Circuit operational amplifiers
pp 237-249
8.2 Basic operational amplifier circuits
The high voltage gain (open loop), the infinite input impedance and zero output impedance of an ideal operational amplifier design allowing wide variety of circuits for performing various computer functions.
Most of the circuits based on the application of negative feedback or anasyzefxis between output and the inverting input of the TE. Since the operational amplifiers have two inputs, the inverting and non - inverting, the combination
application input and feedback of the output signal enables the design of two main categories of circuits, the circuits are non-inverting feedback (non inverting feedback) and feedback circuits with inverting (inverting feedback). Typical amplifiers are manufactured in this way are called their non-inverting amplifier (non inverting amplifier) and inverting amplifier (inverting amplifier) and will be described in detail immediately below.
Figure 8.2.1 Non Circuit - Aol = inverted amplifier open-loop voltage gain (Open loop)
8.2.1 Non-inverting amplifier
The circuit is a non - reverse amplifier is shown in Fig 8.2.1 A key feature of this circuit is applied to the signal gain to the non - inverting input and the feedback rate of the output signal to the inverting input. The feedback is negative because the anatrofodotoumeno signal subtracted from the input signal. The term non - inverting amplifier
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It follows from (8.2.1) and (8.2.2) that the voltage gain of the amplifier will be reversed:
(Non inverting amplifier) that results from the phase difference between output and input signal is 0 °, so the input waveform at the output is reversed. Also, the feedback applied is called non-inverting (non inverting feedback) in this
the feedback phase difference between output and input signal is 0A.
As discussed above, the open-loop voltage gain, ie, without feedback, an operational amplifier is very large (typically, AOL «(100 000). To determine the voltage gain of the non-inverted amplifier, which is equal to the so-called closed-loop gain of the op ACL, we can follow the methodology often used by employees with feedback amplifiers as follows: After the voltage of an input h1 is equal to the input signal (Vi) and the other input voltage is equal to h2 the feedback voltage, the voltage between the two inputs of the operational amplifier is HR = Y |-h2. This tendency is called the error signal voltage (error voltage or signal). The output voltage is analogous to y0.
So the non-inverting input of operational amplifier (Fig. 8.2.1) the voltage will be:
u! = U: 8.2.1
whereas under voltage divider (R1, R2), the inverting input of operational amplifier voltage will be:
h2 = R28.2.2
• Aa = closed-loop voltage gain (Closed Loop) An important conclusion derived from Eq. 8.2.3 is that the non-inverted amplifier that the voltage gain, AV, is larger than the equal of the unit. The AV will approach the unit only if R2 is much greater than the R1 will be equal to the unit if R2 is removed or disconnected from the circuit (R2 = ¥ <shorted the R1 (R1 = 0). When the voltage gain is equal to the unit circuit is called voltage followers (voltage follower), because the output voltage continuously monitors the input voltage value.
Finally we note that the non - inverting amplifier has all the properties due to the non - inverting input. This means:
• Reduce the gain (from AOL to ACL)
• Increase the input impedance
• Reduce the output impedance
• Increased bandwidth
Figure 8.2.2 Response of frequency band and evrss the operational amplifier open loop gain (AOL) and closed-loop (ACL)
The increased price of bandwidth an inverted amplifier due to the influence of the feedback bandwidth operational amplifier as shown in Fig 8.2.2. Note that the open loop voltage gain of an operational amplifier remains constant up to a frequency fOL, and then begins to decrease at about 6dB/oct. This means that the gain-plasiazetai infrastructure for each doubling of frequency. The fOL called threshold or cutoff frequency (cutoff frequency) and is approximately the bandwidth of the open loop gain of the TE. At this frequency the gain is reduced to 0.707 (-3dB) of baseline. The frequency for which the gain is equal to the unit denoted by fT. When applying the feedback closed-loop gain decreases while increasing the corresponding cutoff frequency, fCL, ie the bandwidth of the gain. In each case the product of x gain cutoff frequency is equal to fT, ie
AOL · fOL = ACL · fCL = fT 8.2.4
This rule is assessing the impact of the choice of a gain bandwidth of the circuit.
Example 8.2.1
In the circuit of Fig 8.2.1 the input voltage is 10 mV, while R1 = 9
KO and KO R2 = 1. Calculate the closed-loop voltage gain
and the output voltage of the circuit.
Solution
Since we Ex.8.2.3
Example 8.2.2
In a non-inverted amplifier TE (Fig. 8.2.1) applies
signal at the entrance of 10 mV. If the output of the signal must import
yes 1V, and R2 = 1KO to calculate R1.
Solution
Since we Ex.8.2.3
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Operational AMPLIFIERS
Therefore
and therefore
Example 8.2.3
An operational amplifier (eg 741) is open loop voltage gain of 100000 and frequency fT = 1,5 MHz. If through feedback (a circuit non-inverted amplifier) reduce the gain and made 100, to calculate the cutoff frequency of the open loop gain and the cutoff frequency of the closed loop gain.
Solution
It is known that
and
fT 1,5 x 106 Hz
8.2.2 inverting amplifier
The circuit reversed an amplifier is shown in sch.8.2.3. A key feature of this circuit is applied to the signal inverting input via a resistor R2 (input resistance). The feedback from the output signal and is transferred to the inverting input through another resistance R., (feedback resistor). The feedback and in this case is negative.
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O term inverting amplifier (inverting amplifier) resulting from
that the phase difference between signal and esodou input signal is 180 ° degrees. Applied to the circuit inverting feedback (inverting feedback). O term results from the fact that during this feedback phase difference between output and feedback input signal is 180 ° degrees.
Figure 8.2.3 Circuit reversed amplifier
The reverse voltage gain in amplifier (gain closed-loop) is calculated easily by the observation that the (negative) feedback operational amplifier is trying to reduce to zero the potential difference between the two inputs. So as non - inverting input is grounded, and should the inverting input to behave as if the potential is of the earth.
The gain is calculated as follows: If an input voltage applied, yi between the free end of the R2 and earth will pass through the R2 stream i. The stream will be given by the relationship
8.2.5
Because the input impedance of the operational amplifier is very large (theoretically infinite), the first Kirchhoff rule imposes that current which will pass through the R1 will be equal to that which passes through R2. Thus the output voltage V0 should have such a value to satisfy the equation
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Operational AMPLIFIERS
After the current of i in Eq. 8.2.5 and Eq. 8.2.6 is the same, the latter shall be equal, so that the loop gain kieistou anastrefontos amplifier will be given by the expression
The negative sign indicates 180 ° phase difference between input voltage and output voltage.
Careful observation of Eq. 8.2.7 leads us to conclude that the inverting amplifier has a voltage gain which, in absolute value may be either greater than one or less than one. In other words, the inverting amplifier can enhance the signal or the signal fades, depending on the choice of the resistors R1 and R2. Finally, the voltage gain can be zero if we have R1 = 0O.
This behavior is quite different than the non-inverted amplifier, in which the voltage gain is greater than or equal to the unit, ie the amplifier can not weaken the signal.
Finally we note that the inverting amplifier has all the properties due to inverting input. This means:
• input impedance equal to R2.
• Reduce the output impedance.
• Increase bandwidth.
Example 8.2.4
In the booster circuit of Fig 8.2.3 the input voltage is 10mV,
and we have KO R1 = 10 and R2 = 1 KO. To calculate the gain (closed loop) of the amplifier and the output voltage.
Solution
Since we ex.8.2.7
A = Ri = 10 = 10KO
Acl = R2 = - = 10 1KO
thus the output voltage will be
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y0 = ACL · yi = 10 × 10 mV = 100 mV
with phase change 180 °.
Example 8.2.5
In an inverted amplifier TE (Sch.8.2.3) applies to
input signal of 20 mV. If the output of the signal is 1 V and R2 = 1
KO, to calculate R1.
Solution
From Eq. 8.2.7 we
A = y0 = 1V = 50
Acl = HI = 0.02V = 50
with phase change 180 °. We have also
Consequently
R1 = 50 R2 = 50 x KO
8.2.3 Other circuits with operational amplifiers
In previous sections analyzed the operational amplifier circuit that operates as inverting or non-inverting amplifier. A common feature of these two circuits were using only one input for the implementation of the signal. Beyond these two basic circuit there are many others that perform mathematical operations (operations) and justify the name of this basic unit of analog electronics, etc. As mentioned elsewhere, the operational amplifier can be used to perform mathematical operations such as summation, subtraction, integration, differentiation, etc. In this chapter we only focus on two of these instruments and examine the corresponding circuits. The first case to be considered is that
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Operational AMPLIFIERS
where the closed loop gain is given by the equation
R1
in which the operational amplifier performs subtraction. The second is that in which operational amplifier performs addition. These circuits are called on subtracter and adder, respectively, to state:
Figure 8.2.4 subtracter amplifier circuit
A. The amplifier circuit of the subtracter (difference amplifier) is shown in Fig 8.2.4.
As seen from the figure, the circuit is derived from the combination of an inverted amplifier (upper circuit) and a non-inverted amplifier (lower circuit). As a result, there are two entrances to which respective signals h1 and h2. The output of the subtracter circuit amplifier is given by the equation:
In a subtracter amplifier the two resistors to the inputs (R-) should be the same as the other two (R1), to avoid the problems that arise with the compensation voltage and input current. The voltage gain is not affected if different resistors in each entry, however, sufficient reason to R1/R2 is the same.
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Figure 8.2.5 adder amplifier circuit three inputs
O subtracter amplifier principle operates as a differential amplifier, the output voltage that is proportional to the difference in voltages applied to the inputs. Such is the inner workings of the operational amplifier Eq. 8.1.2, which has been built at the entrance of a differential amplifier. Nevertheless, the amplifier sch.8.2.4 not called a differential amplifier (differential amplifier) but subtracter amplifier (difference amplifier). The difference arises from the same circuit, which is in use and feedback voltage divider at the entrance, and from that input resistance is determined by the resistors used. Another circuit, the so-called instrumental instrumentalist or amplifier (instrumentation amplifier) is used as a common differential amplifier. In this feedback occurs internally and the input resistance of the approaches that of ideal operational amplifiers.
B. The amplifier circuit of the adder (summing amplifier) is shown in Fig 8.2.5. O summing amplifier scheme has three inputs and the operation is based on the inverted amplifier. Adders amplifiers can be constructed with non - reversed amplifiers. O number of entries in both cases is unlimited.
In this circuit we exploit the fact that the inverting input behave as if it has the potential of non-per-turn, ie 0. So the first rule of Kirchhoff
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Operational AMPLIFIERS
that the sum of input currents must be
equal to the current of the feedback loop (i1 + i2 + i3 + ... = if).
Applying these calculated output voltage, which is given by
the following equation:
Because the ratios of resistors determine the above expression
corresponding to the inverting voltage gain feedback,
conclude that the circuit of Fig 8.2.5 is possible to select separate voltage gain for each input separately. So on. 8.2.10
can be written more clearly as follows:
The negative sign simply indicates reverse-phase 180 ° (reverse polarity) to the final result of summation.
Example 8.2.6
In a subtracter amplifier (Fig. 8.2.4) the resistance of
circuit are R1 = 100 KO. If the non-inverting input
applied voltage h1 = 25 mV and the inverting voltage u2 = 10mV
To calculate the voltage at the output.
Solution
Based on Eq. 8.2.9, the required voltage gain of the amplifier will subtracter
A = R1 = 100K = first
Acl = R2 = '0 = 10KO
So, in Eq. 82.8, the output voltage will be
y0 ^ = A · (h1 - h2) = 10 x (25 mV - 10 mV) = 10 × 15 mV = 150 mV Example 8.2.7 <br /> In an adder amplifier (Fig. 8.2.5.) resistors circuit are R1 = 100 KO, R2 = 56 KO, R1 = 22 KO and Rf = 220 KO. If the inputs of applied voltages h1 = 25 mV, h2 = -15 mV and Y3 = 10 mV, respectively, to calculate the output voltage.
Solution
The output voltage of an inverting adder with feedback given by Eq. 8.2.10. So the output voltage will be
so finally: V0 = 96,05 mV
8.2 SUMMARY There are two basic types of negative feedback, which are used in operational amplifiers: The non-inverting and inverting feedback feedback. The respective amplifiers take their name from the kind of feedback that we have non-inverted amplifiers and turned amplifiers. The negative feedback reduces the voltage gain but increases the input impedance and bandwidth of the amplifier. The negative feedback reduces the output impedance of the operational amplifier. In an operational amplifier (with or without feedback) the product of gain x bandwidth is a key parameter (fT) and allows to determine the bandwidth for each value of the closed loop gain. In the non-inverted amplifier output signal is symfasiko on the input signal. In the non-inverted amplifier input impedance of the amplifier is equal to the input impedance of the operational amplifier. In the inverted amplifier output signal is 180 ° phase difference with the input signal. In the inverted amplifier input impedance of the amplifier is equal to the input impedance (resistance which is inserted at the entrance (-) of operational amplifier), so it is usually small compared with the input impedance of the operational amplifier.
QUESTIONS 8.2 EXERCISES-
What is the non - inverting amplifier and what are the characteristics. The non - inverting input of anaktiristika, lessening What is the voltage gain inverting amplifier causes an increase in:
tissue of a voltage output 8.2.3. What is the product of voltage gain b sfalamatos voltage on the bandwidth and
c voltage feedback
what purpose. 8.2.4 d input voltage.
What is the subtracter amplifier 8.2.8. An aggregator is a wave and what are the main characteristics kloma in which: the characteristics.
A. The output voltage is equal to
8.2.5. What is the summing amplifier is the algebraic sum of and what are the main characteristics <br /> trends in the input characteristics. B. The output voltage is equal to
8.2.6 How many types of negative differential across the supply voltages of the inputs used in Dai Operational enhanced c can enter them all: have a different disinfectants
A. one B. two D. The output voltage is always positive three c
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8.2.9. A subtracter is a wave in which kloma: A. The output voltage is equal to the algebraic sum of the voltages of input B. The output voltage is equal to the difference of the input voltages of c is possible to use random combinations of resistors in the circuit D. The output voltage is always negative
8.2.10. The non-inverting amplifier of Fig 8.2.1 is KO R1 = 19,8, R2 = 200 Ohm and RL = 10 KO. If the open loop voltage gain of the TE is 40,000 to calculate the closed loop voltage gain.
8.3.11. The amplifier of the previous exercise to compute the current which flows through the load resistance if applied to the input of the amplifier a voltage 1mV.
8.3.12.
