This is a radio circuit. Since the diode has a voltage drop of 0.7V, that means that the node highlighted in pink must be above 1V or so.
Does that mean that the LC tank can amplify the microvolts detected by antenna into several volts?
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Sign up to join this communityThis is a radio circuit. Since the diode has a voltage drop of 0.7V, that means that the node highlighted in pink must be above 1V or so.
Does that mean that the LC tank can amplify the microvolts detected by antenna into several volts?
First, yes, a tank circuit can amplify a signal's voltage (or, by arranging it differently, its current). An LC circuit can be used as an impedance transformer. It cannot increase the signal's energy, however, so if you're using a passive (i.e., no amplifiers, tubes, or transistors) LC circuit to increase a signal's voltage, you have to drive a high impedance circuit with it.
Second, that tank circuit can't amplify that antenna's voltage -- it's just not arranged right. But that's a schematic for an AM receiver that's operating in the hundreds of kHz, so a typical backyard antenna is going to act like a capacitive probe. The combination of the tank circuit and the capacitive antenna will result in some "voltage amplification", or at least a voltage high enough to make the radio work, if not well.
Third, that circuit is not going to be a very good radio. It won't even be more than middling by crystal radio standards. The reason for that is because the transistor amplifier that follows the tank circuit has a low input impedance; that low impedance will load the tank circuit and broaden its response (make it respond to more than one station at a time) even if it does manage to amplify the signal.
Fourth, don't listen to that Third person. It'll make a radio that'll receive strong local AM stations, so what are you waiting for? Make it work, then make it work better.
"Magnify" is probably a better word than "amplify." The resonant LC circuit can indeed raise the voltage. The LC circuit cannot, however, increase the power. Most folks think of an amplifier circuit with an external power source when they see the word "amplify."
There are two things going on that make it possible for a diode detector to work with just the signal level from the antenna.
People usually talk about the forward voltage of a silicon diode being 0.7V or 0.3V for germanium diode.
The fact of the matter is that the forward voltage depends on the current through the diode.
Take a look at the lower left corner of that chart. At very low current, the forward voltage is very low. You don't have to get a high voltage out of your antenna and LC circuit for the diode to work. You need some voltage, but mostly you need to draw only a tiny amount of current.
Old time crystal radios managed that by using a headset with a very high internal resistance (the coil itself, there was no extra resistor added.) With an internal resistance of tens of thousands of ohms, the headsets pulled very little current through the diode. That allowed the diode to work at a very low voltage.
Germanium diodes are better for that kind of thing than silicon diodes. The forward voltage of germanium diodes is lower than for silicon diodes, more so in the very low current regions needed to make a crystal radio work.
The "0.7V forward voltage of a silicon diode" is a simplification. It works fine when dealing with higher voltage and higher current stuff - you don't generally care a lot if your power supply rectifier has a forward voltage of 0.6V or 1.1V depending on the current.
At low current (radio signals) and high current (high current power supplies,) you have to look at things a little closer.
At low current, you are aided by the lower forward voltage. At high current, you begin to have to pay more attention to the power loss due the the higher voltage drop.
The 0.7V approximation isn't wrong, it just isn't always a good approximation.
In the circuit you are looking at, the base of Q1 is "lifted up" by R1 so that Q1 is always conducting to some extent. The additional current from the LC tank and D1 vary how much Q1 conducts. At the high spots of the radio wave, Q1 conducts a tiny bit more. At the low spot of the wave, Q1 conducts a tiny bit less.
The transistor circuit is a common emitter amplifier rather than the simple switch that you seem to be viewing it as.