circuit theory 1 finals

1. Rojien V. Morcilla, R.E.E.
Circuit Theory 1

2. OPERATIONAL AMPLIFIERS
The op amp is an electronic unit that behaves like a voltage-controlled voltage source.
An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration.
The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes.
Typical op amp package
A typical one is the eight-pin dual in-line package (or DIP), shown above, Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are:
The inverting input, pin 2.
The noninverting input, pin 3.
The output, pin 6.
The positive power supply V+, pin 7.
The negative power supply V, pin 4.
A typical op amp: (a) pin configuration, (b) circuit symbol.
Powering the op amp.
Power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked. By KCL,
The equivalent circuit of the non ideal op amp.
The differential input voltage vd is given by
Thus, the output vo is given by
A is called the open-loop voltage gain because it is the gain of the op amp without any external feedback from output to input.
A practical limitation of the op amp is that the magnitude of its output voltage cannot exceed |VCC|. In other words, the output voltage is dependent on and is limited by the power supply voltage.
op amp can operate in three modes, depending on the differential input voltage vd :
Positive saturation, vo = VCC.
Linear region, VCC vo = Avd VCC.
Negative saturation, vo = VCC.
We will assume that our op amps operate in the linear mode
A 741 op amp has an open-loop voltage gain of 2105, input resistance of 2 M, and output resistance of 50 . The op amp is used in the circuit. Find the closed-loop gain vo/vs . Determine current i when vs = 2 V.
If the same 741 op amp in Example 1 is used in the circuit below, calculate the closed-loop gain vo/vs . Find io when vs = 1 V
8. IDEAL OP AMP
An op amp is ideal if it has the following characteristics:
Infinite open-loop gain, A .
Infinite input resistance, Ri .
Zero output resistance, Ro 0.
An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance.
9. IDEAL OP AMP
Two important characteristics of the ideal op amp are:
The currents into both input terminals are zero:
This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit exists there and current cannot enter the op amp. But the output current is not necessarily zero.
The voltage across the input terminals is negligibly small; i.e.,
Thus, an ideal op amp has zero current into its two input terminals and negligibly small voltage between the two input terminals.
10. IDEAL OP AMP
A 741 op amp has an open-loop voltage gain of 2105, input resistance of 2 M, and output resistance of 50 . The op amp is used in the circuit. Find the closed-loop gain vo/vs . Determine current i when vs = 2 V.
11. IDEAL OP AMP
If the same 741 op amp in Example 1 is used in the circuit below, calculate the closed-loop gain vo/vs . Find io when vs = 1 V.
12. Inverting Amplifier
An inverting amplifier reverses the polarity of the input signal while amplifying it.
An equivalent circuit
for the inverter
The inverting amplifier
In this circuit, the noninverting input is grounded, vi is connected to the inverting input through R1, and the feedback resistor Rf is connected between the inverting input and output. Our goal is to obtain the relationship between the input voltage vi and the output voltage vo.
Applying KCL at node 1,
But v1 = v2 = 0 for an ideal op amp, since the noninverting terminal is grounded. Hence,
or
Notice that the gain is the feedback resistance divided by the input resistance which means that the gain depends only on the external elements connected to the op amp.
Example: Refer to the op amp in Fig. 1. If vi = 0.5 V, calculate: (a) the output voltage vo, and (b) the current in the 10 k resistor.
Solution:
(a) Using the previous equation,
(b) The current through the 10-k resistor is
Figure 1
Example: Determine vo in the op amp circuit shown in Fig. 2.
Solution:
Applying KCL at node a,
But va = vb = 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence,
Figure 2
Notice that if vb = 0 = va, then vo = 12, as expected
Practice Problem
Two kinds of current-to-voltage converters (also known as transresistance amplifiers) are shown in Figures below.
(a) Show that for the converter in Fig. (a),
(b) Show that for the converter in Fig. (b),
17. Noninverting Amplifier
A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain.
In this case, the input voltage vi is applied directly at the noninverting input terminal, and resistor R1 is connected between the ground and the inverting terminal. We are interested in the output voltage and the voltage gain.
Figure 3
Application of KCL at the inverting terminal give:
But v1 = v2 = vi . Hence,
or
The voltage gain is Av = vo/vi = 1 + Rf /R1, which does not have a negative sign. Thus, the output has the same polarity as the input.
Notice that the gain depends only on the external resistors. Notice that if feedback resistor Rf = 0 (short circuit) or R1 = (open circuit) or both, the gain becomes 1. Under these conditions (Rf = 0 and R1 =), the circuit in Fig. 3 becomes that shown in Fig. 4, which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus, for a voltage follower
Figure 4: The voltage
follower.
A voltage follower used to isolate two cascaded stages of a circuit.
Example: For the op amp circuit in Fig. 5, calculate the output voltage vo.
Solution:
We may solve this in two ways: using superposition and using nodal analysis.
METHOD 1:
Using superposition, we let
vo = vo1 + vo2
where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V input. To get vo1, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence it gives,
Figure 5
To get vo2, we set the 6-V source equal to zero. The circuit becomes a noninverting amplifier so that
Thus,
vo = vo1 + vo2 = 15 + 14 = 1 V
METHOD 2: Applying KCL at node a,
But va = vb = 4, and so
or vo = 1 V, as before.
Practice Problem
Calculate vo in the circuit shown
22. Summing Amplifier
A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs.
It takes advantage of the fact that the inverting configuration can handle many inputs at the same time. We keep in mind that the current entering each op amp input is zero.
Applying KCL at node a gives,
i = i1 + i2 + i3 (1)
But
Figure 6: The summing amplifier
(2)
We note that va = 0 and substitute Eq. (1) into Eq. (2). We get
Indicates that the output voltage is a weighted sum of the inputs
Example: Calculate vo and io in the op amp circuit in Fig. 7.
Solution:
This is a summer with two inputs. Using the previous Eq;
The current io is the sum of the currents through the 10-k and 2-k resistors. Both of these resistors have voltage vo = 8 V across them, since va = vb = 0. Hence,
Figure 7
Practice Problem
Calculate vo and io in the op amp circuit shown
25. Difference Amplifier
A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs.
Difference (or differential) amplifiers are used in various applications where there is need to amplify the difference between two input signals.
Applying KCL to node a,
or
(3)
Applying KCL to node b,
Figure 8: Difference amplifier.
(4)
or
But va = vb. Substituting Eq. (3) into Eq. (4) yields
(5)
or
Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that vo = 0 when v1 = v2. This property exists when
Thus, when the op amp circuit is a difference amplifier, Eq. (5) becomes
If R2 = R1 and R3 = R4, the difference amplifier becomes a subtractor, with the output
Example: Design an op amp circuit with inputs v1 and v2 such that vo = 5v1+3v2.
Solution:
The circuit requires that
vo = 3v2 5v1(6)
This circuit can be realized in two ways.
DESIGN 1 If we desire to use only one op amp, we can use the op amp circuit of Fig. 8. Comparing Eq. (6) with Eq. (5),
Also,
DESIGN 2 If we desire to use more than one op amp, we may cascade an inverting amplifier and a two-input inverting summer, as shown in Fig. 9. For the summer,
or
If we choose R1 = 10 k and R3 = 20 k, then R2 = 50 k and R4 = 20 k.
vo = va 5v1 (7)
and for the inverter,
va = 3v2 (8)
Combining Eqs. (7) and (8) gives
vo = 3v2 5v1
Figure 9
which is the desired result. In Fig. 9, we may select R1 = 10 k and R2 = 20 k or R1 = R2 = 10 k.
Practice Problem
Design a difference amplifier with gain 4.
Example: An instrumentation amplifier shown is an amplifier of low level signals used in process control or measurement applications and commercially available in single-package units. Show that
Solution:
Recognize that amplifier A3is a difference amplifier. Thus,
Since the op amps A1 and A2 draw no current, current i flows through the three resistors as though they were in series. Hence,
But; and va = v1, vb = v2. Therefore
Thus;
Practice Problem
Obtain ioin the instrumentation amplifier circuit shown.
33. Cascading op amp Circuit
A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next.
When op amp circuits are cascaded, each circuit in the string is called a stage; the original input signal is increased by the gain of the individual stage. Op amp circuits have the advantage that they can be cascaded without changing their input-output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance.
A = A1A2A3
Example: Find vo and io in the circuit in Fig. 10.
Solution:
This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp,
Figure 10
At the output of the second op amp,
But vb = va = 100 mV. Hence,
The required current io is the current through the 10-k resistor.
Example: If v1 = 1 V and v2 = 2 V, find vo in the op amp circuit of Fig. 11.
Solution:
The circuit consists of two inverters A and B and a summer C as shown in Fig. 11. We first find the outputs of the inverters.
Figure 11
These become the inputs to the summer so that the output is obtained as
Practice Problem
Determine vo and io in the op amp circuit in Fig. A.
If v1 = 2 V and v2 = 1.5 V, find vo in the op amp circuit of Fig. B.
Figure B
Figure A
37. CAPACITORS AND INDUCTORS
38. CAPACITORS
A capacitor consists of two conducting plates separated by an insulator (or dielectric).
A capacitor is a passive element designed to store energy in its electric field.
When a voltage source v is connected to the capacitor, the source deposits a positive charge q on one plate and a negative charge q on the other. The capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that
Figure 13: A capacitor
with applied voltage v.
Figure 12: A typical capacitor.
q = Cv (9)
39. CAPACITORS
Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F).
1 farad = 1 coulomb/volt.
Although the capacitance C of a capacitor is the ratio of the charge q per plate to the applied voltage v, it does not depend on q or v. It depends on the physical dimensions of the capacitor.
where A is the surface area of each plate, d is the distance between the plates, andis the permittivity of the dielectric material between the plates.
40. CAPACITORS
Three factors determine the value of the capacitance:
1. The surface area of the platesthe larger the area, the greater the capacitance.
2. The spacing between the platesthe smaller the spacing, the greater the capacitance.
3. The permittivity of the materialthe higher the permittivity, the greater the capacitance.
Figure 14: Circuit symbols for capacitors:
(a) fixed capacitor, (b) variable capacitor.
41. CAPACITORS
To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides of Eq. (9). Since
differentiating both sides of Eq. (9) gives
The voltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (11). We get
where v(t0) = q(t0)/C is the voltage across the capacitor at time t0.
(10)
(11)
(12)
(13)
or
42. CAPACITORS
The instantaneous power delivered to the capacitor is
The energy stored in the capacitor is therefore
Figure 14: Current-voltage relationship of a capacitor.
(14)
(15)
43. CAPACITORS
We note that v() = 0, because the capacitor was uncharged at t = . Thus,
Using Eq. (9), we may rewrite Eq. (16) as
(16)
(17)
44. CAPACITORS
Important properties of a capacitor:

A capacitor is an open circuit to dc when voltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero.

45. The voltage on a capacitor cannot change abruptly -The capacitor resists an abrupt change in the voltage across it. According to Eq. (11), a discontinuous change in voltage requires an infinite current, which is physically impossible. 46. The ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit. 47. A real, non ideal capacitor has a parallel-model leakage resistance. The leakage resistance may be as high as 100 M and can be neglected for most practical applications.

circuit theory 1 finals

Education

ideal op ampan op amp

op amp circuit diagrams

op amps

typical op amp package

input terminals

output voltage

infinite input resistance

output resistance