AM detector -- direction of signal after tank circuit - Page 1
Forum section: electronics / rf-microwave

author: petert
author_rank: Regular Contributor
timestamp: August 03, 2019, 04:56:27 am
content: Hello,Assuming you have a simple AM detector radio, with an LC tank circuit as filter, as in this link:https://www.falstad.com/circuit/e-amdetect.htmlI don't fully understand how the tank circuit "makes" a difference between the input from the antenna and the output on the right of the tank circuit. Since the input is an AC waveform there is no single direction of current as would be in DC. So what forces a signal to be an input, what what forces it to be an output?Or in other words, what causes the directionality of the filter? I do understand that the tank circuit oscillates because the capacitor gets charged while the inductor discharges, and vice versa. I also get that the tank circuit attenuates all waves that are not of the resonant frequency.But I struggle to understand how this works when used as a filter since I would assume ever wire connected to the tank circuit would be affected equally, and therefore there would be no input or output.Edit: I simplified the above circuit to the minimal one necessary for this question, withjust the antenna and tank circuit:Could someone shed some light on this? Thanks.

author: bob91343
author_rank: Super Contributor
timestamp: August 03, 2019, 05:14:30 am
content: I don't really get the question. The circuit gets stimulated by signals from the antenna. Signals close in frequency to resonance will have least attenuation. They will provide signal for the diode's conduction. The filter on the right eliminates much of the input frequency due to its being in the cutoff range there, but doesn't attenuate the modulation much because that is in its passband.The 3k resistor doesn't accomplish anything, as I see it. It wastes signal power.The way detectors work is largely incorrectly described in the literature. Basically there are two frequency ranges of interest, the carrier (with sidebands) and the modulation. There is no input signal at modulation frequency but due to the diode, there are mixing products between sidebands and carrier such that modulation frequency components appear. The rf is unable to reach the output to any great degree due to the 300 nF capacitor. But that's not large enough to bypass modulating frequencies.Detectors like this are a compromise between the carrier and modulating frequency ranges. The time constant has to be slow enough to prevent rf from reaching the output, but short enough not to interfere with the modulating frequency.

author: @rt
author_rank: Super Contributor
timestamp: August 04, 2019, 08:15:48 am
content: Because the entire bottom half of the circuit is tied to ground, or it can’t work. That’s why.If a crystal radio isn’t grounded, and still works, it’s grounded through capacitance to Earth.The voltage at the antenna, and therefore the top half of the circuit is swinging back & forth with the RF signal relative to ground (considered zero volts).

author: petert
author_rank: Regular Contributor
timestamp: August 04, 2019, 08:40:26 am
content: QuoteSorry, but I am not sure to what part you replied. How does that create an output and input?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 04, 2019, 08:53:57 am
content: Right, ordinary passive filters are reciprocal, they work the same way in either direction. But that's not a problem here, if some signals are transmitted by the diode and out the antenna, it will be much weaker and not a problem for anyone else.Which does indeed happen, you can observe harmonics produced by the diode with a sensitive receiver.Although I suspect, that's a bit more in-depth than what you're asking about?Since the "antenna" is connected right to the tank, it's actually that the tank is reflecting the other frequencies away from the circuit, back up the antenna.Tim

author: petert
author_rank: Regular Contributor
timestamp: August 04, 2019, 09:30:51 am
content: Quote from: T3sl4co1l on August 04, 2019, 08:53:57 amThat's what I was actually wondering about!Is there a way I could somehow observe that in a simulation (something similarly simple as in my second link, which omits anything on the right of the tank circuit)? Since the antenna is modeled as a "perfect" signal generator, the output of the antenna will stay unaffected.Or is there a way to manually compute the effects on such a circuit, besides just accepting that a tank circuit "reflects"?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 04, 2019, 12:05:45 pm
content: Yes, you need to construct a reflectance bridge. This can be a simple equation (which is fine in SPICE, but I don't recall if there's any way to input equations in Falstad), or constructed from transformers and a resistor.Or you can familiarize yourself with the equation, and see it work by inspection. Basically whenever the feedpoint voltage is below nominal, it's reflecting out of phase (impedance less than nominal), or above, in phase (impedance greater than nominal). The open circuit voltage is twice nominal, and nominal voltage falls exactly at nominal (matched) resistance.A definition of the reflectance coefficient:Gamma = ((Vin/Iin) - Zo) / ((Vin/Iin) + Zo)Where Vin is the input voltage, Iin the input current, and Zo the system (source) impedance.It looks like the impedance is quite high to begin with (the ground return resistor) so there may be some freedom implied, as far as what we consider the system impedance to be. Normally we'd pick the default 50 ohms, because pretty much all our test equipment is set to that. If we don't need to connect to these, and we're just doing a few components anyway, we might choose a completely unique system impedance, and work from there.In a real crystal radio, we might use an electrically short antenna (since, at these frequencies, a resonant antenna (having a nice reasonable feedpoint impedance) take a BIG tower), which will have a high source impedance, namely a combination of resistance and capacitance. We tune out the capacitance, and then we're left with the resistance, which we can use to drive our detector diode and speaker, and anything left gets reflected back out into space. Basically the ambient waves waft by our piece of wire, and interact with it. The field is distorted, due to the presence of the wire, and exactly how it's distorted depends on what we have connected to it.Tim

author: coppercone2
author_rank: Super Contributor
timestamp: August 04, 2019, 11:31:02 pm
content: i think the military uses that to detect radios, the reradiation

author: vk6zgo
author_rank: Super Contributor
timestamp: August 05, 2019, 12:01:40 am
content: Quote from: coppercone2 on August 04, 2019, 11:31:02 pmNah! They use local oscillator radiation from Superheterodynes.The re-radiated signal from refections is swamped by the mass of signals already available to receive.During WW2 this was touted as an adbantage of TRF receivers---- no detectable radiation, unlike regenerative, superhet, or direct conversion radios.

author: coppercone2
author_rank: Super Contributor
timestamp: August 05, 2019, 02:24:46 am
content: it won't show up if you beam a powerful signal into a field radio from a electronic warfare airplane? I thought they might be able to do that to direct missiles or something in the general location. Won't the internal signal be ultra weak compared to if they flew over it with a powerful transmitter on a dish or horn?I don't mean like finding bugs, more like finding a truck with an antenna on it at night from far away with clouds under camouflage etc.guess its too weak, maybe in ww2 lol