I am trying to follow this circuit diagram:

enter image description here

Taken from (http://www.techlib.com/electronics/crystal.html)

Using two oscilloscopes I am watching the signal on the tuner circuit and the voltage created by the amplifier. First, before connecting the RF amp circuit, I adjust the tuner to ensure I am getting a good signal, then when I connect the positive terminal from the capacitor to the tuner circuit, the amplified signal does not appear to be modulating--I'm pretty sure it's not the proper signal.

My configuration is very slightly different:


simulate this circuit – Schematic created using CircuitLab

I suspect that the RF amp circuit is affecting my tuning, but I don't know why and I don't know how to fix or compensate for this.

  • \$\begingroup\$ Your configuration is not "slightly" different, it is completely different. Why not copy the schematic? \$\endgroup\$ Commented Oct 15, 2015 at 3:06
  • \$\begingroup\$ @KevinWhite I've updated the diagram to show what I've done. I'm not sure how different it is. I thought it would be nearly the same since I'm still using a tank circuit \$\endgroup\$
    – Klik
    Commented Oct 15, 2015 at 16:56
  • \$\begingroup\$ The design you have there will not work very well as you have the amplifier directly across the tuning element (C1). The original circuit was designed that way for a reason; to reduce the loading on the tuned circuit by the amplifier or detector. Why not implement as the original design? Do you have a specific objective for implementing it like you are? \$\endgroup\$ Commented Oct 15, 2015 at 21:01
  • \$\begingroup\$ The way I have it implemented, the inductor in the tuner circuit acts as a transformer; the signal from the antenna runs through the first few coils and then straight to ground. This way I get a better signal, (which was my issue before). However, I am really curious how this is affecting my tuned circuit, this is the answer I am looking for. I thought the inductor and capacitor were in parallel so the impedance is calculated as such, but I don't know how the amp circuit is affecting it. \$\endgroup\$
    – Klik
    Commented Oct 15, 2015 at 23:38
  • 1
    \$\begingroup\$ You have C4 directly connected to the power supply. The power supply directly connected to the diode. C4 should connect to the diode and R4 BUT not to the power supply. R4 should be 1k, not 100 ohm. The PSU should be 9v. Why are you modifying the circuit given on techlib? \$\endgroup\$ Commented Oct 16, 2015 at 2:50

1 Answer 1


Your RF "amp" is loading the tuned circuit. When the gain pot is at Rmin you have a prospective voltage gain of maybe 100 but the input impedance could be only 500 ohm. So the mismatch is robbing your gain. When the gain pot is at Rmax, Zin goes up but gain is less than 2 anyway. This could be fixed by using a JFET like MPF102 or an emitter follower or tapping down the coil closer to ground.

  • \$\begingroup\$ I have further analyzed the circuit and I believe what you're suggesting is not the correct answer. The reason I think so is that after attaching the RF amp circuit, I must adjust my variable inductor or capacitor to get the signal back. \$\endgroup\$
    – Klik
    Commented Oct 15, 2015 at 16:29
  • \$\begingroup\$ Hold on, you might be right and it's just that I am not understanding your answer. Kevin White also mentioned the amp circuit loading my tuned circuit. Please excuse me, I am very new to circuit design and am only familiar with analyzing DC circuits. I need an answer good enough for a "noob" like me. \$\endgroup\$
    – Klik
    Commented Oct 15, 2015 at 23:41

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