Physics 160 Lecture 15
Physics 160 Lecture 15 R. Johnson May 18, 2015 Negative Feedback Formalism In + Out A B Representative of a non-inverting voltage amplifier A= open loop gain (forward gain with no feedback), typically huge (e.g. 100,000) B= gain of feedback network, typically < 1 AB= loop gain Feedback Vout A (Vin B Vout ) V A 1 1 G out Vin 1 AB B 1 1 AB Ideally, you want the loop gain AB to be large, Ideally large so that G is determined by only the feedback network gain B. May 18, 2015 Physics 160 2 In 3 V+ 8 Non-Inverting Amp, More Complete Analysis + Out - 4 2 1 V- OUT You may want to add a 9.1k series resistor to this leg to equalize effects of non negligible input non-negligible current in the case of bipolar inputs. R2 90k R1 10k 0 Using the not-so-simplified analysis: Suppose the forward gain is A=105 (fairly typical). H Here the h ffeedback db k gain i iis B B=R R1/(R1+R R2)=1/10 ) 1/10 ((voltage l di divider). id ) Then the loop gain is AB=104. V 1 G outt 10 9.999 4 Vin 1 10 May 18, 2015 Physics 160 3 Example R3 3 V1 10k 15Vdc 445.9uV + V+ 4 15.00V U5A - LM324 R2 0V V 11 241.5uV V- OUT 2 -206.1mV 1 9990k V2 1Vac 0Vdc R1 10k V3 15Vdc 0 Vout 1 G 1000 1 1000 Vin A May 18, 2015 Physics 160 4 Non-Inverting Amp Output 90 degree phase change from capacitive roll-off (low-pass filter) filter). The negative feedback turns into positive feedback at this point! The amp will oscillate if the overall loop gain is greater than unity at this frequency. From G=989, instead of 1000, we infer that A=90 A 90,000 000 (not infinite). Gain-bandwidth product is evidently about 10001000=106. So a follower (G=1) would work up to 1 MHz. May 18, 2015 Physics 160 5 LM324 Data Sheet http://www.onsemi.com/pub_link/Collateral/LM324-D.PDF We expect a frequency limit of ~1 kHz G=1000 A follower (G=1) will work up to ~1 MHz. dB 20 log10 G 20 log l (90000) 99 dB May 18, 2015 Physics 160 6 Follower IN OUT Non-inverting amp with 100% feedback. Be aware that the internal compensation capacitor is very important here to avoid high-frequency oscillations! The LM324, like the LM411, has a large enough internal capacitor to enhance the Miller effect that it is stable when operated as a follower. Some op amps require an external compensation capacitor, so you can optimize the operation. May 18, 2015 Physics 160 7 AC Coupled Non-Inverting Amp (The op amp here is assumed to be operated with dual power supplies, so that the signal can oscillate about ground.) This resistor is essential, to supply pp y some bias current to the op amp input transistors. This is no different from what we did with discrete transistors. This extra resistor is a good idea, to make sure that the input bias current gives the same voltage shift to both inputs. Not important for the LM411, but for a BJT op amp like the LM324 it can be very important. May 18, 2015 The input impedance of this circuit now is only 10 kohms! (But if you use the LM411 with JFET inputs inputs, the resistor can be much, much bigger.) Physics 160 8 AC Coupled Follower with Bootstrap IN OUT If necessary, for a follower the bias network can be bootstrapped as shown here, in exactly the same way as we did for the emitter-follower. (Note that this example is for an amplifier with both positive and negative supplies supplies, so the inputs and outputs can be centered on ground potential.) May 18, 2015 Physics 160 9 AC Non-Inverting Amplifier An alternative to using a coupling capacitor, or a good idea even if the input is AC coupled. IN OUT This amp will have a gain of 10 at q above 1/(2R ( frequencies 2C1)). At DC the gain is only unity, which avoids problems with DC offsets getting amplified. May 18, 2015 Physics 160 10 Inverting Amp, More Complete Analysis R2 Vin Vout i R1 R2 R1 2 V- In 4 i - + Out 8 3 1 V+ OUT Without R1 this is a transresistance amplifier. feedback 0 V Vin iR1 Vin (Vin Vout ) B where Vout A (V V ) AV Eliminate V : May 18, 2015 R1 B R1 R2 Forward gain 1 B 1 Vout R2 G Vin R1 1 1 AB Physics 160 11 Inverting Amp, Input Current Compensation R1 2 - V- In 4 R2 Add a resistor here of value R1R2/(R1+R2) May 18, 2015 + Out 8 3 1 V+ OUT 0 A real op amp will have some input bias current (especially for BJT input) input). The current flowing through R1 and R2 will shift the voltage at the inverting input. We should add a resistor to the g input p to compensate: p non-inverting Physics 160 12 Analog Computing Analog computing: – – – – Integrator Differentiator Summing amplifier Log amp Op-amps provide nearly ideal performance for these devices, far, far beyond what can be achieved with passive components. (Note: (N t historically hi t i ll analog l computers t were widely used, but by now digital computers have such high performance and low cost that analog computers are largely obsolete.) May 18, 2015 Physics 160 A 1960 Newmark analog computer, made up of five units. This computer was used to solve differential equations and is currently housed at the Cambridge g Museum of Technology gy (from Wikipedia). 13 Integrator Example A real circuit will have to provide a method to zero the charge on the capacitor at the start of the integration! 15.00V t 1 Vout (t ) Vout (0) Vin dt R1C1 Q C1 100n 0 U4 3 + V+ V Virtual ground 7 -1.321V 40.32nV V1 0V OUT R1 2 V1 = 0 V2 = 0 0.2 2 TD = 0 TR = 0 TF = 0 PW = 1m PER = 10m V3 V 1k B1 5 out 6 1 V LF411 40.32nV R2 1k 0V V2 15Vdc -15.00V May 18, 2015 - V- 15Vdc B2 4 iin dV V d dQ in C1 out dt R1 dt Physics 160 Note DC offset. This is hard to control, since there is no DC feedback. Adding g a large resistor in parallel with C1 can help control this (but also gives some error in the integration) integration). 14 Integration Example Input Square Pulse Perfect Integration (but note the inversion of sign) May 18, 2015 Physics 160 15 Differentiator iin dV dQ C 2 in dt dt Vout iin R1 R1C 2 C1 0.1n dVin dt R1 100k V2 U1 3 2 V3 .01uF V May 18, 2015 B1 6 1 V LF411 For stability, stability it becomes an integrator at high frequency. V1 15Vdc - 5 4 V1 = 0 V2 = .1 1 TD = 0 TR = .5m TF = .5m PW = 0.5m PER = 2m 1k B2 OUT C2 V- R2 + V+ + Virtual ground 7 15Vdc 0 Physics 160 16 Differentiation Example Slow response on edges is needed f stability for t bilit May 18, 2015 Physics 160 17 Stability Issues • Integrator: – voltage gain falls at 6 dB per octave as frequency increases. – inherently stable against oscillations even with an uncompensated op-amp. – the feedback itself looks like Miller-effect frequency compensation. • Differentiator: – voltage gain rises at 6 dB per octave as frequency increases. – inherently unstable against oscillations. – the gain must be rolled off at high frequency by introducing an integration, or oscillation is guaranteed! 1 jRC Differentiator : G jRC Integrator : G May 18, 2015 Physics 160 18 Sum (or Subtraction) Amp What are really being added here are the 3 currents, which combine at the virtual ground. ground Inverse of the sum of the 3 voltages V2, V3 V4 V3, V4. Resistor added in an attempt to minimize offset errors due to input current. May 18, 2015 Physics 160 19 Diode is to protect against baseemitter breakdown in case input goes negative. Log Amp Emitter voltage is logarithm of collector current. D1 D1N914 Q1 Q2N3904 FET inputs for nearly zero current error. R2 OUT 2 10k B1 Q2 Q2N3904 6 U2 3 + 7 5 1 B2 OUT LF411 2 4 VIN - V- R1 0Vdc B2 V+ V1 - V- + V+ U1 7 15k 3 15Vdc Source of ~constant current current. B1 5 6 1 V 4 LF411 R3 Do DC scan of this source. 15Vdc Output will be the log of this input. Constant current. Diode drop here provides temperature compensation. V2 15k R4 1k 16 non non-inverting inverting amp With a log amp you can do analog multiplication and division! Ebers-Moll: VBE VT log( I C I S ) VT log I S I C Vin / R1 May 18, 2015 But I C I S Vout C VBE 16 C 0.96 V log10 I C I S Physics 160 0.026 16 ln(10) 0.96 20 Log Amp DC Bias Example In a real circuit, put a capacitor in parallel with the diode, for stability. 15.00V D1 D1N914 Q1 Q2N3904 119 1 A 119.1pA 1.000uA -1.023uA 23.25nA 987.4uA R2 0V 15k R1 10.00mV 1.000uA 5.329mA 0.01Vdc 10k U2 3 -40.13pA + B2 OUT 2 -40.13pA 40.13pA LF411 B1 5 6 -189.3uA 1 -2.170mA In a real circuit, use pins 1 and 5 of the op amp to trim the input offset to zero. V 3.029V 4 0V V2 2.171mA 7.398uA 1.000uA 15Vdc 7 189.3mV 980.0uA -987.4uA 4 VIN Q2 Q2N3904 V+ 15Vdc V- V1 V- U1 3 2.171mA 5 B2 + -40.32pA 6 OUT 988.4uA 2 1 B1 -40.32pA LF411 -2.170mA V+ -1.831uV 7 -515.9mV R3 189.3mV 189.3uA 189.3uA 15k R4 1k 0V 4.340mA -15.00V 15 00V May 18, 2015 Physics 160 21 Log Amp Simulation Good logarithmic response over about b t 6 or 7 d decades, d but fails for inputs below about 10 microvolts. May 18, 2015 Physics 160 22 Behavior, Limitations of Real Op-amps • • • • • • • • • • Input offset voltage Input current, and input-current mismatch – Design to minimize voltage offsets from this current – Input offset current See the data sheet for the Temperature dependences LF411 (posted on the course Slew rate web page) page). – e.g. correcting crossover distortion Frequency range (gen. purpose op amps are not for high frequencies) Voltage gain and phase shift – Stability! CMRR and Common-mode input range Input impedance with negative feedback is extremely high Output impedance with negative feedback is very low – What is really relevant is the limitation on output current drive Output range (how close can the output get to the supply rails?) – Single-supply op amps; can the output slew all the way to zero? May 18, 2015 Physics 160 23 Offset Voltage and Input Current • Offset voltage – 2 mV max. for the 411 • This is the effective mismatch at the input. At the output it gets multiplied lti li d b by th the gain! i ! – Trimming + V+ U1 3 7 V1 15 OS2 2 - V- OUT OS1 6 1 R2 4 LM741 5 V2 – Temperature and time drifts • 10k 15 I Input t currents t (esp. ( for f BJTs) BJT ) – Balancing resistors on inputs – Input offset current • A/C Amp: reduce DC gain to unity to minimize offset effects May 18, 2015 Physics 160 24 Charge Sensitive Amp Integrate the pulse of current from a detector to produce an output voltage proportional to the total charge. Vout R6 10G C1 0.1p Virtual ground U4 3 + V1 V2 May 18, 2015 VV 2 15Vdc B2 OUT 15Vdc - B1 5 6 1 V LF411 I1 4 I1 = 0 I2 = 1u TD = 50n TR = 1n TF = 1n PW = 98n PER = 100u Q 0.1pC 0 1pC 1V C 0.1pF 7 This is not a practical resistor value in the real world! V+ • C2 10p R1 R3 100Meg 100Meg Detector Model, with 100 fC pulse l off charge. h Physics 160 Even though the feedback capacitor is 100 time smaller than the detector capacitance, it effectively gets multiplied by A (forward gain). Therefore, almost all of the charge ends up on the feedback capacitor. Hence the detector ends up with nearly zero volts on it (virtual ground). 25 Charge Amp Example Short pulse of current. Rise time is non-zero due to the non-zero output impedance of the amplifier. May 18, 2015 Slow decay from the 10Gohm resistor in the feedback network. Physics 160 26 Voltage divider divides the output by 100, correspondingly reducing the feedback current. The DC f db k iis th feedback thus almost l t equivalent to a 10 Gohm resistor. T Network 15.00V R6 R2 4.024mV 100Meg 100Meg R4 1Meg C1 4.028mV V+ U4 3 7 0 1p 0.1p + V1 OUT 2 4.032mV - B1 5 406.4mV 6 1 V LF411 4 I1 V- 15Vdc I1 = 0 I2 = 1u TD = 50n TR = 1n TF = 1n PW = 98n PER = 100u B2 C2 10p R1 R3 100Meg 100Meg 0V V2 15Vdc -15.00V May 18, 2015 Physics 160 27 Charge Amp with T Network Short pulse of current. Slow decay from the T network, acts the same as the 10 Gohm resistor. May 18, 2015 Physics 160 28
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