Presentation is loading. Please wait. # Electronics Principles & Applications Sixth Edition Chapter 7 More About Small-Signal Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A.

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Electronics Principles & Applications Sixth Edition Chapter 7 More About Small-Signal Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler

Amplifier Coupling Voltage Gain FET Amplifier Negative Feedback Frequency Response INTRODUCTION

Dear Student: This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow you to view that segment again, if you want to.

Concept Preview Cascade amplifiers can use capacitive coupling. When dc gain is required, direct coupling is required. The Darlington configuration is an example of direct coupling. Transformer coupling offers the advantage of impedance matching. The impedance ratio is equal to the square of the turns ratio. Tuned transformers provide selectivity.

V CC These two points are at different dc voltages. Capacitive coupling is convenient in cascade ac amplifiers.

V CC Direct coupling is required for dc gain.

V CC The darlington is a popular dc arrangement.

V CC P S 10:1 10  Z RATIO = T RATIO 2 = 10 2 = 100 Z COLLECTOR = 100 x 10  = 1000  Transformer coupling offers the advantage of impedance matching.

V CC Transformer coupling can be used in bandpass amplifiers to achieve selectivity. fRfR Gain

Amplifier Coupling Quiz Capacitive coupling is not useful for _________ amplifiers. dc Dc frequency response requires ________ coupling. direct Transformer coupling offers the advantage of _________ matching. impedance Tuned transformer coupling provides frequency _____________. selectivity A darlington amplifier is an example of _________ coupling. direct

Concept Review Cascade amplifiers can use capacitive coupling. When dc gain is required, direct coupling is required. The Darlington configuration is an example of direct coupling. Transformer coupling offers the advantage of impedance matching. The impedance ratio is equal to the square of the turns ratio. Tuned transformers provide selectivity. Repeat Segment

Concept Preview The input impedance of a C-E amplifier is equal to the equivalent parallel resistance of the base divider and r in of the transistor. r in is  times the sum of the emitter resistances when the emitter resistor is not bypassed. Loading the output circuit changes the clipping points and decreases the voltage gain. The clipping points are shown by the ac load line. The ac load line passes through the same Q- point as the dc load line.

R B1 E B C RLRL V CC R B2 RERE = 12 V 2.7 k  22 k  = 2.2 k  More about solving the practical circuit for its ac conditions: = 220  Z in = ?

R B1 E B C RLRL V CC R B2 RERE = 12 V 2.7 k  22 k  = 2.2 k  Z in is a combination of R B1, R B2, and r in of the transistor. = 220  r in =  (R E + r E ) r in =  (220  + 9.03  ) r in = 34.4 k  Note: r in =  r E when R E is bypassed. Determine r in first:

R B1 E B C RLRL V CC R B2 RERE = 12 V 2.7 k  22 k  = 2.2 k  = 220  Z in = 1 R B2 1 r in 1 + R B1 1 +++ Z in = 1 2.7 k  1 34.4 k  1 22 k  1 Z in = 2.25 k  R B1, R B2, and r in act in parallel to load the input signal.

R B1 V CC R B2 RERE = 12 V 2.7 k  22 k  RLRL = 2.2 k  = 220  Load = 2.2 k  What happens when an amplifier is loaded? R L and the Load act in parallel. R P = 1.1 k 

R B1 R B2 RERE V CC = 12 V 2.7 k  22 k  RLRL = 2.2 k  = 220  Load = 2.2 k  There are two saturation currents for a loaded amplifier. R P = 1.1 k  I SAT(DC) = V CC R L + R E = 4.96 mA I SAT(AC) = V CC R P + R E = 9.09 mA

0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A There are two load lines for a loaded amplifier. DC TEMPORARY AC The DC load line connects V CC and I SAT(DC). A temporary AC load line connects V CC and I SAT(AC).

0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A 5.3 V DC AC TEMP. AC The quiescent V CE is projected to the DC load line to establish the Q-point. The AC load line is drawn through the Q-point, parallel to the temporary AC load line.

0 24 6 8 1012 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA 20  A 0  A 100  A 80  A 60  A 40  A 5.3 V AC The AC load line shows the limits for V CE and if the Q-point is properly located. With loaded amplifiers, the Q-point is often closer to saturation.

R B1 R B2 RERE V CC = 12 V 2.7 k  22 k  RLRL = 2.2 k  = 220  Load = 2.2 k  What about voltage gain for a loaded amplifier? R P = 1.1 k  A V = RPRP R E + r E A V = 1.1 k  220  9.03  = 4.8

V CC Z in of the 2nd stage loads the 1st stage. When analyzing cascade amplifiers, remember: 2nd 1st

Amplifier ac Conditions Quiz Emitter bypassing _________ an amplifier’s input impedance. decreases Loading at the output of an amplifier ________ its voltage gain. decreases A loaded amplifier has two load lines: dc and ___________. ac The clipping points of a loaded amplifier are set by its _______ load line. ac In a cascade amplifier, the Z in of a stage _______ the prior stage. loads

Concept Review The input impedance of a C-E amplifier is equal to the equivalent parallel resistance of the base divider and r in of the transistor. r in is  times the sum of the emitter resistances when the emitter resistor is not bypassed. Loading the output circuit changes the clipping points and decreases the voltage gain. The clipping points are shown by the ac load line. The ac load line passes through the same Q- point as the dc load line. Repeat Segment

Concept Preview A common-source JFET amplifier uses the gate as the input and the drain as the output. The forward transfer admittance (Y fs ) can be determined from the drain family of curves. Voltage gain is equal to Y fs times R L. Source bias produces negative feedback and decreases the voltage gain. The gain with feedback is determined by the feedback ratio and the open-loop gain. The feedback can be eliminated with a source bypass capacitor.

Drain Source Gate V DD = 20 V V GS = 1.5 V RGRG C R L = 5 k  Input signal Common-source JFET amplifier. Fixed bias I SAT = 20 V 5 k  = 4 mA Phase-inverted output

0 2 4 1 V DS in Volts I D in mA 5 10 15 20 25 3 -2.5 -2.0 -1.5 -0.5 0 N-channel JFET characteristic curves V GS in Volts Load line The Q-point is set by the fixed bias. 8 V P-P 1 V P-P A V = 8

0 2 4 1 V DS in Volts I D in mA 5 10 15 20 25 3 -2.5 -2.0 -1.5 -0.5 0 Determining forward transfer admittance: Y fs = IDID  V GS V GS in Volts V DS 1.6 mA = 1.6 mS

D S G V DD = 20 V V GS = 1.5 V RGRG C R L = 5 k  When the forward transfer admittance is known, the voltage gain can be determined using: A V = Y fs x R L = 1.6 mS x 5 k  = 8 This agrees with the graphic solution.

D S G V DD V GS = I D x R S RGRG C RLRL RSRS Source bias eliminates the need for a separate V GS supply. I S = I D This resistor also provides ac negative feedback which decreases the voltage gain.

V in - BV out A(V in - BV out ) BV out A = open loop gain Summing junction V in V out A B Feedback A negative feedback model B = feedback ratio V out = A(V in - BV out ) V out = AV in - ABV out AV in V out 1 = - AB AV in V out AB +1 = V in V out AB +1 A = V in V out AB +1 A = A V in V out A simplified model

D S G V DD RGRG C RLRL RSRS = 5 k  = 800  The feedback ratio (B) for this circuit is easy to determine since the source and drain currents are the same. B = 800  5 k  = 0.16

AB +1 A V in V out Use the simplified model: A (WITH NEG. FEEDBACK) = 8 (8)(0.16) + 1 = 3.51

CSCS D G V DD RGRG C RLRL RSRS The source bypass capacitor will eliminate the ac negative feedback and restore the voltage gain.

JFET Amplifier Quiz In a common-source amplifier, the input signal goes to the _______. gate In a common-source amplifier, the input to output phase relationship is ____. 180 o The voltage gain of a C-S amplifier is equal to Y fs x _________. load resistance Source bias is produced by current flow through the _______ resistor. source An unbypassed source resistor _______ the voltage gain of a C-S amp. decreases

Concept Review A common-source JFET amplifier uses the gate as the input and the drain as the output. The forward transfer admittance (Y fs ) can be determined from the drain family of curves. Voltage gain is equal to Y fs times R L. Source bias produces negative feedback and decreases the voltage gain. The gain with feedback is determined by the feedback ratio and the open-loop gain. The feedback can be eliminated with a source bypass capacitor. Repeat Segment

Concept Preview Dc negative feedback stabilizes the Q-point. Ac negative feedback decreases gain. Ac negative feedback increases bandwidth. Ac negative feedback reduces distortion. Amplifier gain is maximum at mid-band. The break frequencies are where the gain drops by 3 dB. Amplifier bandwidth is found by subtracting the lower break frequency from the upper break frequency.

Amplifier Negative Feedback DC reduces sensitivity to device parameters DC stabilizes operating point DC reduces sensitivity to temperature change AC reduces gain AC increases bandwidth AC reduces signal distortion and noise AC may change input and output impedances

0.707 A max A f The frequency response curve of an ac amplifier Bandwidth The gain is maximum in the midband. A max Midband The bandwidth spans the -3 dB points which are called the break frequencies. -3dB

50  10  F 1 k  100  1k  6.8 k  The emitter bypass capacitor in this amplifier has a significant effect on both gain and bandwidth.

Gain in dB 0 50 Frequency 10 Hz 100 MHz BW 1 BW 2 Gain and bandwidth with and without the emitter bypass

Amplifier Frequency Response The lower break frequency is partly determined by coupling capacitors. It is also influenced by emitter bypass capacitors. The upper break frequency is partly determined by transistor internal capacitance. Both break frequencies can be influenced by negative feedback.

Concept Review Dc negative feedback stabilizes the Q-point. Ac negative feedback decreases gain. Ac negative feedback increases bandwidth. Ac negative feedback reduces distortion. Amplifier gain is maximum at mid-band. The break frequencies are where the gain drops by 3 dB. Amplifier bandwidth is found by subtracting the lower break frequency from the upper break frequency. Repeat Segment

REVIEW Amplifier Coupling Voltage Gain FET Amplifier Negative Feedback Frequency Response

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