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IIP3 measurement (Read 13611 times)
aaron_do
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IIP3 measurement
Sep 01st, 2010, 11:36pm
 
Hi all,


suppose I wish to measure the IIP3 of a circuit, and I have two possible options:

1) I could use a vector signal generator which generates both tones from a single output. I'm not sure how the two tones are generated, but each tone is spread in frequency such that the signal looks much like a wideband FSK signal.

2) I could use two separate signal generators and combine the signals using a coupler.

Should I expect any significant difference in the measurements from the two methods?

I ended up using method 1, and I measured the entire signal bandwidths to get the IIP3 value. However, I found that the measured IIP3 was quite a bit worse (10 dB) than what I got in simulation. The design consists of an LNA, IQ mixers with zero-IF, and a pair of low-pass filters. My next question is, is such a huge discrepancy normal? Could it be poor device modeling? My PDK uses the BSIM3 model. For the simulations, I did not simulate the IIP3 of the LNA and mixers, just the low-pass filters (I simulated the LNA IIP3 separately). I estimated the overall IIP3 by subtracting the LNA+mixer gain from the filter IIP3.


thanks,
Aaron
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vp1953
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Re: IIP3 measurement
Reply #1 - Sep 2nd, 2010, 3:59pm
 
Hi Aaron,

In your method one, can you clarify what the signal looks like spectrally. When you say the signal looks like a wideband FSK signal, is there only one tone at any given instant of time?

For IIP3, is your MixerIP3 - LNAgain << FilterIp3 - Mixergain-LNAgain? If this is the case then you can use the overall IP3 calculation in the manner you have calculated.

However this is still an approximation and the actual numbers need to be simulated (and not very difficult to do so considering that you already have most of what is needed).
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aaron_do
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Re: IIP3 measurement
Reply #2 - Sep 2nd, 2010, 5:43pm
 
Hi,


thanks for the reply.

Quote:
is there only one tone at any given instant of time?


I figured that somebody would ask that. Actually I didn't check that, but I'm curious if it would make a difference. Something to think about.

Quote:
For IIP3, is your MixerIP3 - LNAgain << FilterIp3 - Mixergain-LNAgain?


I guess you mean it the other way around right? i.e. the mixer IIP3 doesn't affect the calculation. Unfortunately, I'm just not sure how accurate such a simulation would be anyway, since I'm using a passive mixer with the BSIM3 model. My understanding, however, is that a passive mixer should be able to achieve pretty good linearity.

I'm hoping for some advice on how accurate these simulations are in terms of linearity. Also, what is the preferred method for measuring IIP3?


thanks,
Aaron
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vp1953
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Re: IIP3 measurement
Reply #3 - Sep 3rd, 2010, 2:44pm
 
Hi Aaron,

Quote:
I figured that somebody would ask that. Actually I didn't check that, but I'm curious if it would make a difference. Something to think about.


I think it would make a big difference.

Quote:
I guess you mean it the other way around right?


Yes you are right, it should be the other way round. If it is a passive mixer then it has negative gain (in db). If Gain of (mixer +lna) is low then even the LNA IP3 has an effect. Without knowing actual IP3 values and gains, it is possible that the formula could be a little bit off.

If it is not possible to do an IP3 simulation, how about a p1db measurement and then get IIP3 from that. In Cadence spectre, this is simple simulation (power sweep in PSS). It is easy to do this with a single tone measurement. IIP3 = P1db + 10db (also an approximation)
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Re: IIP3 measurement
Reply #4 - Sep 3rd, 2010, 2:59pm
 
Quote:
2) I could use two separate signal generators and combine the signals using a coupler.


Two separate signals combined using a power splitter/combiner is commonly done. But i think some of the newer equipment can generate two tones simultaneously, so maybe you dont need to use two separate sources.
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pancho_hideboo
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Re: IIP3 measurement
Reply #5 - Sep 3rd, 2010, 9:03pm
 
aaron_do wrote on Sep 1st, 2010, 11:36pm:
1) I could use a vector signal generator which generates both tones from a single output.
Yes, you can do it very easily.
aaron_do wrote on Sep 1st, 2010, 11:36pm:
I'm not sure how the two tones are generated,

vp1953 wrote on Sep 3rd, 2010, 2:59pm:
But i think some of the newer equipment can generate two tones simultaneously,
Almost all signal generators can generate two tones as DSB-AM with suppressed carrier, as far as tone separation is within 100kHz.

But high performance signal generators which have wide-modulation ability are required for generation of wide separated two tones.

aaron_do wrote on Sep 1st, 2010, 11:36pm:
but each tone is spread in frequency such that the signal looks much like a wideband FSK signal.
Don't use FSK.
IP3 is due to AM/AM characteristics.
However FSK signal is a constant envelope signal.
See http://www.designers-guide.org/Forum/YaBB.pl?num=1190971685/9#9

aaron_do wrote on Sep 1st, 2010, 11:36pm:
2) I could use two separate signal generators and combine the signals using a coupler.

Should I expect any significant difference in the measurements from the two methods?
Currently separtion of two tones is at most MHz order even if high performance signal generators are used as method-(1).
And suppression of carrier could be not good.
Merit of method-(1) is that we don't have to adjust amplitude level of two tones.

aaron_do wrote on Sep 1st, 2010, 11:36pm:
For the simulations, I did not simulate the IIP3 of the LNA and mixers,
just the low-pass filters (I simulated the LNA IIP3 separately).
I estimated the overall IIP3 by subtracting the LNA+mixer gain from the filter IIP3.
Your estimation of IIP3 may be not appropriate.

Study signal level budget for total cascaded system.
Here you have to take care of filtering characteristics of each block as well as IM3 of each block.
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« Last Edit: Sep 4th, 2010, 10:28am by pancho_hideboo »  
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aaron_do
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Re: IIP3 measurement
Reply #6 - Sep 5th, 2010, 4:08am
 
Quote:
But high performance signal generators which have wide-modulation ability are required for generation of wide separated two tones.


that sounds like what i'm using.

Quote:
Don't use FSK.
IP3 is due to AM/AM characteristics.
However FSK signal is a constant envelope signal.


thanks!

Quote:
Study signal level budget for total cascaded system.
Here you have to take care of filtering characteristics of each block as well as IM3 of each block.


I took into account the channel filter's filtering ability. But I guess the front-end filtering has no affect on the overall IIP3.


thanks,
Aaron
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pancho_hideboo
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Re: IIP3 measurement
Reply #7 - Sep 5th, 2010, 4:42am
 
aaron_do wrote on Sep 5th, 2010, 4:08am:
Quote:
But high performance signal generators which have wide-modulation ability are required for generation of wide separated two tones.
that sounds like what i'm using.
If so, why can't you generate two tones ?

aaron_do wrote on Sep 5th, 2010, 4:08am:
I took into account the channel filter's filtering ability. But I guess the front-end filtering has no affect on the overall IIP3.
Show me input frequency, output frequency and cutoff frequency of LPF for measuring IIP3.

RF1? RF2? IF1? IF2? IM3_Upper? IM3_Lower?
Here IF1, IF2, IM3_Upper and IM3_Lower are all not zero frequency even if your receiver's architecture is zero-IF.

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aaron_do
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Re: IIP3 measurement
Reply #8 - Sep 5th, 2010, 8:46am
 
Hi pancho_hideboo,


I am using a zero-IF architecture where the low-pass filter -3dB frequency is 1 MHz. Channel spacing is equal to 5 MHz, and I tested the circuit with one tone in the adjacent channel and one tone in the alternate channel of equal power.

I set the following parameters

LO frequency = 2.45 GHz
First RF frequency = 2.455 GHz
Second RF frequency = 2.4605 GHz

the center frequency of the 3rd order intermodulation signal falls at 500 kHz. So then I took readings of the output intermodulation signal power versus the input power, and selected a point where the slope of output versus input (in dBm) was approximately 3. From the formula,

output power = 3* input power + B

where both power levels are in dBm, I calculated B in dB. I tested the desired gain separately by setting,

LO frequency = 2.45 GHz
RF frequency = 2.4505 GHz

and took the gain as the output power versus the input power. i.e.

output power = input power + A

again A is in dB. So I took the IIP3 as,

0.5 * (A - B) - loss_compensation - 3 dB

The loss_compensation is all of the loss before the DUT, and the 3 dB is because the signal generator generated each tone a -3dB lower than the total power. This is basically the same method I used in my simulations except that for the filter IIP3, I just used two tones at 5 MHz and 10.5 MHz with no LO. I"d appreciate if you would let me know if you think i've done anything wrong here.


thanks,
Aaron




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Re: IIP3 measurement
Reply #9 - Sep 7th, 2010, 4:26pm
 
Hi Aaron,

Quote:
I am using a zero-IF architecture where the low-pass filter -3dB frequency is 1 MHz. Channel spacing is equal to 5 MHz, and I tested the circuit with one tone in the adjacent channel and one tone in the alternate channel of equal power.


I think your RF1, RF2, IM3_upper and IM3_lower should be such that they are all within the same channel and have nearly same gain. Otherwise it gets pretty complicated as to what to do.

At the filter output, plot the fundamental (RF1 after downconversion) vs input RF1 input power (after compensation etc). Plot either IM_upper or IM3_lower also vs. RF1 input power. The point where the two curves meet by extrapolation or by actual measurement is the IIP3.

I could be wrong but I dont think the intermod products go as 3x input power as you have in your formula below. Not sure how you get the 0.5(A-B)
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aaron_do
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Re: IIP3 measurement
Reply #10 - Sep 7th, 2010, 5:55pm
 
Hi,


sorry if I was unclear about my formulae.

Quote:
I think your RF1, RF2, IM3_upper and IM3_lower should be such that they are all within the same channel and have nearly same gain. Otherwise it gets pretty complicated as to what to do.


The reason I ran the simulations/measurements the way I did, was to emulate actual worst case conditions. From my understanding, the system's modulation scheme is constant envelope, and so the actual direct distortion due to nonlinearity is unimportant. What is important is how the interfering signals intermodulate and create undesired products in the desired channel.

If you take the input output relationship as a taylor series, you get something like y = ax + cx2 + bx3 + ...
The IM3 comes from the bx3 component which in dB is 10logB + 30logx which is why the IM3 goes varies as 3*Pin. If I let y1 = 10loga + 10logx = A + Pin and y3 = 10logb + 30logx = B + 3*Pin, then the IIP3 is the point where y1 and y3 intersect, or Pin = 0.5*(A-B).

Quote:
At the filter output, plot the fundamental (RF1 after downconversion) vs input RF1 input power (after compensation etc). Plot either IM_upper or IM3_lower also vs. RF1 input power. The point where the two curves meet by extrapolation or by actual measurement is the IIP3.


I think this method is fine as long as all tones are within the bandwidth of the system. I think if you want to do this in other cases, you have to correct for the filtering.


cheers,
Aaron
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Re: IIP3 measurement
Reply #11 - Sep 7th, 2010, 9:37pm
 
HI Aaron,

Quote:
If you take the input output relationship as a taylor series, you get something like y = ax + cx2 + bx3 + ...


The expansion for Taylor series assumes that the coefficients a,b,c for RF1 and RF2 are same (after downconversion). But when RF1 gets downconverted to in channel and RF2 gets dowconverted to out of channel, then the coefficients are completely different for each signal; very clearly "a" is different for each (if I were to represent each signal separately by a Taylor series)

Now if you were to say that x = signal 1 +attentuating_factor*signal 2, in this case x already represents the output signal and cannot be used again as an input in the Taylor series.
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aaron_do
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Re: IIP3 measurement
Reply #12 - Sep 8th, 2010, 1:35am
 
Hi,


Quote:
I could be wrong but I dont think the intermod products go as 3x input power as you have in your formula below.


my taylor series explanation was simply in response to the above.

Quote:
The expansion for Taylor series assumes that the coefficients a,b,c for RF1 and RF2 are same (after downconversion). But when RF1 gets downconverted to in channel and RF2 gets dowconverted to out of channel, then the coefficients are completely different for each signal


Are you saying that due to the coupling between the filtering and the nonlinearity, the IM3 versus input will not be a cubic relationship outside of the channel bandwidth? If so, then I think you have a good point, but I believe this applies more to strongly nonlinear circuits such as power amplifiers.

From my simulations, the cubic relationship was a reasonable assumption, but I used a PSS+PAC analysis, so one of the tones was treated as a small signal. The measured relationship was approximately cubic, but it was hard to say for sure because the input tones must be fairly large in order for the IM3 component to exceed my circuit's noise level. At such high levels, the relationship may have already strayed from a cubic one.

Regardless, if you take both interfering signals in-band, then you must apply some kind of correction later on (because in reality they are not), and I believe this would be even more inaccurate than my method.


thanks,
Aaron
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Re: IIP3 measurement
Reply #13 - Sep 9th, 2010, 1:26am
 
Hi vp1953,


I thought about what you said, and I went back and re-simulated my circuit using a QPSS analysis with one large tone and one moderate tone. Previously I had been using a PSS+PAC analysis with only one large tone. This time the simulated IIP3 was very close to the measured IIP3. Perhaps the reason was related to what you said about the filtering+nonlinearity not being independent. I need to do some more simulations to make sure...


Thanks for the help,
Aaron

EDIT: Ok simulation results using PSS with two large tones agrees with QPSS as well as measurement. PSS+PAC with one large tone and a small signal does not agree with the others. Perhaps somebody could shed some light on this...maybe i'll post it as a separate question.
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