Razavi - RF Microelectronics的笔记 - Current-Driven Passive Mixer

On page 367, while discussing about current-driven passive mixer, there is this saying:

the switches in Fig. 6.39(b) also mix the baseband waveforms with the LO, delivering the upconverted voltages to node A.

image

I am quite confused on where does this waveform come from. I seems like the output (i.e V1V2) "reflects" back on input (i.e the source current).

I checked the reference [5], which doesn't explain the problem well neither.

image
image

According to reference [5],

at the RF side, the baseband voltage appears as +vBB(t)/2 or vBB(t)/2, depending on which switch is ON.

image

In Fig.2.(a), vBB(t) is the voltage between a differential pair. I agree on the fact that vBB(t) oscillate between positive and negative values, due to commutating of switches. On the other hand, vRF is a single end voltage, and it's continuous in time, because on-time of each switch are complementary and covers the entire time exclusively.

With this doubt, I turn to simulation for better understanding. This time I use Simulink.

image
image

In this simulation, ωRF=550MHz,ωLO=500MHz. Therefore, we can find a down-converted shape at 50MHz corresponds to the voltage of differential pair on the baseband side. To make the shape of ZBB(f) more distinguishable, I model the load as RC combination in series. So the shape of the down-converted signal is defined by ZBB(s)=R+1sC

The input source current has a clean spectrum peaks at 550MHz.

There exists a up-converted ZBB(f) shaped spectrum at ωRF, along with a image around ωLO at 450MHz, which corresponds to the single-ended voltage at input node.

Now the doubt becomes clear. The explanation is quite straightforward. The voltage of the current source is the product of current it supplies and load impedance. The central frequency of the current is ωRF, so the resultant voltage is actually the spectra of load impedance up-converted to ωRF

posted @   湿毛巾堵鼻孔  阅读(23)  评论(0编辑  收藏  举报
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