The Operational Amplifier as a differential voltage-controlled voltage source

Transcription

1 The Operational Amplifier as a differential voltage-controlled voltage source Operational amplifiers (op amps) are high performance differential amplifiers. They have inverting and noninverting inputs and typically 2 DC power leads. The 741 is a good general purpose op amp in a dip Introduction to Operational Amplifiers, Spring We have seen that with a variety of transistors we can build switches, oscillators, amplifiers, modulators etc, but still they are often inconvenient to use. While transistors remain the fundamental building block of most interesting circuit elements, they have been combined into standard configurations and packages that function at a higher level and provide greater performance. This of course leads to a wide variety of devices and even new classes of devices, such as the separation of digital and analog circuits. One of the most useful examples of this higher level analog devices is the operational amplifier, (often shortened to op-amp). The op-amp is a differential amplifier (it has two inputs and the output voltage is an amplified copy to the difference between the inputs). 1

2 A simple op amp model non-inverting input +Vs An op amp is driven by the imbalance of the two inputs. When there is a differential voltage it quickly saturates. output inverting input -Vs Introduction to Operational Amplifiers, Spring This simple water analogy comes close to giving the dynamics of an op-amp. As the difference in force on the two inputs becomes finite, then the blue piece rotates and the output is connected to one of the two supply voltages. The channels are such that the output is quickly sent to either +/- Vsupply. When the balance between the inputs is restored, then the output is once again set to zero. With this analogy I want to make it clear that it is very difficult to hold the inputs in such a state that the output is between +/- the supply voltage. 2

3 The Operational Amplifier as a differential voltage-controlled voltage source red blue green Vs -Vs Introduction to Operational Amplifiers, Spring The example of the operational amplifier as a very high gain differential amplifier makes it clear that the output is driven very quickly to the supply voltages. With a gain of about 1 million it only takes a few micro-volts difference between the two inputs to drive the amplifier into saturation. If the + (non-inverting) input is higher than the - (inverting) then the amplifier saturates to + Vsupply. So for the wave-shapes given, when fed to the inverting input the output is out of phase with the input, and when fed to the non-inverting input the output is in phase with the input. 3

4 The ideal Operational Amplifier = A(V + - V - ); A ª10, 000 R in = The ideal op amp has a very high gain, and an infinite input impedance Introduction to Operational Amplifiers, Spring The ideal op amp has an infinite gain, an infinite input impedance and an arbitrarily low output impedance. None of these are actually true of course, but we can use these ideals in designing circuits and then the design process will be simple. Clearly such an ideal amplifier would be wonderful, with infinite input impedance it would never load any circuit, and it could drive any load without being pulled down. As we will see, with infinite gain we can use feedback to choose any gain we wish. A challenge of the op amp is that A is not a good parameter, it varies from device to device and is temperature and operating condition dependent. Good op amp design avoids dependences on A. 4

5 A model of an ideal Operational Amplifier V V + + V 0 A(V + -V - ) V - V 0 V - V 0 A is of order 500, Introduction to Operational Amplifiers, Spring A model for an operational amplifier that is often given is shown above. This is a somewhat useful concept, the basic idea is that the two input have arbitrarily high input impedances and the difference in voltage of these acts as an independent voltage source with a high gain. It turns out that with a couple of simple rules (like for the BJT transistor) we can design and analyze op amp circuits much more efficiently. So we will not use this model which in my opinion is more useful in circuits than in electronics. 5

6 Real op amps R in R out = A(V + - V - ) ideal op amp open loop gain is infinite, real A ~ 10 4 to R in ~10 6 to W. R out ~ 10 to 1000 W. limited power delivery. input current is very nearly zero, na -> pa. clips at +/- Vs Introduction to Operational Amplifiers, Spring So now we get to real op amps, they are similar to the ideal, with the limitations shown above. In addition you need to be aware the A is dependent on everything. 6

7 Demo of A for an op amps (741) +V s V 0,V ~ +V s V + V - -V s V 0 (V + -V - ) mv ~ -V s V 0 =A(V + -V - ) Introduction to Operational Amplifiers, Spring A nice way of looking at an op amp is to correlate the voltage output to the input. Since we only care about the differential voltage input, the horizontal axis is that, and the vertical axis is the output voltage. The slope of the curve in the center is A (the open loop gain of the op amp), and the flat regions are where the op amp has saturated and is reporting the voltage rails. Notice that since A is large, the slope of the curve is high and therefore the two axes have different units, micro-volts for the input and volts for the output. 7

8 Demo of A for an op amps (741) +15V -15V V 0 attn -80dB V 0 A Ê -80dB =10 log P A ˆ 0 Á Ë P 0 Here we have a constant imedance so P = V 2 R Ê A ( V 0 ) 2 ˆ Á -80dB =10 logá R 2 V Á 0 = 20 log Ê V A ˆ 0 A Á so V 0 =10-4 Ë V Ë R 0 V Introduction to Operational Amplifiers, Spring This first demo includes an attenuator to reduce the scale of the output and make the measurement simpler. This is an 80 db attenuator, and so it reduces the voltage by a factor of 10,000 assuming a constant impedance. 8

9 Comparator Vs V - is a settable voltage between V s and gnd. Since there is no feedback the op amp saturates to +Vs when > V set, and to -Vs when < V set. Use the rule that V + - V - = Introduction to Operational Amplifiers, Spring One use of an op amp with large open loop gain is a comparator. If Vin is greater than the set voltage the output will be at the positive supply rail, and otherwise it will be at the negative. The gain is so high that you will not ba able to set the output in between the rails. 9

10 Negative Feedback R in I in R F I F The op amp is easily kept from saturating through negative feedback. Take part of the output and inject this into the inverting input. For the ideal op amp the current through R F is equal and opposite that through R in. match the currents: - V - R in = - - V - R F resultant gain: = - R F R in Introduction to Operational Amplifiers, Spring Nearly all applications of op amps however involve negative feedback. In this case some of the output voltage is fed back into the inverting input of the op amp. Now as the op amp has more gain, the output increases and since it goes into the inverting input this reduces the output. Very quickly the op amp finds its operating point. Here is the advantage of having a large A, the device is thus very stiff and finds the operating point extremely quickly. Notice that the operating point does not depend on A, provided that A is large. 10

11 Rules for op amps with feedback R in R out When the op amp has found its operating point, 1. The two input voltages are equal. 2. The current into each input is zero. = A(V + - V - ) Introduction to Operational Amplifiers, Spring Here are the promised two rules for op amps with feedback. The first simply states that the voltage difference goes to zero and thus there is no output voltage. This follows from the behavior of the ideal op amp. The current rule can be thought of as, an input current provides an internal voltage that the op amp is going to correct by adjusting the output voltage. So this is again the same rule as above, but written in terms of an effective current at the inputs. 11

12 Using Negative Feedback Ï feedback Ì Ó network Ï feedback Ì Ó network For negative feedback, the op amp adjusts its output so that the feedback current/voltage is that necessary to keep V + - V - equal to zero Introduction to Operational Amplifiers, Spring There are two simple means of using negative feedback. The feedback of course must always go to the inverting input, but the source can either also go to the inverting input (thus having an inverting amplifier), or the source can go to the non-inverting input and result in a non-inverting amplifier. The strength of the feedback (how much of the output is fed back) will determine the gain of the device. 12

13 Buffer or unit gain amplifier For V + - V - to equal zero, then =. Therefore the gain is 1. Recall that the input impedance of an op amp is quite high, so the buffer provides an impedance transformation and protects from the load Introduction to Operational Amplifiers, Spring Here the amplifier is driven in the non-inverting input with the output voltage directly fed to the inverting input. The op amp will amplify until the voltages at the inverting and non-inverting inputs are the same. This of course can only happen if the output is equal to the input. Thus the buffer has a unit gain (the voltage output is the same as the input). Of course the input impedance of the op amp is quite high and the output impedance is low, so there is an impedance transformation that can result in a power gain. In general when both inputs are used, you should try to employ the rule that the voltages must be equal. Here there is no way of using the rule that the currents are zero (although they essentially are since the input impedance is so high). 13

14 Inverting amplifier R 2 I 1 = -I 2 R 1 I 2 R 1 = - R 2 I1 = - R 2 R 1 Use the rule that the current I - is zero. To avoid offset errors place a R = R 1 R 2 to gnd from the noninverting input Introduction to Operational Amplifiers, Spring The inverting amplifier has the non-inverting input grounded and sums the inputs from the input and the output into the inverting input. The way to analyze this is to realize that when I1 and I2 are equal and opposite then there is no current flowing into the op amp input and the two input voltages will be equal. In other words, both the input path and the feedback path are driving the same internal resistance in the op amp. So if the two currents are equal but opposite in sign then the voltages they induce across this internal resistor will also be equal and opposite - so the overall voltage into the inverting terminal is zero. The currents are given above and setting them equal and opposite results directly in the gain. In practice there is a small input current at the inverting input, and so it is normal to place a resistor to ground from the non-inverting input to give the same effective impedance. This is a small effect but observable. 14

15 Demo of an inverting amplifier 1kW 10kW Introduction to Operational Amplifiers, Spring In this demo we show the ouput vs input voltages for an inverting amplifier. By lowering the variable feedback resistor the gain can be made very small, the device essentially becomes an inverting buffer. 15

16 Non-inverting amplifier R 1 R 2 Use the rule that V + - V - equals zero. V + = V - = R 2 R 1 + R 2 = R 2 R 1 + R 2 = R 1 + R 2 R 2 To avoid offset errors arrange that R source = R 1 R Introduction to Operational Amplifiers, Spring The non-inverting amplifier sends the input into the non-inverting terminal. The feedback must of course go to the inverting input. The gain is adjusted by sampling the output voltage through a voltage divider. By adjusting the ratio of R1 to R2 you can vary the strength of the feedback and thus change the gain. Here use the rule that the input voltages must be the same. Again using the rule immediately provides the gain. In this case to avoid an offset due to input current mismatch it is common to place a series resistor inline with the non-inverting input and adjust it to the effective impedance into the inverting input. 16

17 Demo of non-inverting amplifier 1kW 10kW 1kW Introduction to Operational Amplifiers, Spring A demo similar to the previous ones but now for the non-inverting configuration. Notice that in the case of the inverting amplifier as the feedback resistor was varied the gain could go much less than 1, here the gain can only approach 1. 17

18 Non-inverting amplifier R f R f R A(V + -V - ) R A R = A( V + -V - ) = A Vin - R + R f V + = or Ê R ˆ V - = Á Ë R + R out}vout V Ê f A = V 0 1+ A R ˆ Á Ë R + R f Introduction to Operational Amplifiers, Spring

19 Non-inverting amplifier (Cont.) so = A Ê R ˆ, then A Ê R ˆ Á R + R >>1 Ë f 1+ AÁ Ë R + R f Thus, ~ R +R f R Introduction to Operational Amplifiers, Spring