Introduction to Electronics - Part 7 pdf

Introduction to Electronics 159 The Source Follower R sig R G R S R L C in C out v sig v o + + - - V DD v in + - Fig. 228. Source follower circuit. v sig R sig r d R G R L R S g m v gs v o + + - - G S D v in + - i in v gs + - r d || R S || R L = R L ’ Fig. 229. Source follower small-signal equivalent circuit. The Source Follower Small-Signal Equivalent Circuit This follower uses fixed bias : I G = 0 V GSQ = 0 I D = I DSS ⇒⇒ Tremendously large R in is obtained by sacrificing bias stability, which isn’t very important in this circuit anyway, as we shall see. The small-signal equivalent is constructed in the usual manner: Introduction to Electronics 160 The Source Follower v sig R sig r d R G R L R S g m v gs v o + + - - G S D v in + - i in v gs + - r d || R S || R L = R L ’ Fig. 230. Source follower small-signal equivalent circuit (Fig. 229 repeated). vvv v vv in gs o gs in o =+ ⇒ =− (233) () vgviR gv v R Rvg R R o m gs in L m gs gs G Lgsm G L =+ ′ =+       ′ =+       ′ 1 (234) () vvvg R R oinom G L =− +       ′ 1 (235) 1 11 ++       ′       =+       ′ g R Rv vg R R m G Lo inm G L (236) A v v g R R g R R v o in m G L m G L == +       ′ ++       ′ = 1 1 1 05. to 0.8 typically (237) Voltage Gain This one requires a little more algebra. Beginning with: and We replace v gs in eq. (234) with eq. (233), and solve for v o /v in : Introduction to Electronics 161 The Source Follower v sig R sig r d R G R L R S g m v gs v o + + - - G S D v in + - i in v gs + - r d || R S || R L = R L ’ Fig. 231. Source follower small-signal equivalent circuit (Fig. 229 repeated). vvvvvg R R in gs o gs gs m G L =+=+ +       ′ 1 (238) viRiRg R R in in G in G m G L =+ +       ′ 1 (239) () R v i RgRR in in in GmGL ==++ ′ 1 (240) Input Resistance Replacing v o in eq. (233) with eq. (234): But v gs = i in R G : Solving for v in / i in : Because I G = 0, R G can be several M Ω . With the additional multiplying factor of R L ’, R in can become extremely large !!! Introduction to Electronics 162 The Source Follower v test R sig r d R G R S g m v gs + - G S D i test v gs + - Fig. 232. Determining output resistance of the source follower. i v R v r v RR gv test test S test d test Gsig mgs =++ + − (241) v R RR v gs G Gsig test =− + (242) iv RrRR gR RR test test Sd G sig mG Gsig =++ + + +       11 1 (243) Output Resistance This calculation is a little more involved, so we shall be more formal in our approach. We remove R L , apply a test source, v test , and set the independent source to zero. From a KCL equation at the source node: But R G and R sig form a voltage divider: Substituting eq. (242) into eq. (241): Introduction to Electronics 163 The Source Follower v test R sig r d R G R S g m v gs + - G S D i test v gs + - Fig. 233. Determining output resistance of the source follower (Fig. 232 repeated). R v i RrRR gR RR o test test Sd G sig mG Gsig == ++ + + + 1 11 1 (244) () RRrRR RR gR oSdGsig Gsig mG =+ +       || || || (245) Thus: Finally, we recognize this form as that of resistances in parallel: Introduction to Electronics 164 Review of Bode Plots Af j f f j f f j f f v ZZ P () =       +       +       12 1 1 1   (246) Review of Bode Plots Introduction The emphasis here is review . Please refer to an appropriate text if you need a more detailed treatment of this subject. Let us begin with a generalized transfer function: We presume the function is limited to certain features: ● Numerator and denominator can be factored. ● Numerator factors have only one of the two forms shown. ● Denominator factors have only the form shown. Remember: ● Bode plots are not the actual curves, but only asymptotes to the actual curves . ● Bode magnitude plots are not based on the transfer function itself, but on the logarithm of the transfer function - actually, on 20 log A v . ● The total Bode response for A v ( f ) consists of the magnitude response and the phase response . Both of these consist of the sum of the responses to each numerator and denominator factor. Introduction to Electronics 165 Review of Bode Plots 0 dB f z1 20 dB/decade Fig. 234. Bode magnitude response for jf / f Z1 . 0 dB f z2 20 dB/decade Fig. 235. Bode magnitude response for 1 + jf / f Z2 . The Bode Magnitude Response Now, let’s review the Bode magnitude response of each term: The numerator term : j f f Z 1 The magnitude response increases 20 dB per decade for all f . For f = f Z1 the term has a magnitude of 1. Thus the magnitude response has an amplitude of 0 dB at f Z1 . The numerator term : 1 2 + j f f Z For f << f Z2 the imaginary term is negligible; the magnitude is just 0 dB. For f >> f Z2 the imaginary term dominates, thus the magnitude increases 20 db per decade. The denominator term : 1 1 + j f f P For f << f P1 the imaginary term is negligible; the magnitude is just 0 dB. For f >> f P1 the imaginary term dominates, thus the magnitude decreases 20 db per decade (because the term is in the denominator). 0 dB f p -20 dB/decade Fig. 236. Bode magnitude response for 1 + jf / f P1 . Introduction to Electronics 166 Review of Bode Plots +90 O Fig. 237. Bode phase response for jf / f Z1 . 0 O f z2 /10 10 f z2 +90 O 45 O /decade Fig. 238. Bode phase response for 1 + jf / f Z2 . The Bode Phase Response Now, let’s review the Bode phases response of each term: The numerator term : j f f Z 1 The phase response is simply 90 o for all f . The numerator term : 1 2 + j f f Z For f << f Z2 the imaginary term is negligible; the phase is just 0 o . For f >> f Z2 the imaginary term dominates, thus the phase is 90 o . At f = f Z2 , the term is 1 + j 1; its phase is 45 O . The denominator term : 1 1 + j f f P For f << f Z2 the imaginary term is negligible; the phase is just 0 o . For f >> f Z2 the imaginary term dominates, thus the phase is -90 o . At f = f Z2 , the term is 1 + j 1; its phase is -45 O . 0 O f p /10 10 f p -45 O /decade -90 O Fig. 239. Bode phase response for 1 + jf / f P1 . Introduction to Electronics 167 Review of Bode Plots V o (s) R + 1/ sC V in (s) + - Fig. 240. Single-pole low-pass RC circuit, A V V sC R sC sRC v o in == + = + 1 1 1 1 (247) () A jRCf j f f f RC v b b = + = + = 1 12 1 1 1 2 ππ where (248) A f f v b = +       1 1 2 2 (249) Single-Pole Low-Pass RC The review of the details of the Bode response of a single-pole low-pass RC circuit begins with the s-domain transfer function: Note that there is a pole at s = -1/ RC and zero at s = . ∞ For the sinusoidal steady state response we substitute j 2 π f for s : This fits the generalized single-pole form from the previous page, except we’re using “ f b ” instead of “ f P .” The term f b is called the half- power frequency, the corner frequency, the break frequency, or the 3-dB frequency. Gain Magnitude in dB: From: Introduction to Electronics 168 Review of Bode Plots () A f f f f f f f f v dB b b bb = +       =− +       =− +       =− +               20 1 1 20 1 20 1 20 1 10 1 2 2 2 2 2 22 log log log log log (250) () A v dB =− = 10 1 0log dB (251) A f f f f v dB bb =−       =−       10 20 2 log log (252) f b /10 f b 10 f b 100 f b -3 dB -40 dB -20 dB A v , dB f Fig. 241. Bode magnitude plot for single-pole low- pass, in red. The actual curve is shown in blue. We obtain: Bode Magnitude Plot: From eq. (250), at low frequencies ( f / f b << 1): And, at high frequencies ( f / f b >> 1): Note that the latter equation decreases 20 dB for each factor of 10 increase in frequency (i.e., -20 db per decade ). [...]... ckt RC + vx Rin - + A v - vo x Fig 249 Model equivalent to amplifier section Coupling Capacitors 173 Introduction to Electronics (3) redraw the entire circuit (Fig 2 47) as shown: RS Ro Cin + vx Rin - + vs - + Avo vx - Cout RL + vo - Fig 250 Complete circuit redrawn with amplifier section replaced by its model Note that both sides are identical topologically, and are single-pole, high-pass circuits:... Introduction to Electronics 175 Design Considerations for RC-Coupled Amplifiers 1 RC-Coupled amplifiers: Coupling capacitors - capacitors cost $ Direct-Coupled amplifiers: No capacitors - bias circuits interact - more difficult design, but preferable 2 Determine Thevenin resistance “seen” by each coupling capacitor Larger resistances mean smaller and cheaper capacitors 3 Choose fb for each RC circuit to meet... Generic dual-supply common emitter ckt (Let RL’ = RL || RC , RE = REF + REB ) vo - Low- & Mid-Frequency Performance of CE Amplifiers Introduction to Electronics 177 Both common-emitter topologies have the same small-signal equivalent circuit: iin RS + vs - B ib io C + RB vin β ib r RC RL + vo - - E Rin RL’ REF Fig 254 Generic small-signal equivalent of common emitter amplifier Midband Performance vo − βRL... single-pole high-pass, in red The actual curve is shown in blue (258) Coupling Capacitors Introduction to Electronics 172 Coupling Capacitors Effect on Frequency Response VCC RB RC Cout Cin RS + vs - Q1 + + vin RL vo - Amplifier Source Load Fig 2 47 Representative amplifier circuit, split into sections In our midband amplifier analysis, we assumed the capacitors were short circuits, drew the small-signal... -4 5O -9 0O Fig 243 Bode phase plot for single-pole low-pass, shown in red The actual curve is shown in blue Review of Bode Plots Introduction to Electronics 170 Single-Pole High-Pass RC 1/sC The s-domain transfer function: + + Vin(s) R Av = Vo(s) - R 1 +R sC = sRC sRC + 1 (255) Note there is a pole at s = -1 /RC, and a zero at s = 0 Fig 244 Single-pole high-pass RC circuit For the sinusoidal steady state... Performance of the CE Amplifier Introduction to Electronics 189 High-Frequency Performance of CE Amplifier The Small-Signal Equivalent Circuit We now have the tools we need to analyze (actually, estimate) the high-frequency performance of an amplifier circuit We choose the common-emitter amplifier to illustrate the techniques: VCC R1 RC Cout Cin RS Q1 + + vs - + vin R2 CE RE RL vo - - Fig 269 Standard common... only Av = R S’ + v s’ - vo ′ ≈ −gmRL vπ (286) B’ C + vπ - Cπ Cµ (1+gmRL’) g m vπ R L’ + vo - Fig 272 Final (approximate) equivalent after applying the Miller Effect High-Frequency Performance of the CE Amplifier R S’ + v s’ - Introduction to Electronics B’ 191 C + vπ - Cπ Cµ (1+gmRL’) g m vπ R L’ + vo - Fig 273 Final (approximate) equivalent after applying the Miller Effect (Fig 272 repeated) So we have... due to C2 1 2πCeqRThevenin (283) Because C1 + Cπ >> C2 , the pole due to C1 + Cπ will dominate The pole due to C2 is usually negligible, especially when RL’ is included in the circuit B E 188 Introduction to Electronics The Hybrid-π BJT Model C B’ + vπ - rπ C1 Cπ gmvπ C 2 ro E Fig 268 Miller Effect applied to hybrid-π model (Fig 265 repeated) The overall half-power frequency, then, is usually due to. .. meet overall -3 dB requirement Judicious choice can reduce overall cost of capacitors 4 Calculate required capacitance values 5 Choose C values somewhat (approximately 1.5 times larger) larger than calculated Some C tolerances are as much as -2 0%, +80 % Vales can change ±10 % with time and temperature Low- & Mid-Frequency Performance of CE Amplifiers Introduction to Electronics 176 Low- & Mid-Frequency... repeated) Now we use the hybrid-π equivalent for the BJT and construct the small-signal equivalent circuit for the amplifier: Cµ RS + vs - B + vin - rx rµ B’ + RB = R1||R2 vπ E rπ Cπ C g mv π RL||RC ro E RL’ = ro||RL||RC Fig 270 Amplifier small-signal equivalent circuit using hybrid-π BJT model High-Frequency Performance of the CE Amplifier Introduction to Electronics 190 High-Frequency Performance We . Performance of CE Amplifier Introduction We begin with two of the most common topologies of common- emitter amplifier: Introduction to Electronics 177 Low- & Mid-Frequency Performance of CE. determine the its model parameters, and . . . Introduction to Electronics 173 Coupling Capacitors R in R o + - + - v x A vo v x R L C out v o + - R S C in v s + - Fig. 250. Complete circuit redrawn. leads to the following phase equation: This is just the low-pass phase plot shifted upward by 90 o : Introduction to Electronics 172 Coupling Capacitors R S R B R C R L C in C out v s v o + + - - V CC Q 1 v in + - Source Amplifier Load Fig.