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I would like to simulate this schematic in a SPICE AC analysis. A regular switch model only works in a transient simulation. Is there an equivalent impedance that I can use in the place of the switches (and the switched resistor)?

The frequency of the switch is 100kHz 50% duty cycle

A better question would be, is there a way to substitute some kind of impedance in for the switch and resistor if the switching frequency is 100kHz?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ What simulator are you planning to use? Many have a built-in switch primitive. \$\endgroup\$ Commented Aug 2, 2018 at 20:44
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    \$\begingroup\$ Ltspice has switch models \$\endgroup\$
    – EE_socal
    Commented Aug 2, 2018 at 20:56
  • \$\begingroup\$ @EE_socal The switch models only work in a transient simulation, I want to run an AC sweep. \$\endgroup\$
    – Voltage Spike
    Commented Aug 2, 2018 at 21:50
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    \$\begingroup\$ You should basically use an averaged equivalent circuit. E.g. R2 acts as a resistor while its switch is on and as open circuit otherwise, so for a given voltage its average current is D (duty cycle) times the DC one. In simpler terms R2(D)=R2/D. The same obviously goes for R1 and its own switch. Of course for the above to be meaningful you have to keep your AC analysis well below switching frequency. \$\endgroup\$
    – carloc
    Commented Aug 2, 2018 at 23:01
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    \$\begingroup\$ I would guess the switches act like a regular sampler, with the usual sinc frequency response, but the aliasing here occurs volens-nolens. For example, a .TRAN freq sweep from 100 to 1k, with fs=1k, will not give the usual sampled symmetry, but something mingled. I don't know how to represent such an aliasing difference in .AC. Furthermore, as carloc mentions, it is dependent on the duty cycle, if this would be the case. \$\endgroup\$ Commented Aug 4, 2018 at 6:37

2 Answers 2

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Your circuitry looks like a non-ideal switched-capacitor circuit (due to the 10k resistors). Hence, it can be treated mathematically using digital signal processing tools - provided that

  • the clock period Tcl is much larger than the time constant of the RC-combination (seems to be fulfilled with Tcl=1E-5>>1E-7=RC),

  • The period Ts of the input signal is much larger than the clock period Tcl (Ts>>Tcl, fs<<100kHz).

In this case, your circuit with two switches and the grounded capacitor can be treated as a simple combination of a grounded capacitor and a "toggled" switch (non-ideal with an on-resistance of 10k), which allows capacitor charging during one switch period and discharging during the other period. However, if both above mentioned requirements are fulfilled, the capacitor can be assumed as fully charged/discharged in each clock period - and the value of the resitor (10k) plays no major role.

For such a combination a time continuous linear ac equivalent does exist (based on z-transformation) which was described by B.D. Nelin in 1983 (IEEE Transactions Circuits and Systems, vol CAS-30, pp 43-48).

Such a toggled-switch capacitor block (called "storistor") has the following transfer function in the z-domain: Hz=z*exp(-1/2).

In each simulation programm (ac analysis) this storistor model can be realized using an analog delay element (delay line) with a delay of Td=(1/2)Tcl . The delay line should be terminated properly with a resistor Rl which equals the characteristic impedance of the delay line.

Fazit: For low frequency signals (fs<<100kHz) , the shown combination produces nothing else than a pure delay of the signal.

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That's not impossible if your V1 has much lower frequency than 100kHz. I have built an adjustable audio frequency notch filter which was made adjustable by switching resistors ON and OFF having a variable duty cycle. It worked. The states of the capacitors were freezed (except leakages and OPAMP input currents) when the resistors were OFF.

If the resistors are ON P% of the time, the effective resistance of resistor R is =100*R/P

You can use this only to those voltages which are held by the capacitors during the OFF states of the resistors. The method has no sense if the voltage over R1 or R2 is your output.

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