I have a switched capacitor circuit that looks like below:

It operates at 80MHz to realize a 625k resistor. It works fine in the circuit. I tested with VDAC(input)=1.8V and the voltage at the right is 1.5V. Average current of 0.46uA flows in the circuit.

However, I am very much concerned about the practical implementation problems (noise). When I found the noise voltage of this circuit at 100kHz bandwidth, it comes out as:

$$\frac{K.T}{C}.100kHz = 0.0207$$ which is -33dB.

The input referred noise requirement of my circuit is around -110 dB. Am i doing something wrong or its just that the switched capacitor is not feasible?

Even if I try to add bigger capacitor in parallel, it can improve the accuracy, however, the noise voltage still cannot match to -110 dB.

Any guidance would be helpful.

Thank you.

## Best Answer

You did not calculate the noise correctly.

\$\frac{kT}{C}\$ is the integrated noise power which includes the Bandwidth of the circuit. So, you do not multiply it by bandwidth, bandwidth is multiplied when noise power spectral density is considered.

^{simulate this circuit – Schematic created using CircuitLab}To calculate the noise in your circuit, consider the above schematic. In one phase of the switching period, the capacitor C is connected (via \$SW_1\$) to the input voltage while it is connected to the output (\$C_{out}\$) in the second phase (via \$SW_2\$).

In \$\phi_1\$ (\$SW_1\$ closed and \$SW_2\$ open):A noise charge will be sampled on the \$C_{out}\$ capacitor. From thermodynamics, we know that each degree of freedom has energy of \$\frac{KT}{2}\$ in thermal equilibrium. Since voltage/charge is the only state of the capacitor, noise energy stored should be: $$\frac{\overline{q_{o_n}^2}}{2C_o} = \frac{kT}{2}$$ $$\overline{q_{o_n}^2} = kTC_o$$

In \$\phi_2\$ (\$SW_2\$ closed and \$SW_1\$ open):From similar reasoning as above, a noise charge will be sampled on the capacitor \$C\$ given by: $$\overline{q_{n}^2} = kTC$$

Since in this phase both the capacitors are in parallel, a charge redistribution will happen, which after a simple math gives the following final charge on the output capacitor \$C_o\$: $$Q_{o_n} = \frac{C_o}{C_o+C}(q_n+q_{o_n})$$. Since the noise charges on the capacitors are uncorrelated, we get: $$\overline{Q_{o_n}^2} = (\frac{C_o}{C_o+C})^2(\overline{q_{n}^2}+\overline{q_{o_n}^2})$$ Substituting the values, we get the noise charge of: $$\overline{Q_{o_n}^2} = \frac{kTC_o^2}{C_o+C}$$ Thus, the output voltage noise becomes: $$\overline{v_{o_n}^2} = \frac{kT}{C_o+C}$$ Assuming, you have \$C=C_o=20fF\$, the output noise power comes out to be \$-69.8dB\$.