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Message started by aaron_do on Aug 21^st, 2008, 9:12pm

Title: passive mixer linearity
Post by aaron_do on Aug 21^st, 2008, 9:12pm

Hi all,

generally for a passive mixer, we can improve the conversion gain and to a certain degree the noise performance by lowering the duty cycle. THis is generally done by reducing the DC gate bias of the switching FETs.

Supposedly this leads to poorer linearity as the high order LO harmonics will be more prevalent. However, IMO it seems that the harmonic generation of the input signal should be smaller since when the DC gate bias of the FET is reduced, the device is ON for a shorter time and hence covers a smaller operation region.

Any thoughts on this?

Title: Re: passive mixer linearity
Post by nxing on Aug 26^th, 2008, 5:27pm

Hi aaron_do,

Can you tell some details why reduce the cycle can actually increase the gain? actually, I am thinking that if the duty cycle is reduce, the gain will be less than 2/pi.

Regards,

nxing

Title: Re: passive mixer linearity
Post by aaron_do on Aug 27^th, 2008, 2:18am

Sure. You can find this in several papers on passive mixers btw.

The switch is a variable resistor. Its resistance can be expressed in a fourier series, so it has a DC component and an AC component.

Supposed a voltage is applied to the input at RF. This voltage will be converted into an IF signal by the first order component of the resistance. It will then reach the output and the load will be the zeroth order component. So overall you will get an equation which has the term G₁/G₀, where G₁ is the first order component of the conductance and G₀ is the zeroth order component. G₁/G₀ is dependent on the duty cycle, where a smaller duty cycle results in a high G₁/G₀. Note that G is zero when the transistor is off. Therefore a longer off time results in a smaller G₀.

The gain can reach a maximum value of 1 (0 dB). You will find a similar result in "A 12-mW Wide Dynamic Range CMOS Front-End for a Portable GPS Receiver" by Arvin R. Shahani, Derek K. Shaeffer, and Thomas H. Lee.

If you actually try and write out the equations you will find it a little more difficult than what i've tried to explain.

cheers,
Aaron

Title: Re: passive mixer linearity
Post by RFICDUDE on Aug 27^th, 2008, 6:58pm

You asked for thoughts ..

It seems as the duty cycle reduces the mixer is approaching an ideal impulse sampler as opposed to the natural sampling you get with 50% duty cycle. This explains the increased LO harmonic content as the switching aliases approach unity in respect to the input for ideal impulse sampling.

I suspect, but don't know for sure, that the degraded linearity originates from distortion during the rise/fall times of the switches. This becomes a greater percentage of the overall signal passed through the mixer, so the significance of the rise/fall and settling time impacts the desired signal more.
Just a guess/thought.

Title: Re: passive mixer linearity
Post by aaron_do on Aug 27^th, 2008, 7:39pm

thanks. I hadn't even considered that at all...

Aaron

Title: Re: passive mixer linearity
Post by Poojan Wagh on Aug 28^th, 2008, 5:17pm

aaron_do wrote on Aug 27^th, 2008, 2:18am:

If you're doing a double-balanced switch-mode mixer, the usual analysis assumes you are multiplying with a +/- 1 square/rectangular signal. The first order harmonic of this pulse signal is given by:

2/pi * sin(pi * d)

where d is the duty ratio. The maximum conversion gain is given by d=50% (conversion gain of 2/pi, or -4 dB).

Title: Re: passive mixer linearity
Post by aaron_do on Aug 28^th, 2008, 7:36pm

Hi,

thanks. Actually i am aware of that analysis too. Perhaps you could read the paper I mentioned (very good passive mixer analysis) in the previous post or maybe run some simulations with changing bias point. Personally, my own simulations agree that the maximum conversion gain is 1.

The paper I mentioned also concluded that a sine-wave drive gives a better conversion gain than a square-wave drive...

cheers,
Aaron

Title: Re: passive mixer linearity
Post by kelly on Jan 20^th, 2011, 10:05pm

Hi Aaron,

I looked up the paper by TLee as you mentioned. I didn't see the analysis, Perhaps, I missed it?

A lot of the passive mixer paper analysis is for the I/Q mixers (by having non-overlapping clocks, you actually don't lose current or no two voltages are on at the same). But I assume the same principle apply even if you just have a single voltage mode passive mixer, right?

My problem is, I actually see this in the simulation (I have only one mixer in voltage mode), but I can't explain it.

Could you explain your G1/Go ananlysis again?

Thanks much.

Title: Re: passive mixer linearity
Post by aaron_do on Jan 21^st, 2011, 5:06am

Hi Kelly,

I'm pretty sure there's an analysis there. Anyhow, if you're interested in my own analysis, then you could read the following paper:

A. V. Do et. Al, "An Energy-Aware CMOS Receiver Front End for Low-Power 2.4-GHz Applications", IEEE Transactions on Circuits and Systems I, vol. 57, no. 10, Oct 2010, pp. 2675-2684.

Read section IVB and the Appendix. You could also refer to a book (I can't exactly remember the title) by Stephen A. Mass about nonlinear microwave design.

cheers,
Aaron

Title: Re: passive mixer linearity
Post by kelly on Jan 23^rd, 2011, 2:18pm

got it, thanks. I must have gotten the pre-journal (short paper) version of it the 1st time I searched (the text was only 1 page).

I guess maybe another way to see why IIP3 degrades is that the mixer has a higher average impedance for the 25% LO duty cycle vs the 50% case.

Thanks.

Title: Re: passive mixer linearity
Post by kelly on Feb 3^rd, 2011, 11:55pm

Hi All,
I have a stupid question, won't the 25% duty cycle causes a huge RF feedthtrough? Am I not thinking this right?

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 5^th, 2011, 11:15am

Not if the architecture remains double balanced.

Title: Re: passive mixer linearity
Post by kelly on Feb 6^th, 2011, 1:53am

oh, yeah, thanks. I forgot that the 25% duty cycle is non-overlapping, so the DC is still 0. Which is different from the normal duty cycle distortion case.

Title: Re: passive mixer linearity
Post by kelly on Feb 8^th, 2011, 2:18pm

Hi,

I am looking at the RF feedthrough as a function of the DCD on LO. It seems that it's very sensitive (-35dBc for a 49.5% or55.05% DCD). Does anyone see the same thing? By the way, this is just a single mixer not a IQ modulator.

Thanks.
Kelly

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 9^th, 2011, 5:25pm

Is the single mixer operating from a 1/4 duty cycle LO?
If so, is there a RF LC tank circuit on the RF side?
What is the load on the baseband side?

Title: Re: passive mixer linearity
Post by kelly on Feb 10^th, 2011, 10:42am

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.

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 10^th, 2011, 5:55pm

Hmmm, if it is double balanced and 50% duty cycle LO then I'm not sure why there is excessive LO leakage.

By "DCD" do you mean duty cycle or is this some reference to DC gate bias relative to the LO swing (I'm not exactly sure what you mean by "DCD on the LO")?

Title: Re: passive mixer linearity
Post by kelly on Feb 10^th, 2011, 6:36pm

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.

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 16^th, 2011, 3:24am

I understand now.

If the fundamental of the LO is balanced, then you could AC couple the LO to the gates of the mixer devices and bring in a common mode gate bias voltage through high valued resistors. This should prevent DC offset on the LO drive from creating LO feedthrough.

You may still have some Vth mismatch, but it should be much less than 9mV of error.

Title: Re: passive mixer linearity
Post by kelly on Feb 16^th, 2011, 10:52pm

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.

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 19^th, 2011, 5:05am

For your results above, are the LO signals AC or DC coupled to the gates?

Title: Re: passive mixer linearity
Post by kelly on Feb 19^th, 2011, 5:09pm

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

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 20^th, 2011, 8:32am

Ok, then just AC couple LO and bring in a common mode bias. It should, at least, eliminate switching errors due to LO DCD.

Have you attempted to AC couple the LO?

Reverse mixing is a characteristic of passive mixers. I suppose it is a concern if there are emissions requirements at the mixer input.

Title: Re: passive mixer linearity
Post by kelly on Feb 20^th, 2011, 3:09pm

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.

Title: Re: passive mixer linearity
Post by RFICDUDE on Feb 20^th, 2011, 7:38pm

You may benefit a lot from carrying out the analysis yourself.

The LO signal with DCD has some DC imbalance because of the DCD, but if you AC couple the LO then the DC imbalance is not passed on to the gates. Thus the LO DCD contribution to carrier feedthrough should be eliminated (i.e. you should be able to increase DCD without seeing a strong sensitivity to LO feedthrough).

Title: Re: passive mixer linearity
Post by kelly on Feb 21^st, 2011, 1:10am

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.

Title: Re: passive mixer linearity
Post by aaron_do on Feb 21^st, 2011, 5:34am

Hi Kelly,

can I confirm a few things with you?

1) this is a double-balanced design.

2) you have a square wave drive.

3) by 0.5% DCD, you mean that the positive LO is driven with 50.5% duty cycle, and the negative LO is driven with 49.5% duty cycle (or vice-versa).

In a double-balanced design, the DCD on the LO would cause a common-mode LO leakage to the RF ports, so I'm not sure if it would propagate to the IF ports (as a differential signal). In a single-balanced design, I would definitely expect to see all the tones that you are seeing with any DCD.

Also, I'm not so sure if your use of a DC offset to represent the DCD is correct. If it is a square wave, then the switches will take on only two states: ON or OFF. A 9 mV DC signal is not likely to have that much effect on whether a switch is ON or OFF (when it is driven by a large square wave). However, a change in the duty cycle directly affects how long the switch is ON/OFF.

I'm sorry if I have misunderstood your circuit. Perhaps it would be more clear with a schematic.

cheers,
Aaron

Title: Re: passive mixer linearity
Post by kelly on Feb 21^st, 2011, 10:09am

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.

Title: Re: passive mixer linearity
Post by kelly on Feb 21^st, 2011, 10:12am

sorry when I say IF output I mean IFP-IFN, so it's a diff output

Title: Re: passive mixer linearity
Post by vp1953 on Feb 21^st, 2011, 12:29pm

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.

Title: Re: passive mixer linearity
Post by kelly on Feb 21^st, 2011, 1:04pm

Hi VP1953,

OK, good. I guess I am not going crazy.
But I have a question then, all the 2nLO+RF at the IF outputs seem to track the even harmincs on the LO, says around -36dBc with DCD of 49.5 or 50.5%. Isn't that pretty big? I assume you can never have perfect 50% duty cycle (due to mismatch and differernt rise and fall times). 50.5% seems pretty good to me, how will one take out this sensitivity then?
Thanks.

Title: Re: passive mixer linearity
Post by aaron_do on Feb 21^st, 2011, 7:47pm

Hi Kelly,

I assume your dBc is with respect to the RF signal. In that case -36 dBc should be ok since it means you can achieve an SNR of 36 dB. However, as you say, it depends on whether you can actually get that kind of DCD.

Just one last check which I guess you are doing correctly anyway, but you are checking your IF differentially right? Another thing, I believe you will get better cancellation using a quadrature design.

regards,
Aaron

Title: Re: passive mixer linearity
Post by kelly on Feb 22^nd, 2011, 10:52pm

Hi Aaron,

Yes, IF is differential. Thanks (everyone) for all your help.

Kelly

Title: Re: passive mixer linearity
Post by cherry girl on Jul 29^th, 2011, 2:44am

Hi kelly,

As you mentioned , you IP3 curves seem a little weird below 0 dBm.I have encoutered the same problem before, I think maybe two reasons causeing this situation. At first, it is possible that you mixer's noise performance is too bad , so in the low level of input power , the desired sigal is easily distoried by nois;For another reason ,you simulation is not accurate,you needn't use QPSS , it seems take too long time to run one time,just take harmonic balance (hb)with two tones,just remember the harmonic numbers should be large enough to ensure your simulation accuracy.