Electronic PRINCIPLES
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1 MALVINO & BATES Electronic PRINCIPLES SEVENTH EDITION
2 Chapter 9 AC Models
3 Topics covered in Chapter 9 Base-biased amplifier Emitter-biased amplifier Small-signal operation AC beta AC resistance of the emitter diode Two transistor models Analyzing an amplifier AC quantities on the data sheet
4 Base-biased amplifier The reactance of a coupling capacitor is much smaller than the resistance AC input into base Amplified and inverted output at the collector AC output coupled to the load
5 The coupling capacitor X C R R R Good coupling: X C < 0.1 R SHORT OPEN 1. For ac analysis, the capacitor is a short. 2. For dc analysis, the capacitor is open.
6 +30 V 1 MΩ 5 kω β dc = kω 100 μv A base-biased amplifier with capacitive coupling A dc analysis reveals I B = 30 μa, I C = 3 ma and V C = 15 V.
7 I B 30 μa Base current t I C 3 ma Collector current V C t 15 V Collector voltage t
8 +30 V 1 MΩ 5 kω + 15 V V β dc = kω 100 μv The base-biased amplifier with voltage waveforms
9 The voltage gain of an amplifier is the ac output divided by the ac input. v in A V = 200 v out A V = v out v in
10 The bypass capacitor R Ac ground X C Good bypassing: X C < 0.1 R Note: The bypass capacitor appears open to dc and shorted to ac
11 VDB and TSEB amplifiers Dc voltages and currents are calculated mentally by opening capacitors The ac signal is coupled via a coupling capacitor The bypass capacitor causes an ac signal to appear across the base-emitter junction and provides higher gain
12 A VDB amplifier with voltage waveforms +10 V +1.8 V 10 kω 3.6 kω V V 100 kω 100 μv 2.2 kω 1 kω
13 A TSEB amplifier with voltage waveforms +10 V 3.6 kω V V kω 100 μv 2.7 kω 1 kω -2 V
14 Distortion The stretching and compressing of alternate half cycles Undesirable in high-fidelity amplifiers Can be minimized by keeping the ac input small
15 I E Q V BE Input signal Large-signal operation produces distortion
16 The 10 percent rule Total emitter current consists of dc and ac To minimize distortion, i e must be small compared to I EQ The ac signal is small when the peak-topeak ac emitter current is less than 10 percent of the dc emitter current
17 I E i e I EQ = 10 ma Less than 1 ma V BE Total emitter current: I E = I EQ + i e Small-signal operation: i e(pp) < 0.1I EQ
18 The dc current gain is given as: β dc = I C I B The ac current gain is given as: β ac = i c i b Use CAPITAL letters for dc quantities and lowercase letters for ac.
19 I E V BE The size of the ac emitter current depends on the Q point.
20 Total emitter current: I E = I EQ + i e Total base-emitter voltage: V BE = V BEQ + v be The ac resistance of the emitter diode is defined as: r e = v be i e The ac resistance of the emitter diode decreases when the dc emitter current increases
21 Ac resistance of the emitter diode Equals the ac base-emitter voltage divided by the ac emitter current The prime ( ) in r e indicates that the resistance is inside the transistor
22 r e = v be i e I E Larger i e Smaller i e V BE Note that r e varies with the operating point. This implies that r e is a function of the dc emitter current.
23 Formula for ac emitter resistance Derived by using solid-state physics and calculus: r e = 25 mv I E Widely used in industry because of its simplicity and it applies to almost all commercial transistors
24 Transistor model Ac equivalent circuit for a transistor Simulates how a transistor behaves when an ac signal is present Ebers-Moll (T model) and π type models are widely used
25 The T model of a transistor: z in(base) = v be i b i c v be = i e r e z in(base) z in(base) = i er e i b z in(base) = βr e i b r e i e
26 The π model of a transistor is based on z in(base) = βr e : i b z in(base) βr e i c i e Clearly shows the input impedance of Br e will load the ac voltage source driving the base
27 Amplifier analysis Perform a complete dc analysis Mentally short all coupling and bypass capacitors for ac signals Visualize all dc supply voltages as ac grounds Replace the transistor by its π or T model Draw the ac equivalent circuit
28 Data sheets The four h parameters are a mathematical approach h fe is the ac current gain h ie is equivalent to input impedance β ac = h fe r e = h ie /h fe h re and h oe are not needed for basic design and troubleshooting The h parameters give useful information when translated into r parameters
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