# Miniproject: AM Radio

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2 3 BACKGROUND 2 to 20 khz range (this is the range of frequencies audible by a human being with good hearing). AM radio in the United States is broadcast in the 520 khz to 60 khz, so let s choose a carrier frequency ω c in this range that will be used to modulate our audio signal. We ll represent our audio signal s(t) and carrier c(t) by the following waveforms: s(t) = M cos(ω s t) c(t) = cos(ω c t) M represents the amplitude of the pure cosine that we re modulating. Modulating these signals simply involves multiplying them to produce the modulated signal, like so: = (A + s(t))c(t) = (A + M cos(ω s t))cos(ω c t) () We ve added a constant A to our audio signal before modulating in this case. There are actually a number of different types of amplitude modulation, but our version requires A M and is called double-sideband full carrier (A3E) modulation. It will become apparent in a moment why it carries this name. It s difficult to see how modulation has helped us until we note the following trigonometric identity: cos(a)cos(b) = [cos(a b) + cos(a + b)] (2) 2 If we apply this trigonometric identity to our modulated signal, we get: = Acos(ω c t) + M 2 [cos(ω st ω c t) + cos(ω s t + ω c t)] = Acos(ω c t) + M 2 [cos((ω s ω c )t) + cos((ω s + ω c )t)] = Acos(ω c t) + M 2 [cos((ω c ω s )t) + cos((ω c + ω s )t)] where the last step uses the fact that cosine is an even function. From this result, you can see that our modulated signal actually contains two sinusoids with half the amplitude of the audio signal one with frequency ω c + ω s and one with frequency ω c ω s and one sinusoid that is a scaled version of our carrier signal. Effectively, we ve shifted our signal up to the carrier frequency, but split it into two parts that surround the carrier frequency. These two parts are called sidebands (as in double-sideband ). To see why this method is called amplitude modulation, consider Equation ; you ll see that it s simply a high frequency cosine whose amplitude is dictated by a lower frequency cosine. Graphically, it appears that the amplitude of the high frequency cosine oscillates at the frequency of the low frequency signal, meaning that the amplitude of the carrier is modulated to carry the audio signal, as shown in Figure. 3.2 Demodulating a pure sine wave Now let s look at how we can retrieve s(t) if we know. Although there are a number of ways to accomplish the detection, the easiest involves multiplying by itself: = (A + M cos(ω s t)) 2 cos 2 (ω c t) = A 2 cos 2 (ω c t) + 2AM cos(ω s t)cos 2 (ω c t) + M 2 cos 2 (ω s t)cos 2 (ω c t) = A2 2 ( + cos(2ω ct)) + AM cos(ω s t)( + cos(2ω c t)) + M2 4 ( + cos(2ω st))( + cos(2ω c t)) where we ve used the trigonometric identity cos 2 θ = 2 ( + cos2θ). Although this expression appears complicated, recall that we only care about signals within the audio frequency range. We can eliminate any signals up near the carrier frequency because we can filter them out using a low-pass filter. Furthermore, any constant terms are also irrelevant, since they provide no audio information and can be filtered out with capacitive coupling. Thus, we can write our demodulated signal as: = AM cos(ω s t) + M2 4 cos(2ω st)

3 3 BACKGROUND 3 Voltage (arb. units) t s(t) c(t) Figure : Audio signal s(t), carrier signal c(t), and modulated signal depicted graphically for ω s = 2π khz, ω c = 2π 5 khz, M = 0.5, and A =. We re using a low frequency carrier here for demonstration only. Note our original signal is present (though scaled by the constant A), along with an attenuated harmonic at twice the frequency. Luckily, a simple ( diode can perform the demodulation. Recall that a diode has an I-V characteristic defined by I D = I S e V D/V th ). Let s look at what happens when V D = : ) I D = I S (e /V th Note that for small x, e x + x + 2 x2. We expect the voltage across the diode to be relatively small, so we can use this approximation to write: ( I D I S + ( ) ) 2 V th 2 V th Thus, we ve produced a signal containing () 2, our demodulated signal. Although we have an extra term added in, we can simply filter it out with a low-pass filter, since all of its signals are near the carrier frequency. Figure 2 shows how the demodulated signal d(t) looks prior to low-pass filtering and removal of the DC component. You can compare this to s(t) and see that they are identical aside from the high frequency components and DC offset present in the demodulated signal. 3.3 Demodulator circuit A simple demodulator is shown in Figure 3. The antenna (often just a long wire) picks up any radio waves that happen to be passing through the air. The LC oscillator composed of L and C pick out the radio frequency (RF) signals around its resonant frequency, given by ω = L C. The diode demodulates the RF signal, and the low-pass filter composed of R 2 and C 2 filters out the remnants of the high-frequency signals produced during modulation and demodulation. Note: Think of the antenna and diode as a current source and the low-pass filter as a transresistance amplifier (current input and voltage output) if it isn t clear why this is a low-pass filter. There are two demodulator schematics shown in Figure 3. The only difference between the two is that the one shown in Figure 3(b) provides better signal reception. You are free to build either one. (Note: This semester, the inductor we re using only has two taps, so you ll need to build the demodulator in Figure 3(a)). For the two demodulators, use a nf capacitor for C 2 and a 00 kω resistor for R 2. If you are building the one shown in Figure 3(b), use a 00 pf capacitor for C 3. As for the inductor, the inductor given to you in the kit has several inductor coils inside the same unit. You will only have to use the terminals as shown in Figure 3.

4 4 BIAS VOLTAGE 4 Voltage (arb. units) t s(t) d(t) Figure 2: Audio signal s(t), modulated signal c(t), and demodulated signal d(t) depicted graphically for ω s = 2π khz, ω c = 2π 5 khz, M = 0.5, and A =. C 3 d(t) White D L 2 White D C 2 d(t) Black C 2 R 2 L Y ellow C R 2 L Y ellow C (a) (b) Figure 3: (a) Simple demodulator schematic (b) A better demodulator schematic 4 Bias Voltage In previous labs, you were instructed to use the function generator offset function to bias the base/gate of a transistor. However, in the case of amplifying the small signal produced by the demodulator, there is no offset function to set the bias! How do we add a DC offset to the signal coming out of the demodulator to bias the transistor in the proper operation region? In this section, we will provide you with a starting point on biasing your transistor. In Lab 6, you learned how to build a voltage source with a diode connected transistor, such as the one shown in Figure 4(a). If you want to add this DC voltage to the AC signal received from the demodulator, you will have to do a little trick to add them together. Figure 4(b) shows how this can be done. In the large signal s perspective, the capacitor is an open circuit to the DC voltage, so the DC voltage has to go into the amplifier. In the small signal s perspective, the DC voltage source is ground to the AC signal. Therefore, you need to put a LARGE resistor (around 00 kω) in front of the DC voltage source to prevent AC signal from flowing out to AC ground. So what resistance should R BIAS be? This depends on the voltage output of the voltage source. For example, if you want the voltage source shown in Figure 4(a) to produce a voltage V OUT, you can use KVL

5 5 BIAS CURRENT 5 V DD V DD R BIAS I BIAS R BIAS I BIAS + V OUT R DC C AC To Amplifier (a) AC Signal Source (b) Figure 4: (a)voltage source built with diode connected transistor (b) Schematic for adding the DC voltage produced by the voltage source to an AC signal and the current equation for the transistor to find the required resistance: V OUT = V DD I BIAS R BIAS V OUT = V DD W 2 L k (V OUT V TN ) 2 R BIAS V DD V OUT R BIAS = W 2 L k (V DC V TN ) 2 You can also build a voltage source with BJT. When calculating the required resistance, you just need to replace the MOSFET current equation with the BJT equivalent. 5 Bias Current Besides voltage biasing, sometimes you may want to current sources to bias your circuit. For example, the common source amplifier shown in Figure 5 uses a current mirror instead of a resistor for biasing. The advantage of this design is to increase the output resistance of the amplifier to increase the gain of the amplifier. Let s think about how much current we should bias the transistors. Recall the gain of a common source amplifier is: A v = g m R out In this particular example, the output resistance R out is equal to r o r o2. For simplicity, let s assume that r o = r o2. This assumption is valid when the channel length modulation parameters λ and λ 2 are equal, which is the case most of the time when using matching pair of transistors. Then, you can simplify the gain equation to: If we substitute A v = g m (r o r o2 ) = g m r o 2 2I D V GS V T N for g m and λ I D for r o, then we get: 2I D A v = V GS V TN λ I D 2 = (V GS V TN ) λ

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