Dr. Ahmed Heikal – Lecture 2

Electronic Devices Ninth Edition Floyd

Chapter 6

AC Quantities

V AC quantities are indicated

with a italic subscript; rms rms avg values are assumed unless Vce V Vce otherwise stated. ce VCE Vce The figure shows an example of vce a specific waveform for the collector-emitter voltage. Notice

the DC component is VCE and 0 t 0 the ac component is Vce. Resistance is also identified with a lower case subscript when analyzed from an ac standpoint.

Linear Amplifier

A linear amplifier produces an replica of the input signal at the output. +VCC

Ic Vb ICQ V BQ R1 RC

Vce C2 VCEQ Rs C1 Ib IBQ

Vs R2 RE RL

For the amplifier shown, notice that the voltage waveform is inverted between the input and output but has the same shape.

AC Load Line

Operation of the linear amplifier can be illustrated IC Q I B

using an ac load line. b I

The ac load line is different Ic I than the dc load line because CQ Q a capacitor looks open to dc but effectively acts as a

short to ac. Thus the 0 VCE collector resistor appears to Vce

be in parallel with the load VCEQ resistor.

Transistor AC Model

The five resistance parameters (r-parameters) can be used for detailed analysis of a BJT circuit. For most analysis work, the simplified r-parameters give good results. The simplified r-parameters are C C shown in relation to the transistor model. βacIb βacIb An important r-parameter is r '. It e B B r ′ appears as a small ac resistance e Ib between the base and emitter. r e′

' 25 mV re = E E IE

The Common-Emitter Amplifier

In the common-emitter (CE) amplifier, the input signal is applied to the base and the inverted output is taken from the collector. The emitter is common to ac signals.

VCC

RC C3 R1 Vo ut C1

Vin

RE C2

The Common-Emitter Amplifier

What is re' for the CE amplifier? Assume stiff voltage-divider bias.

VCC 27 kΩ +15 V VB = 15 V = 4.26 V 68 kΩ+ 27 k Ω RC C3 R1 3.9 kΩ V = 4.26 V – 0.7 V = 3.56 V 68 kΩ E C 1 10 µF

VE 3.56 V = = = 1.0 µF RL IE 1.62 mA 3.9 kΩ R 2.2 kΩ R2 E Ω 27 k RE C2 2.2 kΩ 100 µF ' 25 mV 25 mV re = = = 15.4 Ω IE 1.62 mA

The Common-Emitter Amplifier

Notice that the ac resistance of the collector circuit is RC||RL. What is the gain of the amplifier?

VCC +15 V V R RR A =out = c = C L v '' RC C3 Vrin e r e R1 3.9 kΩ 68 kΩ 3.9 kΩΩ 3.9 k C1 A = = 127 10 µF v Ω 15.4 1.0 µF RL Ω R2 3.9 k 27 kΩ C The gain will be a little lower if the RE 2 2.2 kΩ 100 µF input loading effect is accounted for.

The Common-Emitter Amplifier

Greater gain stability can be achieved by adding a swamping resistor to the emitter circuit of the CE amplifier. The gain will be lower as a result. VCC +15 V

RC What is the gain with the addition C3 R1 3.9 kΩ 68 kΩ of the swamping resistor? (Ignore C 1 10 µF the small effect on re'.) 1.0 µF R Vout Rc RRC  L E1 R A = = = 33 Ω L v ''++ 3.9 kΩ Vin rR e E1 rRe E1 R 2 Ω 27 k RE2 C2 3.9 kΩΩ 3.9 k 2.2 kΩ A = = 38.2 100 µF v 15.4 Ω+33 Ω

The Common-Emitter Amplifier

Multisim is a good way to check your calculation. For an input of

10 mVpp, the output is 378 mVpp as shown on the oscilloscope display for the swamped CE amplifier.

input

output

The Common-Emitter Amplifier

In addition to gain stability, swamping has the advantage of increasing

the ac input resistance of the amplifier. For this amplifier, Rin(tot) is given by Rin(tot) = R1||R2||βac(re' + RE1) VCC +15 V

RC C3 R1 3.9 kΩ What is Rin(tot) for the amplifier if 68 kΩ C βac = 200? 1 10 µF

1.0 µF Rin(tot) = R1||R2||βac(re' + RE1) R E1 R 33 Ω L 3.9 kΩ = 68 kΩ||27 kΩ||200(15.4 Ω + 33 Ω) R 2 Ω 27 k RE2 C2 = 6.45 kΩ 2.2 kΩ 100 µF

The Common-Collector Amplifier

The common-collector amplifier (emitter-follower) has a voltage gain of approximately 1, but can have high input resistance and current gain. The input is applied to the base and taken from the emitter.

+VCC

R1 C1

Vin C2 Iin Vo ut

R2 RE RL

The Common-Collector Amplifier

The power gain is the ratio of the power delivered to the input resistance divided by the power dissipated in the load. This is approximately equal to

the current gain. That is, Ap ≈Ai. V You can also write power gain CC as a ratio of resistances: 2 VL R1 PRR C1 LL2 in() tot V AApv= =2 = in PRin Vin L C2 Vout Rin() tot R2 RRin() tot in () tot RE RL ≅=1 RRLL The next slide is an example…

The Common-Collector Amplifier

Calculate the power gain to the load for the CC amplifier using a ratio of

resistances. Assume Av = 1 and βac = 200. Use re' = 2 Ω.

VCC +15 V

Rin(tot) = R1||R2||βac(re' + RE||RL) R Ω Ω Ω Ω 1 = 39 k ||220 k ||200(2 + 500 ) C1 39 kΩ Vin

= 24.9 kΩ C2 µ Vout Ω 0.22 F RL = 1.0 k R2 3.3 µF 220 kΩ R RL R E in() tot 24.9 kΩ 1.0 kΩ 1.0 kΩ Ap = = = 24.9 RL 1.0 kΩ

The Common-Collector Amplifier

The input voltage-divider in the previous example is not “rock-solid” but the overall power gain is good. A “rock solid” stiff voltage-divider is not always the best design. Can you spot the problem illustrated here?

Rin(tot) = R1||R2||βac(re' + RE||RL) VCC +10 V = 10 kΩ||10 kΩ||200(25 Ω + 3.0 kΩ)

= 4.96 kΩ R1 C1 10 kΩ β = 200 V RL = 10 kΩ in C2 R 4.96 kΩ Vout in() tot R Ap = = = 0.496! 2 Ω R RL 10 kΩ 10 k RE L 4.3 kΩ 10 kΩ The problem is the power gain is less than 1!

The Darlington Pair A Darlington pair is two transistors connected as shown. The two transistors act as one “super β” transistor. Darlington transistors are available in a single package. Notice there are two diode drops from base to emitter. VCC VCC

R1 RC C1

Vin Q1

Q2 C2 Vo ut R2 R E RL

CE Amplifier Darlington CC amplifier Load

The Sziklai Pair Another high β pair is the Sziklai pair (sometimes called a complementary Darlington), in which a pnp and npn transistor are connected as shown. This configuration has the advantage of only one diode drop between base and emitter.

+VCC

What is the relation between IE2 and IB1?

Vin βDC1

IB1 βDC2

The DC currents are: IC1 IE2 IC1 is βDC1 x IB1 and is equal to IB2 RE IE2 is approximately equal to βDC2 x IC1 Therefore, IE2 ≈ βDC1βDC2IB1

The CB Amplifier The common-base (CB) amplifier is used in applications where a low input impedance is acceptable. It does not invert the signal, an advantage for higher frequencies as you will see later when you study the Miller effect.

What is the purpose of C2?

+VCC

RC C3 C forces the base R1 2 C V to be at ac ground. 2 o ut

RL C1

Vin

R2 RE

Multistage Amplifiers To improve amplifier performance, stages are often cascaded where the output of one drives another. This an example of a two-stage direct-coupled amplifier in which the input and VCC output signals are capacitively +12 V coupled.

R1 RC RE3 C3 10 kΩ 1.0 kΩ 330 Ω Vo ut 10 µF R Q2 L 2N3906 330 Ω C1 Vin Q1 2N3904 VS 1.0 µF 100 mV pp R2 R Ω E1 1.0 kHz 4.7 k 100 Ω

RE2 C2 330 Ω 47 µF

Differential Amplifiers A differential amplifier (diff-amp) has two inputs. It amplifies the difference in the two input voltages. This circuit is widely used as the input stage to operational amplifiers. Differential-mode inputs are illustrated.

+VCC

Vo ut 1 Vo ut 2 RC1 RC2 1 2

Q1 Q2 1 2 Vin1 Vin2

–VEE

Differential Amplifiers The same amplifier as in the last slide now is shown with common-mode inputs. Diff-amps tend to reject common-mode signals, which are usually due to noise. Ideally, the outputs are zero with common-mode inputs.

+VCC

Vo ut 1 Vo ut 2 RC1 RC2 1 2

Q1 Q2 1 2 Vin1 Vin2

–VEE

r-parameter One of a set of BJT characteristic parameters

that include αac, βac, re', rb', and rc'.

Common- A BJT configuration in which the emitter is emitter the common terminal to an ac signal.

ac ground A point in a circuit that appears as a ground to ac signals only.

Input The resistance seen by an ac source resistance connected to the amplifier input.

Output The ac resistance looking in at the amplifier resistance output.

Common- A BJT configuration in which the emitter is collector the common terminal to an ac signal.

Differential An amplifier in which the output is a function amplifier of the difference between two input voltages.

Common-mode A condition where two signals applied to differential inputs are of the same phase, frequency and amplitude.

1. The equation for finding the ac emitter resistance of a BJT is

' 25 mV a. re = IB

' 25 mV b. re = IE

' 0.7 V c. re = IB

' 0.7 V d. re = IE

2. For a CE amplifier, a swamping resistor will a. increase the input resistance b. increase the gain c. both of the above d. none of the above

3. A well-designed CC amplifier has a. voltage gain > 1 b. current gain > 1 c. both of the above d. none of the above

4. In a CC amplifier, the power gain is approximately a. one b. equal to the voltage gain c. equal to the current gain d. none of the above

5. The amplifier shown is a a. differential amplifier

b. CE amplifier +VCC

RC c. CC amplifier C3 R1 C V d. CB amplifier 2 o ut

RL C 1 Vin

R2 RE

6. An advantage to this amplifier is that it a. has high current gain

b. has high input resistance +VCC

RC c. is non-inverting C3 R1 C V d. all of the above 2 o ut

RL C 1 Vin

R2 RE

7. Together, Q1 and Q2 form a a. Swamped amplifier

b. Differential pair VCC c. Sziklai pair R1 C1

d. none of the above Q1

Q2 C2 Vo ut R2 R E RL

8. A CC amplifier with a power gain less than 1 is a. a buffer b. an inverting amplifier c. unstable d. an example of poor design

9. An npn and a pnp transistor acting together as a single high β transistor is a a. Darlington pair b. Sziklai pair c. Differential pair d. cascaded amplifier

10. If identical signals are applied to both inputs of a differential amplifier, ideally the output will be a. zero b. equal to one of the signals c. equal to the sum of the two signals d. very large

Answers: 1. b 6. c 2. a 7. d 3. b 8. d 4. c 9. b 5. d 10. a