Operational Amplifiers Magic Rules Application Examples Tom Murphy, UCSD

Winter 2012 UCSD: Physics 121; Op-Amp Introduction Op-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematicOp-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematic There are two inputsThere are two inputs –inverting and non-inverting And one outputAnd one output Also power connections (note no explicit ground)Also power connections (note no explicit ground)  divot on pin-1 end inverting input non-inverting input V+V+ VV output

Winter 2012 UCSD: Physics 121; The ideal op-amp Infinite voltage gainInfinite voltage gain –a voltage difference at the two inputs is magnified infinitely –in truth, something like 200,000 –means difference between + terminal and  terminal is amplified by 200,000! Infinite input impedanceInfinite input impedance –no current flows into inputs –in truth, about  for FET input op-amps Zero output impedanceZero output impedance –rock-solid independent of load –roughly true up to current maximum (usually 5–25 mA) Infinitely fast (infinite bandwidth)Infinitely fast (infinite bandwidth) –in truth, limited to few MHz range –slew rate limited to 0.5–20 V/  s

op amp: modello 4

Winter 2012 UCSD: Physics 121; Op-amp without feedback The internal op-amp formula is:The internal op-amp formula is: V out = gain  (V +  V  ) So if V + is greater than V , the output goes positiveSo if V + is greater than V , the output goes positive If V  is greater than V +, the output goes negativeIf V  is greater than V +, the output goes negative A gain of 200,000 makes this device (as illustrated here) practically uselessA gain of 200,000 makes this device (as illustrated here) practically useless  + VV V+V+ V out

Winter 2012 UCSD: Physics 121; Infinite Gain in negative feedback Infinite gain would be useless except in the self- regulated negative feedback regimeInfinite gain would be useless except in the self- regulated negative feedback regime –negative feedback seems bad, and positive good—but in electronics positive feedback means runaway or oscillation, and negative feedback leads to stability Imagine hooking the output to the inverting terminal:Imagine hooking the output to the inverting terminal: If the output is less than V in, it shoots positiveIf the output is less than V in, it shoots positive If the output is greater than V in, it shoots negativeIf the output is greater than V in, it shoots negative –result is that output quickly forces itself to be exactly V in  + V in negative feedback loop

Winter 2012 UCSD: Physics 121; Even under load Even if we load the output (which as pictured wants to drag the output to ground)…Even if we load the output (which as pictured wants to drag the output to ground)… –the op-amp will do everything it can within its current limitations to drive the output until the inverting input reaches V in –negative feedback makes it self-correcting –in this case, the op-amp drives (or pulls, if V in is negative) a current through the load until the output equals V in –so what we have here is a buffer: can apply V in to a load without burdening the source of V in with any current!  + V in Important note: op-amp output terminal sources/sinks current at will: not like inputs that have no current flow

Winter 2012 UCSD: Physics 121; Positive feedback pathology In the configuration below, if the + input is even a smidge higher than V in, the output goes way positiveIn the configuration below, if the + input is even a smidge higher than V in, the output goes way positive This makes the + terminal even more positive than V in, making the situation worseThis makes the + terminal even more positive than V in, making the situation worse This system will immediately “rail” at the supply voltageThis system will immediately “rail” at the supply voltage –could rail either direction, depending on initial offset  + V in positive feedback: BAD

feedback 9 A0A0 B V in V out V’V’

Winter 2012 UCSD: Physics 121; Op-Amp “Golden Rules” When an op-amp is configured in any negative- feedback arrangement, it will obey the following two rules:When an op-amp is configured in any negative- feedback arrangement, it will obey the following two rules: –The inputs to the op-amp draw or source no current (true whether negative feedback or not) –The op-amp output will do whatever it can (within its limitations) to make the voltage difference between the two inputs zero

Winter 2012 UCSD: Physics 121; Inverting amplifier example  + V in V out R1R1 R2R2

Winter 2012 UCSD: Physics 121; Inverting amplifier example Applying the rules:  terminal at “virtual ground”Applying the rules:  terminal at “virtual ground” –so current through R 1 is I f = V in /R 1 Current does not flow into op-amp (one of our rules)Current does not flow into op-amp (one of our rules) –so the current through R 1 must go through R 2 –voltage drop across R 2 is then I f R 2 = V in  (R 2 /R 1 ) So V out = 0  V in  (R 2 /R 1 ) =  V in  (R 2 /R 1 )So V out = 0  V in  (R 2 /R 1 ) =  V in  (R 2 /R 1 ) Thus we amplify V in by factor  R 2 /R 1Thus we amplify V in by factor  R 2 /R 1 –negative sign earns title “inverting” amplifier Current is drawn into op-amp output terminalCurrent is drawn into op-amp output terminal  + V in V out R1R1 R2R2

Winter 2012 UCSD: Physics 121; Non-inverting Amplifier  + V in V out R1R1 R2R2

Winter 2012 UCSD: Physics 121; Non-inverting Amplifier Now neg. terminal held at V inNow neg. terminal held at V in –so current through R 1 is I f = V in /R 1 (to left, into ground) This current cannot come from op-amp inputThis current cannot come from op-amp input –so comes through R 2 (delivered from op-amp output) –voltage drop across R 2 is I f R 2 = V in  (R 2 /R 1 ) –so that output is higher than neg. input terminal by V in  (R 2 /R 1 ) –V out = V in + V in  (R 2 /R 1 ) = V in  (1 + R 2 /R 1 ) –thus gain is (1 + R 2 /R 1 ), and is positive Current is sourced from op-amp output in this exampleCurrent is sourced from op-amp output in this example  + V in V out R1R1 R2R2

Ref:080114HKNOperational Amplifier15 Noninverting amplifier Noninverting input with voltage divider Voltage follower Less than unity gain

Ref:080114HKNOperational Amplifier16 Ideal Vs Practical Op-Amp IdealPractical Open Loop gain A  10 5 Bandwidth BW  Hz Input Impedance Z in  >1M  Output Impedance Z out 0   Output Voltage V out Depends only on V d = (V +  V  ) Differential mode signal Depends slightly on average input V c = (V + +V  )/2 Common-Mode signal CMRR  dB

Winter 2012 UCSD: Physics 121; Summing Amplifier Much like the inverting amplifier, but with two input voltagesMuch like the inverting amplifier, but with two input voltages –inverting input still held at virtual ground –I 1 and I 2 are added together to run through R f –so we get the (inverted) sum: V out =  R f  (V 1 /R 1 + V 2 /R 2 ) if R 2 = R 1, we get a sum proportional to (V 1 + V 2 ) Can have any number of summing inputsCan have any number of summing inputs –we’ll make our D/A converter this way  + V1V1 V out R1R1 RfRf V2V2 R2R2

Winter 2012 UCSD: Physics 121; Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider:  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

Winter 2012 UCSD: Physics 121; Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider: –V node = V + R 2 /(R 1 + R 2 )  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

Winter 2012 UCSD: Physics 121; Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider: –V node = V + R 2 /(R 1 + R 2 ) So I f = (V   V node )/R 1So I f = (V   V node )/R 1  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

Winter 2012 UCSD: Physics 121; Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider: –V node = V + R 2 /(R 1 + R 2 ) So I f = (V   V node )/R 1So I f = (V   V node )/R 1 –V out = V node  I f R 2 = V + (1 + R 2 /R 1 )(R 2 /(R 1 + R 2 ))  V  (R 2 /R 1 )  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

Winter 2012 UCSD: Physics 121; Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider: –V node = V + R 2 /(R 1 + R 2 ) So I f = (V   V node )/R 1So I f = (V   V node )/R 1 –V out = V node  I f R 2 = V + (1 + R 2 /R 1 )(R 2 /(R 1 + R 2 ))  V  (R 2 /R 1 ) –so V out = (R 2 /R 1 )(V   V  )  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

23 Differentiator (high-pass) V in V out C R

Winter 2012 UCSD: Physics 121; Differentiator (high-pass) For a capacitorFor a capacitor So we have a differentiator, or high-pass filterSo we have a differentiator, or high-pass filter  + V in V out C R

Winter 2012 UCSD: Physics 121; Low-pass filter (integrator) I f = V in /R, so C·dV cap /dt = V in /RI f = V in /R, so C·dV cap /dt = V in /R –and since left side of capacitor is at virtual ground: –and therefore we have an integrator (low pass)  + V in V out R C

esempio:

Ref:080114HKNOperational Amplifier27 Frequency-Gain Relation Ideally, signals are amplified from DC to the highest AC frequency Practically, bandwidth is limited 741 family op-amp have an limit bandwidth of few KHz. Unity Gain frequency f 1 : the gain at unity Cutoff frequency f c : the gain drop by 3dB from dc gain G d GB Product : f 1 = G d f c 20log(0.707)=3dB

Ref:080114HKNOperational Amplifier28 GB Product Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f 1 = 10 MHz and voltage differential gain G d = 20V/mV Sol: Since f 1 = 10 MHz By using GB production equation f 1 = G d f c f c = f 1 / G d = 10 MHz / 20 V/mV = 10  10 6 / 20  10 3 = 500 Hz 10MHz ? Hz

Gain-Bandwidth product: GBP = A o BW