09_op-amps.ppt
- 1. Operational Amplifiers Magic Rules Application Examples
- 2. Winter 2012 UCSD: Physics 121; 2012 2 Op-Amp Introduction • Op-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematic • There are two inputs – inverting and non-inverting • And one output • Also power connections (note no explicit ground) + 2 3 4 7 6 divot on pin-1 end inverting input non-inverting input V+ V output
- 3. Winter 2012 UCSD: Physics 121; 2012 3 The ideal op-amp • Infinite 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 impedance – no current flows into inputs – in truth, about 1012 for FET input op-amps • Zero output impedance – rock-solid independent of load – roughly true up to current maximum (usually 5–25 mA) • Infinitely fast (infinite bandwidth) – in truth, limited to few MHz range – slew rate limited to 0.5–20 V/s
- 4. Winter 2012 UCSD: Physics 121; 2012 4 Op-amp without feedback • The internal op-amp formula is: Vout = gain(V+ V) • So if V+ is greater than V, the output goes positive • If V is greater than V+, the output goes negative • A gain of 200,000 makes this device (as illustrated here) practically useless + V V+ Vout
- 5. Winter 2012 UCSD: Physics 121; 2012 5 Infinite Gain in negative feedback • Infinite 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: • If the output is less than Vin, it shoots positive • If the output is greater than Vin, it shoots negative – result is that output quickly forces itself to be exactly Vin + Vin negative feedback loop
- 6. Winter 2012 UCSD: Physics 121; 2012 6 Even under load • 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 Vin – negative feedback makes it self-correcting – in this case, the op-amp drives (or pulls, if Vin is negative) a current through the load until the output equals Vin – so what we have here is a buffer: can apply Vin to a load without burdening the source of Vin with any current! + Vin Important note: op-amp output terminal sources/sinks current at will: not like inputs that have no current flow
- 7. Winter 2012 UCSD: Physics 121; 2012 7 Positive feedback pathology • In the configuration below, if the + input is even a smidge higher than Vin, the output goes way positive • This makes the + terminal even more positive than Vin, making the situation worse • This system will immediately “rail” at the supply voltage – could rail either direction, depending on initial offset + Vin positive feedback: BAD
- 8. Winter 2012 UCSD: Physics 121; 2012 8 Op-Amp “Golden 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
- 9. Winter 2012 UCSD: Physics 121; 2012 9 Inverting amplifier example • Applying the rules: terminal at “virtual ground” – so current through R1 is If = Vin/R1 • Current does not flow into op-amp (one of our rules) – so the current through R1 must go through R2 – voltage drop across R2 is then IfR2 = Vin(R2/R1) • So Vout = 0 Vin(R2/R1) = Vin(R2/R1) • Thus we amplify Vin by factor R2/R1 – negative sign earns title “inverting” amplifier • Current is drawn into op-amp output terminal + Vin Vout R1 R2
- 10. Winter 2012 UCSD: Physics 121; 2012 10 Non-inverting Amplifier • Now neg. terminal held at Vin – so current through R1 is If = Vin/R1 (to left, into ground) • This current cannot come from op-amp input – so comes through R2 (delivered from op-amp output) – voltage drop across R2 is IfR2 = Vin(R2/R1) – so that output is higher than neg. input terminal by Vin(R2/R1) – Vout = Vin + Vin(R2/R1) = Vin(1 + R2/R1) – thus gain is (1 + R2/R1), and is positive • Current is sourced from op-amp output in this example + Vin Vout R1 R2
- 11. Winter 2012 UCSD: Physics 121; 2012 11 Summing Amplifier • Much like the inverting amplifier, but with two input voltages – inverting input still held at virtual ground – I1 and I2 are added together to run through Rf – so we get the (inverted) sum: Vout = Rf(V1/R1 + V2/R2) • if R2 = R1, we get a sum proportional to (V1 + V2) • Can have any number of summing inputs – we’ll make our D/A converter this way + V1 Vout R1 Rf V2 R2
- 12. Winter 2012 UCSD: Physics 121; 2012 12 Differencing Amplifier • The non-inverting input is a simple voltage divider: – Vnode = V+R2/(R1 + R2) • So If = (V Vnode)/R1 – Vout = Vnode IfR2 = V+(1 + R2/R1)(R2/(R1 + R2)) V(R2/R1) – so Vout = (R2/R1)(V V) – therefore we difference V and V + V Vout R1 R2 V+ R1 R2
- 13. Winter 2012 UCSD: Physics 121; 2012 13 Differentiator (high-pass) • For a capacitor, Q = CV, so Icap = dQ/dt = C·dV/dt – Thus Vout = IcapR = RC·dV/dt • So we have a differentiator, or high-pass filter – if signal is V0sint, Vout = V0RCcost – the -dependence means higher frequencies amplified more + Vin Vout C R
- 14. Winter 2012 UCSD: Physics 121; 2012 14 Low-pass filter (integrator) • If = Vin/R, so C·dVcap/dt = Vin/R – and since left side of capacitor is at virtual ground: dVout/dt = Vin/RC – so – and therefore we have an integrator (low pass) + Vin Vout R C
- 15. Winter 2012 UCSD: Physics 121; 2012 15 RTD Readout Scheme
- 16. Winter 2012 UCSD: Physics 121; 2012 16 Notes on RTD readout • RTD has resistance R = 1000 + 3.85T(C) • Goal: put 1.00 mA across RTD and present output voltage proportional to temperature: Vout = V0 + T • First stage: – put precision 10.00 V reference across precision 10k resistor to make 1.00 mA, sending across RTD – output is 1 V at 0C; 1.385 V at 100C • Second stage: – resistor network produces 0.25 mA of source through R9 – R6 slurps 0.25 mA when stage 1 output is 1 V • so no current through feedback output is zero volts – At 100C, R6 slurps 0.346 mA, leaving net 0.096 that must come through feedback – If R7 + R8 = 10389 ohms, output is 1.0 V at 100C • Tuning resistors R11, R7 allows control over offset and gain, respectively: this config set up for Vout = 0.01T
- 17. Winter 2012 UCSD: Physics 121; 2012 17 Hiding Distortion • Consider the “push-pull” transistor arrangement to the right – an npn transistor (top) and a pnp (bot) – wimpy input can drive big load (speaker?) – base-emitter voltage differs by 0.6V in each transistor (emitter has arrow) – input has to be higher than ~0.6 V for the npn to become active – input has to be lower than 0.6 V for the pnp to be active • There is a no-man’s land in between where neither transistor conducts, so one would get “crossover distortion” – output is zero while input signal is between 0.6 and 0.6 V out in V+ V crossover distortion
- 18. Winter 2012 UCSD: Physics 121; 2012 18 Stick it in the feedback loop! • By sticking the push-pull into an op-amp’s feedback loop, we guarantee that the output faithfully follows the input! – after all, the golden rule demands that + input = input • Op-amp jerks up to 0.6 and down to 0.6 at the crossover – it’s almost magic: it figures out the vagaries/nonlinearities of the thing in the loop • Now get advantages of push-pull drive capability, without the mess + Vin out V+ V input and output now the same
- 19. Winter 2012 UCSD: Physics 121; 2012 19 Dogs in the Feedback • The op-amp is obligated to contrive the inverse dog so that the ultimate output may be as tidy as the input. • Lesson: you can hide nasty nonlinearities in the feedback loop and the op-amp will “do the right thing” + Vin dog inverse dog “there is no dog” We owe thanks to Hayes & Horowitz, p. 173 of the student manual companion to the Art of Electronics for this priceless metaphor.
- 20. Winter 2012 UCSD: Physics 121; 2012 20 Reading • Read 6.4.2, 6.4.3 • Pay special attention to Figure 6.66 (6.59 in 3rd ed.)