    |        |
  GND -o/o--+    +-o/o-  +----+  |\    |
       phi2        | phi1     +--|-\   |
                  GND            |  >--+--> Vo
                              +--|+/
                              |  |/
                             GND

(Left plate: $V_i$ on $\phi_1$, ground on $\phi_2$; right plate: ground on $\phi_1$, op-amp input on $\phi_2$.) In the z-domain: $H(z) = +\frac{C_1}{C_F}\cdot\frac{z^{-1}}{1 - z^{-1}}$. A non-inverting integrator removes the need for an extra inverter in biquads, which is a big advantage of SC over active-RC.