à 10. DC (DIRECT-COUPLED) AMPLIFIERS
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1 0.DC-Amps-X.nb à 0. DC (DIRECT-COUPLED) AMPLIFIERS ü AC COUPLED SMALL SIGNAL AMPLIFIERS ADVANTAGES:. Signal, load and the amplifier bias are separate. One can work on the bias calculations stage by stage w/o worrying about an interaction with the signal source or the load. 2. No dc current flows through the load or through the signal source. (Magnetic saturation of the load or the signal source is prevented. Also, their dc resistances do not interfere with the bias.) 3. Stages can easily be cascaded. Design of a stage involves only the ac loading effects of its neighbors, dc conditions of the adjacent stages are completely independent of each other. DISADVANTAGES: For dc isolation and stability AC coupled amplifiers depend on the isolating capability of capacitors while expecting the capacitors to be fully transparent (almost like a short circuit) at the signal frequency. a. At low frequencies capacitors fail to act like short circuit. Therefore, the AC coupled amplifier behaves like a high pass filter and becomes useful only above a certain cut-off frequency. b. 3 capacitors are needed for each amplifier stage. Capacitors are bulky and costly and cannot be integrated on a silicon chip. c. "DC-like" slowly varying signals cannot be amplified because of impractically high values of capacitors required. ü DIFFERENTIAL AMPLIFIERS 0. THE DIFFERENCE AMPLIFIER (THE DIFFERENTIAL AMPLIFIER) COMMON-MODE RESPONSE DIFFERENTIAL - INPUT RESPONSE A. The Tail Current Source B. The BJT Current Source 0. 2 BJT DIFFERENCE AMPLIFIER with ZENER DIODE REGULATED CURRENT SOURCE 0. 3 OUTPUT RESISTANCE OF A BJT CURRENT SOURCE EXAMPLE ELE 343 Notes Prof. M.G. Guven
2 2 0.DC-Amps-X. ü 0. THE DIFFERENCE AMPLIFIER (THE DIFFERENTIAL AMPLIFIER)* BJT VERSION The circuit is made symmetrical. I O is a DC-current source. Q = Q 2 (matched transistors). R C = R C2 = R C and R = R 2 Typically,» V EE» =» V CC» = ±0 V ±2 V ±5 V and v O = v 02 I CQ = I CQ2. I BQ = I BQ2 I EQ = I EQ2 I EQ + I EQ2 = I O I EQ = I EQ2 = I O ÄÄÄÄÄÄÄÄ 2 î I BQ = I O ê 2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Hb EFF + L b EFF and I CQ = I O ê 2 * ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Hb EFF + L
3 0.DC-Amps-X.nb For bias v v 2 = 0 V X = 0 - I BQ.R - V - V BEQ For proper bias For both transistors V CEQ = V CQ - V EQ = V CC - R C.I CQ - V X where V EQ = V X V CEQ > V CESAT V CC - R C.I CQ - V X > V CESAT The Difference Amplifier with v and v 2 applied: Two extreme cases:. Pure Differential Input 2. Pure Common-Mode Input v ª-v 2 = v INDM ê 2 v ª v 2 = v INCM ELE 343 Notes Prof. M.G. Guven
4 4 0.DC-Amps-X..» v» ª» v 2» but signs are opposite. 2.» v» ª» v 2» and signs are the same. The outputs :. v OUT = v O "Single ended" 2. v OUT = v O2 "Single ended" outputs are measured wrt the common (ground). 3. v OUT = v O - v O2 "Differential output" outputs are measured wrt the each other. ü COMMON-MODE RESPONSE Right ª Left Differential Output v O - v O2 ª 0 always in common mode Therefore,. Zero response for common-mode input--differential output operation What if single ended output, v O or v O2? Since i E = i E2 = I O / 2 and fixed, ï i C = i C2 = fixed ï v O = v O2 = fixed, Therefore, 2. Zero response for common-mode input--single-ended output operation. Note that, in the single-ended output case, the output voltage contains a bias voltage since it is measured wrt the ground. However, the measured voltage does not change with (does not respond to) the common mode input. In the differential output, the bias voltages are also canceled leaving an absolute zero output.
5 0.DC-Amps-X.nb Summary: The difference amplifier circuit shown above does not respond to common mode signals applied to its input. It "rejects" common-mode signals ü DIFFERENTIAL - INPUT RESPONSE I will assume the signals are small so that I can do a small-signal (linear) analysis. v = Hv INDM Lê2 v 2 = -Hv INDM L ê 2 v = v 2 v - v 2 = v INDM SSAC Equivalent: The pivot point behaves as if it is tied to ground. Its potential (voltage) cannot change. Therefore it behaves like a "Virtual Ground". ELE 343 Notes Prof. M.G. Guven
6 6 0.DC-Amps-X. So, equivalently: Dv O =-R C h fe Di B î Di B = i j V indm ê 2 y ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ z k R + h ie { Single - Ended Output Differential Gain» A vdm» = D i j Dv O y ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ z = k V indm { R C h fe ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 HR + h ie L If R is small then A vdm = R C h fe ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = R C h fe ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 h ie 2 h fe. ÄÄÄÄÄÄÄÄÄÄÄ ktêq I CQ = ÄÄÄÄÄ 2 R C I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ kt ê q Then :. Gain for Differential Input ê SingleEnded Output : A vdm = ÄÄÄÄÄ 2 R C I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ kt ê q where I CQ > I O ê 2 2. Input Impedance R indm = V indm ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ I indm where V indm ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = HR + h ie L.Di B and I indm =Di B 2 Therefore, R indm = 2 h ie since R &R 2 are external. 3. Output Impedance HR out L single-ended output = R C In conclusion: The design of differential amplifier is very simple. Given (A vdm & R indm ) or (A vdm & R out ) or (R indm & R out ) use appropriate pairs of equations (two equations) to determine the two unknowns, I CQ & R C. ï I O > 2 I CQ
7 0.DC-Amps-X.nb Differential Gain Hfor Single Ended Output operationl A vdm = ÄÄÄÄÄ 2 R C I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ kt ê q A vdm = ÄÄÄÄÄ HDC bias on R C L ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 2.59 V ÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 kt ê q mv = 50» R C.I CQ»H» VL CC - V CBQ» where V CBQ Zero or reverse biasing Potential Max. Differential Gain achievable HA vdm L max possible ª ÄÄÄÄÄ V CC ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 kt ê q ~ 250 for V CC = 2 VDC. à THE TAIL CURRENT SOURCE ü A. SIMPLE V EE R E CIRCUIT Assuming v and v 2 do have a common DC component. Then, Hv L DC - R.I BQ - V BEQ = V EQ I O =- -» V EE» -V EQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄ Ä =» V EE» +V EQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄ and 2 I EQ = I O R E R E Therefore, I O & I BQ & I CQ depend on what Hv L DC is. This creates a disadvantage for this simple implementation of the tail current source that has to be kept in mind. ELE 343 Notes Prof. M.G. Guven
8 8 0.DC-Amps-X. a. Differential Mode Gain: SSACC: Because of virtual ground behaviour of point "X" (the tail) for pure differential input, 2 R E 's get shorted to ground with no effect on the differential response. Therefore, the equations derived earlier with an ideal current source biasing the tail should be applicable to this circuit.. Gain for Differential Input ê SingleEnded Output : A vdm = ÄÄÄÄÄ 2 R C I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ kt ê q where I CQ > I O ê 2 2. Input Impedance R indm = 2 h ie since R &R 2 are external. 3. Output Impedance : HR out L single-ended output = R C b. Common Mode (Response) Gain: SSAC:
9 0.DC-Amps-X.nb SSAC analysis on the left half: ELE 343 Notes Prof. M.G. Guven
10 0 0.DC-Amps-X. V incm = R Di B + h ie.di B + 2 R E H + h fe L Di B Amplitute of small signal of "v o " HV o L cm =Dv o =-R C.h fe.di B Common Mode HSSACL Gain A vcm = HV ol cm -R C.h fe.di B ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ V incm HR + h ie + Hh fe + L 2R E Di B If R E is large enough, A vcm ô R C ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 R E A vcm ô R C I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄ 2 R E I CQ I ÄÄÄÄÄÄÄÄÄÄ ktêq M ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Ä, I ÄÄÄÄÄÄÄÄÄÄÄ ktêq M I O > 2 I CQ» -R C h fe»» A vcm» = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ R + h ie + Hh fe + L 2 R E A vdm A vdm» A vcm» = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Ä = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ R I o E ÄÄÄÄÄ 2 I ÄÄÄÄÄÄÄÄÄÄÄ ktêq M DC voltage drop on R ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ E 2kTêq Ultimate low A vcm will be attained with values approaching A vdm >» A vdm» ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ»V EE» ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Make ê Pick V EE much larger than kt ê q 2kTêq Common - Mode - Rejection - Ratio = C.M.R.R = D» A vdm» ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ» A vcm» > V EE ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 kt ê q Conclusions :. Differential Response is same as the differential amplifier with an ideal current source. 2. Common mode response is not zero. CMMR is not infinite. 3. Common mode rejection can be improved by increasing V EE but it will always be inferior. ü B. BJT CURRENT SOURCES. Simple BJT Current Source
11 0.DC-Amps-X.nb Comparison of BJT and resitor characteristics where r O = ê h oe = / slope = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ V CEX+V ÅÅÅÅÅÅÅÅ AX I CQX is much larger than an R X. 2. BJT Current Source with Emitter Degeneration ELE 343 Notes Prof. M.G. Guven
12 2 0.DC-Amps-X. If a unbypassed R EX is present the equivalent small signal resisitance of the resulting current source becomes much higher than h oe -. Therefore, do not bypass R EX of the current source circuit. (See next section for proof.) à OUTPUT RESISTANCE OF A BJT CURRENT SOURCE WITH EMITTER DEGENERATION It's small signal equivalent circuit to calculate r o using the I test, V test method.
13 0.DC-Amps-X.nb Rearranging the circuit for the loop equations: ELE 343 Notes Prof. M.G. Guven
14 4 0.DC-Amps-X. -R EX 0 J N = i j R BX + h ie + R EX y -V Test k-r EX + h - - z J I N oe.h fe R EX + h oe { I 2 -R EX LinearSolveA i j R BX + h ie + R EX y 0 k-r EX + h - - z, J NE oe.h fe R EX + h oe { -V Test R EX V Test 99- ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ I ÅÅÅÅÅÅÅ h oe.h fe - R EX M R EX + I ÅÅÅÅÅÅÅ h oe + R EX MHh ie + R BX + R EX L =, Hh ie + R BX + R EX L V Test 9- ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ I ÅÅÅÅÅÅÅ h oe.h fe - R EX M R EX + I ÅÅÅÅÅÅÅ h oe + R EX MHh ie + R BX + R EX L == Hh ie + R BX + R EX L V Test I 2 =-ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ I ÄÄÄÄÄÄÄ.h h fe - R EX M R EX + I ÄÄÄÄÄÄÄ + R EX MHh ie + R BX + R EX L oe h oe r o == V Test ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ I Test = V Test ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ -I 2 I ÅÅÅÅÅÅÅ h oe.h fe - R EX M R EX + I ÅÅÅÅÅÅÅ h oe + R EX MHh ie + R BX + R EX L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ h ie + R BX + R EX r 0 =+ i j ÄÄÄÄÄÄÄÄÄÄ k h oe y + R EX z + { R EX i ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ j y ÄÄÄÄÄÄÄÄÄÄ.h fe - R EX z h ie + R BX + R EX k h oe { a. For small R EX (approaching Common-Emitter)
15 0.DC-Amps-X.nb r 0 = ÄÄÄÄÄÄÄÄÄÄ + R EX ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ. ÄÄÄÄÄÄÄÄÄÄ.h fe ô r 0 = ÄÄÄÄÄÄÄÄÄÄ h oe R BX h oe b. For large R EX and small R BX (i.e. Common-Base) h oe r 0 = ÄÄÄÄÄÄÄÄÄÄ + R EX ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ. ÄÄÄÄÄÄÄÄÄÄ.h fe ô r 0 = h fe ÄÄÄÄÄÄÄÄÄÄ h oe R EX h oe Implement V BBX.R BX R B2X with as small as R BX as possible. Solution: Use a zener diode with a small r Z to subsitute for R B2X. Then, R BX > r Z // R R BX > r Z h oe ELE 343 Notes Prof. M.G. Guven
16 6 0.DC-Amps-X. ü 0.2 BJT DIFFERENCE AMPLIFIER with ZENER DIODE REGULATED CURRENT SOURCE Note that ( % S b L IO requires R BX << Hb + L R EX. The Zener diode circuit is ideal for good stability.
17 0.DC-Amps-X.nb V REX = V Z - V BEX and I EX > I CX > I O Pick V Z,R EX such that R EX > V Z - V BEX ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Pick R X such that D Z avoids operating in its soft breakdown. I O I R - I BX = I D > I ZMIN V R =» V EE» -V Z Pick R X =» V EE» -V Z ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ <» V EE» -V Z ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ I D + I BX I ZMIN + I BX Then, one can get output resistance as high as, r o > h fex.h oex - I-V chs. of BJT Current Source With Zener where: Slope = ÅÅÅÅÅ r o = ÅÅÅÅÅÅÅÅ hoex h fex ELE 343 Notes Prof. M.G. Guven
18 8 0.DC-Amps-X. Simple BJT Tail Current Source Slope = ÅÅÅÅÅ r o = h oex Example : Given the specs of a BJT Difference Amplifier, design its current source, calculate its common - mode and differential mode gains, input impedances, CMRR and CMR range. V Z = 6V V CC =+5 V V EE =» 5 V» V A = 00 R C = 7 K b > 00 I O = 2mA I ZMIN =.5 ma. Design of Zener Diode biased BJT Current Source : R EX = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄ 2mA = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 5.3 = 2.65 KW 2mA R X = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ Ä = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄ > 3KW HI DMIN I ZMIN L I BX + I DMIN I ÄÄÄÄÄÄÄÄÄÄÄÄ 2 ma 00 M + 3mA 2. Find A vd,a vcm, R indm Upper common - mode range If R and R 2 > 0 A vd = ÄÄÄÄÄ 2 Hv L max = Hv 2 L max R C.I CQ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ kt ê q = ÄÄÄÄÄ H7KL I ÄÄÄÄÄÄÄÄÄÄÄÄ 2 ma M 2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 3.5 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 35 A vcm > R C ÄÄÄÄÄÄÄÄÄÄÄÄ 2 r o = R C R C ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄ > ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 Hh oex L -.h fex I 2 I ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ cx V ÄÄÄÄÄÄÄÄ M -.h AX +V fex CEX where r o is the small signal resistance of the current source and is equivalent to R E of simple current source in the equation above. Calculating V CEX, V CEX = V X - H-» V EE» +V Z - V BEQ L
19 0.DC-Amps-X.nb For HV CEX L Hv L DC = 0, Hv 2 L DC = 0 V X =-V BEQ, therefore, HV CEX L Hv L DC = 0, Hv 2 L DC = 0 = H L > 9 V > V CESAT O.K. Since I EX > I O, H h oex L - = i - I cx y j ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ z = i j ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 ma k V AX + V CEX { k 00 V + 9Vy { - z > 55 K 7KW A vcm = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄ Ä > 6 * * 55 KW*00 I + ÄÄÄÄÄÄÄ 9 00 M CMRR = A vdm ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 35 > = 07 db of rejection HVery Good!L A -4 vcm 6.0 This will be true as long as Hv CM L MIN < Hv L CM = Hv 2 L CM < Hv CM L MAX, the common - mode range of the amplifier. Calculating the common - mode range minimum, Since the current source behavior collapses for V X -V EE + V REX + V CESAT Then the common - mode voltage applied to bases of Q and Q 2 can not go below, Hv MIN = v 2 MIN L V BEQ Hor 2L + HV X L MIN V BEQ - V EE + V REX + V CESAT HV X L MIN =-5 V + HV Z - V BEX L + V CESAT =-5 V + H6-0.7L + V =-8.7 V Hv CM L MIN = V XMIN + V BEQ = > -8V Calculating the common - mode range maximum, ELE 343 Notes Prof. M.G. Guven
20 20 0.DC-Amps-X. Condition for proper operation of Q and Q 2 : HV CC - R C I CQ L > V X + V CESAT Substitute V X = v MAX - V BEQ v MAX V CC - R C I CQ + V BEQ - V CESAT v MAX = 5 V - 7 * ma > 7.7
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