LAB 2: BJT CE Amplifier Record waveforms from the oscilloscope to a USB stick and paste these in your report. NO WAVEFORM = NO SCORE. You must use lab voltmeters

(no simulation). 2.3.1 VB (Volts): VE (Volts): Vc (Volts): Emitter Current IE here: As ß >100, lc must approximate: Rc value here: Voltage at Collector with new Rc value here: [7 Marks, 1 ea] 2.3.2 You must use the LTSPICE simulator. Your Simulated Bode Plot (Gain vs Frequency) here: [5 Marks] 2.3.3 You must use lab oscilloscopes (no simulation). VACIN = VACOUT = AC voltage gain calculated = AC voltage gain and measured (insert waveform below) = Difference in phases = Clipped waveform (insert waveform below): Max voltage swing without clipping is = ALWAYS SHOW YOUR CALCULATIONS. NO CALCULATION = NO SCORE. 2.4 2.3.4 You must use lab oscilloscopes (no simulation). New calculated undistorted gain = Show calculations... New measured undistorted gain = [14 Marks, 2 ea] Read the notes... Demonstrate your circuit to the Lecturer in the Lab. Use theory. NO THEORY = NO SCORE. [10 Marks] Conclude discussing other effect/s of adding an emitter bypass capacitor.: [Max Total Score = 50 Marks] [4 Marks] [10 Marks]/n LAB 2: BJT Common Emitter Amplifier Design Aim: To investigate the design process of a simple common-emitter amplifier. 1 Introduction The common-emitter amplifier is one of the most useful and fundamental circuits for which a BJT is typically used. In this experiment, the standard circuit configuration is constructed and the DC biasing conditions are explored. Then, an AC signal is applied to experimentally determine the gain of the circuit. Lastly, an emitter bypass capacitor is added to the circuit to increase the gain. This lab combines calculations with practical measurement and simulation. 2 Background Examine Figure 1 below. The resistors R1 and R2 are establishing a DC biasing condition for the base of the BJT. The emitter resistor RẺ is then effectively setting the emitter current and hence base current, making the circuit less dependent on ẞ and ambient temperature. The collector resistor, Rc, should be chosen to divide the supply voltage approximately in half. This is so that when an AC voltage is applied to the input of the amplifier the output voltage at the collector can swing in equal amounts both up and downwards (often called "mid-point" biasing). You will investigate the DC properties of the amplifier and subsequently the AC properties. 5 C1 HH 10uF R1 56kQ R2 10kQ RC C2 HH Q1 10uF BC337 12 RE 1kQ 1 6 .V1 - 15 V Figure 1: Common Emitter Amplifier with Emitter Resistor 3 Procedure 3.1 DC quiescent biasing conditions To construct the circuit in Figure 1, initially use Rc = 1 k and then measure the DC voltages at the Base, Emitter and Collector (relative to ground) and populate the table in the Results section. You must use lab instruments for this (no simulation). 1. From the voltage across the emitter resistor, R₁, determine the emitter current, IE. 2. Knowing that ß> 100 for the BC337-25 (from the previous experiment) what does Ic approximate? 3. Given this information, determine a more suitable value for Rc seeing that we wish to ensure the quiescent (DC) output voltage is about half of the supply voltage? 4. Alter Rc to be your calculated value and recheck the voltage at the Collector now. 3.2 AC Bode plot by simulation We can now consider the AC behaviour of the circuit by taking a full bode plot. In order to do this practically, the signal generator frequency is swept and at different points, the gain, as the amplitude relationship between the two channels, and the phase shift are measured and plotted on a gain vs frequency graph. This is difficult and time consuming to do and so we will do it by simulation only using LTSPICE. The circuit is as shown in Figure 2. A signal generator (AC signal voltage source) must be connected to the input capacitor C₁ and to ground. Simulate and click a voltage probe to the output end of capacitor C₂ and another voltage probe at capacitor C₁'s input node with the signal generator to create your AC Bode plot from 1 Hz to 100 MHz. 1Vpk at 1 kHz C2 HH V2 10μ SINE(0 1 1k 0000) R1 56k R2 10k RC C1 HH 10μ. Q1 BC337-25 RE 1k V1 20/05 OUT 15 R3 100k Figure 2: Measuring AC Gain of the Common Emitter Amplifier 3.3 AC gain at 5 kHz We will now measure the circuit's real gain at a frequency where we know it will be close to maximum, so lets use 5000 Hz for this section. You must use lab instruments for this (no simulation). 5. Measure the AC peak-to-peak voltage of the input source and the output source. 6. Now determine the AC voltage gain (the ratio of Vout/Vin); calculate it also. 7. Compare the phases of each signal- what is the main difference? 8. Increase the input voltage source to 2 Vpk, the output should start to saturate, can you see the AC waveform is "clipping"? draw a clipped waveform. 9. What is the maximum output voltage swing without clipping? 3.4 Increasing max AC gain by adding an emitter bypass capacitor You should find the AC gain of the amplifier is quite low. This is because the emitter resistor is applying a large amount of negative feedback to the amplifier and this has reduced the gain. However, if a capacitor is connected in parallel with the emitter resistor then the AC small signal gain will increase dramatically. You must use lab instruments for this (no simulation). 10. Add a 10 μF capacitor across RE and retest the gain at 5 kHz ensuring the output doesn't clip (reduce input voltage to 20 mV). Calculate new gain too. 4 Conclusions A common-emitter amplifier has been simulated and compared with theory. The DC bias points where first established to determine the quiescent collector- emitter current. Then a suitable value of Rc was selected to achieve mid-point biased operating point. The AC voltage gain was measured and was found to increase dramatically when an emitter bypass capacitor was added to the circuit. 11. What other effect/s does adding the bypass capacitor have on the circuit?