Hi RFICDUDE,

It's 50% duty cycle and down conversion voltage mode mixer.  There is no LC tank on the RF side.  RF side  (matched to 50ohms with real resistors) goes off chip to an attenuator, balum, and a saw.  IF side goes to the gates of an on chip VGA.  I plan to have some caps (~5p) at the mixer output (VGA input) just to have some light filtering.

Hi RFICDUDE,

DCD means duty cycle distortion on the LO, sorry about not being clear about that.  

The way is see it is that the DCD of the LO waveform, produce a DC offset of the LO port, therefore, causes the RF feedthrough.  

With 0.5% DCD (49.5% duty cycle), I found that the DC of the differential LO is around 9mV.  So I put a DC offset of 10mV on one of the LO sides (with 50% duty cycle), the RF feedthrough is 30dB less that the 49.5% duty cycle with no offset applied.  I wouldn't expect that result.

Is the passive mixer somehow more sensitive to the DCD than just a pure DC offset on the LO?  Or I am looking at this wrong?

Thanks.

Hi RFICDude,

So I looked at the LO spectrum for 2 cases, LO offset =10mV and LO DCD =0.5%  

LO offset =10mV case (no DCD):
LO has a DC component of -40dBC, and odd harmonic LO (LO is a square wave)
RF feedthrough=-60dBc

LO DCD = 0.5% case (no addition offset):
LO has a DC component of -40dBC, now you have both even and odd harmonics of LO (ok, so I guess you get evens 'cause now it's not exactly 50% square wave?)
RF feedthrough=-40dBC (20dB bigger than jus the DC offset case)

So, do the even harmonics introduces addition offset that's why the RF feedthrough is much bigger in the DCD=0.5% case?


Thanks.

Hi RFICDUDE,

It's DC.  So the LO goes from 0 to 1V at the gate of the switching devices, and the source/drain are biased 200mV above ground.

When I say I get RF feedthrough, I don't just get RF, I also get what I thought is LO-IF (reverse mixing) and all the higher order ones.  Now I think what I get is actually all the 2nLO+/-RF, eg, LO-IF=LO-(RF-LO)=2LO-RF.  

I tried it with the active mixer (which shouldn't have the everse mixing term), but I get the same thing when DCD=0.5% is applied.

Kelly

Kelly

Hi RFICDUDE,

No, I haven't.  What switching error does AC couple eliminate? Is it just the DC, won't you still have a non-50% duty cycle after the AC couple?

Thanks.

hi RFICDUDE,

So I AC coupled the LO signal.  If I looked at the dft of LO before and after the AC, the DC component decrased by >30dB.  But the even harmonics are still present in both signals.  So as a result, I still get all the 2nLO+/-RF components at the IF outputs including RF for n=0.  

Do you expect the duty cycle distortion causes the even harmonics on the LO ?

Thanks.

Hi Aaron,

Yes, you understand the circuit correctly.  What you and RFICDUDE describing is what I expect to see but I don't from my simulations.  Maybe I am doing something wrong, just can't figure it out.  To answer your question,
1.  yes. I use vcvs to do the single to diff on both LO and RF.  This is just a single mixer,  no I/Q path.
2. yes.
3. yes.  specifically, I use a Vpulse source, the pulse width is 1/flo/2.02-rtime ( flo=lo frequency, rtime= rise time=fall time) for DCD=49.5.  It's 2 for 50% duty cycle.  I then use the two vcvs to do the single to diff for both LO ports.

I agreed with what you said about the LO/RF leakage.

I am not using DC offset to model DCD.  In fact, if I just use the DC offset=10mv , the RF feedthrough is not that much (as you pointed using the square wave) and there is no 2nLO+/-RF at the IF output.
 
I use the vpulse decribed above to model DCD and I see both DC and even harmonics tones on LO.  What I don't understand is why I am seeing all the 2nLO+/-RF on the IF output when I model the DCD the way I did.  

Thanks.

Hi Kelly,

If i got this right, you have a double balanced mixer with complementary LO signals(call it LO+, LO-) driving the LO ports. When the duty cycle of LO+ and LO- are changed say to 50.5 and 49.5, you are seeing 2n.LO+RF?

I would absolutely expect to see even harmonics of LO for the above scenario; when LO+ and LO- are perfectly complementary, the DC and the even harmonics get cancelled out. When there is slight difference in duty cycles, there is incomplete cancellation of even harmonics and a direct RF feedthrough.