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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\$\begingroup\$ You don't use regular silicon diodes for radios. The AA119 is a germanium diode with very low voltage drop. An LC tank circuit might resonate, but that would only amplify the carrier frequency, not all the frequency components carrying the data. That LC is a filter to tune the station. \$\endgroup\$– DKNguyenOct 7, 2022 at 14:22
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\$\begingroup\$ @DKNguyen,hi, thanks for replying, "amplify the carrier frequency", yes the radio actually detect the carrier frequency (MHz) so this means that it amplifies the MHz microvolts into several volts? :) \$\endgroup\$– jessica smithOct 7, 2022 at 14:27
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1\$\begingroup\$ You DC bias the transistor above the cutoff and the AC RF signal rides on top of that so you are always above the cutoff. \$\endgroup\$– DKNguyenOct 7, 2022 at 14:54
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1\$\begingroup\$ @jessicasmith you already have the answer to that. The resistors bias the transistor appropriately. The signal gets amplified by the transistor, not before it. \$\endgroup\$– Marcus MüllerOct 7, 2022 at 14:54
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1\$\begingroup\$ @jessicasmith: You will find that the transistor "gets turned on" whether there is an input signal or not. The input signal just varies how "on" the transistor is. \$\endgroup\$– JREOct 7, 2022 at 15:03
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2 Answers
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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.
answered Oct 7, 2022 at 14:40
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\$\begingroup\$ HI, so this ""voltage amplification", has to be at least several volts right because it has to pass through a diode and even turn on the transistor ? \$\endgroup\$ Oct 7, 2022 at 14:49
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2\$\begingroup\$ @jessicasmith No, you bias the transistor at a DC level above its cutoff voltage and that C3 couples the AC radio signal to ride on that DC bias. I can't comment much about the diode though but earlier I did say it was a very low forward voltage germanium diode that needs to be used. \$\endgroup\$– DKNguyenOct 7, 2022 at 14:51
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1\$\begingroup\$ Not just a low forward voltage diode -- it needs to be a leaky low forward voltage diode, or the junction of C2 and C3 will just charge up to the peak tank voltage (minus diode drop) and stay there. Fortunately for this circuit, germanium diodes also tend to have a lot of reverse leakage. \$\endgroup\$ Oct 7, 2022 at 16:19
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"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.
- Resonance in the LC tank increases the amplitude of the signal.
- The forward voltage required is much less than you think it is.
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.