LNA topology - Page 1
Forum section: electronics / rf-microwave

author: arildj78
author_rank: Regular Contributor
timestamp: June 12, 2019, 01:54:34 pm
content: I am designing a 500kHz receiver. The signal is extremely weak and I’m throwing money and engineering at it to maximize the range. The signal is just a carrier wave without any extra modulation and my task is to measure signal strength to estimate distance.The antenna is a loopstick with Q factor around 200 and I believe it is best to put on an extra coil on the ferrite rod to tap the signal from the tank circuit.Now, what would be a sensible topology for the Low Noise Amplifier (LNA)? Should I go for a transimpedance design or a voltage amplifier? With such a high antenna impedance (more than 100k), will Johnson noise be a big factor and what for which design will it be of least significance?Arild

author: HB9EVI
author_rank: Frequent Contributor
timestamp: June 12, 2019, 05:51:57 pm
content: that sounds all a bit unclear...first I guess you should study the propagation conditions you meet on 500kHz; maybe you get an idea what it means to pinpoint a signal source on MW according to its signal strengthwhat do you mean with throwing money in it? 500kHz are not millimeter waves; I used for my 472kHz transverter mostly normal AF components like BC550 or BD139.weak signal tracing requires small pass bandwidths to reach better s/n ratios; just amplification is not going to do; additionally there's a lot of man made noise on those frequencies, which make reception of small signals trickier.somehow that whole undertaking sounds quite pointless; I'd say you have to go more into detail if you're serious about it.

author: arildj78
author_rank: Regular Contributor
timestamp: June 12, 2019, 09:37:57 pm
content: Avalanche tranceivers transmit a burst of CW energy at 457kHz for 80ms with repetition every second. Estimating the distance from the transmitting beacon to the receiver is done by checking the signal strength. Since the TX is low power, normal detection ranges is about 50m (well within the near field of the TX).I’m trying to build a RX with better detection range. The plan is a high Q antenna, an awesome LNA, downconversion, bandpassfiltering and detection.To get better range I need both low noise and a lot of gain. Now I’m asking for help on choosing a topology for the LNA.The loopstick antenna has more than 100k impedance. Which topology is most appropriate? Would I gain something from inductive coupling to the loopstick tankcircuit or should I tap the signal straight from the tankcircuit? Would a standard negative feedback opamp design with e.g. 20dB gain give a better SNR than than a transimpedance design? Are there other topologies that is even better?Feel free to suggest literature if you don’t want to answer yourself.

author: Odysseus
author_rank: Regular Contributor
timestamp: June 13, 2019, 02:56:56 am
content: The signal/noise ratio is going to be dominated by your ferrite rod antenna, not the following amplifier. The available signal power is determined by the rod's size, and the noise power is set by the copper/magnetic losses in the antenna. You can play with the transformation ratio to get a lower/higher source impedance, but the SNR will not change.A JFET input opamp with a low noise current density, like a jellybean TL081, in a non-inverting configuration will easily be below the noise generated by a 100kOhm or higher source impedance.

author: arildj78
author_rank: Regular Contributor
timestamp: June 13, 2019, 03:43:52 am
content: The 100k source impedance is calculated with the formula R=1/Q * sqrt(L/C). Is this a “real” resistance and hence affected by Johnson noise or is it some sort of equivalent resistance? The DC resistance of the coil is aprx 750mOhm.So how would I increase the SNR? Accept lower Q? That would also lower the antenna gain?For educational purposes, if the antenna were quiet enough, what would be the most quiet topology for such an application? I’m looking at the data sheet for a OPA847 and there are several different designs in the application section, but none is marked as narrowband or high impedance source.Arild

author: Odysseus
author_rank: Regular Contributor
timestamp: June 13, 2019, 05:06:07 am
content: That equation is valid for a series resonant circuit. A tank resistance of 100 kOhm almost certainly meant you were considering a parallel resonant circuit, in which case the equation would be R=Q*sqrt(L/C).Yes, that resistance is real and as noisy as any other 100 kOhm resistor.And no, you do not want to decrease Q. The lower the losses, then the higher the Q, and the higher the SNR at resonance.I would just use a non-inverting configuration with the positive input driven by the parallel output voltage from the resonant tank. The amount of voltage gain is up to you.And the OPA847 would be a terrible choice compared to the TL081. Here's why:Start with the voltage noise density of your antenna. I'm going to assume you correctly calculated a 100kOhm parallel resistance.The voltage noise density of a 100k Ohm resistance will be sqrt(4kTR)=sqrt(4*1.38e-23J/K*300K*100kOhm) = 40.7 nV/√Hz.Now lets compare that to the noise specs of a TL081 and the OPA847:The TL081 has a voltage noise density of 16nV/√Hz and a current noise density of 0.01 pA/√Hz.The current noise flows through the source resistance, so it contributes a negligible 0.01pA/√Hz*100kOhm = 1nV/√Hz.The uncorrelated noise from the source resistor and opamp sum in RMS fashion only, so the resulting total noise is sqrt(40.7² + 16²)= 43.75nV/Hz.Thus the TL081 barely increases the noise level. Specifically, the noise figure would only be 20*log10(43.75/40.7) = 0.6dB, which is excellent.The OP847 has a voltage noise density of only 0.85nV/√Hz, a magnitude better than the TL081.BUT, it's current noise density is at best 2.5pA/Hz, over two magnitudes worse, thus contributing 2.5pA/√Hz*100kOhm = 250nV/√Hz!Now the total noise is sqrt(250²+40.7²) = 253nV√Hz and so the noise figure is a miserable 20*log10(250.5/40.7) = 15.9dB! This is basically garbage.The OPA847 sucks so bad in this application because is has bipolar inputs, which have high bias currents and high current noise densities.It is much better suited to lower source impedances. I'll leave it as an execise to you to figure out what it's optimum source resistance and resulting noise figure would be.

author: arildj78
author_rank: Regular Contributor
timestamp: June 13, 2019, 07:41:29 am
content: Thanks! That was a very helpful calculation exercise! One thing though (for my own understanding of the subject), should the noise voltage contributed by noise current through the source also be added in the equation below?QuoteQuoteIt didn't affect the sum in this example, but I'm just checking that my understanding of the subject is correct.You mention that I can connect the tankcircuit directly to the noninverting input of the amplifier. What difference would it make to use a transformer (extra coil on the ferrite rod) to translate the impedance and signal amplitude up or down? In what case would it be better to use a transimpedance amplifier? I have seen example circuits used for amplification of weak signals from e.g. photodiodes, but I don't know which factors to consider before choosing such a design?Arild

author: Odysseus
author_rank: Regular Contributor
timestamp: June 13, 2019, 09:07:03 am
content: Yup, you got it. The correction you made would be the correct solution, in general.Let's say you used a tap on the coil at 1/10th of it's length, in an auto-transformer fashion.There the signal voltage would also be 1/10th of the full tank voltage, and the noise voltage is reduced by a factor of 10 as well, keeping the SNR unchanged.The impedance gets reduced by a factor of 100, however, so in your case would become 1kOhm.The true invariant that is left unchanged by impedance transformation is the available signal or noise power.A trans-impedance amplifier appears as a short circuit to the source, so is typically used for amplifying current signals.This is in contrast to the open circuit presented by a non-inverting amplifier, which is convenient for amplifying voltages.A trans-impedance amplifier would not work for amplifying the voltage signal of the parallel resonant circuit in your tank, as this would short out the resonance.Instead, you could reconnect your tank in a series resonant configuration. You could then re-interpret the loss in your tank as a series resistance equal to 100kOhm/200² = 2.5Ohm.Combined with a trans-impedance amplifier with a gain of 2.5Ohm*200=500 Ohm, this would give you the same voltage level as the parallel circuit originally did.Unfortunately, the noise voltage of a 2.5Ohm resistance is only 0.2nV, and you'll be hard-pressed to find ANY amplifier with noise voltage that low. Maybe it could be done with discrete transistors.Keep in mind that the noise of the 500Ohm resistor doesn't matter too much, as this appears in the output after the 200x amplification of your signal, and is only sqrt(200) times noisier than the noise of your source. Ultimately, it contributes 14 times less noise than the antenna does.Noise is just one of many parameters to consider when evaluating an amplifier; there's gain, bandwidth, linearity, input and output impedance, and many more. There's no simple answer for every situation, but the underlying math doesn't really change.

author: arildj78
author_rank: Regular Contributor
timestamp: June 15, 2019, 06:03:01 pm
content: Still working through your numbers to understand this subject and have two questions.When connecting to an antenna with a weak signal like this, what would be more beneficial, designing an amplifier with a high impedance input or with a matched impedance? The way I understand it, high impedance will give me the highest voltage on the input while a matched impedance will maximize the power transfer from antenna to amplifier? In my application the wavelength is very long compared to the wires. Would this choice affect the signal to noise ratio?QuoteWhat is general form of the equation2.5Ohm*200=500 Ohm. I can see 2.5ohm from series resistance, 200 from Q factor, and 500ohm from 1/200 of 100kOhm - but how did you come up with that relationship?Arild

author: Odysseus
author_rank: Regular Contributor
timestamp: June 16, 2019, 12:26:37 am
content: At low frequencies, it's practical to design high or low input impedance amplifiers (using opamps).At high frequencies, this isn't really practical, and transmission line effects makes matched impedances easier to work with, with the added benefit of maximized power gain/transfer, minimized reflections, etc.For a given amplifier, you can calculate the optimum source impedance that minimizes the amplifiers noise figure, and then transform the impedance of your source (sensor. transducer, photodiode, antenna, etc) to match.Often, however, this noise optimized impedance is close enough to just use a high input impedance voltage amplifier, or a low input impedance current amplifier, or an impedance matched power amplifier.Let me elaborate on those numbers I threw out. Here I'm using lowercase 'p' and 's' to refer to different quantities in the parallel and series circuits.The series resistance is derived by combining Q=Rp*sqrt(C/L) for a parallel RLC circuit and Q=1/Rs*sqrt(L/C) for a series RLC circuit to get Rs=Rp/Q² .Then equate the power in the parallel and series circuits to get P=Vp²/Rp=Is²*Rs and rearrange to get Vp/Is = sqrt(Rp*Rs) = sqrt(Rp²/Q²) = Rp/Q.Hence, the trans-impedance amplifier should have a gain of Rp/Q so that it converts Is into the same voltage as Vp.

author: arildj78
author_rank: Regular Contributor
timestamp: June 16, 2019, 09:43:03 pm
content: I have now tried to simulate two different configurations. Feel free to comment if I'm totally off track or if this makes sense.I have based the design on the following parallel tank circuit. L=303.213µH C=400pF Q=250 (BW 1.828kHz)From this i calculated the resonance frequency to be 457kHz and an equivalent circuit to be an ideal voltage source in series with 193.594k\$\Omega\$. I also assumed that this would be equivalent to a current source in series with a 3.48\$\Omega\$ resistance if I rewired the tankcircuit to be a series resonant circuit.To test this I simulated the circuit in LTSpice. The ideal sources have been designed to dump 1pW (-90dBm) into their series resistance if short circuited. The op amp used in the simulation is the LT1028 just because it is a low noise op amp available in LTSpice. It does not have high enough gain bandwidth product so for the output voltage simulation I dropped the frequency to 10kHz.From the noise plots I can see that with these configurations and the LT1028 there are more than two orders of magnitude lower noise density in the transimpedance amplifier. I will continue to test with other op amps but for now it looks as there is a good idea to use the transimpedance amplifier.Attached is screenshots from the simulation.

author: Odysseus
author_rank: Regular Contributor
timestamp: June 16, 2019, 11:32:52 pm
content: I'm calculating your Rp would be 217.6kOhm.In the trans-impedance circuit, you should have the current source in parallel with R1, and C1 isn't necessary.Otherwise, looks great!Just a tip: In LTSpice, you can calculate noise factor by dividing the total output noise by the output noise contributed from the source impedance, so you can add an expression like V(onoise)/V(R1).Then the noise figure is just the log representation of that: 20*log10(V(onoise)/V(R1)).

author: arildj78
author_rank: Regular Contributor
timestamp: June 17, 2019, 01:30:48 pm
content: When I connect R1 in series with Isource I get 33mV amplitude on the output, same as the non-inverting amplifier. This reasoning is based on aYouTube clip(screenshot attached) where a parallel RLC circuit is compared to an ideal voltage source with a series resistance and a current source with parallel resistance. For my series RLC resonant configuration I assumed that everything was opposite... Voltage source with parallel R and current source with series R.I have googled quite a lot and read what I believe is the relevant paragraps in The Art of Electronics, but I have not found anything to back up my assumption.When I put R1 i parallel with the current source I get Vout (amplitude) 132mV. This could be because I have calculated the ideal current source to fit with my initial assumption...Are you able to enlighten me?

author: Odysseus
author_rank: Regular Contributor
timestamp: June 18, 2019, 04:36:39 am
content: For a given source impedance, you can use either a voltage source in series, or a current source in parallel as an equivalent circuit, just like your screenshot shows.This applies for both the series and parallel RLC circuit. Seehttps://en.wikipedia.org/wiki/Source_transformationfor more info.It is incorrect to use a voltage source in parallel or a current source in series with the source impedance. If you think about it, you will realize these circuits would eliminate the effect of the source impedance from the circuit.You are probably getting the wrong result because it was actually 132uV and the Gain-BW of the LT1028 is too low to be used at a gain of 12500 and a frequency of 10kHz. That would need a GBW in excess of 125MHz.In reality, we have been incorrectly referring to the second circuit as a trans-impedance amplifier. Actually, it is just an inverting amplifier with a very small input resistor, and hence a very high gain.Your calculation of 43.5kOhm is correct.

author: radiolistener
author_rank: Super Contributor
timestamp: June 30, 2019, 06:41:46 pm
content: Quote from: arildj78 on June 12, 2019, 01:54:34 pmresonant frequency will be f = 457000 Hz and 3 dB bandwidth will be df = f / Q = 1828 Hz.Thermal noise depends on active resistance, bandwidth and temperature:E = sqrt( 4 * k * T * R * df )wherek - Boltzmann constant, k = 1.380649e-23 [J/K]T - temperature in Kelvin degrees (T = °C + 273.15)R - resistancedf - bandwidthLC circuit impedance at resonant frequency is the following:Z = 2 * pi * f * L * QIn your case Z = 2 * pi * 457000 * 0.000303213 * 250 = 217663 And thermal noise of your LC circuit (antenna) for t = 20 °C and 1828 Hz bandwidth will be:E = sqrt( 4 * 1.380649e-23 * (20+273.15) * 217663 * 1828 ) = 2.54  = -135

author: Kleinstein
author_rank: Super Contributor
timestamp: June 30, 2019, 07:53:09 pm
content: It was already noted before: the LT1028 is not suitable for a high source impedance, because it has way to much current noise.If it needs to be an OP, it would be like an OPA141. The more typical choice would be a discrete JFET like J309 or 2Sk3557.There is no real need for high end JFETs - quite a few get a noise temperature well below room temperature or a noise figure well below 2 dB.

author: bson
author_rank: Unknown
timestamp: July 03, 2019, 10:36:26 pm
content: Since your carrier is fixed you could use a crystal bandpass filter to get the bandwidth down to a few Hz or less. This will push the noise floor way down. Then adjust the transmitter to where it peaks on the receiver. Calibrate the readings to a couple of known distances and done.

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 12:18:32 pm
content: I've tried to read up on crystal filters and it looks like there is quite a bit of time consuming work to find matching crystals. I'm challenging myself to design a device for manufacture and because of this I'm a little reluctant to using the procedures I've found to build a crystal filter.My present plan is to use a transimpedance amplifier as the LNA, then mix down the signal to low audio frequencies and a apply something like an 8-pole active bandpass filter to the IF to achieve the required selectivity. After that rectify and integrate the audio before feeding it to an ADC and a micro.Any thoughts?

author: Kleinstein
author_rank: Super Contributor
timestamp: July 06, 2019, 12:52:31 pm
content: Why a trans-impedance amplifier ? A resonant ferrite antenna in high impedance, so that it would be more like a FET based amplifier, possibly just a follower.With a ZF in the audio range, one could directly go to the µCs ADC - so no hardware rectification and some of the close in filtering could also be done in software.With a downwards mixer one has to care about mirror frequencies, so that one should have quite some filtering before the mixer. Depending on the exact frequency this could be a cheap ceramic filter, usually made for AM radio. They are not as narrow as a crystal, but still narrow band and usually easier to use. One could reduce the filtering requirements by using a Tayloe mixer and going all the way with digital I/Q.

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 01:01:35 pm
content: I'm trying to figure out how to get the absolute maximum possible SNR out of this thing. As mentioned earlier the transmitter is fixed at 457kHz at quite low power. Now I find myself in the near field trying to receive the magnetic field with a resonant LC circuit. The goal is to pick up the transmitter at the greatest possible distance. As a minimum 150ft but I hope to achieve up to 500ft.I am free to configure the LC as a series or parallel circuit and was hoping that a series circuit would have a lower ESR so that there would be less Johnson noise. Furthermore I figured that a series circuit would be the equivalent to a current source and hence the transimpedance amplifier.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 06, 2019, 02:41:28 pm
content: The trans-impedance amplifier together with the antenna in series mode would close the resonant circuit. So much of the same Q and damping would apply. The difference is either having some loading by an amplifier at a higher voltage point or having the not perfect TIA carrying the current. Usually it is easier to get a high impedance amplifier near perfect than a TIA. JFETs get a noise figure well below 1 dB for a high impedance source and are thus already quite close to perfect. The high signal impedance of the parallel resonance is not a bad thing - it about matches the best SNR point of a JFET amplifier. If needed one can chose a lower impedance by tap at a different part of the coil, or using less turns / a larger cap.The more important point than the amplifier noise is likely keeping out other frequency signals, e.g. mirror frequencies. Chances are man made interference is way higher than the background noise from the amplifier and antenna. In the near field and such a low frequency there is little chance to get a more directional antenna. So all that is left is really good filtering.For a 457 kHz signal a 455 kHz ceramic ZF filter is a rather good match - they are dirt cheap, stable and about the right bandwidth.

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 03:46:32 pm
content: If connecting the LC in parallel, would you filter it before the LNA or after?Would it make sense to mix the signal down to a say 500 Hz and then create a band pass with BW=200Hz to achieve good selectivity?

author: radiolistener
author_rank: Super Contributor
timestamp: July 06, 2019, 05:04:53 pm
content: there is no sense to mix it down to 500 Hz, because noise level of environment at 500 Hz will be much higher than at 200 kHz and receiver will be affected with 50/60 Hz interference from mains.

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 05:06:54 pm
content: I was hoping that it would be easier to create a brickwall bandpass at such low frequency. Is this wrong?

author: Kleinstein
author_rank: Super Contributor
timestamp: July 06, 2019, 05:16:14 pm
content: A possible configuration could be like antenna - JFET amplifier (possibly just a follower) - ceramic filter - amplifier - mixer(e.g. Tayloe type) .Than 2 channels with a little filtering, but not much and going to 2 ADC channels / ADCs. The fine filtering and final demodulation / detection could than be done in the digital domain.The ZF would be more like in the 10-50 kHz range - so not too low.

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 05:52:57 pm
content: I'm really on deep water here. Do you know of any good online literature that explains why you suggest that design? I don't want to dive to deep, just understand why my original thought was wrong.First of all, I was hoping that I could find some really high end integrated solutions.Secondly, designing a JFET amp is scary. I am far more comfortable using a reference design from a low noise op amp datasheet.Thirdly, what factors should be considered when choosing the intermediate frequencyFourthly, How could do I get the really weak signal through a narrow bandwidth filter for high selectivity and what would be the best filter design.Lastly, where do I source these components??? I have browsed through DigiKey, but the amount of 455kHz ceramic filters on stock is not to great.

author: bson
author_rank: Unknown
timestamp: July 06, 2019, 06:53:17 pm
content: Quote from: Kleinstein on July 06, 2019, 12:52:31 pmA JFET follower would also nicely bias the signal for an ADC. But I think it would be unrealistic to expect a high SNR without a pre-selection filter. Which in this case, since the carrier is fixed, can be a high Q filter.As for crystals, even a single crystal stage is probably adequate, followed by a 2 or 4 order Cauer to tighten up the transition bands. Remember old 27MHz RC? Channel selection used to be done by swapping transmitter and receiver crystals... so a single crystal is definitely not a bad starting point. (Although those were for the LO, but same deal.)

author: arildj78
author_rank: Regular Contributor
timestamp: July 06, 2019, 07:02:56 pm
content: Plan was some sort of LNA to avoid loading down the tank, then a high Q-filter, but to get the bandwidth narrow, and the transition bands tight, I figured that it would be benificial to mix down the frequency to a really low one. If not I get into trouble with gain bandwidth product of the op-amps in my band pass filter.Does anyone have a link to some good online material on this subject and the factors affecting this design?

author: iMo
author_rank: Super Contributor
timestamp: July 06, 2019, 09:52:12 pm
content: The 457kHz is the international agreed frequency standard for the avalanche beacons/transceivers (carried by the people buried under snow) - it could be a special ceramic filters are available for that frequency.Quote from: arildj78 on June 12, 2019, 01:54:34 pmThe best way to start is to study how the competitors are doing that - there are many avalanche beacons/transceivers on the market available.