Why do we need a diode in an AM demodulation circuit

amplitude modulationdemodulationradioRF

The incoming signal is a modulated carrier wave, right? If we only apply a low-pass filter to it, we will get an AC signal that can be passed to a speaker directly, right?

Using a diode, we end up with DC that must then be turned to AC again for the speaker to work, so why bother with the diode? What does adding it actually do?

Best Answer

If we only apply a low-pass filter to it, we will get an AC signal that can be passed to a speaker directly, right?

No, not at all. The low-pass filter would just block the RF signal (your modulated carrier) completely. So, you'd get nothing.

Using a diode, we end up with DC

Not really!

A diode alone doesn't smooth anything. Even assuming an ideal diode (which seems to be the diode model you have in your head, which is pretty far from correct for RF signals), you'd get a signal that is 0 when the input signal is negative (i.e., half of the time), and the original signal when it's positive.

So, your signal after the diode is kind of "twice as high" in frequency than your original RF signal; I really can't call that "DC"; I'd call it "high-frequency AC with a positive voltage offset".

The trick here is while the average of an RF modulated carrier is 0, no matter its amplitude, the average of the RF modulated carrier with the negative halves "cut off" suddenly becomes positive, and proportional to the amplitude of the carrier.

That's why we call it amplitude modulation!

that must then be turned to AC again for the speaker to work

Not really, a speaker would still work with a DC offset, but it's a non-issue anyways:

You're not putting the signal coming out of the diode directly to the speaker – you still need the low-pass filtering step, which is actually doing the "average" mentioned above.

"Adding the negative halves back to make things look like AC" is inherent to the way you then amplify the demodulated signal; you can just add a series capacitor.

What does adding it actually do?

Without it you don't get a receiver, so I'd argue the diode in a diode detector is the central part.

The diode fulfills the purpose of applying a non-linear function to input signal, if you need to think about it in more mathematical terms:

Say we start with a sine carrier of frequency \$f_c\$ with amplitude \$A_m\$. This \$A_m\$ is always positive, \$A_m>0\$. (so, it's the audio plus a constant offset such that it's never "dipping" into negative values.)

Because that amplitude contains our message, I'm giving it the subscript \${}_m\$.

For example, \$f_c=100\,\text{kHz}\$ or \$f_c=144\,\text{MHz}\$, so much much "faster" than our message changes. That's why I'm not giving \$A_m\$ an "of time" \$(t)\$; for all we care on short times, \$A_m\$ is unknown, but stays the same.

Let's call our receive signal \$r(t)\$ (so, it's a function of time \$t\$: at different times \$t\$, it has a different values. So, here's our reception of our modulated carrier:

$$r(t)= A_m \cdot \sin(2\pi f_c t)$$

So, we want to get the \$A_m\$ from that formula, and get rid of the \$\sin(\ldots)\$! A filter can't do that, because all a filter ever does is multiply each frequency with its own gain, so you might get a "dampened" \$A_m \cdot \sin(2\pi f_c t)\$, but not anything without the high-frequency sine. A filter can't shift in frequency!

Any non-linearity can. To illustrate that, you need to understand that a diode does not actually do the "off when negative, fully on when positive" thing you've learned. A diode has a diode curve, which is an exponential function that is very close to zero in the negative inputs, and very quickly grows for positive inputs.

Because radio signals are by and large pretty weak, we can concentrate on values a few microvolts to millivolts below and above 0 V (depends on your antenna and your receive amplifier, different story). In a pretty good approximation, for such small inputs, we ignore the complicated exponential function of that diode and go just with "yeah, it's quadratic. It takes the input and squares it!" (and everything that it doesn't square stays at \$f_c\$, so our low-pass filter throws it out anyways. We ignore it.)

So, yeah, our diode squares our input. We get:

\begin{align} (r(t))^2 &=(A_m \cdot \sin(2\pi f_c t))^2\\ &= A_m^2 \cdot (\sin(2\pi f_c t))^2 \end{align}

Here, we need to ask our friend Wikipedia for help (unless we had a fresh calculus course, and know how to prove such things ourselves). There's things we can say about squares of sines, and they say that $(\sin (x))^2 = \frac{1-\cos(2x)}{2}$. So, in our formula:

\begin{align} &= A_m^2 \cdot (\sin(2\pi f_c t))^2\\ &= A_m^2 \cdot \frac{1- \cos(2\cdot2\pi f_c t)}{2}\\ &= A_m^2 \cdot \left(\frac{1}{2}-\frac{\cos(2\pi (2\cdot f_c) t)}{2}\right)\\ &= \frac{A_m^2}{2}-A_m^2\frac{\cos(2\pi (2\cdot f_c) t)}{2} \end{align}

Waaaait, that last line is strange: now we have two added components (from our multiplicative \$A_m\cdot\sin\ldots\$), and while the first one is just a constant value \$A_m/2\$, the second is at twice the carrier frequency!

OK, we throw our low-pass filter at that, which we designed to cut off anything above what is audible. That especially cuts away anything at twice the carrier frequency! We get

$$ m(t) = \frac{A_m^2}{2}. $$

Neat. There's our \$A_m\$, without the modulation onto the carrier (note: the carrier frequency didn't even matter! Just that it's much higher than our message signal's frequency, so that our low-pass filter can cut it off.)

Yeah, and as you said, that's not a great signal to put through a speaker, so we feed it to our audio amplifier through a DC-blocking capacitor, which removes the DC offset that the original \$A_m\$ had.

(In real AM radio a bit more is going on, to make the \$A_m^2\$ not be a mess compared to the original \$A_m\$, but that's honestly just "improvements of audio quality"; not really necessary.)

Related Topic