Crystal radio antenna impedance matching

[Colin]

Because, once you have turned on the transistor, you don't have to overcome the junction-drop and the transistor will "diode detect" the signal. All the diode does is prevent the voltage on the capacitor (say 10n) from leaking back to the low impedance of the tuned circuit.

 

[unclejed613]

A lot of the things you are saying, are wrong.

example?

 

[Colin]

"In a very high impedance tank circuit, a standard 1N34A diode just won't cut the mustard. The diode will load the circuit"

The diode does not load the circuit at all. It merely passes the signal to the output once the amplitude is above the natural characterist voltage-drop across the diode. Any diode can be used. They all have different voltage drops.

 

[unclejed613]

Because, once you have turned on the transistor, you don't have to overcome the junction-drop and the transistor will "diode detect" the signal. All the diode does is prevent the voltage on the capacitor (say 10n) from leaking back to the low impedance of the tuned circuit.

once you include active devices, such as a transistor, then it's a transistor radio and no longer a "crystal" radio. in a crystal radio, the diode doesn't isolate the tuned circuit from loading (and the tuned circuit should be a high impedance node, not a low impedance node, that's what maintains good selectivity). that's the reason for the coupling cap and inductor in the circuit i posted, to isolate the high impedance of the RF circuit from the low impedance (at RF) node of the audio portion. the "low impedance" portion of the RF section is the series resonant antenna circuit. the purpose of the diode is to rectify the RF signal, and demodulate the audio from the carrier wave.

 

[JimB]

A lot of the things you are saying, are wrong.

You should read the following article to let you understand how the crystal set works:
The Crystal Set

Still using ETO to promote your own website I see.

JimB

 

[unclejed613]

"In a very high impedance tank circuit, a standard 1N34A diode just won't cut the mustard. The diode will load the circuit"

The diode does not load the circuit at all. It merely passes the signal to the output once the amplitude is above the natural characterist voltage-drop across the diode. Any diode can be used. They all have different voltage drops.

yes, it does... there's some junction capacitance across the diode, and there's current through the diode, both of which contribute to loading of the RF high impedance node. try it on LTSpice, or try it on real hardware.

 

[Colin]

"After the diode, there's a filter capacitor, to remove RF component from the audio signal. This makes a short circuit for RF, and thus, a very low impedance to ground. As the diode, while conducting, is a low impedance, the RF signal loads heavily the tank circuit, no matter what you do."

You have to read my Crystal Set article, but the only signal that appears across the RESERVOIR capacitor is the audio component. There must be a resistor across the capacitor to discharge it so the audio is created. A crystal earpiece is effectively a 22n capacitor and it needs a resistor across it to discharge the "capacitor"

You can't say the antenna loads the tuned circuit because it is delivering the microscopic voltages and signals.
It is the earpiece that loads the circuit.

 

[unclejed613]

if you really want to delve into crystal radio sets, and get a perspective of people who used them PROFESSIONALLY, https://ia802308.us.archive.org/12/items/wirelesstele00buchrich/wirelesstele00buchrich.pdf Practical Wireless Telegraphy
is a must read. the circuits in that book are well explained, and not tinkerer/experimental designs, but actual radios that were in use, and well designed.

 

[Colin]

"Still using ETO to promote your own website I see."

Listen, I get 10,500 visitors a day to my website and have had over 24 million visitors. I don't think this website would have any effect on my readership.

The only reason I reply to your website is to get comments for my SPOT THE MISTAKE pages as practically none of you have the faintest idea about analysing a circuit.

 

[unclejed613]

"Still using ETO to promote your own website I see."

Listen, I get 10,500 visitors a day to my website and have had over 24 million visitors. I don't think this website would have any effect on my readership.

The only reason I reply to your website is to get comments for my SPOT THE MISTAKE pages as practically none of you have the faintest idea about analysing a circuit.

so, what you are telling us is that you are trolling...

 

[Colin]

"yes, it does... there's some junction capacitance across the diode, and there's current through the diode, both of which contribute to loading of the RF high impedance node. try it on LTSpice, or try it on real hardware."

The diode is not considered as part of the load.
When you talk about "loading" the crystal set, you only consider the earpiece. A crystal earpiece is a very light load and a set of 2,000 ohm headphones is a heavier load. Anything less than 1,000 ohms will not work.

 

[Colin]

"in a crystal radio, the diode doesn't isolate the tuned circuit from loading "

Actually it does. Because it allows the tuned circuit to rise to about 400mV before it delivers its voltage to the rest of the circuit.

 

[unclejed613]

"yes, it does... there's some junction capacitance across the diode, and there's current through the diode, both of which contribute to loading of the RF high impedance node. try it on LTSpice, or try it on real hardware."

The diode is not considered as part of the load.
When you talk about "loading" the crystal set, you only consider the earpiece. A crystal earpiece is a very light load and a set of 2,000 ohm headphones is a heavier load. Anything less than 1,000 ohms will not work.

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.

 

[Colin]

"once you include active devices, such as a transistor, then it's a transistor radio and no longer a "crystal" radio."

Who said the one and two transistor sets were crystal sets?

I simply went from a crystal set to more advanced designs because a crystal set needs such a long antenna that it is practically impossible for most people to get a crystal set to work.

 

[unclejed613]

btw, i looked at your "spot the mistake page", and i guess you don't realize a lot of those "mistakes" are intentional. they're called "easter eggs" to get people to buy the kit after they try to build it without buying the kit (and obviously doesn't work).

 

[JimB]

You can't say the antenna loads the tuned circuit because it is delivering the microscopic voltages and signals.

Really?
What about the inductive coupling of the resistive part of the antenna impedance into the tuned circuit?

The only reason I reply to your website is to get comments for my SPOT THE MISTAKE pages as practically none of you have the faintest idea about analysing a circuit.

A quote, thought to be made by Mrs Colin...
..."Nobody knows anything about electrics except my little boy"

JimB


(This thread is going down hill rapidly)

 

[Colin]

The diode is not considered as part of the load. Remove it and the amplitude of the tuned circuit would be reduced by 300mV.
It's a bit like saying the leads to a motor form part of the load.
There are losses in the leads but we don't talk about them as being part of the load.

 

[Colin]

"btw, i looked at your "spot the mistake page", and i guess you don't realize a lot of those "mistakes" are intentional. they're called "easter eggs" to get people to buy the kit after they try to build it without buying the kit (and obviously doesn't work)."

Where do you get this tripe from?

 

[unclejed613]

we were talking about crystal sets here. you popped in with talk of transistors etc...

don't feed the trolls...

 

[audioguru]

Because, once you have turned on the transistor, you don't have to overcome the junction-drop and the transistor will "diode detect" the signal. All the diode does is prevent the voltage on the capacitor (say 10n) from leaking back to the low impedance of the tuned circuit.

But the transistor is already biased to be turned on. The RF signal simply causes it to turn on a little more and turn off a little less at the RF frequency that you cannot hear.

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.