Register to reply

Confusion About Simple AM Radio Receiver/Tuner

by Glenn900
Tags: confusion, radio, receiver or tuner, simple
Share this thread:
Glenn900
#1
Apr15-12, 02:02 PM
P: 4
I am confused by the design of this AM receiver/tuner circuit.

http://imgur.com/f9n05

I don't understand how the inductor and capacitors on the left design a circuit that resonates at a certain frequency, and I also don't understand why the resistor and capacitor in parallel after the diode create a low-pass filter.

Are the inductor and two capacitors not in parallel with each other? This circuit makes more sense to me if I consider the two capacitors to be in parallel with each other, and in series with the inductor. Then the inductor and capacitors should exhibit resonance at a given frequency and act as a band-pass filter, passing this specific frequency to the rest of the circuit. Is this right?

Can anyone help me out with this circuit?
Phys.Org News Partner Engineering news on Phys.org
Creative adaptation of a quadcopter
Filling up could cost less thanks to biomolecular engineering breakthrough
Tesla says decision on battery factory months away
skeptic2
#2
Apr15-12, 04:03 PM
P: 1,803
In this circuit it doesn't matter if you consider the two capacitors in parallel are in series or in parallel with the inductor, the effect is the same.

Imagine the inductor as a mass that is being moved at a regular frequency by the antenna. The capacitors are like a spring that is alternately compressed and stretched by the oscillating mass. Since the spring keeps the mass oscillating, the antenna only has to supply enough input power to offset the losses due to friction. The oscillation of the mass will increase until the losses equal the power input from the antenna.

Likewise in the circuit, the voltage in the inductor-capacitor circuit increases until the losses equal the power supplied by the antenna. If the voltage is high enough, the positive peaks of the oscillation will pass through the diode and charge the capacitor. The 10K resistor allows the capacitor to discharge when the peak of the signal is less than the voltage the capacitor is charged to. In this way the voltage across the final capacitor follows the peak voltage of the signal. When amplified it provides an audio output.
vk6kro
#3
Apr15-12, 05:21 PM
Sci Advisor
P: 4,016
Quote Quote by Glenn900 View Post
I am confused by the design of this AM receiver/tuner circuit.

http://imgur.com/f9n05

I don't understand how the inductor and capacitors on the left design a circuit that resonates at a certain frequency, and I also don't understand why the resistor and capacitor in parallel after the diode create a low-pass filter.

Are the inductor and two capacitors not in parallel with each other? This circuit makes more sense to me if I consider the two capacitors to be in parallel with each other, and in series with the inductor. Then the inductor and capacitors should exhibit resonance at a given frequency and act as a band-pass filter, passing this specific frequency to the rest of the circuit. Is this right?

Can anyone help me out with this circuit?
There is a transformer at the left. This carries a signal from the antenna in the primary and the secondary is tuned with the capacitors so that it resonates with them at one frequency.

At this frequency, there is a step-up in voltage and this voltage appears across the secondary of the transformer.

This voltage is then rectified by the detector diode and the resulting waveform is filtered to remove the high frequency components, leaving only audio. This is amplified in the amplified speaker.

The capacitor C is variable and allows tuning of the parallel tuned circuit to receive different frequencies.


Register to reply

Related Discussions
How does charge pass through a capacitor in a radio tuner Classical Physics 1
First Project - Simple FM Radio/Receiver made Complex Electrical Engineering 1
Series radio tuner Electrical Engineering 1
Radio Tuner without an inductor coil Electrical Engineering 1
Analog Radio Tuner Electrical Engineering 2