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I've been trying to make an extremely simple AM receiver (without even a tuner), however, it doesn't seem to be working.

As shown in my schematic, what I did was connect the antenna wire to the base of a BC547 transistor. The collector of that is connected to a 9V battery and the emitter is connected to the base of another BC547.

This BC547's collector is connected to the same 9V battery and the emitter is connected to a germanium diode (so that only positive voltage radio signals are received), and this diode is connected to a piezo buzzer, which is then connected to ground.

So basically, I'm using the transistors to amplify the antenna's signal so that the piezo buzzer can make noises as it receives AM radio signals.

When I connect the power, I hear a chirp initially, but nothing else. Even if it's extremely low power, I should still hear some small chirps/noises right? What am I doing wrong? Thanks.

enter image description here

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  • \$\begingroup\$ How long is the antenna? Can you connect an earphone instead of the buzzer? \$\endgroup\$
    – Chu
    Apr 13, 2019 at 16:33
  • \$\begingroup\$ Thanks for your reply. The antenna is about 6 feet. Also, I have a low-power AM transmitter I made, and I put its antenna right next to the receiver antenna, so I know that power probably isn't an issue. I don't have an earphone, so unfortunately, no. \$\endgroup\$
    – F16Falcon
    Apr 13, 2019 at 16:50

3 Answers 3

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Seems an amplifier is needed, as well as a rectifier and current buffer.

schematic

simulate this circuit – Schematic created using CircuitLab

This is a low-input-capacitance gain of 300x amplifier (gm of Q1 at 0.1mA, times 47Kohm). Then we add a diode lightly biased on with 4uA, with the collector of Q3 sitting at 4 or 5 volts across the piezo.

Experiment with adding a small capacitor across base-emitter of Q3, for more sensitivity (like a PEAK_HOLD capacitor).

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  • \$\begingroup\$ Thanks! So the "Pick LC" part that you have on the left side can replace the antenna in the main circuit to make this a "specific-channel" radio, right? \$\endgroup\$
    – F16Falcon
    Apr 14, 2019 at 13:59
  • \$\begingroup\$ Also, if you're able to make a simplified version (as simple as you can... I'm a beginner haha) of a channel-specific radio for my second question, I'd appreciate it and give you my vote there! \$\endgroup\$
    – F16Falcon
    Apr 14, 2019 at 14:00
  • \$\begingroup\$ you want to detect some (weak, remote) AM signal, right? not just detect that AM transmitter you can hold near the circuit. Unless you have a huge antenna (many many meters long and high up), you'll need to amplify, and then detect, and then boost the current. Here the first 2 transistors are a cascode (low input capacity) amplifier (which does not waste the RF energy in MillerEffect charging of Cob); then the diode detects, and finally Q3 boosts the current. \$\endgroup\$
    – analogsystemsrf
    Apr 14, 2019 at 16:57
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The diode really isn't necessary (transistor base-emitter junction is sufficient).
But there are a few flaws....

  • The antenna has no ground reference. So much of the detected voltage goes to the antenna, not to the piezo.

  • The piezo is an AC device, like a capacitor. It would be better to provide a DC path from emitter to ground.

The inductor may be hard-to-find. You might try substituting a large-value resistor (one Megohm or larger).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Even the inductor isn't necessary. I know because I've done it by accident. \$\endgroup\$
    – Robert Endl
    Apr 13, 2019 at 17:08
  • \$\begingroup\$ I tried this, and I also tried removing the inductor/resistor. It still doesn't work for some reason. Any other suggestions? \$\endgroup\$
    – F16Falcon
    Apr 13, 2019 at 17:45
  • \$\begingroup\$ Also, I should clarify that my buzzer takes in DC voltage (it has some circuit inside that converts the DC voltage to make the piezo vibrate). \$\endgroup\$
    – F16Falcon
    Apr 13, 2019 at 17:46
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    \$\begingroup\$ I'd assumed a simple, raw piezo, not one that adds an active oscillator driven with DC. I don't think the DC-driven buzzer can be made to work in this circuit. \$\endgroup\$
    – glen_geek
    Apr 13, 2019 at 20:08
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That circuit has a capacitive load: the piezo. Once charged up, there is no discharge path.

I'd make about 1milliSecond discharge time constant. Assuming 10,000pF for the piezo (or 1e-8 farads), place 100,000 ohms across the piezo.

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  • \$\begingroup\$ I tried doing this, but it's not working still. To clarify, I have one of those buzzers that can take in DC voltage (it has some sort of circuitry inside to make the piezo move using DCV). Any suggestions? \$\endgroup\$
    – F16Falcon
    Apr 13, 2019 at 17:44
  • \$\begingroup\$ Does your buzzer need about 5 volts, to make a sound? \$\endgroup\$
    – analogsystemsrf
    Apr 13, 2019 at 18:12
  • \$\begingroup\$ Nope, just 1.5V. Here's the datasheet: taydaelectronics.com/datasheets/A-5019.pdf \$\endgroup\$
    – F16Falcon
    Apr 13, 2019 at 18:23

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