Why simulate resistors with switched capacitors in MOS

Need for switched capacitors in MOS technology

Active-RC filters for audio and voice frequencies need resistors of hundreds of kΩ to MΩ, which MOS ICs cannot provide well:

• Chip area: diffused or poly resistors have low sheet resistance, so a 1 MΩ resistor occupies a very large area; a switched capacitor giving 10 MΩ needs only about 1 pF.
• Accuracy: absolute values of on-chip R and C vary by about ±20 %, and the errors are independent, so the RC time constant may be off by ±40 %. With SC, the time constant becomes $\tau = C/(f_cC_R)$, which depends only on a capacitor ratio (accurate to about 0.1 % in MOS) and a crystal clock.
• Temperature and linearity: capacitor ratios track with temperature and voltage; diffused resistors do not.
• Tunability: changing the clock frequency moves all pole and zero frequencies together.
• MOS gives near-ideal switches (very high off resistance) and op-amps with high input impedance that hold charge, so SC circuits are natural in MOS.

Simulating a resistor with a switched capacitor (parallel type)

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

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

So a capacitor switched at rate $f_c$ behaves like a resistor $R_{eq} = 1/(f_cC_R)$. Example: $C_R = 1$ pF, $f_c = 100$ kHz gives $R_{eq} = 10$ MΩ.

Other forms: the series switched capacitor gives the same $R_{eq} = 1/(f_cC_R)$, and the bilinear type gives $R_{eq} = 1/(4f_cC_R)$.

Conditions: two-phase non-overlapping clock (so the nodes are never shorted together), $f_c \gg$ highest signal frequency (typically 50–100 times), and the nodes are driven by low-impedance sources such as op-amp outputs or virtual ground.