I have a switched capacitor circuit that looks like below: enter image description here

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.

  • 1
    \$\begingroup\$ What are you using for switches? \$\endgroup\$
    – Voltage Spike
    Commented May 29, 2019 at 20:17
  • 1
    \$\begingroup\$ What is the \$K\$ in your noise equation? What is the \$-33\mathrm{dB}\$ with respect to? \$\endgroup\$
    – TimWescott
    Commented May 29, 2019 at 20:23
  • 1
    \$\begingroup\$ And how did you derive your noise equation? \$\endgroup\$
    – TimWescott
    Commented May 29, 2019 at 20:24
  • \$\begingroup\$ @TimWescott I used the noise power, vn^2(rms)=KT/C, where $$K$$ is boltzmann constant and C is the capacitance. Now, the noise power = 20*log10(vn^2(rms)). Well, I guess -33dB is with respect to the signal level (0dB) (full scale of 1V). \$\endgroup\$
    – sundar
    Commented May 29, 2019 at 22:58
  • 1
    \$\begingroup\$ for switch-caps, using 10 pF value, the RMS noise for ANY VALUE OF RESISTOR will be 20 microVolts RMS. The resistor+capacitor set the timeconstant and thus the bandwidth, but sqrt(K*T/C) defines the noise. \$\endgroup\$ Commented May 30, 2019 at 2:09

1 Answer 1


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\$.


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