Operational Amplifier and Digital Electronics Dr. M Ramegowda Associate Professor of Physics Govt. College (Autonomous), Mandya Dr. M Ramegowda 2 Dr. M Ramegowda Chapter 1 The Differential Amplifier 1.1 The basic differential amplifier Circuit diagram of the emitter coupled differential amplifier is shown in figure 1.1. It consists of two identical transistors Q1 and Q2. The emitters of the transistors are connected to −VEE through a common resistor RE. Collectors are connected to +VCC through equal collector s resistances RC . Two inputs v1 and v2 are applied to the bases of Q1 and Q2 through resistances R1 and R2, and the output vo is measured across two collectors as shown in the circuit diagram. VEE and RE form an approximate constant current source for the amplifier. This amplifier is very useful because ó +VCC ÿ RC ¡ RC ¡ ¨vo Q1 Q2 ÿ R1 ¡ ¡ R2 RE v1 ∼ ∼ v2 ¡ ý© ©ý ñ −VEE Dr.Figure M 1.1: Basic Ramegowda difference amplifier • it operates without input capacitors, means that it can work as a DC amplifier • it provides voltage gain for differential signals on the inputs, vid = v1 − v2 and attenuating interfering common-mode signals, vcm = (v1 + v2)=2 • it provides the inverting and non-inverting inputs needed for operational amplifiers. The zero signal emitter current is given by VEE = IE = I1 + I2 (1.1) RE 3 4 CHAPTER 1. THE DIFFERENTIAL AMPLIFIER ÿý RC ¡ RC ÿ ¨¡ ÿ vo # ic ic # v1 1 2 v2 ð ©ÿ ©ÿ ò re ¡ re ¡ i i 1 ù ÿ2 ù RE ¡ ý Figure 1.2: Small signal model of Basic difference amplifier A more complete analysis can be done using the small signal AC model for the circuit which is shown in the Fig. 1.2. When both inputs are at 0V , the current splits equally in the two branches. If v1 is raised slightly while holding v2 = 0V , i1 start to increase, and the voltage drop across RC of transistor Q1 increases. Hence vo = vc1 − vc2 becomes positive, so v1 is called non-inverting input. In the same way, raising v2 slightly with v1 grounded, increases the value of i2. This now increases vc2 and vo = vc1 − vc2 becomes negative, so v2 is called inverting input. When both v1 and v2 are applied, the amplifier works in the differential mode with input vid = v1 − v2. Then the gain of the amplifier is v A = o vid 1.2 Configurations of differential amplifier Differential amplifier can be used in four modes 1. Dual input, balanced-output differential amplifier 2. Dual input, unbalanced-output differential amplifier 3. Single input, balanced-output differential amplifier 4. Single input,Dr. unbalanced-output M differential Ramegowda amplifier When both inputs v1 and v2 are applied, it is said to be dual-input and the output across the collectors of two transistors is taken as balanced-output. When one of the input is applied, it is said to be single-input and the output across the collector of one of the transistor with respect to ground is taken as unbalanced output. All the components in the two emitter biased circuits, which constitute a differential amplifier, must be matched in all respects for proper operation and the magnitude of the supply voltages +Vcc and −VEE must be equal. A multistage amplifier with a desired voltage gain can be formed using a direct connection between successive stages of differential amplifiers. The beauty of the direct connection between stages is that it removes the lower cutoff frequency imposed by the coupling capacitors. In instrumentation systems, differential amplifiers are widely used to compare two input signals. 1.3. ANALYSIS OF DUAL INPUT AND BALANCED OUTPUT DIFFERENTIAL AMPLIFIER5 1.3 Analysis of Dual input and balanced output differen- tial amplifier The circuit of the dual input balanced output differential amplifier is shown in the Fig 1.3. The two input signals (dual input), v1 , and v2 are applied to the bases of transistors Q1 and Q2. The output vo is measured between the collectors of two transistors, which are at the same dc potential. Because of the equal dc potential at the two collectors with respect to ground, the output is referred to as a balanced output. ó +VCC ÿ RC RC ¡ ¡ ¨vo Q1 Q2 ÿ R1 ¡ ¡ R2 RE v1 ∼ v2 ∼ ¡ ý© ý© ñ −VEE Figure 1.3: The dual input balanced output differential amplifier In the analysis, r-parameters are used. The advantage of r-parameters are, 1. AC analysis of differential amplifiers with r-parameters is simpler, more straight forward, and less cumbersome. 2. There is no need to manipulate the r-parameters at different operating levels except for the re value. 3. The performance equations obtained are easy to remember since they are not complex or lengthy 4. ResultsDr. obtained using r-parameters M Ramegowda compare favorably with the actual results. 1.3.1 DC analysis In the dc analysis it is customary to determine the operating point values (ICQ and VCEQ). The dc equivalent circuit can be obtained simply by reducing the input signals v1 and v2 to zero. The dc equivalent circuit thus obtained is shown in Fig. 1.4 Since both emitter-biased sections of the differential amplifier are symmetrical (matched in all respects), it require to determine ICQ and VCEQ for only one section. Then these ICQ and VCEQ values can be used for other transistor. By applying KVL to the base-emitter loop of the transistor Q1, we get R1IB + VBE + 2IERE − VEE = 0 6 CHAPTER 1. THE DIFFERENTIAL AMPLIFIER ó +VCC ÿ RC RC ¡ ¡ ¨vo Q1 Q2 IB û û IB ÿ R1 ¡ ¡ R2 ý RE ý ¡ ñ −VEE Figure 1.4: DC equivalent circuit of the dual input balanced output differential amplifier IC Since IB = β and IC = IE, R1IC /β + VBE + 2IC RE − VEE = 0 IC (R1/β + 2RE) = VEE − VBE VEE − VBE IC = 2RE + R1/β Generally R1/β << RE, the above equation can be written as VEE − VBE IC = (1.2) 2RE Thus the quiescent current of transistors Q1 and Q2 are independent of the collector resistances. VBE = 0:6V for silicon transistors and VBE = 0:2V for germanium transistors. The collector-emitter voltage of the transistor Q1 is given by VCE = VC − VE From the Fig. 1.4 we can write, VC = VCC − IC RC and VE = −VBE. Therefore, VCE = VCC − IC RC + VBE (1.3) The quiescent collector current ICQ = IC and collector-emitter voltage VCEQ = VCE can be calculated by using the equations 1.2 and 1.3. 1.3.2 ACDr. analysis M Ramegowda The ac equivalent circuit can be obtained by reducing VCC and VEE to zero. The ac equivalent circuit of Fig. 1.3 is shown in Fig. 1.5. Applying KVL to loop I and loop II in the Fig. 1.5, we get v1 = R1ib1 + reie1 + RE(ie1 + ie2) (1.4) v2 = R2ib2 + reie2 + RE(ie1 + ie2) (1.5) Since ib is very small compared to ie, the equations 1.4 and 1.5 can be written as (re + RE)ie1+ REie2 = v1 (1.6) REie1+(RE + re)ie2 = v2 (1.7) 1.3. ANALYSIS OF DUAL INPUT AND BALANCED OUTPUT DIFFERENTIAL AMPLIFIER7 C1 E1 E2 C2 ÿ ð ú ÿ ò ð ø ÿ ð re ie1 ie2 re ó B1 ó B2 i i b1 û b2 û I II RC R1 RE ¡ R2 RC ¡ ¡ ¡ ¡ v1 ∼ v2 ∼ ý ©ý ý ©ý ý Figure 1.5: AC equivalent circuit of difference amplifier The above equations can be written in the form of matrix equation as re + RE RE ie1 v1 = (1.8) RE RE + re ie2 v2 Then we can write, ∆ i = 1 (1.9) e1 ∆ ∆ i = 2 (1.10) e2 ∆ where re + RE RE ∆ = RE RE + re 2 2 = (RE + re) − (RE) = (RE + re − RE)(RE + re + RE) = re(2RE + re) (1.11) v1 RE ∆1 = = v1(RE + re) − v2RE (1.12) v2 RE + re re + RE v1 ∆2 = = (RE + re)v2 − REv1 (1.13) RE v2 It is considered that the voltage at the collector of Q2 is greater than the voltage at the collector of Q1. Therefore theDr. output voltage is M given by Ramegowda vo = −vc2 − (−vc1) = vc1 − vc2 = ic1RC − ic2RC = (ic1 − ic2)RC Since ic ' ie, vo = (ie1 − ie2)RC On substituting the values for ie1 and ie2 by using the equations 1.9 and 1.10, we get ∆ − ∆ v = 1 2 R o ∆ C 8 CHAPTER 1. THE DIFFERENTIAL AMPLIFIER On substituting the values of ∆; ∆1 and ∆2 by using equations 1.11, 1.12 and 1.13, we get v1(RE + re) − v2RE − [(RE + re)v2 − REv1] vo = RC re(2RE + re) v1(2RE + re) − (2RE + re)v2 = RC re(2RE + re) (2RE + re)(v1 − v2) = RC re(2RE + re) RC = (v1 − v2) (1.14) re RC vo = vid (1.15) re where vid = v1 − v2 The gain of the amplifier is v R A = o = C (1.16) vid re Thus a differential amplifier amplifies the difference between the two input signals and gain of the amplifier is independent of RE. 1.3.3 Differential input resistance It is the equivalent resistance measured at either input terminal with other terminal grounded. v1 Ri1 = (1.17) i b1 v2=0 v1 = (1.18) i /β e1 ac v2=0 v1βac = (1.19) i e1 v2=0 Substituting the values of ie1 from equation 1.9, we get v1βac∆ Ri1 = ∆1 Substituting the values of ∆ and ∆1 from equations 1.11 and 1.12, we get v1βacre(2RE + re) Ri1 = vDr.1(RE + re) − v2RE M Ramegowda Since v2 = 0, v1βacre(2RE + re) Ri1 = v1(RE + re) Generally, RE re, 2RE + re ' 2RE and RE + re ' RE.