hi,

I am designing a double balanced CMOS passive mixer.
Input to the mixer is a LCR tank circuit ( i.e. LNA load ).
Output of mixer is a 100f F cap .
I am using a sinusoidal LO drive.

My problem is this:

how should i get the input impedance ( equivalent capacitance  in parallel with equivalent resistance) offered by the mixer?
I need the equivalent capacitance  to choose the inductance value at the LNA output to resonate at the desired frequency.

I am asking this because, the input impedance will be time varying with a period of  one half of the  LO period.

I need help in this regard.

regards,
Prabhu

I agree with TT.
A small signal sweep (ac or sp) centred on the LO switching point will give you a good enough idea of the value of the capacitance to add, as non-linear large signal changes in the capacitance tend (in most cases!) to cancel over a cycle.
You can confirm the small signal load capacitance you derive for loading your LNA (i.e mixer Cin) is correct by comparing the ac gain reponse when resonating with your inductor against a frequency swept transient large signal simulation. In the past I have often found they match quite well.

By having a direct connection MixerLNA, you probably have a too high impedance to really drive the mixer
switches in voltage mode but rather current mode.

As far as I can imagine your design, it seems to me that the LNA output impedance is probably to high.
Before any conclusion: can you check by introducing a VCVS with gain = 1 if this effect diappear ?

TT

how exactly are you stimulating your circuit and in what simulation ?
are you using large signal (transient) for both RF and LO. What are the RF and LO freq are you using ? Is everything differental and balanced ?

anyway assuming there isn't a simple bobo somewhere i suggest you sweep the LO drive power to see the effect on the tone. You can repeat with the RF tone if required.

reducing your LO power by 1dB should give you a larger (ideally 3dB) reduction in any 3rd order distortion introduced by that stage.

Hi,

you have a balanced passive mixer right? If you downconvert frf by high side injection for example you will get flo-frf. But i believe because of the symmetry of the passive mixer (S21=S12) your flo-frf will get upconverted (from output to input) and appear at the mixer input port (remember your input port impedance is high). The upconverted signal will have the signal + an image: flo-(flo-frf)=frf and flo+(flo-frf)=2flo-frf which is close to your original frf. Naturally the higher the output impedance from the LNA, the higher the image signal...

The solution to this problem is to use a quadrature balanced passive mixer. That way the image when re-upconverted will be cancelled. Remember the receiver mixer is basically opposite to a transmitter mixer which inputs quadrature IF and outputs differential RF.

Hope that helps!

Aaron

I am using BSIM3 model for 0.18um technology.
I am using Spectre simulator to do QPSS analysis with LO being the large tone and RF being the medium tone.
fLO = 2.44 GHz, fRF = 2.45 GHz.

Circuit Setup:

Drain and Source of the 4 transistors are biased at 0.7 V through large biasing resistors (100k).
Gates are biased at 1.2 V .
LO drive is square / sine wave of ±0.5 Vp around 1.2 V.
Mixer input ( ac coupled ) is differentialy driven with a sine wave source ( 500 uVdifferential ) and 1600 ohms differential source resistance.
Mixer output has a load capacitance of 50 fF on either side.

Observations:

At the mixer input I get a comparable ( one third wrt RF ) tone at 2.43 GHz.
Transient analysis and DFT of the mixer input also gives similar result.
Circuit is fully differential and balanced.

I sweeped the LO amplitude and found that there is not much change in the amplitude of 2.43 GHz ( 2LO - RF ) signal at the mixer input.
So this means the problem is not due to any third order non-linearity but due to the switching action of the passive mixer itself.

As Aaron says,
Is it be possible for the downconverted signal at the mixer output to mix backwards and add to the actual mixer input signal?
How can that happen?
1) There is no active source at the mixer output.
2) Also there will  be no reflections since we are using lumped components in simulation.

Hi Prabhu,

I guess there's more than one way of looking at it...if you simply observe a transient analysis then the current will only be flowing in one direction since as you say there is no source at the output...however if you split it into frequency components and use the principal of superposition it is possible to observe currents flowing in both directions on the same wire in a nonlinear/time varying element.

I don't think your point on lumped elements and reflections is valid since even if you have assumed wave equations it doesn't physically mean current is flowing in both directions. Wave equations always apply and treating each element as a lumped element is a simplifying tool...just like a transmission line can be treated as a network of Ls, Rs and Cs.

cheers,
Aaron