Imagine the following situation: You recently notice that one of your grandparents is getting hard of hearing in one ear. However, they are reluctant to purchase a hearing aid because they don’t want to be seen wearing one. Then it strikes you, after completing the Opamp topic in EE2002, that you could easily design a DIY microphone amplifier that would serve as a hearing
aid, using a standard hands-free headset with a 3.5mm audio jack. Your battery-powered amplifier could also be hidden in a pocket, and then your grandparent would look like they are listening to music with a headset.
So you investigate. Using a function generator to drive the headset, you discover that the output waveform needs to reach ~400mVpk (0.8Vpp) before your grandparent can hear it at 1kHz. But you also discover that your average microphone puts out only ~20mVpk signal for soft sounds. So you decide to amplify it by a gain of 20x to reach 400mVpk using a rail-to-rail LT1880 opamp, running on a rechargeable 2xAA supply (Vs = ±1.25V) as shown below.
1) Observe that Vic = Vi = 40mVpp as expected. Design the LT1880 opamp to act as a microphone preamplifier with a gain of 20x such that the output is 400mVpk (0.8Vpp).
- Note that a typical condenser (electret) microphone is best modeled as a Norton equivalent current source. A standard part like the AUM-5047L requires 0.2 mA of DC bias current and hence drops 1V across the 5 k load; Vic, therefore, has a DC value of 0.25V which is filtered out using Comic, hence Vi above will have no DC component (centered at 0V). The audio signal from the mic is represented as 4 uA of peak AC current, and hence produces 20mVpk (40mVpp) of AC voltage across the 5 k load. Right-click on I1 in LTspice to see these parameters.
2) But if you try to connect the output of the opamp to both your earbuds, which each have a load resistance of 32 (hence setting RL = 32 // 32 = 16 as highlighted below), you should discover that the opamp output Vo becomes heavily distorted:
- This is because the LT1880 is rated to only drive a maximum load current of 20mA. The load current for driving 400mVpk into 16 = 25 mA and exceeds this max limit.
A BJT amplifier can now come to the rescue! A common-collector amplifier has the potential to drive more than 25mA from the supply and can serve as a more ideal voltage buffer if you bias the Q-point with IC > 25mA. But what should it be exactly? This assignment will guide you to figure that out. For ease of calculation, you may treat the most negative voltage as 0V (Ground) rather than using Vs- = -1.25V. Hence VCC = 2.5V as shown in Figure 1.\
