ECEN 1400 - Introduction to Digital and Analog Electronics

Peter Mathys, Spring 2014


Lab 3: Low Power Audio Amplifier, Introduction to Transistors

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Goals of this Lab


Prelab

P1. Transistor Amplifier. Transistors are used for amplification (in analog circuits) and switching (in digital circuits). There are two basic types of transistors, bipolar junction transistors (BJT) and field-effect transistors (FET). Here we are interested in using a BJT for amplification. The physical structure, the schematic symbol, and the current-controlled current source (CCCS) equivalent circuit of a NPN BJT are shown in the following figure.

Physical structure, schematic symbol, and CCCS equivalent circuit for NPN BJT

Most BJTs are made from silicon that has been altered either by adding impurities with additional electrons (N-type) or by adding impurities with less electrons (P-type) than pure silicon. NPN refers to a BJT that is made of two N-type silicon regions with a P-type layer sandwiched in-between. An important parameter of a BJT is its current gain beta. For the small signal transistors in your kit (2N2222 or 2N3904) beta is typically about 200. Note that a BJT has three terminals, labeled base (B), collector (C), and emitter (E). The main current flow is between the collector and the emitter and the base is used to control this current flow. The base emitter voltage vBE that is needed to enable the collector emitter current flow is typically 0.7 V for silicon transistors. The simple CCCS model works well for such purposes as setting the dc operating point. For small signal ac analysis the transconductance model using a voltage-controlled current source (VCCS) as shown below works well.

Transconductance model for NPN BJT

In this problem the goal is to simulate and analyze the following simple transistor amplifier whose schematic is shown below on the left. Vcc denotes the voltage that is used to power the circuit. The corresponding voltage source is not shown explicity in the schematic. On the right side the CCCS equivalent circuit is drawn.

Simple transistor amplifier with NPN BJT, dc analysis

Using the CCCS model, the base current iB, the collector current iC, and the collector voltage vC can be computed for the dc operating point as shown above. Note that for dc the capacitor acts like an open circuit. Compute iB, iC, and vC for Vcc = 9 V, beta = 200, vBE = 0.7 V, RB = 1 megaohm, and RC = 10 kiloohm.

For ac the voltage the dc sources (vBE and Vcc) act like short circuits and the corresponding transconductance model for small ac signals is shown next.

Simple transistor amplifier with NPN BJT, transconductance model

The analysis of this equivalent circuit is shown below. First i2(t) is computed and then the ouptut voltage vC(t) as a function of the input voltage vS(t) is obtained. The ac gain G of the circuit is computed as G = vC(t)/vS(t).

Simple transistor amplifier with NPN BJT, ac analysis

Compute the tranconductance gm and the ac gain for the same numerical values of RB, RC, and iC as specified found above.

To be able to simulate this circuit, build the following schematic in Multisim.

Simple transistor amplifier with NPN BJT in Multisim

Measure the base current iB, the collector current iC, and the collector voltage vC using a multimeter or an instantaneous voltage and current measurement probe in Multisim. Compare the results from the simulation with the values you computed above.

To measure the ac gain G for sinusoidal voltages use the following schematic in Multisim.

Simple transistor amplifier with NPN BJT in Multisim, with ac input

Note the voltage divider at the input which is necessary because the lowest amplitude that the waveform generator can produce is 50 mVpp. This value is too large for the linear range of this transistor amplifier. Measure G at frequencies f = 1 kHz, 10 kHz, 100 kHz, and 1 MHz. Compare the measured values of G with the value that you computed above. Make sure that you make your measurements using sinusoidal waveforms that are not distorted. Choose the amplitude from the waveform generator accordingly.

Next, change RB to smaller values (e.g., 470 kiloohms, 100 kiloohms, and 47 kiloohms), leaving RC and Vcc unchanged. For which value of RB do you obtain the largest ac gain G at f = 10 kHz? Use both analytical computations and Multisim simulations to verify your results.

P2. Basic LM386 Audio Amplifier. In this problem the goal is to build the following circuit in Multisim using the LM386 low power audio amplifier.

LM386 Minimum Parts Schematic in Multisim

There is one catch: The LM386 is not directly available in Multisim. You will have to create this part first before you can use it. The detailed instructions for doing this are here.

In the schematic above, can you spot any RC lowpass or highpass filters? If so, which? What are their cutoff (or -3 dB) frequencies?

P3. Performance of the LM386 Audio Amplifier. The goal of this problem is to measure the following performance parameters of the LM386 circuit in P2:

Devise a measurement scheme for each of these items. The power PL delivered to a load resistor RL by a sinusoidal waveform with peak-to-peak voltage Vpp is computed as

Formula to compute PL for RL driven by sinusoidal voltage

To determine the input resistance of the LM386 try the following strategy. Set the amplitude of the waveform generator to a specific value, e.g., 100 mVpp and measure the resulting output voltage vO. Then insert a resistor like R3 in the schematic below between the waveform generator and the LM386. Change the value of R3 until vO drops to one half of its value (without changing anything else). The value of R3 for which this happens is equal to the input resistance of the LM386.

Schematic with resistor R3 to determine input resistance of LM386

Lab Experiments

E1. Transistor Amplifier. Build the amplifier whose schematic was given in prelab problem P1. Repeat the measurements specified for P1 (dc currents iB, iC, dc voltage vC, ac voltage gain G for f = 1 kHz, 10 kHz, 100 kHz, 1 MHz, and value of RB that yields highest G at f = 10 kHz) on the real transistor. To identify the pins of the transistor use the datasheet, either for the 2N2222 transistor or the 2N3904 transistor. In the labe report specify which type of transistor you used.

E2. Basic LM386 Audio Amplifier. Build the LM386 audio amplifier according to the schematic given in prelab problem P2 (repeated here for convenience) on your breadboard.

LM386 Minimum Parts Schematic in Multisim

Note capacitors C3 and C4. They should be mounted close to the LM386 integrated circuit. The function of C4 is to act as short term energy storage for the time-varying instantaneous power needs of the amplifier. Capacitor C3 acts as a short circuit for high frequencies across the power supply (for ac purposes a dc voltage source needs to be a short circuit).

Use a 9 V battery to power the amplifier. The 9 V battery connector and the speaker, both with wires for breadboard connection attached, are shown below.

8 ohm speaker and 9 V battery connector with wires for breadboard connection

You should use the 9 V battery as power supply for all measurements. Make a note in your lab report what the actual voltage from the battery was for each set of measurements. Use the speaker to convince yourself that you indeed built an audio amplifier. But for the dc bias and the ac gain measurements you have to use a resistive load of approximately 8 ohms. Note that this power amplifier may produce more than 1/4 W of power. Combine resistors accordingly to obtain 8 ohms +-10%. In your lab report state what resistance you used to emulate the speaker.

E3. Performance of the LM386 Audio Amplifier. Make the performance measurements you devised in prelab problem P3 on the audio amplifier you built in E2. Explain differences between the simulation and the real circuit.

When you make your maximum output power measurement, connect the oscilloscope probe across the power supply terminals. What do you expect to see on the oscilloscope for a dc voltage. What do you actually see on the oscilloscope (you may need to use ac coupling)? Why is there a discrepancy?