ElDev 04
ElDev 04
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Introduction… …Introduction
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20 transistors
11 resistors The two inputs are υ1 and υ 2. A differential voltage between them causes
current flow through the differential resistance Rd. The differential voltage
1 capacitor is multiplied by A, the gain of the op amp, to generate the output-voltage
source. Any current flowing to the output terminal vo must pass through the
output resistance Ro.
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Ideal Characteristics Two Basic Rules
• A = ∞ (gain is infinity)
• Vo = 0, when v1 = v2 (no offset voltage) • Rule 1
• Rd = ∞ (input impedance is infinity) – When the op-amp output is in its linear range, the two input terminals
are at the same voltage.
• Ro = 0 (output impedance is zero)
• Rule 2
• Bandwidth = ∞ (no frequency response limitations) and no
– No current flows into or out of either input terminal of the op amp.
phase shift
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• Inverting and non-inverting amplifiers • When looking at feedback we derived the circuit of an
amplifier from ‘first principles’
• Normally we use standard ‘cookbook’ circuits and select
component values to suit our needs
• In analysing these we normally assume the use of ideal op-
amps
– in demanding applications we may need to investigate the
appropriateness of this assumption
– the use of ideal components makes the analysis of these
circuits very straightforward
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…Basic operational amplifier circuits… …Basic operational amplifier circuits…
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Therefore choose R2 = 1 kΩ and R1 = 25 kΩ • For more details of these circuits see the relevant section
(we will consider the choice of values later) of the course text (as shown on the slides)
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Vo = −I i R
Thus the circuit has a gain of unity
• At first sight this might not seem like a very useful circuit,
however, it has a high input resistance and a low output
resistance and is therefore useful as a buffer amplifier
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…Other useful circuits… …Other useful circuits…
R1 R1
Vo = (V1 − V2 ) Vo = −(V1 + V2 )
R2 R2
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• An integrator • A differentiator
1 t d Vi
Vo = −
RC 0∫Vi d t Vo = − RC
dt
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Instrumentation Amplifiers Active Filters
Adding capacitors to op-amp circuits provides external control of the
cutoff frequencies. The op-amp active filter provides controllable cutoff
frequencies and controllable gain.
• Low-pass filter
• High-pass filter
Differential Mode Gain • Bandpass filter
v3 − v4 = i ( R2 + R1 + R2 ) • Bandstop filter
v1 − v2 = iR1
v3 − v4 2 R2 + R1 The filter transfer function T(s) ≡ Vo(s)/Vi(s).
Gd = =
v1 − v2 R1
Advantages: High input impedance, High CMRR, Variable gain
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Low-Pass Filter—Second-Order High-Pass Filter
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…Real operational amplifiers… …Real operational amplifiers…
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Selecting component values Effects of feedback on op-amp circuits…
• Our analysis assumed the use of an ideal op-amp • Effects of feedback on the Gain
• When using real components we need to ensure that our – negative feedback reduces gain from A to A/(1 + AB)
assumptions are valid – in return for this loss of gain we get consistency, provided
• In general this will be true if we: that the open-loop gain is much greater than the closed-loop
– limit the gain of our circuit to much less than the gain (that is, A >> 1/B)
open-loop gain of our op-amp – using negative feedback, standard cookbook circuits can be
– choose external resistors that are small compared with the used – greatly simplifying the design
input resistance of the op-amp – these can be analysed without a detailed knowledge of the
– choose external resistors that are large compared with the op-amp itself
output resistance of the op-amp.
• Generally we use resistors in the range 1 to 100 kΩ
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• Example • Example
– determine the input and output resistance of the following – determine the input and output resistance of the following
circuit assuming op-amp is a 741 circuit assuming op-amp is a 741
Open-loop gain (A) of a 741 is 2 × 105 Open-loop gain (A) of a 741 is 2 × 105
Closed-loop gain (1/B) is 20, B = 1/20 = 0.05 Closed-loop gain (1/B) is 20, B = 1/20 = 0.05
(1 + AB) = (1 + 2 × 105 × 0.05) = 104 (1 + AB) = (1 + 2 × 105 × 0.05) = 104
Feedback senses output voltage therefore it Feedback senses output voltage therefore, it
reduces output resistance of op-amp (75 Ω) by reduces output resistance of op-amp (75 Ω) by
104 to give 7.5 mΩ 104 to give 7.5 mΩ
Feedback subtracts a voltage from the input, Feedback subtracts a current from the input,
therefore it increases the input resistance of therefore it decreases the input resistance. In
the op-amp (2 MΩ) by 104 to give 20 GΩ this case the input sees R2 to a virtual earth,
therefore the input resistance is 1 kΩ
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Key points
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