Why switched capacitors in IC; 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.

SC resistor. A capacitor $C$ switched between nodes 1 and 2 by non-overlapping clocks transfers $\Delta q = C(V_1 - V_2)$ each period $T_c$, so $I = f_cC(V_1 - V_2)$ and $R_{eq} = 1/(f_cC)$.

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

Realization of summer and integrators

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 integrator. The active-RC integrator has $T(s) = -\frac{1}{sRC_F}$. Replacing $R$ by a switched capacitor $C_1$:

$$T(s) = \frac{V_o}{V_i} = -\frac{f_cC_1}{sC_F} = -\frac{1}{s}\cdot\frac{C_1}{C_F}f_c$$

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

The integrator time constant $C_F/(f_cC_1)$ is set by a capacitor ratio. (In the z-domain, the stray-insensitive version gives $H(z) = -\frac{C_1}{C_F}\cdot\frac{1}{1 - z^{-1}}$.)

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.