Hi,
this circuit derives from that of the old crystal radio and is a simple LC circuit. The chosen frequencies are example. I guess the concept is the same for everyone. To simulate a variable capacitor I used the ".step" command with the minimum and maximum capacity values of the condenser itself. R2 and L1 are the parameters of an inductance and are a single component. If we look at the Lspice circuit, I had to add the R1 resistor because obviously I would have the voltage even out. The problem is that this resistor in the original scheme is not there and it works anyway. So why does the simulator need R1 when the original scheme is not there and works equally? I want to understand if in case of real assembly of the circuit, I must also mount R1 or not.

Thank you!

 

The voltage source V1 is modeled as an ideal voltage source.
An ideal voltage source has zero source resistance. The voltage measured at the terminals of V1 will always be constant regardless of the load presented on V1.

The application of R1 adds source resistance to the voltage source. Now you are measuring the voltage output, not at V1, but after the source resistance R1. The voltage will change depending on the current drawn by the load. This is because there is now a voltage across R1 that must be subtracted from V1.

 

Hi D,
V1 signal source is an ideal voltage source needs an internal resistance for that circuit.
E

 

You added a resistor for nothing. The impedance of a short compared to the wavelength of the received signal frequency is capacitive in nature. This capacitance depends on the thickness of the antenna and other factors. But the shorter the shorter the smaller the equivalent antenna capacitance. Try using a dummy capacitor with a capacitance of 10 to 30 pF instead of a resistor. This will be more correct than using a resistor.

 

And also there is a missing duode

 

A real antenna has an impedance, which may be inductive or capacitive depending on the antenna dimensions in relation to the wavelength handled.

 

resistor is used just as a crude way of coupling signal with the circuit. this could be done differently. one method mentioned is to use a small capacitor. another is to add another coil to ferrite antenna with couple of turns (an inductive coupling). both are commonly used in real circuits. it is just that detector circuit is the absolutely bare bone design where everything that could be removed is actually removed. as a result you get poor selectivity, need for big antenna and proximity to a powerful AM radio station. btw your signal only simulates carrier, you can add low frequency signal to it, and see demodulation working.

 

First of all thank you all for your kind replies. So if you want to use an antenna formed by a coil, there is no need to add any additional components in series to the signal input? I made the modification suggested by @Bordodynov and it works however the tuning adjustment with the variable capacitor no longer works as it should. It seems that the capacitor phase out the signal by about 90 ° whereas it did not happen before.

 

In the real world R3 is the earpiece which,depending on type could be resistive, inductive/resistive or capacitive.

R2 represents the resistance of the inductor. 5ohm feels instinctively too large; 80 turns of 20awg on a 25mm former would be 50uH and 6.3m of wire giving 0.2ohm and much more selectivity (higher Q).

 

See

 

The circuit shown above is illiterate. The low impedance of the detector shunts the circuit greatly and results in a not very large signal. Also very low selectivity is obtained. To eliminate this you need to use autotransformer or transformer switching. This is what I did. See below.

 

Hi @Bordodynov thanks for schematic. Is 0.98 the ratio between primary (L1) and secondary (L3) . I have some RF transfomers (Coilcraft) and i think are 1:1. It is correct?

 

No. I took a transformation ratio of ~7:1. An additional transformer is not necessary and would be harmful. If your loop coil has 50 turns, take the signal for the diode off the seventh turn counting from the common point. Transformers and autotransformers allow for impedance matching.
I showed the transformer because it is easier to understand and easy to calculate the inductance of the primary. For the autotransformer the two inductances are harder to calculate and I was too lazy to do it. Especially to explain to you where these inductances come from.

 

Thank you @Bordodynov

 

In the real world R3 is the earpiece which,depending on type could be resistive, inductive/resistive or capacitive.

R2 represents the resistance of the inductor. 5ohm feels instinctively too large; 80 turns of 20awg on a 25mm former would be 50uH and 6.3m of wire giving 0.2ohm and much more selectivity (higher Q).

Hi, @Irving do you have have some calculator to achieve this data?

 

Hi, @Irving do you have have some calculator to achieve this data?

From this site:



7.8 - 8cm long and 80 turns = approx 0.9mm dia wire, or 20awg enameled "magnet" wire

 

@Irving ,i seen too but i wasn't sure what to put in the relative permeability folder. Thanks.

 

@Irving ,i seen too but i wasn't sure what to put in the relative permeability folder. Thanks.

air = 1
ferrite rod = 100 - 600 typically

 

The ferrite rod is not a ring and the effective magnetic permeability is less than 100.
https://coil32.net/

 

The ferrite rod is not a ring and the effective magnetic permeability is less than 100.

Good point, yes I was quoting values for ferrite rings from memory; I'd forgotten the rod is 'open (magnetic) circuit' so to speak...