Common-Emitter Amplifier Biasing and GainExperiment 14

Alexandra Mutty12/19/2008

IntroductionIn this lab, I will be connected a common emitter amplifier circuit to observe the effects

on the circuit and circuit gain with a bypass capacitor in the circuit and without. I will calculate the gain of the circuit both with and without a bypass capacitor to see the effect the component has on the circuit, as well as take drawings of what the waveforms I see.

Figure 1.Schematic used to build our common emitter amplifier circuit in the protoboard. A function generator was later placed into the input shown in the figure.

Procedure

MaterialsPower supplyOscilloscopeDMMFunction Generator560 Ω resistor1000 Ω resistor8.2 k Ω resistor18 k Ω resistorTwo 25 µF 50 V capacitors100 µF 50 V capacitor2N2222 NPN transistor

Part 1 Bias StabilizationWe first connected the circuit of Figure 1, a common emitter

amplifier circuit. We used a DMM to measure the IC current, shown in the schematic of the circuit. Also using the DMM, we measured 3 other points in the circuit, the base to emitter voltage (VBE), the collector to emitter voltage (VCE), and the emitter voltage (VE). The values were recorded into Table 1, in the row of Step 2.

Table 1Bias Stabilization

StepCurrent, mA

Ic

Voltage, V Waveform

VBE VCE VE Input Vp-p Output Vp-p

2 3.37 0.65 3.75 1.88 x x x x

4 3.31 0.65 3.91 1.86 30m 3

5 3.34 0.65 3.9 1.85 5 m .47

After taking down the information, we next calculated the value of IE, using two different methods. One consisted of Ohm’s Law, and the other used Kirchhoff’s Law.

Calculating IE Using Ohm’s Law: VE / RE

VE / RE = 1.88 / 560 = 3.357 mA

Calculating IE Using Kirchhoff’s Law: IE = VTh – VBE / RE + RTh.

To find Vth (Thevenin voltage) we used the formula:

VTh = ( VCC ) ( R2 ) / R1 + R2

= ( 9 V ) ( 8.2k V ) / (18k V + 8.2k V)

= 2.81 V

To find Rth (Thevenin resistance) we used the formula:

RTh = R1R2 / R1 + R2

( R1R2 are added together in parallel 1 / 1/ 18k Ω + 1/ 8.2k Ω = 5.6k Ω )

= 5.6k Ω / (18k Ω + 8.2k Ω)

= 0.21374

IE = VTh – VBE / RE + RTh

IE = 2.81 – 0.65 / 560 = 3.274 mA

So, our final calculations for IE come out to be 3.357 mA and 3.274 mA for the two different formulas used to calculate it. The results are extremely accurate, and only differ from each other by a small margin, less than a tenth of a milliamp.

Next, we connected function generator across the input of our circuit (shown in Figure 1). The frequency was set at 1k Hz. The oscilloscope was then connected across the output of the circuit (also shown in Figure 1).

We increased the amplitude on the function generator, hoping to the sine wave to the maximum undistorted output. We couldn’t get an accurate reading, so we added a potentiometer into the circuit, between the function generator and the input, it’s resistance at maximum. This lowered the voltage going into our circuit to a point low enough that we would actually read an undistorted output.

We increased the amplitude until just before the output became distorted (or flattened out, looking more like a square wave), and kept it there. We recorded the output of the oscilloscope (both the waveform and the peak to peak value) into Table 1, along with the input voltage (base to ground, VB), the base to emitter voltage (VBE), the collector to emitter voltage (VCE), and the emitter voltage (VE). The waveforms we saw were drawn into the Table with their voltages, to better show the phase shifts between the different leads of the transistor. All of the values were recorded into the Step 4 row.

Next, we decreased the amplitude to the lowest possible value while still remaining an undistorted output. Once again, we couldn’t get the function generator’s output to go low enough to read it, so we went as low as possible on the generator and recorded the same information as we had in the previous step. The data was recorded into Table 2, step 5.

Part 2 Effect of Emitter Bypass Capacitor on Gain

We connected the oscilloscope across the output of the amplifier and again found the maximum undistorted output, as we had for part 1. We then decreased the amplitude until it was about half of the original undistorted output, and recorded it into Table 2 for Steps 7,8 under collector voltage. We then measured the input voltage (base to ground) and emitter voltage, as well as record the waveforms shown into Table 2. We noticed that for Steps 7,8, the emitter was only 90 degrees out of phase (behind the base), while the collector was a full 180 degrees out of phase from the base.

Table 2Emitter Bypass Capacitor and Gain

StepWaveform (to ground)

Base Vp-p Collector Vp-p Emitter Vp-p Gain

7, 8 34m 3.8 6.5m 111.76

9, 10 100m 138m 2 Vdc 1.38

11, 12 4m 10m x 2.5

For Steps 9 and 10, we removed C3 from the circuit (the bypass capacitor), and re-measured all the previous values. Out input voltage increased, and out output voltage decreased dramatically. There was no ac voltage through the emitter, and all we read was about 2 volts DC. We recorded all of this into Table 2.

For steps 11 and 12, we placed a hook up wire across RE, which shorted the resistor. We measured and recorded the same things as the previous step, and put the data into Table 2. One thing we noticed about shorting RE was that we saw no ac voltage, but no real DC voltage either. There was just a sort of unfocused blurry line across the screen that we couldn’t seem to focus. We also noticed that the IC (shown in Figure 1) increased from the previous constant of around 3.3mA to 9mA exactly.

Next, we calculated the gain of each of the altered circuits. To do this, we used the formula: Av = Vout / Vin. The output voltage was the collector to ground measured in Table 2, and the input was base to ground. We calculated the gain using this formula for each of the rows in Table 2, and recorded it into the Gain column. We noticed that the highest gain came from the circuit with the bypass capacitor in, and the lowest gain came from the circuit without the bypass capacitor. The circuit with the short across RE had a slightly higher gain than the one without the bypass capacitor, but it was still far less than the circuit using the bypass capacitor, which was about 50 times greater.

DiscussionIn a common emitter amplifier circuit, the gain of the circuit is relatively good for all

aspects like current, voltage, and power, if not great at any one of those. To boost the voltage gain of the circuit, we can put in a bypass capacitor, like the one we put into our circuit, shown in Figure 1. When we calculated the gain of the first circuit we made, we got 111.76. Many of the more common amplifiers have a gain of around 100, which makes are circuit fairly average in aspects of gain.

Without the bypass capacitor, though, the gain of the common emitter amplifier is relatively low, only a gain of about 1 ½ , as we can see in Table 2. We also will read no ac voltage without the capacitor in, as is also shown in Table 2. We can see a small amount of DC voltage from the oscilloscope, but no waveform is shown. With a short in, we don’t even get a voltage of dc or ac, just a sort of unfocused trace across the screen.

Another thing we noticed in the lab was the phase shifts between the different points of measure. Between the incoming voltage (base to ground) and the output voltage (collector to ground), we noticed an exact 180 degree phase shift. We know this to be correct because as we’ve learned in related, the input and output should be exact opposites of each other. This is because before the transistor turns on, the current builds up at the transistor, causing a decrease in voltage, which gives you the bottom half of the sine wave. When the transistor turns on, the current goes through and decreases, which increases the voltage, giving us the top half of our sine wave. So when the input voltage is low, the output is high, and vice versa, which gives us the 180 degree phase shift.

Questions

1. What is the purpose of CE (C3) in Figure 14-5? What happens to the ac amplifier gain when it is removed?

The purpose of C3 is to be used as a bypass capacitor, which increases the gain of a common emitter amplifier circuit. When the capacitor is removed from the circuit, the gain decreases. As shown in Table 1, our gain went from 111.76 to 1.38.

2. What is the effect on the amplifier performance of omitting RE?Shorting RE in the circuit causes an overall decrease in gain from when there was a bypass capacitor in the circuit. It is, however, an improvement from just having RE on the emitter lead. The gain increased by about 1 from the previous 1.38 gain value when the capacitor was taking out, but it far from as effective as when the bypass capacitor was in the circuit, which had a gain of 111.76. The current through IC also increased when RE was shorted

ConclusionIn this lab, I’ve learned about common emitter amplifiers, and how the bypass capacitor

works in the circuit, and what it does for the gain. I now know that the bypass capacitor improves the gain by almost 100 times that of the amplifier circuit without the capacitor. This allows someone to put the least amount of voltage into the circuit and get the most out.

I’ve also learned the effects of taking the capacitor out of the circuit. The voltage gain decreased dramatically, as we saw from calculating the gains in Table 2. The capacitor has the effect of needing more input voltage while getting less output voltage. This makes the circuit much less effective than if we had left the capacitor in.

We also noticed the effects of replacing RE with a short from the emitter to ground. The gain is still extremely low, but not so much as the previous circuit. Shorting RE will give a slightly larger gain in the circuit without a capacitor.

I’ve also learned how to calculate the gain of a circuit using the formula of Gain = Output / Input. This works for power gain, voltage gain, and current gain. So long as I have both the input and outputs of each, I can calculate each in turn using this formula.

Chapter Summary

The common-emitter (CE) is unique in that its input and output signal voltages are 180° out of phase. (See the input and output voltage waveforms in Figure 9.1.) This phase shift can be explained as follows:

The input voltage and current are in phase. The input and output currents are in phase.

Therefore, output current is in phase with the input voltage.

An increase in output current results in a decrease in output voltage, and vice versa (as given by Vout = VC = VCC – ICRC).

Therefore, output voltage is 180° out of phase with output current. Since the output current is in phase with the input voltage, the input and output voltages are 180° out of phase.

FIGURE 9.1 Common-emitter input and output voltages.

The ac emitter resistance (r'e) of a transistor is a dynamic value (like zener impedance) that is used only in ac calculations. For a small-signal amplifier, the value of r'e can be approximated using

The process used to determine the value of r'e is demonstrated in Example 9.1 of the text.

The ac current gain of a transistor is different than its dc current gain. This is because the two values are measured differently, as illustrated in Figure 9.4 of the text. As shown in that figure,

Note that transistor spec sheets list dc as hFE and ac as hfe.

Coupling and Bypass Capacitors

Amplifiers are often cascaded (connected in series) to increase gain. Each amplifier within the cascade is referred to as a stage, and the overall circuit is referred to as a multistage amplifier.

Coupling capacitors are used to provide a combination of ac coupling and dc isolation between the stages of an amplifier. They are positioned in series between amplifier stages as shown in Figure 9.2. The coupling capacitors shown ensure that the dc biasing of each stage is not affected by the

biasing of another stage (or the presence of the source and load). Coupling capacitors are normally high-value components that provide little reactance at the lowest anticipated operating frequency of the circuit.

FIGURE 9.2 Coupling capacitors.

The bypass capacitor shown in Figure 9.3 is connected in parallel with the emitter resistor. The capacitor provides an ac ground at the transistor emitter terminal. This ac ground allows the circuit to provide the highest voltage gain possible (as explained in Section 9.4 of the chapter).

FIGURE 9.3 Emitter bypass capacitor.

The typical common-emitter waveforms are illustrated in Figure 9.10 of the text. Note that there is no change in VE, due to the presence of the bypass capacitor (which effectively shorts the ac component of the emitter voltage to ground).

AC Equivalent Circuits

The ac equivalent of any amplifier is derived by:

Shorting the coupling and bypass capacitors Replacing all dc sources with a ground symbol

The complete process is demonstrated in Example 9.2 of the text.

Amplifier Gain

Voltage gain is the factor by which ac signal voltage increases from the amplifier input to the amplifier output. Stated mathematically,

The voltage gain of a CE amplifier also equals the ratio of ac collector resistance to ac emitter resistance. As shown in Figure 9.12 of the text:

The total ac resistance in the collector circuit equals RC || RL.

The total ac resistance in the emitter circuit equals r'e.

These relationships hold true for a single-stage CE amplifier with a bypassed emitter resistance (and must be modified for other circuits). The voltage gain of a basic CE amplifier can be found as:

where fC = RC || RL. The complete procedure for calculating the voltage gain of a CE amplifier is demonstrated in Example 9.4 of the text.

The ac emitter resistance of the transistor (r'e) can be affected by changes in temperature. As a result, the voltage gain of a CE amplifier may also be affected when there is an increase (or decrease) in operating temperature. A circuit construction technique,

known as swamping, is sometimes used to stabilize voltage gain. (Amplifier swamping is addressed in Section 9.6 of the text.)

The current gain (Ai) of a CE amplifier is lower than the value of hfe for the transistor, because of the current dividers in the base and collector circuits. The power gain of a CE amplifier equals the product of current gain (Ai) and voltage gain (Av).

Gain and Impedance Calculations

The voltage gain of an amplifier is affected by the resistance of its load (as demonstrated in Figure 9.16 of the text). As load resistance increases, so does amplifier voltage gain. Voltage gain reaches its maximum value when its load opens (all other factors being equal).

The input impedance of a CE amplifier equals the parallel combination of the base biasing resistors and the input impedance of the transistor base (see Figure 9.17 of the text). For a voltage-divider biased CE amplifier,

Zin = R1 || R2 || Zbase

where Zbase = hfer'e. The value of Zbase for a transistor is listed as hie on its spec sheet. The calculation of Zin

for a CE amplifier is demonstrated in Example 9.9 of the text.

When the values of Zin and Zbase for a base CE amplifier are known, the current gain of the circuit can be found using

The calculation of Ai for a CE amplifier is demonstrated in Example 9.10 of the text.

To calculate the total power gain of a multistage amplifier, you must first determine the values of Ai and Av for each stage. Then, the total current gain (AiT) is found as the product of the individual stage gains. The same holds true for the total voltage gain (AvT). Finally, the overall power gain can be found using

ApT = AvTAiT

Multistage voltage gain calculations are demonstrated in Examples 9.11 and 9.12 of the text.

Swamped (Gain-stabilized) Amplifiers

A swamped amplifier reduces variations in voltage gain by increasing the ac resistance of the emitter circuit. The swamped amplifier is also referred to as a gain-stabilized amplifier.

A swamped amplifier is shown in Figure 9.4. The resistor labeled rE is not bypassed, so it is a part of the ac equivalent circuit for the amplifier. For the circuit shown, the total ac emitter resistance is (r'e + rE).

FIGURE 9.4 A swamped (gain-stabilized) amplifier.

When the amplifier is designed so that rE > r'e, the effects of any change in r'e on voltage gain are minimized. This principle is demonstrated in Example 9.14 of the text. Note that the addition of a swamping resistor also has the effect of increasing

the base input impedance of the transistor, as given in the following relationship:

Zbase = hfe (r'e + rE)

The effect of swamping on amplifier input impedance is demonstrated in Examples 9.15 and 9.16 of the text.

H-parameters

Hybrid parameters, or h-parameters, are transistor characteristics that are measured under specific conditions. The four h-parameters are listed in Table 9.1 and measured as shown in Figure 9.23 of the text.

TABLE 9.1

Parameter label

Parameter Measurement condition

hie Base input impedance

Output shorted

hfe Base-to-collector current gain

Output shorted

hoe Output admittance Input openhre Reverse voltage Input open

feedback ratio

The values of hie and hfe can be used to provide a more accurate value of r'e, as follows:

When the parameters are listed as maximum and minimum values, the geometric average of the two is used. In many cases, h-parameters are provided using graphs like those shown in Figure 9.26 of the text. When using these graphs:

Determine the minimum and maximum h-parameter values at the value of ICQ.

Use the geometric average of the values obtained in any circuit analysis problems.

The h-parameter analysis of a CE amplifier is demonstrated in Examples 9.18 and 9.19 of the text.

Amplifier Troubleshooting

There are several appoaches that can be taken to troubleshooting multistage amplifiers. One such approach is to begin at the final stage output and work back toward the input. (This approach makes sense when you consider that many electronic

systems contain two or more signal paths that are combined at later stages. By working your way back from the final output, you can trace through multiple signal paths without having to test them all.)

One problem that is common to CE amplifiers is nonlinear distortion. Nonlinear distortion is characterized by asymmetrical alternations of the amplifier output, as shown in Figure 9.29 of the text. Nonlinear distortion is most commonly caused by driving the transistor in an amplifier into the nonlinear operating region of its base curve.