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The answers in this article are spontaneously discussed and replied by RF practitioners, and there are inevitably errors and omissions in the answers, and everyone is welcome to criticize and correct.
01 Discussion on the deterioration of AM-PM during PA multi-cascade imitation
Q: Have you ever encountered a single imitation PA, the AM-PM performance is quite good, but when you take the pre-stage driver stage, and then look at the AM-PM input and output of the PA module (not cascaded whole), the performance deteriorates a lot?
A: The pre-driver psat output? How do you set the signal in front of you when you imitate PA?
Q: When the pre-stage outputs Psat, from the corresponding power of the single imitation PA, the AM-PM of the PA should be quite good. Excitation source set with port.
A: It may be the harmonic influence of the pre-stage.
Q: Oh! Indeed, remind me, thanks, I'll take another look!
02 Discussion on noise test calibration
Q: Why does de-embedding only remove half of the line loss when testing noise?
A: Because the output cable loss has little effect.
A: The insertion loss of the passive device in front of the amplifier directly increases the noise figure. The noise probe is your calibration surface. Later insertion loss has little effect on noise figure.
Q: Okay, thank you.
03 Discussion of Class B PA bias voltage and current
Q: There is a problem with the IV curve of class B in this article "High Efficiency PA Design: From Class A to Class J": Class B, that is, when the drain current is 0, why is the drain voltage not zero? And when the drain current is distorted, why is the drain voltage not distorted?
For AC signals, isn't that RF Choke inductor separated from the power supply? So the drain current is 0, is the small signal model loop current also 0?
A: You can study the book of cripps, which talks about the theory of power amplifier waveforms, and there is a hypothesis in waveform analysis: a harmonic short in output termination, that is, the output current Ids contains Idc, fundamental, harmonics, but the output matching is short circuit to all harmonics, so when Ids reaches the load end, only the fundamental current will generate voltage, and the voltage Vds is always a sine wave, and the peak-to-peak is 2Vdc.
This is true for both classa AB b c waveform theory analysis.
A: "For AC signals, isn't that RF Choke inductor separated from the power supply?" So the drain current is 0, is the small signal model loop current also 0? I didn't really understand what this meant.
Q: The voltage vds is always a sine wave with a peak-to-peak of 2VDc, which is incomprehensible. The rationality of this assumption is not understood.
A: Transistors are not resistors, why should the current be 0 if the current is zero.
Q: When the tube is closed, the drain current is 0. It stands to reason that the loop between the tube and the load is broken, why is the drain voltage still changing.
A: Simply understood, you close the faucet, the water flow is gone, but the water pressure in the faucet is still there, and your drain always has voltage, which has nothing to do with whether you clip it off or open.
Q: But this drain voltage should be fixed, why does it change?
The tube is closed, the channel is disconnected, and the positive fundamental current and negative harmonic current can still flow? This is a bit of a physical off-the-box, right?
A: Physically it is the total current in a time domain, and you expand it to get the fundamental frequency component and harmonics.
You said that this harmonic current flow does not flow or stands in the perspective of the time domain, but in fact, it is already in the frequency domain after unfolding.
A: If you look at the output characteristic curve, you will find that ideally, when the voltage is 0, the current is the maximum.
Q: The main thing is that the channel is clipped off, and neither holes nor electrons can pass through, so you can't violate this premise.
A: Also, is the gain you changed the gain inside the transceiver? What level is it?
Q: "If you look at the output characteristic curve, you will find that ideally, when the voltage is 0, the current is the maximum" I understand this IV curve (there is current), but I can't understand the IV curve when the current is 0.
A: Does it just mean that the phase difference between voltage and current is 180 degrees?
Q: I see that the diagram on the book of CRIPPS is also drawn like this, the current is distorted, but the voltage is a complete sine wave; So it should be right, but I don't understand. The diagram of the CRIPPS god, even the peak drain current is distorted, but the voltage is still complete.
A: The first image is not distorted, it is because of Vknee.
Q: Eun, I used the wrong picture; But my doubts remain;
A: The voltage is sinusoidal because it assumes that all harmonics are filtered, and the current waveform is only related to the bias, under normal circumstances, if the harmonics are not filtered out, the voltage is also a half sinusoid.
Q: Yes.
A: This first is current voltage inversion, if there is no harmonic network, you fold the current up is the voltage waveform, and then with the harmonic network filtered out, it becomes the above one.
You're just because the current and voltage have an inverting relationship
Q: Makes sense, is that the reason for this?
A: This should be a harmonic control network with a harmonic impedance of 0, and at the high q tank of the fundamental wave, the other harmonic impedances are 0
A: cripps are all default to no voltage harmonics, only the fundamental wave, the fundamental voltage and the current invert, the current waveform is only related to your bias state, so that's it.
Q: It is this harmonic control network that turns the drain voltage of normally half a sine wave into a complete sinusoidal fundamental wave.
Q: Thank you, sure enough, there are many gods in the group! I figured this out before going to bed, thank you so much for getting a good night's sleep.
04 Discussion of the memory effect
Q: Guys, two-tone test, why does the signal amplitude of the same order not equal represents the memory effect?
A: cripps that book has said, the essence is AM-PM distortion, Huizhiwei has an article about this "5G PA "memory effect" phenomenon, formation and elimination" The original text please search on WeChat public account.
Q: Thank you.
05 Discussion of the relationship between sensitivity and thermal noise
Q: Please ask a question, according to the sensitivity formula 100M, the best sensitivity can only be -94dB, because the thermal noise is already -94dB. Why can the actual test EIS go up to -108dB? Can sensitivity still be less than thermal noise? The feeling is related to the gain in the maximum direction of the antenna, but I don't know how to explain it. The antenna gain is 20dB.
A: Spread spectrum Take a look.
WCDMA is often in a noisy floor communication situation.
A: That 20dB spread spectrum gain is also too powerful
Q: Is spread spectrum gain an algorithm gain?
A: CDMA spread spectrum gain, the signal has been under the noise floor, the sensitivity is smaller than the thermal noise is normal, the smaller the string added after -174, the higher the sensitivity, -140~-150 receivers are also very common.
Q: How to understand it intuitively? How can signals that are lower than thermal noise be distinguished?
A: Didn't the big guys say, go to see the spread spectrum communication, there is a spread spectrum gain, and it comes out directly.
A: "How to understand it intuitively?" How can signals that are lower than thermal noise be distinguished? "Shannon's formula, in information transmission, increasing the signal bandwidth can lower the signal-to-noise ratio threshold.
Q: Thank you!
06 Discussion on intermodulation and intermodulation
Q: What is the difference between intermodulation and intermodulation?
A: Intermodulation: When weak signals and strong interference pass through a nonlinear system at the same time, interference interferes with the modulation of the signal.
Intermodulation: Two interfering signals with frequencies of w1 and w2 act on the nonlinear system, and the output will produce harmonic components.
Q: Can it be understood that intermodulation produces changes in amplitude and intermodulation produces changes in frequency?
A: I think so.
A: According to Radio Frequency Integrated Circuits and Systems, intermodulation is a two-tone signal input system with the same amplitude, while intermodulation is the modulation of interference signals onto useful signals.
Content in the course
Q: Thank you.
07 Discussion of PA current derivation
Q: Excuse me, when I read page P41 of the cripps book, I don't understand here, how do these two formulas come out? Thank you;
A: Today theta=alpha/2, at this time the id is 0 (it is recommended to read Zhang Yuxing's book, you will understand)
Q: The first red frame is understood; But the second one does not know how to deduce it? How do you eliminate both Iq and Ipk?
A: IQ and ipk are represented by cos(alpha/2), so aren't there two relationships there, the line in the red box.
Q: Understood; To synchronize the two equations; Thank;
08 Discussion of the principle of DPD
Q: Why does DPD improve AM-PM and AM-AM effects from the physical principle?
A: This should be a math problem.
Q: Why can predistortion signals improve the efficiency and linearity of the same PA? Is the power of the predistortion signal greater than the original power?
A: A signal passing through a nonlinear device causes changes in amplitude and phase. Relative amplitude changes are described by AM and changes in relative phase are described by PM. It's just that DPD needs to assume that the amplitude and phase changes of signals of different frequencies are consistent after passing through PA, that is, it needs to be a narrowband signal, not a wideband signal. Mathematically, it is to use a higher-order matrix to digitize the amplitude-frequency response and phase-frequency response, right?
Q: I recommend understanding this problem from the perspective of semiconductor physics. In this way, DPD cannot be deeply understood, and the possible adverse consequences of DPD cannot be well used and prevented.
DPD improves the AM-AM effect is well understood, and it must be that the amplitude of the signal (PA input signal) becomes larger after predistortion to improve the gain compression of the original signal, so that the amplitude gain of the large signal is consistent with the small signal.
A: By default, the device characteristics are accepted, and DPD is just a matter of modeling, and it has nothing to do with semiconductor physics.
A: It's a bit like analyzing diode circuits from the frequency domain. After going around a complex mathematical analysis, the same conclusion was reached
A: Generally speaking, DPD is a method of fighting poison with poison. Personally, I understand that DPD is also a distortion treatment, except that the amplitude and phase are completely opposite to the signal distortion.
Q: This is what predistortion means by "pre-distortion".
A: Therefore, DPD needs to run char in advance, and it is enough to directly check the table call parameters when actually using it.
Q: This is in the frequency domain. How to see DPD signals in the time domain? What is the difference between the time domain characteristics and the original signal? If the original input signal is an OFDM signal, is the peak-to-average ratio of the signal after passing through the DPD module improved? Has the average power changed?
A: The distortion noise amplitude is very small, if this can produce a change in the peak-to-average ratio, can this signal still be used?
Q: Why can't it be used, AM-AM distortion is the signal peak amplitude compression, DPD to make up for this amplitude compression, only the signal amplitude input to the PA is larger, in order to ensure that the original amplitude gain and the amplitude gain of the small signal are consistent. This is also the principle of DPD correcting AM-AM distortion! What is not distorted? It is the same as the long before and after signals
A: Using MATLAB tools to simulate it, I feel that I will understand it deeper.
A: The more PA compression, the higher the peak-to-average ratio expansion after DPD, so a threshold should be added.
Q: Understand people! And what about the average power?
A: The peak-to-average ratio of the signal input to PA is equal to the compression threshold of the modulated signal peak-to-average ratio plus DPD. If the compression threshold is too large, it is possible to burn the PA. As for the average power of the input PA, because the output power of the pa does not change, the gain does not change (the peak-to-average ratio may be slightly reduced when it becomes larger, ignored), DPD is because the loop will control the input signal to ensure that the output power does not change.
Q: There is a bit of a problem with the logic. It should be said that the small and medium-sized signals in the average power are linearly amplified, so there is no need to change, then the output power is basically unchanged.
A: The DPD algorithm is still mainly calibrating adjacent channels for the current large bandwidth, or multi-C calibration, rather than the third order. If it is RF direct mining, multi-C scenario, IBW=400M, if FB ADC sampling 3 times the bandwidth, then the data rate is too high, hardware links and FPGA implementation are difficult.
Q: It is possible that this is a segmented synthesis.
A: How about segmentation? The ADC rate is 983M and you can't see anything 3*400M on three times the bandwidth. And from the application point of view, it seems that the third order does not matter, it is far away, and it will be dried up by the filter.
Q: 400M divided into 4 pieces of 100M.
A: OKOK, thank you.
A: DPD is more critical to fitting accuracy
[Note: The discussion of this issue has not converged, welcome to join the Wisdom RF discussion group to continue to participate in the discussion of this issue]
09 Discussion on LNA design
Q: After the layout of 0.5-2G wideband LNA is drawn, the imitation gain and the pre-imitation ratio are 0.3-0.4dB lower, and the noise figure is about 0.2dB higher.
A: It should be okay, it is recommended to check the changes in IN gmax. The entire input matches the path.
Q: Input matching is achieved by adding feedback to the circuit, and the final design done by myself does not have the hard indicators of the project, so I can only compare it to see the layout before.
A: Calculate the changes in IMN gmax before and after to see how much loss the layout brings.
A: The noise figure is abnormal and the gain is normal, mainly depending on whether the delay and summation operation of beamforming is performed in the analog domain or the digital domain.
Q: The noise figure will also change when the gain changes, and it will affect a little input matching, because it is matched with feedback, and I still understand that the NF deteriorates a little.
A: I feel that he means that the noise deterioration is too small, maybe the simulation is a bit problematic, and it should not only increase by 0.2dB. I generally increase NF by more than 0.8dB in the mmWave band.
Q: Whether it is good or not depends on the relative value, if the total is large, isn't it normal to deteriorate by 0.2?
A: 0.5-2G noise figure deteriorates too much.
Q: 3.2 deteriorates to 3.4.
A: It's already good, if possible, measure a few more samples to average.
Q: No, it's not me, this is just after mentioning the parasitic parameters to do the post-simulation and pre-imitation comparison.
A: The minimum pre-imitation NF in his frequency band is 3.2dB, which should be untuned.
Q: Broadband generally cannot be very low.
A: What process do you use? I feel that 65 and below should not be adjusted.
Q: After the feedback, NF seems to be very low, NFmin has a fast 3.
A: What's the craft? This frequency band is three points.
Q：65nm。
A: You can check the circuit again, there is still room for improvement.
Q: I don't read the literature very low.
A: You have a lot of more than 2 points in this band. Power consumption 3~5mW. You can find more TCAS articles, a lot.
Q: Thank you. I'll take a look again, the broadband NF I found in IEEE before is not very low, 2. There are definitely many, but the structure may be more complicated.
10 Discussion of noise simulation and measured deviations
Q: Why is the s-parameter simulation test similar to the low-noise amplifier chip of DC-1GHz, but the noise test result is much worse than the simulation? Specifically, 6MHz-1GHz, the test and simulation of noise in the full working frequency band are much worse
A: Is it tested in a shielded room?
Q: It is tested in a high and low temperature box with a box.
A: How much worse are you actually than the post-imitation?
Q: Here's what it looks like:
Some students said that there must be a mistake in simulating low frequencies, and there should be no model that considers flicker noise in the PDK used (so I want to ask if I have had a similar experience?). )。
A: It is possible that the noise model is not included in the model, so you can check with Fab.
Q: Okay, thanks!
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