Resistance of mixer/switched capacitors

A switched capacitor behaves like a resistor with resistance Req=1/(C*f) (f being the switching frequency).

Based on this many papers describe the input resistance of the mixer by this which is clearly frequency dependent:

• A Resistively Degenerated Wideband Passive Mixer With Low Noise Figure and High IIP2 (Kim, Larson et.al.)
• 1/f Noise in Passive CMOS Mixers for Low and Zero IF Integrated Receivers (Redman-White et Leenaerts)
• A wideband current-commutating passive mixer for multi-standard receivers in a 0.18um CMOS (Kuan et.al.)

However, another paper derives the resistance as being independent of the LO frequency:

• Implications of Passive Mixer Transparency for Impedance Matching and Noise Figure in Passive Mixer-First Receivers (Andrews and Molnar)

This is totally confusing. How does it fit together?

You should refer to an actual mixer circuit and real diode specifications.

Analyzing a diode mixer as switched capacitor circuit seem to miss the point. There's some switch capacitance involved but parallel connected to forward biased diodes. In case of an all-sides 50 ohm terminated mixer, I would expect that the frequency independent real load impedance in combination with on-state switch resistance dominates the input impedance.

Yes sorry I should have been more concrete. I talk about the mixers from the referenced papers which are all passive mixers (i.e., switching transistors), i.e. no diodes et.al.

Passive mixers have the property of transparency (same as N-path filters) which means that they down- or upconvert the impedances at the mixer ports.

I don't see much difference between diode and transistor mixer regarding RF port impedance, but LO impedance would be different.

Unfortunately I have no knowledge about diode mixers.
But it's not only about mixers but also sample & hold (hence I wrote "switched capacitor") in the title.

Imagine a switch with a C afterwards. What is its resistance looking into it? 1/(fsw*C). If you add R in parallel it gets more complicated. I show the circuit of one of the referenced papers:



What is the resistance of (c) at point Vx looking into the mixer?

The last paper derived something that is independent of flo at flo! The capacitor C just tells something about the bandwidth around flo.


The other papers argue (for R=0) that the resistance is given by 1/(flo*C) (or a more complex variation of it). This is dependent on flo.


Now one could argue that the first one described the case of a resistive load whereas the second one as capacitive.


However, also the first case has a significant capacitance in parallel which does not change the resistance at flo but only changes impedance shape around it (bandwidth).

Specifically what part of the Andrews paper are you referring to? It's likely that they're assuming your IF frequency is constant, meaning that FLO and Frf are changing together.

Diode mixers are passive mixers, btw.

If a switch has an ESR of Rs and its equivalent switched resistance Req = 2x at 50% where 50% d.f. means 50%d.f

and f= 1/(τoff + τon)
or d.f. =(τon) * f

Req= . Rs * τon*f

Thus Req*C= Rs* τon *f * C

hmm I did something wrong. ( grey cells evaporating)

I suppose if load RL>> Rs it becomes a C divider ratio * d.f.

When I designed S&H mixers , they performed as flawless as the incoming signal. In my case a square to sawtooth 100kHz signal was used for counting Doppler range and the ratio of rise to fall time >10us/10ns=1000:1) was maintained on the mixer output at DC. or shifting with velocity with an imperfect sawtooth 1000:1. The gate sample was a 1 shot of 10ns triggered by an analog demod signal zero-crossing, so no sample when loss of signal and it would hold output with low drift.