Following voltage summing amplifier integrator differentiator difference
pp 90-99
4.5 following voltage
As shown by Eq. (4.3.2) is less gain can be achieved with non - reverse amplifier is equal to 1. This event, which occurs when R2 = 0, but this amp has the ability to make good adjustment resistors.
When the amplifier wiring to support unit, as in Sch.4.8, called voltage following, because the output voltage is then equal and symfasiki the input voltage, ie the output voltage "follows" continuously and accurately the input voltage.
Figure 4.8. The TE syndesmologimenos voltage as follows
It turns out that applying the following relations:
Operational AMPLIFIERS
91
Af = 1 + A0 1 (4.5.12)
R.f = (1 + A0) R.AR
R0f = R0 R0 1 + A0 A0 (4.5.13)
(4.5.14)
f1f = (1 + A0) f1 A0f1 (4.5.15)
Observe from Eq. (4.5.13) and (5.4.15) that (always after R. >> R0) the input impedance of the following voltage is much larger than the output impedance. Therefore, the following voltage isolating a large resistance from a small pool. Therefore the following voltage called buffer (buffer). So, when inserted between two circuits, eliminates the high output impedance of the first circuit and allows loading with very low resistance load.
Example 4-3
For the TE 741 syndesmologimeno as follows voltage, find the tire size.
Solution
Af = 1
Rif = A0Ri = 2 x 105 x 2 = 400 G0 MO
75
A0 2 x 105 = 0.38 mQ
f1f = A0f1 = 2 x 105 x 5Hz = 1 MHz
(4.6.16)
92
Analogue Electronics
4.6 Adder
The independence between the two inputs, (+) and (-), a TE allows the use of a circuit that performs aggregation tendencies and therefore called aggregator.
The Sch.4.9 shows the TE syndesmologimeno as adder with three inputs,
u1, u2 and u3. Apodeikny
Un =-B 1 B 2 B R1 R2 R 3
If R1 = R2 = R3 = R, this formula becomes:
Rf
(4.6.17)
Hence, the output voltage is the sum of negative (the - shows just the phase difference 180 °) of input signals to enhance Rf / R, and therefore called the amplifier summing amplifier. If you are not Rf = R follows:
u0 = - (u1 + u2 + u3) (4.6.18)
Hence, the circuit is then a simple adder.
Example 4-4
The circuit has Sch.4.9 u1 = 0.25 V, u2 = 0.5 V, u3 = 0.75 V R1 = R2 = R3 = 10 KO and KO Rf = 6.8. Find the output voltage.
Operational AMPLIFIERS 93
Solution
From Eq. (4.6.17), we get:
u0 = - M (0.25 + 0.5 +0.75 @ - 3.1 V
4.7 difference amplifier
If, instead of the sum of two voltages, we need to remove them, use the TE in subtracter circuit. The Sch.4.10 shows the subtracter circuit. It turns out,
that the output of this circuit we have:
u0 =-R2 (u2-ui) =-K (u2-u. |) (4.7.19)
R1 R2
where, K = (4.7.20)
R1 is to enhance the closed loop. That is, the output circuit that generates enhanced the difference of two input voltages, so it acts as a subtracter amplifier.
Of course, if R2 = R1, K = 1 and Eq. (7.4.19) gives:
u0 = - (u2 - u1) (4.7.21)
that is, the subtracter circuit is simple.
Figure 4.10. Subtracter circuit
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Integrator 4.8
In some cases, such as in analog computers, need to be (mathematically) a complete waveform. The circuit in which the input voltage waveform is the integral of the waveform of the output voltage of the integrator is called integrator or amplifier. This circuit is created by the main amplifier circuit of Sch.4.7 if we replace the feedback resistor R2 to capacitor C. Sch.4.11.
Completion Completion sine square pulse signal
Figure 4.11. Integrator circuit
The Eq. (4.8.22) shows that the output voltage is proportional to the time integral of the input voltage and inversely proportional to the time constant T. The minus is due to 180 ° phase difference caused by the inverting configuration of the TE.
At very low frequencies omega = 2pi 0) integrator functions as an open-loop amplifier. This is because then the capacitor C acts as an open circuit (XC = 1 / oC = <*>). For this reason, the practical integrator, to counteract this disadvantage, connect a resistor R2 in parallel feedback capacity C, Sch.4.12. This resistance limits the low frequency amplification in value R2 / R1 and thus minimizes the output voltage changes at reasonable levels.
It turns out that the output voltage of the integrator are considering, is given by:
(4.8.22)
where
t = R2C (4.823)
is the integrator time constant and c is the constant of integration.
To make the proper completion of the circuit input waveform has a duration of input pulse tp be much less than the time constant T of the integrator. In practice, t must satisfy the relation:
tp 10 (4.8.24)
Figure 4.12. Practical integrator circuit with TE
The integrator used in analog computers, the analogue to digital (ADC) and pulse shaping circuits.
Example 4-5
We calculate integrator when he is excited by the symmetric square pulse width of 5 V and frequency 1 kHz.
Solution
The period of the input signal is: 11
T = 1CT3 sec = 1 msec f 103
Hence, the duration of the pulse input, because it is symmetric, are:
T1
tP = 2 = 2 msec = 0.5 msec
According to Eq. (4.8.23) and (4.8.24), the required time constant is:
t = 10tp = R1C Assuming R1 = 10 KO, we have:
Besides, we accept R2 = 100R1 = 1 MO, so the strengthening of the integrator on multi HS will be:
which is a reasonable price.
The Sch.4.13 shows the calculated final integrator.
10 K
Figure 4.13. A practical example of the integrator
4.9 differentiator
Sometimes we need a circuit which can perform
differentiation, and hence differentiation of the input waveform and is therefore called differentiator. One such circuit is shown in Sch.4.14. As the differentiator formed from the inverted amplifier, if the position of R1 put a capacitor C.
It turns out that in such a circuit, the output voltage waveform is given by the time derivative of the input waveform, ie
Figure 4.14. Differentiator circuit with TE
(4.9.25)
where t is the time constant
t = R2C (4.9.26)
To get good differentiation needs the input of the pulse duration tp be much greater than the time constant of the circuit, ie tp >> T. In practice must satisfy the condition:
tp = 10 t (09/04/27)
In practice, the differentiator with TE is not exactly as shown in Sch.4.14 because it is sensitive to noise in it due to transients by opening and closing switches. Although the amplitude noise can be very small, the rate of change (du: / dt) is often large, so the output can have a very large and undesirable fluctuations.
To avoid this, connect a small resistor R1 (typical value between 1 to 100 O KO) in series with the capacitor C and a small capacitor C2, 100 pF typical price at the ends of R2.
Example 4-6
We compute circuit with TE differentiation stimulated by symmetrical pulse amplitude 1 V and frequency 1 kHz.
Solution
T = 1 msec, so tp = - = 0.5 msec f 103 p 2
From Eq. (4.9.26) and (4.9.27) we have:
tp
tp = 10t = 10 R2C, so C = J0R-
We accept R2 = 10 KO, so
C = ° .5 x 10-3 = 0.5 x 10-8, so C = 4.7 nF.
10 x 104
Finally, make a R1 = 100 Ohm and C2 = 100 pF.
Operational AMPLIFIERS
99
The differentiator Sch.4.15 shows the calculated and output waveforms for both cases the waveform input.
Differentiation differentiation sine square pulse signal
Figure 4.15. Practical differentiation circuit example and waveform output
(4.10.28)
