I've been building an AM radio transmitter and had a few questions about my diagram. I built this using components on a breadboard. Having the resistor there is critical otherwise it doesn't work. The first diagram works great.

AM radio schematic diagram

What is this resistor doing? Why is it so important? If I remove it the transmitter basically stops working. The level of resistance in ohms doesn't matter too much oddly. In my mind, all the resistor is doing is lowering the voltage to the crystal. Just having it there makes it work. Adjusting the battery voltage doesn't have the same effect as the resistor. What's so special about it? Is it separating the MP3 player output from the battery somehow? I have no clue. It's dead without the resistor no matter the battery voltage.

Also, how is my MP3 player not shorting out due to a couple of volts going straight to the audio jack? The antenna and Vcc are both positive so I'm not sure how that affects the MP3 player output.

Basically, I'm trying to find out why this resistor is so critical.

AM radio with bypassed resistor schematic diagram

  • \$\begingroup\$ It's a transmitter. I have a Sony ICF-P27 Portable Radio to pick up the signal. \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 2:19
  • \$\begingroup\$ The output pin is the 1Mhz signal from the crystal oscillator. \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 2:42
  • \$\begingroup\$ The crystal ocillator module requires 5 V (or so) to operate. It is obtaining that supply through the resistor. If the resistor value is too high, it will drop too much voltage for the oscillator to work. \$\endgroup\$ Jan 8, 2023 at 2:58
  • \$\begingroup\$ @PeterBennett That makes sense, but why do I need it in the first place? I don't see it doing anything special. Varying the battery voltage doesn't do anything. When I insert the resistor there it magically works. I wish there was some way to simulate this online. I haven't found one that has a crystal. They are all pretty simple. \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 3:05
  • \$\begingroup\$ Random thought. I wonder if the resistor is working as an AC filter. Keeping the DC battery separate from the AC MP3 player output. I'm just guessing at this point. \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 3:09

1 Answer 1


It's a bizarre design. Looks like the MP3 audio output is causing the Vcc input to the crystal oscillator to vary, between 5V and 3V at the peaks and troughs of the MP3 audio signal output. As long as the oscillator's output is sensitive to Vcc, it will give you an AM signal.

The resistor is necessary to allow Vcc to vary with the MP3 signal. Without it, Vcc would be solidly clamped to the battery's positive voltage. A good value of the resistor would roughly match the designed load of the MP3 player, i.e. headphones which might run from 10's to 100's of ohms.

It also seems odd that the MP3 negative is connected to the oscillator output pin, rather than ground, but that's to allow the audio signal to drive Vcc both higher and lower than the median oscillator output voltage.

It's more of an odd hack than an educational circuit.

  • \$\begingroup\$ "than an educational circuit" well, it's certainly reminding me of just how oddball something can be and still work as an AM transmitter! \$\endgroup\$
    – TimWescott
    Jan 8, 2023 at 3:33
  • \$\begingroup\$ That's the part I'd like to know. "solidly clamped" What is the resistor doing to allow the MP3 player to vary the voltage so well? \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 5:49
  • \$\begingroup\$ Can you point me in the right direction for a proper AM circuit? I've been trying for a few years without much luck. Most of the designs online don't work or as you said are "odd hacks". I did have an early design using a transformer but I didn't like having so many bulky components. \$\endgroup\$
    – Derpy
    Jan 8, 2023 at 5:51

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