I get it. The first RF transistor compresses the top of the RF waveform each time it gets near cutoff and the even harmonics distortion causes the modulation to be rectified a little that is amplified by the audio second transistor.
So, does the demodulation occurs because the transistor is near cut-off, as is the case with JFET infnite-impedance detectors?
How can the bias be precisely controlled/adjusted using a single base-collector resistor? (in this case, it seems to be a parallel-parallel negative feedback amplifier, isn't it?).
All the signals hitting the antenna are fighting each other and cancelling each other out. But one signal is reaching the antenna at say one million times a second with very little voltage but each time it runs down the antenna to the coil and tuning capacitor it appears at exactly the right time to add a little more voltage. The voltage gradually rises and falls according to the frequency of the audio being sent by the radio station and this is the only voltage that is appearing at the tapping of the tuned circuit. This voltage simply passes through the 100n to the base of the transistor and the transistor simply amplifies the audioguru signal.
The radio signal might only be a few microvolts in amplitude but it keeps hitting the tuned circuit at exactly the right time and it has one million opportunities every second. The end result is a voltage as high as 1v or more and the ratio of the incoming voltage to the outgoing voltage is what we call the "Q" of the tuned circuit. It is the "Quality" or the amplification provided by the coil and capacitor combination.
Biasing is simple. Before I used a 2N3904 that has a fairly low hFE so it was close to cutoff. Now I am using a BC547C that has a high hFE and it is close to saturation.
You want the collector voltage to be about half the supply voltage so it can swing the most up and down, then with the 4.5V battery the collector is 2.25V and the collector current is 2.25V/1k ohms= 2.25mA.
The datasheet of the BC548C shows that its hFE is typically 520 at 2mA so the base current will be 2.25mA/520= 4.3uA. The base voltage is shown on a graph as 0.65V so the base current is (2.25V- 0.65V)/100k= 16uA which is too high. The transistor is biased almost at saturation.
Confirming too much bias we assume that the collector voltage is 1.3V. Then the collector current is 3.2mA. The base current is (1.3V - 0.67V)/100k= 6.3uA, and it will cause a collector current of 6.3uA x 520= 3.2mA.
Then since the transistor is biased near saturation it rectifies the RF like an amplified diode. The hFE of a BC548C has a wide range from 420 to 800 at about 2mA of collector current. Ones with an hFE of 800 will be even closer to saturation.
If the supply voltage is reduced as the battery voltage runs down, then the transistor will become linear and not rectify the RF so there will be no audio. Then when the supply voltage drops to near 1.5V the transistor will become cutoff and again it will rectify the RF. Here is a BC547C in this circuit showing saturation with a 4.5V battery:
The classic 'crystal set' used 2000 ohm headphones (a standard value back then), more 'modern' ones used crystal earpieces which are even higher impedance.
Would it be better to work near saturation or cut-off?
With JFET I bias near cut-off, and it works fine.
I think that was the case with most valve radios which didn't use demodulator diode (except those using grid-leak approach, which are near saturation, that is not possible with transistors, unless adding a diode gate to ground).
Going back to first page, I just built three regenerative MW receivers: one bipolar, two JFET; one with capacitive throttling, another one with potentiometer. None of them use diode.
All of them are very selective receivers, and have decent gain.
They have minimal parts, although the coil and the final values tweaking has to be done carefully.
"the diode passes current for 1/2 of the cycle of the RF carrier wave, and if there's a low impedance like a capacitor to filter out the RF) on the other side of it does indeed load down the circuit."
The RF (the 1MHz) signal never gets to the diode.
The high frequency is passed to the tuned circuit where just one frequency is constantly increasing in amplitude and it adds a microscopic increase across the tuned circuit each time it arrives. Over a period of time the voltage on the top of the tuned circuit is 350mV and when it gets to 351 mV, the diode passes this 1mV to the load. The voltage keeps increasing and the voltage above 350mV is the voltage that will appear across the headphone (earpiece). Once it reaches a maximum, it starts to gradually decrease. The waveform it produces is the audio being transmitted by the radio station.
The diode does not have to be high-speed but it must be able to following a waveform that may be as high a 10kHz. This is about the highest frequency you will hear on an AM radio.
The diode is merely passing the voltage on the top of the tuned circuit and the only time when "speed" comes into it, is when the signal is changing from a HIGH to a LOW. The diode must be able to turn off quickly enough so the decreasing signal does not taken any current from the load side of the diode. You do not need a capacitor across the load but if one is present (such as a crystal earpiece - 22n) you need a resistor across the capacitor to discharge it.
Now, when it comes to the argument that the antenna poses a load on the tuned circuit, we find the antenna is the "delivery vehicle" and it never poses a load.
When the signal from the antenna increases the voltage on the tuned circuit it produces EXPANDING FLUX and when the 1MHz signal goes from a HIGH to a LOW, the flux collapses slightly but the value of inductance of the coil and the value of the 415pF capacitor contain so much energy that the drop in voltage is microscopically small before the next cycle of the 1MHz signal appears. That is why each cycle of the incoming signal increases the voltage on the tuned circuit to a point where it is sufficient to drive a headphone.
Thus it is not important where the antenna is connected, although many crystal sets have antenna tapping points. A change in antenna connection will alter the overall value of the tuned circuit but not actually put a load on it.
Experimenting with crystal radio, I've reach a point where I'm not able to understand the explanations about antenna and tune coils coupling, and impedance matching.
This topic has veered way off the initial question, with lots of pseudo science and Hocus Pocus, but if the original question is still of interest, and you're willing to handle some fairly technical discussion, then have a look at Ben Tongue's crystal radio analysis. He deals with the whole impedance matching issue from the antenna, through the detector diode, to the headphones. https://kearman.com/bentongue/xtalset/xtalset.html
BTW, Ben Tongue was one of the founding partners of Blonder & Tongue, a US manufacturer of RF equipment.