6.Experiments

5.11. VOLTAGE FOLLOWER

233

(I=E/R). This calculation is particularly easy with resistors of 1 k- value: there will be 1 milliamp of current for every volt of drop across them. For best precision, you may measure the resistance of each resistor rather than assume an exact value of 1 k-, but it really doesn't matter much for the purposes of this experiment. When resistors are used to take current measurements by "translating" a current into a corresponding voltage, they are often referred to as shunt resistors.

You should expect to ¯nd huge di®erences between input and output currents for this ampli¯er circuit. In fact, it is not uncommon to experience current gains well in excess of 200 for a small-signal transistor operating at low current levels. This is the primary purpose of a voltage follower circuit: to boost the current capacity of a "weak" signal without altering its voltage.

Another way of thinking of this circuit's function is in terms of impedance. The input side of this ampli¯er accepts a voltage signal without drawing much current. The output side of this ampli¯er delivers the same voltage, but at a current limited only by load resistance and the current-handling ability of the transistor. Cast in terms of impedance, we could say that this ampli¯er has a high input impedance (voltage dropped with very little current drawn) and a low output impedance (voltage dropped with almost unlimited current-sourcing capacity).

COMPUTER SIMULATION

Schematic with SPICE node numbers:

		1							1								1
								Rpot1		5 kΩ	Rbase 3
	V1							2								Q1
											1 kΩ
								Rpot2		5 kΩ							4
													Rload				1 kΩ
		0
									0							0
	Netlist (make a text ¯le containing the following text, verbatim):
Voltage follower
v1	1	0
rpot1 1 2 5k
rpot2 2 0 5k

rbase 2 3		1k
rload 4 0		1k
q1 1 3	4 mod1
.model	mod1 npn bf=200
.dc v1	12	12	1
.print	dc	v(2,0) v(4,0) v(2,3)
.end

When this simulation is run through the SPICE program, it shows an input voltage of 5.937 volts and an output voltage of 5.095 volts, with an input current of 25.35 ¹A (2.535E-02 volts

234	CHAPTER 5. DISCRETE SEMICONDUCTOR CIRCUITS

dropped across the 1 k- Rbase resistor). Output current is, of course, 5.095 mA, inferred from the output voltage of 5.095 volts dropped across a load resistance of exactly 1 k-. You may change the "potentiometer" setting in this circuit by adjusting the values of Rpot1 and Rpot2, always keeping their sum at 10 k-.

5.12. COMMON-EMITTER AMPLIFIER

235

5.12Common-emitter ampli¯er

PARTS AND MATERIALS

²One NPN transistor { model 2N2222 or 2N3403 recommended (Radio Shack catalog # 2761617 is a package of ¯fteen NPN transistors ideal for this and other experiments)

²Two 6-volt batteries

²One 10 k- potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)

²One 1 M- resistor

²One 100 k- resistor

²One 10 k- resistor

²One 1.5 k- resistor

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

²Design of a simple common-emitter ampli¯er circuit

²How to measure ampli¯er voltage gain

²The di®erence between an inverting and a noninverting ampli¯er

²Ways to introduce negative feedback in an ampli¯er circuit

SCHEMATIC DIAGRAM

	10 kΩ
6 V	Vout
	100 kΩ
10 kΩ	Q1
6 V	Vin

ILLUSTRATION

236	CHAPTER 5. DISCRETE SEMICONDUCTOR CIRCUITS

- -

++

CBE

INSTRUCTIONS

Build this circuit and measure output voltage (voltage measured between the transistor's collector terminal and ground) and input voltage (voltage measured between the potentiometer's wiper terminal and ground) for several position settings of the potentiometer. I recommend determining the output voltage range as the potentiometer is adjusted through its entire range of motion, then choosing several voltages spanning that output range to take measurements at. For example, if full rotation on the potentiometer drives the ampli¯er circuit's output voltage from 0.1 volts (low) to 11.7 volts (high), choose several voltage levels between those limits (1 volt, 3 volts, 5 volts, 7 volts, 9 volts, and 11 volts). Measuring the output voltage with a meter, adjust the potentiometer to obtain each of these predetermined voltages at the output, noting the exact ¯gure for later reference. Then, measure the exact input voltage producing that output voltage, and record that voltage ¯gure as well.

In the end, you should have a table of numbers representing several di®erent output voltages along with their corresponding input voltages. Take any two pairs of voltage ¯gures and calculate voltage gain by dividing the di®erence in output voltages by the di®erence in input voltages. For example, if an input voltage of 1.5 volts gives me an output voltage of 7.0 volts and an input voltage of 1.66 volts gives me an output voltage of 1.0 volt, the ampli¯er's voltage gain is (7.0 - 1.0)/(1.66 - 1.5), or 6 divided by 0.16: a gain ratio of 37.50.

You should immediately notice two characteristics while taking these voltage measurements: ¯rst, that the input-to-output e®ect is "reversed;" that is, an increasing input voltage results in a decreasing output voltage. This e®ect is known as signal inversion, and this kind of ampli¯er as an inverting ampli¯er. Secondly, this ampli¯er exhibits a very strong voltage gain: a small change in input voltage results in a large change in output voltage. This should stand in stark contrast to the "voltage follower" ampli¯er circuit discussed earlier, which had a voltage gain of about 1.

Common-emitter ampli¯ers are widely used due to their high voltage gain, but they are rarely used in as crude a form as this. Although this ampli¯er circuit works to demonstrate the basic concept, it is very susceptible to changes in temperature. Try leaving the potentiometer in one position and heating the transistor by grasping it ¯rmly with your hand or heating it with some other source of heat such as an electric hair dryer (WARNING: be careful not to get it so hot that your plastic breadboard melts!). You may also explore temperature e®ects by cooling the transistor: touch an ice cube to its surface and note the change in output voltage.

5.12. COMMON-EMITTER AMPLIFIER

237

When the transistor's temperature changes, its base-emitter diode characteristics change, resulting in di®erent amounts of base current for the same input voltage. This in turn alters the controlled current through the collector terminal, thus a®ecting output voltage. Such changes may be minimized through the use of signal feedback, whereby a portion of the output voltage is "fed back" to the ampli¯er's input so as to have a negative, or canceling, e®ect on voltage gain. Stability is improved at the expense of voltage gain, a compromise solution, but practical nonetheless.

Perhaps the simplest way to add negative feedback to a common-emitter ampli¯er is to add some resistance between the emitter terminal and ground, so that the input voltage becomes divided between the base-emitter PN junction and the voltage drop across the new resistance:

	10 kΩ
6 V	Vout
	100 kΩ
10 kΩ	Q1
6 V	Vin
	1.5 kΩ

- -

++

CBE

Repeat the same voltage measurement and recording exercise with the 1.5 k- resistor installed, calculating the new (reduced) voltage gain. Try altering the transistor's temperature again and noting the output voltage for a steady input voltage. Does it change more or less than without the 1.5 k- resistor?

Another method of introducing negative feedback to this ampli¯er circuit is to "couple" the output to the input through a high-value resistor. Connecting a 1 M- resistor between the transistor's collector and base terminals works well:

238 CHAPTER 5. DISCRETE SEMICONDUCTOR CIRCUITS

	10 kΩ
6 V	1 MΩ
	Vout
	100 kΩ
10 kΩ	Q1
	Vin

6 V

- -

++

CBE

Although this di®erent method of feedback accomplishes the same goal of increased stability by diminishing gain, the two feedback circuits will not behave identically. Note the range of possible output voltages with each feedback scheme (the low and high voltage values obtained with a full sweep of the input voltage potentiometer), and how this di®ers between the two circuits.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

5.12. COMMON-EMITTER AMPLIFIER

239

	1															1
												Rc				10 kΩ
										2							Vout
Vsupply										100 kΩ
			3												Q1
	0				Vin						Rb 4					0
			0

Netlist (make a text ¯le containing the following text, verbatim):

Common-emitter amplifier vsupply 1 0 dc 12

vin 3 0 rc 1 2 10k

rb 3 4 100k q1 2 4 0 mod1

.model mod1 npn bf=200

.dc vin 0 2 0.05

.plot dc v(2,0) v(3,0)

.end

This SPICE simulation sets up a circuit with a variable DC voltage source (vin) as the input signal, and measures the corresponding output voltage between nodes 2 and 0. The input voltage is varied, or "swept," from 0 to 2 volts in 0.05 volt increments. Results are shown on a plot, with the input voltage appearing as a straight line and the output voltage as a "step" ¯gure where the voltage begins and ends level, with a steep change in the middle where the transistor is in its active mode of operation.

240	CHAPTER 5. DISCRETE SEMICONDUCTOR CIRCUITS

5.13Multi-stage ampli¯er

PARTS AND MATERIALS

²Three NPN transistors { model 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of ¯fteen NPN transistors ideal for this and other experiments)

²Two 6-volt batteries

²One 10 k- potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)

²One 1 M- resistor

²Three 100 k- resistors

²Three 10 k- resistors

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

²Design of a multi-stage, direct-coupled common-emitter ampli¯er circuit

²E®ect of negative feedback in an ampli¯er circuit

SCHEMATIC DIAGRAM

1 MΩ

		10 kΩ	10 kΩ	6 V
	Vin	10 kΩ
				Vout
	10 kΩ
	100 kΩ	100 kΩ	100 kΩ	6 V

ILLUSTRATION

5.13. MULTI-STAGE AMPLIFIER

241

- -

++

CBE

CBE

CBE

INSTRUCTIONS

By connecting three common-emitter ampli¯er circuit together { the collector terminal of the previous transistor to the base (resistor) of the next transistor { the voltage gains of each stage compound to give a very high overall voltage gain. I recommend building this circuit without the 1 M- feedback resistor to begin with, to see for yourself just how high the unrestricted voltage gain is. You may ¯nd it impossible to adjust the potentiometer for a stable output voltage (that isn't saturated at full supply voltage or zero), the gain being so high.

Even if you can't adjust the input voltage ¯ne enough to stabilize the output voltage in the active range of the last transistor, you should be able to tell that the output-to-input relationship is inverting; that is, the output tends to drive to a high voltage when the input goes low, and visa-versa. Since any one of the common-emitter "stages" is inverting in itself, an even number of staged common-emitter ampli¯ers gives noninverting response, while an odd number of stages gives inverting. You may experience these relationships by measuring the collector-to-ground voltage at each transistor while adjusting the input voltage potentiometer, noting whether or not the output voltage increases or decreases with an increase in input voltage.

Connect the 1 M- feedback resistor into the circuit, coupling the collector of the last transistor to the base of the ¯rst. Since the overall response of this three-stage ampli¯er is inverting, the feedback signal provided through the 1 M- resistor from the output of the last transistor to the input of the ¯rst should be negative in nature. As such, it will act to stabilize the ampli¯er's response and minimize the voltage gain. You should notice the reduction in gain immediately by the decreased sensitivity of the output signal on input signal changes (changes in potentiometer position). Simply put, the ampli¯er isn't nearly as "touchy" as it was without the feedback resistor in place.

As with the simple common-emitter ampli¯er discussed in an earlier experiment, it is a good idea here to make a table of input versus output voltage ¯gures with which you may calculate voltage gain.

Experiment with di®erent values of feedback resistance. What e®ect do you think a decrease in feedback resistance have on voltage gain? What about an increase in feedback resistance? Try it and ¯nd out!

An advantage of using negative feedback to "tame" a high-gain ampli¯er circuit is that the resulting voltage gain becomes more dependent upon the resistor values and less dependent upon the characteristics of the constituent transistors. This is good, because it is far easier to manufacture

242	CHAPTER 5. DISCRETE SEMICONDUCTOR CIRCUITS

consistent resistors than consistent transistors. Thus, it is easier to design an ampli¯er with predictable gain by building a staged network of transistors with an arbitrarily high voltage gain, then mitigate that gain precisely through negative feedback. It is this same principle that is used to make operational ampli¯er circuits behave so predictably.

This ampli¯er circuit is a bit simpli¯ed from what you will normally encounter in practical multi-stage circuits. Rarely is a pure common-emitter con¯guration (i.e. with no emitter-to-ground resistor) used, and if the ampli¯er's service is for AC signals, the inter-stage coupling is often capacitive with voltage divider networks connected to each transistor base for proper biasing of each stage. Radio-frequency ampli¯er circuits are often transformer-coupled, with capacitors connected in parallel with the transformer windings for resonant tuning.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

	3	1 MΩ	6
	1	Rf	1			1
	10 kΩ				R6
	R2	10 kΩ	R4
2				10 kΩ		Vout
	4	5			6
R1 3		R3 7		R5	8
Vin	Q1		Q2			Q3
100 kΩ	100 kΩ		100 kΩ
0	0		0			0

1

Vsupply

0

	Netlist (make a text ¯le containing the following text, verbatim):
Multi-stage				amplifier
vsupply 1 0				dc 12
vin 2 0
r1	2	3	100k
r2	1	4	10k
q1	4	3	0 mod1
r3	4	7	100k
r4	1	5	10k
q2	5	7	0 mod1
r5	5	8	100k
r6	1	6	10k
q3	6	8	0 mod1
rf 3 6			1meg
.model			mod1	npn bf=200

.dc vin 0 2.5 0.1

.plot dc v(6,0) v(2,0)

.end