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I am trying to make a radio receiver for 800 kHz AM radio signal.

I connected an antenna with an LC filter circuit which is then connected to a voltage divider biased BJT amplifier whose output is connected with a diode and speaker.

Will this reciever work? Will I be able to listen to 800 kHz radio signal if all the components are connected and their values are chosen properly?

Will you please give me any suggestions for this circuit to make it work?

schematic
(My drawing)

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    \$\begingroup\$ No, R1 & R3, DC shorted to ground via L ... Add a capacitor (10-100 nF) between the two points (common R1 & R3 and C). \$\endgroup\$
    – Antonio51
    Commented Oct 16 at 11:17
  • \$\begingroup\$ Will it work if I add a capacitor between C and mid point of R1 and R3? \$\endgroup\$
    – Alex
    Commented Oct 16 at 13:08
  • \$\begingroup\$ You also have a "short" through the diode and the speaker of the transistor's collector. Your envelope detector will not work well either. Have you already looked at any references for an AM receiver on the internet? \$\endgroup\$
    – GNA
    Commented Oct 17 at 9:11

4 Answers 4

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Will this reciever work?

No, not as shown...because the transistor is not biased properly.
However, it is a good start:

  • Antenna signals feed a bandpass filter that selects one frequency preferentially.
  • A transistor gain stage boosts signal before a diode detector

Another answer suggests re-arranging the order so that the diode detector comes first, followed by an audio amplifier that drives an audio transducer. This has the merit that an audio amplifier would probably have more gain than would a radio-frequency amplifier. However, a diode detector deals poorly with tiny signals from an antenna, providing little audio signal to the following audio amplifier.

Here's a suggested way to bias the transistor so that its base sees the 800 kHz signal on top of a DC bias from R1 & R3 (leftmost circuit). The transistor amplifies 800 kHz signal first, then drives a diode detector through a coupling capacitor to headphones. A loudspeaker has such low resistance that transistor gain would be difficult to achieve...high impedance headphones would allow more gain:

schematic

simulate this circuit – Schematic created using CircuitLab


On the right, a ferrite rod is used as RF selector AND antenna. This is often done in transistor radios. A link winding on the rod having fewer turns than the main winding provides a lower impedance signal to the transistor, which gives higher gain.
It would be convenient to take the ferrite rod from another radio, along with its variable tuning capacitor...everything is already optimized so that the transistor sees a good RF signal between base and the bias resistors R8, R6. A bypass capacitor C9 provides an AC ground reference for radio-frequency signals.
DC base bias current flows through the small link winding on the ferrite rod.


Edit:
The left circuit requires antenna wire connected to C2 & L1 and a connection to GROUND as well. At 800 kHz frequency, most any length wire antenna will add capacitance which adds to C2, affecting resonant frequency.
The ferrite rod transformer of the rightmost circuit needs neither antenna nor ground.

Bias resistors R1 & R3 at left circuit should be chosen on the high side so that they don't load down L1,C2 resonance...more than tens of kilohms.
To achieve high gain, both Rheadphone and R2 ideally would be above 1000 ohms. D1 would best be germanium or schottky rather than silicon. This also applies to right circuit.

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  • \$\begingroup\$ Thanks alot really. I am interested in the diagram you posted on the left. Will the reciever work if choose the values shown in the diagram and then choose suitable values for R1, R2, R3 and R4. \$\endgroup\$
    – Alex
    Commented Oct 17 at 5:03
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    \$\begingroup\$ You may be disappointed at weak headphone signals unless your 800 kHz signal source is a local high-power broadcaster. Capacitor values shown should be roughly correct for 800 kHz. You might choose bias resistors so that Q1 has 2-5 mA DC current. This would require a DC voltage source of about 9-15V. \$\endgroup\$
    – glen_geek
    Commented Oct 17 at 13:01
  • \$\begingroup\$ And will it work if it is near the broadcaster? I can use multistage amplifier and power amplifier aswell of required. All I need to know is that will it work under ideal strength of the signal? \$\endgroup\$
    – Alex
    Commented Oct 17 at 14:16
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    \$\begingroup\$ Work well?...component selection & construction technique influence success greatly when you attempt radio-frequency amplifiers. When I simulated this in LTspice, some subtle influences showed up. I could get the transistor to oscillate by changing some values. Not necessarily a bad thing, but near-oscillation can be confusing if you don't have an oscilloscope to see waveforms and then know what components to adjust. This dead-simple AM receiver i.sstatic.net/Ye9QT.png can work very well with local AM broadcast signal with good headphones (2kohm) \$\endgroup\$
    – glen_geek
    Commented Oct 17 at 18:14
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    \$\begingroup\$ R1 & R3 have large latitude to achieve a desired transistor \$I_C\$. You'll often see these in the 10's of kilohms, perhaps as high as 100's of kilohms, I'd suggest choosing R1 arbitrarily at 100k, then calculate R3, R4 to get \$I_E \approx 3mA\$. \$\endgroup\$
    – glen_geek
    Commented Oct 18 at 12:55
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The one in the photo is a draft project of an am- receiver (m is the modulation index) ... it should work .... once all the components are sized ...

enter image description here

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  • \$\begingroup\$ Is this just Impedence matching network between C1 and Ci? \$\endgroup\$
    – Alex
    Commented Oct 16 at 15:57
  • \$\begingroup\$ The receiver consists of the antenna magnetically connected with a circuit tuned to the frequency of the carrier fc followed by the detection group with the diode, peak detector, connected to the low-pass RC filter that attenuates the unwanted RF components making up the ripple. The capacitor Cc blocks the continuous component as well as Ci (necessary to not modify the polarization of the BJT) . The potentiometer is used to vary the audio volume. Finally the signal is amplified by the common emitter BJT circuit. See this Schematic \$\endgroup\$ Commented Oct 17 at 6:58
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Both the input and output have DC bias problems. On the input the coil is a DC short across R3 and the transistor base so the transistor would be biased below cutoff. The diode and speaker on the output will limit the collector voltage to around one diode drop, approximately 0.6 V.

When amplifying an AC signal such as audio or RF the transistor bias needs to be isolated from any DC paths in the input and load circuits, typically with series ‘blocking’ capacitors or transformers.

Also the input impedance of a CE amp with voltage divider bias is relatively low while an LC tank circuit has a high impedance, so there needs to be impedance matching. This can be done a number of ways such as tapping the amplifier input down on the coil (connected to a tap close to the grounded end of the coil).

Likewise the amplifier will have a relatively high output impedance compared to a speaker, so in addition to DC blocking the output you need to either use a higher impedance load such as high impedance headphones or add a power amplifier to drive a speaker.

There are other improvements that could be made, I would recommend doing some research on receivers and amplifiers, amateur radio books are often a good resource for this and many can be found online.

Addendum:

Trying to design a receiver from scratch is going to be problematic, if it doesn't work you don't know if it's because it's a poor design or if you've built it wrong. I would start with a known working design, get that working and go on from there. One thing you might look into is reflex receivers, they use one transistor to amplify both RF and audio. Eventually you can work your way up to superheterodyne receivers or even software defined radio.

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  • \$\begingroup\$ Really Thanks alot. Will this circuit work if I add a capacitor between lc circuit and the amplifier as you suggested and if the Impedence matching is made. Actually I am a beginner I understand about those compoments and I want to use them to make a reciever. \$\endgroup\$
    – Alex
    Commented Oct 17 at 5:33
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    \$\begingroup\$ @Alex I added a bit to my answer, have a look at it please. \$\endgroup\$
    – GodJihyo
    Commented Oct 17 at 15:58
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You keep using the term "speaker" - it will not work, because common speakers are like an 8-36Ω resistor. They will be unnecessarily quiet. High impedance headphones are needed for "listening" to the demodulator diode - or an amplified speaker or headphones with an amplifier (good bluetooth models that support wired operation are fully amplified when turned on).

On the RF side, you'd use a double variable capacitor for tuning and have one section tune the antenna tank, and the other section tune a collector tank. That way the amplifier would be selective. Or you can lower the Q of the antenna tank and make it wideband - then it won't need tuning, but you lose selectivity.

You'd want another section as an output amplifier. Connecting the demodulator diode directly to the speaker or headphone driver coil is a pessimization that unnecessarily cripples the performance of the circuit.

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