Operational Amplifiers Lesson #5 Chapter 2 BME 372 Electronics I – 138 J.Schesser Operational Amplifiers • An operational Amplifier is an ideal differential with the following characteristics: – Infinite input impedance – Infinite gain for the differential signal – Zero gain for the common-mode signal Inverting input – Zero output impedance v 1 v – Infinite Bandwidth - o v2 + Non-Inverting input BME 372 Electronics I – 139 J.Schesser Operational Amplifier Feedback • Operational Amplifiers are used with negative feedback • Feedback is a way to return a portion of the output of an amplifier to the input – Negative Feedback: returned output opposes the source signal – Positive Feedback: returned output aids the source signal • For Negative Feedback – In an Op-amp, the negative feedback returns a fraction of the output to the inverting input terminal forcing the differential input to zero. – Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit – Since the input impedance is infinite the current flowing into the input is also zero. – These latter two points are called the summing-point constraint. BME 372 Electronics I – 140 J.Schesser Operational Amplifier Analysis Using the Summing Point Constraint • In order to analyze Op-amps, the following steps should be followed: 1. Verify that negative feedback is present 2. Assume that the voltage and current at the input of the Op-amp are both zero (Summing- point Constraint 3. Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest. BME 372 Electronics I – 141 J.Schesser Example: Inverting Amplifier 1. Verify Negative Feedback: Note that a portion of vo is fed back via R2 to the R2 inverting input. So if vi increases and, i 2 therefore, increases vo, the portion of vo fed i1=vin/R1 0 R1 back will then have the affect of reducing vi - - v (i.e., negative feedback). + i + + + RL vin 2. Use the summing point constraint. - vo - 3. Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output vin i1R1 0 since vi is zero due to the summing - point constraint vin i1R1 Zin R1 i1 i2 due to the summing - point constraint i1 i1 v0 i2 R2 0 since vi is zero R2 - vin which is independent of RL (note that the output is opposite to the input : inverted) R1 BME 372 Electronics I – 142 J.Schesser Op-amp • Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is: 1. Independent of the load 2. Dependent only on values of the circuit parameter 3. We can choose the gain of our amplifier by proper selection of resistors. BME 372 Electronics I – 143 J.Schesser Another Example: Inverting Amplifier 1. Verify Negative Feedback: i 4 2. Use the summing point constraint. R2 R4 i2 3. Use KVL at the inverting input node for the i3 R3 i1=vin/R1 0 R1 branch connected to the source and KCL & - KVL at the node where the 3 resistors are vi + R vo L connected vin vin i1R1 0 since vi is zero due to the summing - point constraint i1 i2 due to the summing - point constraint vi i2 R2 i3R3 0 i2 R2 i3R3 since vi is zero i4 i3 i2 vo R4i4 R3i3 BME 372 Electronics I – 144 J.Schesser Another Example: Continued R i R i R i 2 i i 2 2 3 3 3 2 4 R3 R2 R4 vin vin i1R1 0; i1 i2 i2 i2 i3 R3 R1 i =v /R 0 1 in 1 R R R 1 1 i i i ( 2 1)i ( 2 )v - 4 3 2 2 in R3 R3R1 R1 vi + R R2 1 R2 vin v L v R i R i R ( )v R v o o 4 4 3 3 4 in 3 in R3R1 R1 R3 R1 R2 R4 R4 R2 vin ( ) R3R1 R1 R1 vo R2 R4 R4 R2 Av ( ) vin R3R1 R1 R1 v v i R 0 in in 1 1 Zin Rin R1 i1 i1 i2 Why would we use this design over the simpler one? i2 R2 i3R3 1. Same gain but with smaller values of resistance i4 i3 i2 2. Higher gain vo R4i4 R3i3 BME 372 Electronics I – 145 J.Schesser Non-inverting Amp 1. First check: negative feedback? I = 0 2. Next apply, summing point constraint in + io 3. Use circuit analysis + + vi - -- RL vvv 0 vv + R2 in i f f f vin vo -- + R1 v R vvvfoin; f 1 -- RR12 -- v R RR A o 21 1 2 v vR R in 11Note: v Since iZ0; in 1. The gain is always greater than one in in i in 2. The output has the same sign as the input BME 372 Electronics I – 146 J.Schesser Non-inverting Amp Special Case What happens if R2 = 0? Iin= 0 + io vin vi v f 0 v f v f + v + R i - 1 -- RL v f vo vo vin ; + vin vo R1 0 -- v 0 R Av o 1 1 vin R1 -- vin Since iin 0;Zin iin This is a unity gain amplifier and is also called a voltage follower. BME 372 Electronics I – 147 J.Schesser Some Practical Issues when Designing Op-amps • Since ratios of resistor values determines the gain, choosing the proper resistor values is crucial – Too small means large currents drawn – Too large yields another set of problems • Open Loop Gain is not constant but a function of frequency • Non-linearities of the amplifier – Voltage clipping – Slew rate • DC imperfections – Offsets – Bias Currents BME 372 Electronics I – 148 J.Schesser Selecting Resistor Values • Let say we want a gain of 10. This means that R2 = 9R1. Iin= 0 + io • If we chose R1=1Ω, then for a 10 volt + + output, there will be 1 A flowing threw vi - R and R . -- 1 2 + R2 vin vo -- • THIS IS DANGEROUS!!!! + vf R1 • On the other hand if R1=10 MΩ, then -- there may be unwanted effects due to -- R pickup of induced signals Av 1 2 R • Therefore choosing values between 1 100Ω and 1MΩ is optimum BME 372 Electronics I – 149 J.Schesser Frequency Issues • Let’s assume that the open loop gain of our Op-amp is a function of frequency. 20 log |AOL( f )| dB 20 log |A0OL| AoOL AfOL () 1(/ jf fBOL ) where AfoOL is the open-loop gain for 0, fBOL is called the open-loop break frequency since when ff BOL then AfOL( ) A oOL / 2 or at the half power point. fBOL Note that when fAf oOL BOL , AOL ()f 1 This is will define an important relationship for the amplifier when feedback is used BME 372 Electronics I – 150 J.Schesser Frequency Issues Iin= 0 + io + A + oOL Vi - AfOL () 1(/ jf fBOL ) -- + R2 Using phasors: Vin Vo -- RR + VVV11 ; where fOO Vf R1 RR12 RR 12 -- -- V O VVin i V O V O; since V O V iA OL AoOL AOL Af() Af () 1 jff ( / ) V A Af()OL OL BOL A OOL CL A CL 1()1()AfOL Af OL oOL Vin1 A OL 1 1(/ jf fBOL ) A (Note for the ideal Op-amp A oOL OL 1(/ jf f ) A 1 RR R BOL oOL and A 12 1 2 1(/ jf f ) A 1(/) Ajff CL BOL oOL oOL BOL RR11 1(/)1(/)jf fBOL jf f BOL which what we expected.) AoOL 1 AA oOL oCL ff 1(jj )1( ) fABOL(1 oOL ) f BCL BME 372 Electronics I – 151 J.Schesser Iin= 0 + io Frequency Issues + + Vi - -- 20 log |A0OL| 20 log |AOL( f )| dB + R2 Vin Vo -- + Vf R1 -- 20 log |AOL( f )| dB 20 log |A0CL| -- fBOL fBOL(1+AoOL) A interesting factor now comes to light: Where: A oOL AoOL AoCLffAAf BCL BOL(1 oOL ) oOL BOL AffA and (1 ) 1 A oCL BCL BOL oOL oOL 1 AoOL The gain bandwidth product is constant!!! BME 372 Electronics I – 152 J.Schesser Example • For an amplifier with Open-loop dc gain of 100 dB with Open-loop breakpoint of 40 Hz the Bandwidth product for β=1, 0.1, 0.01 β AOCL AOCL(dB) fBCL 1 0.9999 0 4 MHz 0.1 9.9990 20 400 kHz 0.01 99.90 40 40 kHz BME 372 Electronics I – 153 J.Schesser 20 10 0 Iin= 0 0 5 10 15 20 Voltage clipping -10 + io + -20 + Vi - -- • The Op-amp has a limit as to how + 3k Vin Vo large an output voltage and current -- can be produced. (See figure 2.28) + V 1k • For example for the circuit shown it f -- has the following limitations: ±12 V -- and ±25 mA a) If the load resistor is 10k Ω, what is Vomax Vomax 12 12 the maximum input voltage which can b)Iomax 123mA be handled without clipping RL R1 R2 100 3000 1000 b) Repeat for 100 Ω For this case, maximum output current will occur AvCL 1 3/1 4 before maximum output voltage is reached. V V 12 12 V V V V omax omax omax omax 25mA omax omax a)Iomax 4 Iomax RL R1 R2 10 3000 1000 RL R1 R2 100 3000 1000 4.2mA Vomax 2.44V For this case, maximum output voltage will occur Vinmax 2.44 4 0.61V before maximum output current is reached.