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Op Amps I
Due date: Monday, September 21 (midnight)
Reading: HH Sections 4.01–4.10 (pgs. 175-188)
The operational amplifier (op amp) is the single most important active circuit component
for analog electronics work. We will spend several labs on it.
4.1
Installation
The pin-out designations for a standard 8-pin dual-inline package (DIP) op amp are shown
in Fig. 1. When counting pins on a DIP device, always start by locating the “top” of the
package. This can be identified by a semi-circular indentation, a small circle, or both, as
illustrated in Fig. 2. The pins are labled starting in the top left corner and proceeding
counter-clockwise, as shown.
Place an LF411 op amp on your breadboard, top up, straddling one of the central
columns. With the board power off, wire the vertical ‘+’ and ‘-’ columns with ±15 V
respectively, and then connnect these power buses to the +VS and −VS pins of the package.
This is sufficient, but it is always good practice to filter the power supplies connections using
a capacitor to ground. This helps maintain a steady voltage if the input current changes
quickly; without it, the inductance of the line from the power supply can lead to instability.
Here, use a 1 µF tantalum capacitor, one from the +VS pin to ground and another from
the −VS to ground. These capacitors have a polarity, with the lead that must be positive
marked with a +. Be careful to install them correctly.
We will not be using the “offset null” pins today.
4.2
Open-Loop Test
An op-amp is nothing more than a high gain amplifier with high input impedance. You can
observe this amplification with the circuit of Fig. 3. Is the behavior consistent with the gain
specification of 200 V/mV? Can you see why an op-amp is essentially never used in this
“open loop” configuration?
offset null
inverting
non-inverting
1
8
no connection
2
7
3
6
4
5
out
offset null
Figure 1: Pin-out diagram for a standard 8-pin
op-amp package.
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Figure 2: How to count pins on an integrated
circuit chip.
out
Figure 3: Op amp as an open loop amplifier.
4.3
Inverting Amplifier
Assemble the inverting amplifier circuit of Fig. 4. Drive it with a 1 kHz sine wave. What is
the gain (in dB)? What is the maximum output voltage range? Try varying the frenquency.
Drive the amplifier with a triangle, and see if the amplifier is linear. Drive it with a
square wave and see how it responds to a sharp step.
Replace the 10k feedback resistor with a 100k pot, and vary the gain. Can you take the
gain all the way to zero (-∞ dB)? If you kept increasing the feedback resistance, what do
you think is the maximum gain you could achieve?
Measure the input impedance of the amplifier circuit by adding another 1k resister in
series with the input. Does the result make sense?
Verify that the output impedance of the circuit is very low by driving a 10 Ω load. You’ll
need to use a small input signal, because the op-amp can only supply about 25 mA of current
before saturating. This is an important limitation, but the output impedance is properly
defined for small signals, where saturation is not occuring. Don’t try too hard to get a value
for the output impedance. The best answer in the end is that the impedance is low enough
to treat as zero in practice.
4.4
Non-inverting Amplifier
Assemble the non-inverting amplifier shown in Fig. 5. What is the voltage gain? Replace
the 10k resistor with a pot, and observe the range of gains achievable.
Try to measure the circuit’s input impedance by putting a 1M resistor in series with the
in
out
in
out
Figure 4: Inverting amplifier
Figure 5: Non-inverting amplifier
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Vin
Vin
LF411
Vout
LF411
Vout
(b)
(a)
Figure 6: The follower (a) and an application (b).
input. (Note that if you get 1 MΩ or 10 MΩ, you’re making a mistake.) Don’t work too
hard at this either, because eventually it’s not clear whether the impedance will be limited
by the op amp or the breadboard.
You should easily be able to convince yourself that the output impedance of this amplifier
is the same as that of the inverting amplifier.
4.5
Follower
You may be surprised to discover that the single most common op-amp circuit is probably
the follower of Fig. 6(a). By applying the op-amp rules, you can show that this circuit
apparently does nothing: Vout = Vin . What, however, about the input and output currents?
The point of the follower is that its input impedance is very high, while it’s output
impedance is very low. We’ve seen already that when you hook a load to a source, you need
to make sure the load input impedance is large compared to the source output impedance,
if you want to avoid degrading your signal. If you need to hook a low impedance load up
to a high impedance source, you have a problem. The follower solves that problem, as in
Fig. 6(b).
As a simple example, repeat an experiment you did in Lab 1, where you cascaded two
voltage dividers. There, you had to use small resistors in the first divider and large resistors
in the second, in order to avoid loading effects. Build a circuit now that uses a 1:3 divider
driving a follower, driving an identical 1:3 divider. By eliminating the problem of loading,
the follower makes circuit design much easier.
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1
out
2
Figure 7: Summing amplifier.
4.6
Summing Amplifier
Another useful trick is shown in Fig. 7. This circuit takes two inputs V1 and V2 and produces
the (inverted) sum −(V1 + V2 ). What if you don’t want the inversion? Simply follow it by
an inverting amplifier with gain -1.
To get some practice with the idea, design and build a circuit to add a ±5 V variable
DC offset to your function generator signal. The offset should be set using a 10k pot. (Does
the circuit need to include a follower after the pot?)
4.7
Current Sources
We have mostly thought of power supplies as voltage sources: a ‘good’ source is one with a low
output impedance, that maintains its voltage regardless of what load you apply. However,
current sources can also be useful. This describes a supply that produces a constant current,
no matter what load you apply.
The obvious use is in applications that depend specifically on current. For instance, you
might want a good current source to produce a stable magnetic field from an electromagnet
coil. As the coil heats up, its resistance will change, so a voltage source would not be as
desireable. Another common application is to generate a linear voltage ramp by applying a
current source to a capacitor.
Of course, a real voltage source can’t produce unlimited current when you apply it to
a short circuit (= zero resistance) load. Similarly, no current source can generate infinite
voltage in response to an open circuit (= infinite resistance) load. So if you turn on a current
source with no load attached, it will simply ramp up to it’s maximum possible voltage and
sit there. But if you turn it on with a short circuit load, the current source will drive exactly
the current specified, for as long as you like.
Interestingly, the output impedance of a perfect current source is infinite: the Thevenin
equivalent circuit is a very large voltage driving a very large internal resistance. Any finite
load resistance you apply is neglible compared to the internal resistance, and thus has no
effect on the current. More generally, a supply of finite impedance Z acts as a current source
for loads with impedance much less than Z, and a voltage source for loads with impedance
much greater than Z.
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100k
V-
in
100k
R
LF411
V+
Iout= (V+ - V- )/R
100k
100k
LF411
Figure 8: Current source with
floating load.
Figure 9: Current source with arbitrary load.
An op amp can be used to make a simple current source using the circuit of Fig. 8.
Wire this up with a 10k pot as a load, and with your DMM ammeter in series with the
pot. You can use the varible power supply to set Vin . Does the circuit accurately maintain
Iload = Vin /R as you change the pot?
In Fig. 8, the load is floating with respect to ground. Unfortunately, loads that must
be referenced to ground are fairly common. (An oscilloscope is one example.) An op amp
circuit that works as a current source to ground is shown in Fig. 9; you get to analyze it in
your homework assignment.
Note that this circuit is quite flexible: it will deliver the specified current to a load
referenced to any (reasonable) voltage, not just ground. Wire it up and test it with a few
different input voltages and loads. Verify that it doesn’t matter whether the load is returned
to ground or to the +5 V supply.
4.8
Current to Voltage Converter
Another function at which op amps excell is converting a current to a voltage. At first
glance, this seems trivial, since that is just the job of a resistor: V = IR. However, if you
want a large ‘gain’ (ie, a large voltage for a given current), then you need a large resistor.
This would, of course, have a large input impedance, and for a current source, large load
impedances are difficult to handle.
An example where this comes up is the detection of light with a photodiode. Photodiodes
have the property that, under illumination, they produce a current propotional to the light
intensity. The output impedance of a photodiode, however, is very low. (In fact, the output
impedance is nonlinear, since the photodiode is, of course, a diode. What is most relevant,
however, is that the output voltage starts to saturate at just a tenth of a volt or so.)
Figure 10 shows how an op amp can boost the voltage produced without presenting a
large impedance to the photodiode. In fact, since the inverting input of the op amp is at
ground, the photodiode ‘sees’ a short circuit, the perfect load for a current source. The
feedback resistor can be made very large to produce a large signal. Wire up the circuit using
a PNZ335 photodiode, and use it to observe the room lights.
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If you attach the photodiode to the end of a cable and hold it up to the screen of a
scope displaying the output signal, you can observe some interesting feedback effects. Try
reversing the diode orientation.
For reference, the formal name of a current to voltage converter is a transimpedance
amplifier. The gain of a transimpedance amplifier is specified in ohms.
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Figure 10: A photodiode amplifier.
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