Is it possible to design this circuit of crystal radio reciever on bread board?

After tuning the LC tank circuit to the required frequency the output voltage from the parallel LC circuit will be fed to the diode and ear phones.

Can this circuit be made as it is on the bread board so that the output voltage is enough to drive the diode and earphones?

It will also be appreciated if one likes to share the steps it takes to design that type of radio on the bread board.

• Don't know if such a carrier powered radio circuit can work on breadboard (parasitics might be a problem), but know that such circuits typically require a crystal earphone. May 2, 2021 at 7:21
• I suppose the diode VF will be important... also the coil construction... I remember doing such exotics in the 60's. For example, I used a graphite contact with a razor blade as a diode. I also melted sulfur and lead in a test tube for some such "diode"... Primitive but I guess with low VF it was. The inductor was flat with a large surface or cylindrical... The earphone was electromagnetic ... May 2, 2021 at 7:30
• stem-supplies.com/science/circuits/build-a-crystal-radio-kit This seems to imply that it is possible. Keep the connections far apart to minimize the capacitance. May 2, 2021 at 7:48
• (but please be careful if using an outlet: under no circumstances should you connect your radio to the live wire) May 2, 2021 at 14:03
• @Alex: You have a design - that's the schematic diagram. You seem to want to build the circuit on a bread board.
– JRE
May 2, 2021 at 15:23

Is it possible to design this circuit of crystal radio reciever on bread board?

Assuming that you have designed it correctly and assuming you are interested in radio frequencies that are sub 10 MHz, then it can certainly be implemented on a breadboard. But, to design it you need to have: -

• An antenna that is around one-tenth of one-quarter of the wavelength of the carrier wave.
• A germanium detector diode because it has a low forward volt-drop and therefore makes it easier to amplitude demodulate weaker local radio transmissions.
• High impedance earphones so that the energy taken from the tuned circuit does not damp it too much and cause loss of channel selectivity.

After tuning the LC tank circuit to the required frequency.....

It is the antenna's self-capacitance and the tuning capacitor together that form the tuned circuit with the inductor - I'm mentioning this because when using a "short" antenna (as per my note above about the length being one-tenth of one-quarter), the antenna acts as a capacitive source and allows quite selective tuning. Making an antenna that is quarter wave matched will not work effectively at all with this circuit. Yes you might get more signal, but the tuning capability will be very poor and you'll hear several broadcast stations simultaneously (not a desirable effect). It's all down to the projected antenna impedance and an effective crystal radio does require an electrically short antenna.

The picture above comes from this Q and A. I've drawn a red area that represents the monopole length being between one-tenth and one-fifth of a standard quarter wavelength. This is the optimal length of the crystal radio antenna. In other words it should project a capacitive impedance of around 1000 to 2000 ohms. At 1 MHz, that would be around 100 pF.

Also note that the resistive element of the antenna is now a fraction of what it is at standard quarter wave frequencies and this does indeed boost channel selectivity by significantly increasing tuned circuit Q factor. The counter-side to this is that the received signal power is significantly smaller when the antenna is "short" but, as with most simple circuit implementations you can't simultaneously eat your cake and have it.

Given that a breadboard might introduce an extra 5 or 10 pF per node, you can see that it won't be a big deal.

• What's the advantage of Ge compared to a modern Schottky diode?
– mmmm
May 2, 2021 at 9:41
• @mmmm probably not much. Maybe it might be harder to choose a Schottky with quite low reverse capacitance that has the low forward drop of a Ge diode. May 2, 2021 at 10:37
• infact I want to know that what value of inductance, capacitance may i use to get the desired rasonance frequency and to drive the diode and ear phones at the same time? About that design I asked for. I will appreciate if you can give me an example of designing such a circuit. @Andy aka?
– Alex
May 2, 2021 at 16:24
• There isn't anything all that special about any of the values in a crystal radio except the headphone impedance being high and using a germanium signal diode. Everything is so dependent on the operating frequency and length of the antenna it's probably best just playing with a few values in a simulator @Alex May 2, 2021 at 17:43
• You can only go so far before you get all voltage and no current (no current means no headphone acoustic output). It's not really related to exceeding the diode drop but in not reducing the Q of the tuned circuit. Lower resistance = lower Q = less channel selectivity. May 2, 2021 at 19:21

In my experience the crystal radio receiver is quite simple to build, but its low sensitivity requires two conditions:

1. A powerful local radio station in that area.
2. A long antenna, possibly four or five meters.

The radio frequency involved is not critical, the medium wave signals can be treated as audio signals with long wires and long signal path.

• Note that 1. can be a real problem, depending on where you are: long-wave stations are weak in most places of the earth (also, antennas for these will be unwieldy), and medium wave has been shot down throughout Europe.
– mmmm
May 2, 2021 at 9:44
• Yeah, I made a crystal radio once and the only way I could get it to work was by using a 6 meter piece of bare copper wire running up a tree as an aerial. I think the aerial is still up there in my back yard :) . May 2, 2021 at 9:45