Convert LNA 433MHz Signal (21dB) to DC Voltage - Ryan's Question

[RLovelett]

My scenario:
LNA 433MHz signal (21dB) from antenna and the difference between the two signals with a 4.1dB gain. I would like to take the result of the difference amplifier an produce a DC value that is proportional to the output.

Question 1:
What type of circuit would lower the frequency to something like 50MHz?

Question 2:
What type of circuit would take the amplitude of a 50MHz sine wave signal and convert it to a DC voltage that is proportional?

Many thanks!
Ryan

 

[Corneo]

Answer 1:
A mixer

Answer 2:
A detector diode.

 

[oswaler]

Hello Ryan,

It sounds like you are working on a project involving a Low Noise Amplifier (LNA) and a difference amplifier to convert a 433MHz signal to a DC voltage. To answer your questions:

1. To lower the frequency from 433MHz to 50MHz, you would need to use a frequency down-converter circuit. This type of circuit typically involves a mixer, a local oscillator, and a bandpass filter. The mixer combines the incoming signal with the local oscillator signal to produce an output at the difference frequency, in this case 383MHz (433MHz - 50MHz). The bandpass filter then filters out any unwanted frequencies, leaving only the desired 383MHz signal.

2. To convert the amplitude of the 50MHz sine wave signal to a proportional DC voltage, you would need to use a circuit known as a demodulator. A demodulator works by separating the amplitude variations of a modulated signal, in this case the 50MHz sine wave, from the carrier frequency. This results in a DC voltage that is proportional to the amplitude of the original signal.

I hope this helps with your project. Best of luck!