S ELECTED E QUATIONS

Transcription

1 D EIVAIONS OF S ELECED E QUAIONS Equation 2 3 he average value of a half-wave rectified sine wave is the area under the curve divided y the period (2p). he equation for a sine wave is v Vpsin u p Vp area Vpsin u du (-cos u) p0 2p 2p L0 2p Vp Vp Vp 3-cos p - (-cos 0)4 3-(-) - (-)4 (2) 2p 2p 2p Vp p VAVG VAVG Equation 2 2 efer to Figure B. vc Vp(rect)e t LC Vp(rect) 0 Vr( pp) tdis 䊱 FIGUE B When the filter capacitor discharges through L, the voltage is vc Vp(rect)e - t>lc Since the discharge time of the capacitor is from one peak to approximately the next peak, tdis when vc reaches its minimum value. vc(min) Vp(rect)e - >LC Since C W, /LC ecomes much less than (which is usually the case); e - >LC approaches and can e expressed as e - >LC - LC Appendix B

2 B-2 APPENDIX B herefore, he peak-to-peak ripple voltage is V r(pp) a f L C V p(rect) v C(min) V p(rect) a - L C V r(pp) V p(rect) - V C(min) V p(rect) - V p(rect) + V p(rect) L C V p(rect) L C Equation 2 3 o otain the dc value, one-half of the peak-to-peak ripple is sutracted from the peak value. Equation 6 he Shockley equation for the ase-emitter pn junction is where I E the total forward current across the ase-emitter junction I the reverse saturation current V the voltage across the depletion layer Q the charge on an electron k a numer known as Boltzmann s constant the asolute temperature At amient temperature, Q>k 40, so Differentiating yields V DC V p(rect) - V r(pp) 2 V DC a - 2f L C V p(rect) V p(rect) - a 2f L C V p(rect) I E I (e VQ>k - ) I E I (e V40 - ) Since I e V40 I E + I, Assuming I 66 I E, di E dv 40I e V40 di E dv 40(I E + I ) he ac resistance r e of the ase-emitter junction can e expressed as dv>di E. Equation 6 4 di E dv 40I E r e dv di E 40I E 25 mv I E he emitter-follower is represented y the r parameter ac equivalent circuit in Figure B 2(a).

3 DEIVAIONS OF SELECED EQUAIONS B-3 C β ac I β ac I B s 2 s r e E r e V s 2 E I e E V e (a) () FIGUE B 2 Conventional current direction shown. By thevenizing from the ase ack to the source, the circuit is simplified to the form shown in Figure B 2(). V e, I out I e, and I in I. With V s 0 and with I produced y, and neglecting the ase-to-emitter voltage drop (and therefore r e ), Assuming that 77 s and 2 77 s, Looking into the emitter, appears in parallel with s > ac. herefore, E I out I V e s out V e I e I V e I e I e ac I I out I e acv e s V e 2 s V e ac V e > s out a s ac E s ac Midpoint Bias (Chapter 8) he following proof is for the equation on page 400 that shows I D 0.5I DSS when V GS V GS(off ) /3.4. Start with Equation 8 : Let I D 0.5I DSS. I D I DSS a - 0.5I DSS I DSS a - V 2 GS V 2 GS

4 B-4 APPENDIX B Cancelling I DSS on each side, 0.5 a - V 2 GS We want a factor (call it F) y which can e divided to give a value of V GS that will produce a drain current that is 0.5I DSS. 0.5 J - a V 2 GS(off) F K Solving for F, a F - F herefore, I D 0.5I DSS when V GS /3.4. Equation 9 2 I D I DSS a - earranging into a standard quadratic equation form, he coefficients and constant are In simplified notation, the equation is he solutions to this quadratic equation are Equation 9 0 I DSS a - 2I D S F F F I 2 D S I V DSS a - I D S a - GS(off) a I DSS 2 S V 2 I 2 D - a + 2I DSS S I GS(off) V D + I DSS 0 GS(off) I D -B ; 2B2-4AC 2A A general model of a switched-capacitor circuit, as shown in Figure B 3(a), consists of a capacitor, two voltage sources, and V 2, and a two-pole switch. Let s examine this circuit V I2 D 2 S V 2 GS(off) A 2 SI DSS V 2 GS(off) B -a + 2 SI DSS C I DSS I DSS - 2I DSS S I V D + I DSS 2 S GS(off) AI 2 D + BI D + C 0 I D S V 2 I 2 D GS(off)

5 DEIVAIONS OF SELECED EQUAIONS B-5 I 0 /2 2 Position Position Position V C V 2 Position 2 Position 2 0 /2 (a) () FIGUE B 3 for a specified period of time,. Assume that V and V 2 are constant during the time period and V 7 V 2. Of particular interest is the average current I produced y the source V during the time period. During the first half of the time period, the switch is in position, as indicated in Figure B 3(). he capacitor charges very rapidly to the source voltage V. herefore, an average current I due to V is charging the capacitor during the interval from t 0 to t /2. During the second half of the time period, the switch is in position 2, as indicated. Because V 7 V 2, the capacitor rapidly discharges to the voltage V 2. he average current produced y the source V over the time period is Q (0) is the charge at t 0 and Q (/2) is the charge at t /2. herefore, Q (/2) - Q (0) is the net charge transferred while the switch is in position. he capacitor voltage at /2 is equal to V, and the capacitor voltage at 0 or is equal to. By sustituting CV for Q in the previous equation, V 2 Since and are assumed to e constant during, the average current can e expressed as V V 2 I (avg) Q (/2) - Q (0) I (avg) CV (/2) - CV 2(0) I (avg) C(V - V 2 ) CV (/2) - V 2(0) 2 Figure B 4 shows a conventional resistive circuit with two voltage sources. From Ohm s law, the current is I I V - V 2 he current I (avg) in the switched-capacitor circuit is equal to I in the resistive circuit. I (avg) C(V - V 2 ) By solving for and canceling the V - V 2 terms, (V - V 2 ) C(V - V 2 ) V - V 2 V V 2 FIGUE B 4 C As you can see, a switched-capacitor circuit can emulate a resistor with a value determined y the time period and the capacitance C. ememer that the two-pole switch is in each position for one-half of the time period and that you can vary y varying the frequency at which the switches are operated.

6 B-6 APPENDIX B Since /f, the resistance in terms of frequency is Equation 0 An inverting amplifier with feedack capacitance is shown in Figure B 5. For the input, Factoring V out, fc I V - V 2 I V ( - V 2 > V ) he ratio V 2 >V is the voltage gain, -A v. I V ( + A v ) V > ( + A v ) FIGUE B 5 C I I 2 A v V V 2 he effective reactance as seen from the input terminals is or Cancelling and inverting, 2pfC in(miller) in(miller) + A v 2pfC( + A v ) C in(miller) C(A v + ) Equation 0 2 For the output in Figure B 6, Since V >V 2 ->A v, I 2 V 2( + >A v ) I 2 V 2 - V V 2( - V >V 2 ) he effective reactance as seen from the output is out(miller) 2pfC out(miller) V 2 >( + >A v ) V 2 >[(A v + )>A v ] (A v + )>A v 2pfC[(A v + )>A v ]

7 DEIVAIONS OF SELECED EQUAIONS B-7 Cancelling and inverting yields C out(miller) Ca A v + A v Equations 0 29 and 0 30 he total gain, A v(tot), of an individual amplifier stage at the lower critical frequency equals the midrange gain, A v(mid), times the attenuation of the high-pass C circuit. A v(tot) A v(mid) a A X 2 v(mid) a C 2 + X 2 C> 2 f cl 2pC Dividing oth sides y any frequency f, f cl f (2pfC) Since > 2pfC, f cl f Sustitution in the gain formula gives he gain ratio is A v(tot) A v(mid) a 2 + ( f cl >f ) 2 For a multistage amplifier with n stages, each with the same of the gain ratios is and gain ratio, the product he critical frequency f cl of the multistage amplifier is the frequency at which A v(tot) 0.707A v(mid), so the gain ratio at f cl is herefore, for a multistage amplifier, So Squaring oth sides, A v(tot) A v(mid) A v(tot) A v(mid) n 2 c 2 + ( f cl >f cl ) d 2 a n 2 + ( f cl >f ) /2 (2 + ( f cl >f cl ) 2 ) n 2 + ( f cl >f ) 2 2 ( + ( f cl >f cl ) 2 ) n f cl (2 + ( f cl >f cl ) 2 ) n

8 B-8 APPENDIX B aking the nth root of oth sides, A similar process will give Equation 0 30: Equations 0 3 and 0 32 he rise time is defined as the time required for the voltage to increase from 0 percent of its final value to 90 percent of its final value, as indicated in Figure B 6. Expressing the curve in its exponential form gives When v 0.V final, 2 >n + ( f cl >f cl ) 2 a f 2 cl 2 >n - f cl a f cl f cl 22 >n - f cl f cu f cu 22 >n - v V final ( - e - t>c ) 0.V final V final ( - e -t>c ) V final - V final e -t>c f cl 22 >n - FIGUE B 6 V final e -t>c 0.9V final e -t>c 0.9 ln e -t>c ln (0.9) - t C -0. t 0.C V final 0.9 V final t V final e C 0. V final 0 t r t When v 0.9V final, 0.9V final V final ( - e -t>c ) V final - V final e -t>c V final e -t>c 0.V final ln e -t>c ln (0.) - t C -2.3 t 2.3C

9 DEIVAIONS OF SELECED EQUAIONS B-9 he difference is the rise time. he critical frequency of an C circuit is Sustituting, In a similar way, it can e shown that Equation 2 2 he formula for open-loop gain in Equation 2 9 can e expressed in complex notation as Sustituting the aove expression into the equation A cl A ol >( + BA ol ) gives a formula for the total closed-loop gain. Multiplying the numerator and denominator y + jf>f c(ol) yields Dividing the numerator and denominator y + BA ol(mid) gives he aove expression is of the form of the first equation where f c(cl) Equation 4 A cl t r 2.3C - 0.C 2.2C A cl A cl f c 2pC C 2pf c t r pf cu f cu f cu 0.35 t r A ol A ol(mid) >( + jf>f c(ol) ) + BA ol(mid) >( + jf>f c(ol) ) A ol(mid) >( + BA ol(mid) ) + j[ f>( f c(ol) ( + BA ol(mid) ))] A cl f cl 0.35 t f A ol(mid) + jf>f c(ol) A ol(mid) + BA ol(mid) + jf>f c(ol) A cl(mid) + jf>f c(cl) is the closed-loop critical frequency. hus, f c(cl) f c(ol) ( + BA ol(mid) ) In Figure B 7 the common-mode voltage, V cm, on the noninverting input is amplified y the small common-mode gain of op-amp A. ( A cm is typically less than.) he total output voltage of op-amp A is a + - a 2 + V cm

10 B-0 APPENDIX B A similar analysis can e applied to op-amp A2 and results in the following output expression: 2 a a 2 + V cm + V cm + A A3 A cl (2 ) 2 + V cm A FIGUE B 7 Op-amp A3 has on one of its inputs and 2 on the other. herefore, the differential input voltage to op-amp A3 is a a V cm - V cm For 2, Notice that, since the common-mode voltages (V cm ) are equal, they cancel each other. Factoring out the differential gain gives the following expression for the differential input to op-amp A3: Op-amp A3 has unity gain ecause and A v 5 / 3. 6 / 4. herefore, the final output of the instrumentation amplifier (the output of op-amp A3) is he closed-loop gain is Equation a a V cm - V cm 2 - a + 2 (2 - ) (2 - ) a + 2 (2 - ) A cl 2 - A cl + 2 (-jx)>( - jx) ( - jx) + (-jx)>( - jx) (-jx) ( - jx) 2 - jx

11 DEIVAIONS OF SELECED EQUAIONS B- Multiplying the numerator and denominator y j, For a 0 phase angle there can e no j term. ecall from complex numers in ac theory that a nonzero angle is associated with a complex numer having a j term. herefore, at f r the j term is 0. hus, Cancelling yields X j( - jx) 2 + X X X + j 2 + 2X - jx X 2 0 X 3X 3 X X + j( 2 - j2x - X 2 ) X 3X + j( 2 - X 2 ) Equation 6 2 From the derivation of Equation 6, Since X 2pf r C, 2 - X X 2 f r X 2pf r C 2pC Equations 6 3 and 6 4 he feedack circuit in the phase-shift oscillator consists of three C stages, as shown in Figure B 8. An expression for the attenuation is derived using the mesh analysis method for the loop assignment shown. All s are equal in value, and all Cs are equal in value. ( - j>2pfc)i - I 2 + 0I 3 -I + (2 - j>2pfc)i 2 - I 3 0 0I - I 2 + (2 - j>2pfc)i 3 0 C C C I I 2 I 3 FIGUE B 8

12 B-2 APPENDIX B In order to get, we must solve for using determinants: I 3 I 3 3 ( - j>2pfc) - - (2 - j>2pfc) ( - j>2pfc) (2 - j>2pfc) (2 - j>2pfc) I 3 I 3 2 ( - j>2pfc)(2 - j>2pfc) 2-2 (2 - j>2pfc) - 2 ( - >2pfC) Expanding and comining the real terms and the j terms separately. For oscillation in the phase-shift amplifier, the phase shift through the C circuit must equal 80. For this condition to exist, the j term must e 0 at the frequency of oscillation f r. Since the j term is 0, 5 a - 4p 2 f 2 2 C 2 - ja 6 2pfC - (2pf ) 3 3 C 3 6 2pf r C - (2pf r ) 3 3 C 3 0 6(2p) 2 f 2 r 2 C 2 - (2p) 3 f 3 r 3 C 3 0 6(2p) 2 fr 2 2 C 2-0 fr 2 6(2p) 2 2 C 2 f r 2p6C 5 5-4p 2 fr 2 2 C 2-2 a 6C 2 C 2 ( - j>2pfc) (2 - j>2pfc) 2 - (3 - j>2pfc) 3 3 ( - j>2pfc) (2 - j>2pfc) 2-3 (3 - j>2pfc) he negative sign results from the 80 inversion. hus, the value of attenuation for the feedack circuit is 3 ( - j>2pfc)(2 - j>2pfc) 2-3 (2 - j>2pfc) - 3 ( - >2pfC) 3 3 ( - j>2pfc)(2 - j>2pfc) 2-3 [(2 - j>2pfc) - ( - j>2pfc)] B 29