Lecture 4: Feedback and OpAmps


 Gavin Thornton
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1 Lecture 4: Feedback and OpAmps Last time, we discussed using transistors in smallsignal amplifiers If we want a large signal, we d need to chain several of these small amplifiers together There s a problem, though: at each stage of the amplification, distortion (noise) is introduced along with the signal we want Eventually the amplified noise becomes comparable to or bigger than the signal This can be avoided by designing amplifier circuits with feedback that means that part of the output from the amplifier is routed back to the input Concept first developed by Bell Labs engineer Harold Black This made longdistance calls (which must be amplified many times en route) possible
2 Feedback example The smallsignal amp from last lecture would go here. From amp gain, we know that v = Av out i Negative output sent back to positive input negative feedback
3 We want to find the gain for the entire circuit (feedback included) i.e., want to know v out /v in We start with what we already know: v out i in f f Av v = v v v = = R 1 i R1 + R F v out
4 From which we find: R1 v = v v R + R i in out 1 F v = A v R v Note that the gain of the feedback circuit is less than that of the main amplifier Seems counterproductive! 1 out in out R1 + RF AR 1 vout1 + Avin R1 R = + F A v = v A v AR 1+ R + R out in f in 1 1 F
5 Benefits to feedback The general form of the equation from the previous slide is: A Af = 1 ( ± 1) β A where A is the main amplifier gain (often called the opencircuit gain), β is the fraction of the output fed back to the input, and the ± indicates whether the feedback is positive or negative The quantity L = ± βa is called the loop gain for the circuit Now let s assume A = 100, and β = 0.25, and the feedback is negative Then 100 A f = =
6 What happens if the main amp degrades such that A is reduced to 50? 50 A f = = So a drastic change in A results in a tiny change in A f As long as β is constant But β can be determined by resistors (as in our example circuit) and is therefore very stable Thus we see that negative feedback improves the stability of our amplifier
7 A f Frequency response Let s assume that A is bandwidthlimited i.e., it depends on ω, with large frequencies amplified less: 0 ( ) = A ω A ω 1+ ω In a negativefeedback system, the overall gain is: A0 A( ω ) 1 + ω / ωc ω = = 1+ β A( ω ) A 0 1+ β 1 ω / ω + c Ao Ao 1+ β Ao = = 1 + ω / ω + β A 1+ β A 1 + ω / ω + β A ( ) c o o c o Ao 1 1 = = Af ( 0) 1+ β Ao 1 + ω / ( 1+ β Ao ) ωc 1 + ω / ( 1+ β Ao ) ωc c
8 So we see that the cutoff frequency is increased by a factor of 1+βA o that s the same factor by which the amplifier s gain is reduced by the feedback circuit we re trading reduced gain for increased bandwidth
9 Noise reduction Assume now that the amplifier generates some random noise on top of the output signal, so that: vout = Avi + vn without feedback, the output depends on the noise as: dvout = 1 dvn With negative feedback, we have: ( 1 β ) ( ) v = v β Av + v i in i N v + A = v β v i in N
10 This means that: vin β vn vout = A + v 1+ β A dv dv out N ( 1 β β ) Av + + A A v Av + v = = 1+ β A 1+ β A 1 = 1 + β A N in N in N
11 Differential Amplifiers So far we ve talked about amplifying a signal, but often what we really want is to amplify the difference between two signals A circuit that does this is called a differential amplifier, and the following is an example: It s basically two commonemitter amplifiers placed backtoback Point A
12 Let s calculate the differential gain for this circuit: G diff V = out ( V V ) 1 2 to calculate this, let s see what happens when equal but opposite signals are input to V 1 and V 2 : note that V A is unchanged by this, since the two emitter resistors have equal value V = I R I R diff out C C E C V = V = I R = V G 1 E E E 2 I R R = = I R I R 2R E C C E E E E E
13 We now compare this to the common mode gain that occurs when both V 1 and V 2 move in the same direction: G CM V = + out ( V V ) 1 2 Let s see what happens when we increase both voltages by the same amount: V = I R I R out C C E C V = V = I R 2 I R = V G 1 E E E E 1 2 CM I E RC RC = = 2 I R + 2R 2R + 4R ( ) E E 1 E 1
14 If we choose R 1 >> R E, signal differences will be amplified much more than common signal changes We define the common mode rejection ratio (CMRR) as: for our circuit, it s: CMRR CMRR = G G CM diff 2RE + 4R R + 2R = = 2R R E 1 E 1 E
15 Operational Amplifiers (OpAmps) An opamp is an integrated circuit (chip) whose behavior approximates an ideal amplifier: high input impedance low output impedance large gain (factors of a million aren t uncommon) These are used rather than transistors in circuits that require an amplifier Internally, an opamp might look something like this:
16 We re not really going to care too much about the innards of the opamp we just need to be familiar with how it behaves An opamp looks like: on a schematic: inverting input in real life: noninverting input Requires external power input (typically ±12 or ±15V) these connections often not shown on schematic
17 Opamp rules An ideal opamp behaves as follows: 1. Inputs draw no current (infinite input impedance) 2. Output tries to adjust itself so that inputs are at the same voltage Real opamps come pretty close to these ideals Limitations appear as: nonzero input currents finite slew rate for output limitations in bandwidth finite CMRR
18 Inverting amplifier: Opamp circuits Note the negative feedback The voltage at the () input is: V = V i R1 I = 1 V o I2R2 Since the opamp wants its inputs at the same voltage, V = 0 Since the inputs draw no current, I 1 = I 2
19 Putting it all together, we find: Note that the gain depends totally on the resistors, and not at all on the properties of the opamp except of course that the opamp is assumed to behave ideally! Input impedance is just R 1, so we can choose the value but raising input impedance lowers gain not an optimal feature Output impedance is small (<1Ω) V V i V o V o i = R I 1 = R I = R R 2 1 2
20 Follower We can remove the resistors from the previous circuit to get: From opamp rule #1, we see that V o will adjust itself to equal V i the output follows the input This is an amplifier with a gain of 1 but does offer very high input impedance and low output impedance
21 Comparator Consider what happens with the following circuit: The voltage at the + input changes due to the variable resistance Since there s no feedback, the opamp can t do anything to make the inputs equal any difference between V i and V + will be subject to the opamp gain of a million or so
22 Does that mean that a difference of 1V at the inputs will result in an output of ~1 million V? No! The opamp can t exceed the voltage of it s power supply What does happen is that the output swings all the way to the limits (upper or lower) of the power supply whenever the inputs are different in other words, the output tells us whether V + is greater than or less than V i it gives one bit of information about the analog input voltages This circuit is at the interface between analog and digital electronics We ll go into the digital realm starting next lecture
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