Lab 3 - EE 420L: Engineering Electronics II
This lab will utilize the LM324 op-amp
(LM324.pdf).
Review the data sheet for this op-amp.
For the following questions and experiments assume VCC+ = +5V and
VCC- = 0V.
Knowing the non-inverting input, Vp, is at the same potential as the inverting input, Vm, (called the common-mode voltage, VCM) what are the
maximum and minimum allowable common-mode voltages?
The datasheet for the LM324 indicates
the common mode voltage, VCM, has a maximum rating of (Vcc-1.5) and a minimum
of 0V at room temperature (+25
C). This is seen in the image below
directly from the LM324 datasheet. The Vcc in
the experiments is Vcc+ = +5V, thus the maximum
VCM is 5V - 1V = 3.5V. As seen on the datasheet and indicated below in figure 1
in the footnote, Vcc+ is rated for a maximum of
30V.
Figure 1
What is a good estimate for the
op-amp's open-loop gain? Support your answer with a plot from the
datasheet and an entry from the electrical characteristics table.
The open-loop gain of an op-amp is the
gain achieved when the circuit has no feedback loop. An ideal op-amp has
infinite open-loop gain, which does not exist in practical situations. The
LM324 datasheet has several locations indicating open-loop gain. Below in figure 2 is the large signal voltage
gain indicating a gain of 100V/mV from the data sheet.
Figure 2
What is a good estimate for the offset
voltage? For worst case design what value would you use?
The datasheet snip below in figure 3
lists a typical offset voltage of 2mV with a maximum of 5V at room temperature,
thus 2mv is a good estimate for the offset voltage. For worst case design, the
9mV offset will be used to account for high temperature variations and provide
a large range for worst case scenarios.
Figure 3
Build, and test, the following
circuit. Note that a precise value for the 5k resistors isn't important. You
can use 4.7k or a 5.1k resistors.
Below in figure
4 is the schematic of the circuit being tested.
Figure 4
What is the common-mode voltage, VCM?
Does VCM change? Why or why not?
The circuit above in figure 4 includes
a voltage divider at DC, as both capacitors act as opens at DC. Theoretically,
VCM should be +2.5V as indicated below. Figure 5 below will demonstrate the
hand calculations used to determine VCM.
The waveform below in figure 6 shows the VCM measures at approximately
2.5V as indicated.
Figure 5
Figure 6
What is the ideal closed-loop gain?
The circuit above in figure 4 has an
inverting op-amp topology and thus has an ideal closed-loop gain of 
.
This results in a unity gain 1800 out of phase with the
input.
What is the output swing and what is
it centered around? What happens if the input isn't
centered around VCM, that is, 2.5 V? Provide a detailed
discussion illustrating that you understand what is going on.
The output swing is the range of
voltage that an op-amp physically provides at its output. From the calculation
above, the output will swing at the same amplitude as our input centered at
2.5V, but 1800 out of phase. The waveform below in figure 7 displays
the output on channel 4 out of phase, but with the same amplitude as our input
on channel one. The image was taken by
setting the coupling on the oscilloscope to AC.
With DC coupling the offset can be seen but the wave form cannot as the
scale is 100mV and the input is centered at 2.5V
Figure 7
If the input is not
centered around VCM, the output signal changes according to the
difference between the two inputs. The image below in figure 8 is the result of
centering Vm at 0.6V.
The same power supply was used to power VCC and provide a constant DC
voltage to the LM324, but the result was VCM did not match Vm and the difference was larger than the recommended
common node voltage of the op-amp. As
expected the output of the op-amp did not match the input wave form. Again the image was taking with the coupling
of the oscilloscope set to AC.
Figure 8
What is the maximum allowable input
signal amplitude? Why?
The maximum allowable input signal is
2.5V due to the maximum 5V VCC. Any signal larger than 2.5V will be clipped
because it will exceed the supply voltage.
The rail voltage determines the range in which the signal can operate,
in our case that range is from +5V to -5V.
What is the maximum allowable input
signal if the magnitude of the gain is increased to 10? Why?
Increasing the gain by 10 means the
input signal must be decreased by a factor of 1/10th to
250mV. This is to account for the clipping that occurs with respect to the
rails. When the voltage is centered at
2.5V and our rail voltage is +5V to -5V the maximum voltage we can input is
2.5V before we reach the limit of our rail voltage. As 250mV * 10 = 2.5V our maximum input
voltage on an op-amp with a gain of ten is 250mV.
What is the point of the 0.01 uF capacitors from VCC and VCM to ground? Are these
values critical or could 0.1 uF, 1,000 pF,
1 uF, etc. capacitors be used?
The capacitors act as decoupling capacitors in this circuit.
The decoupling capacitors separate the AC and DC components of a signal as a
capacitor acts as an open to DC and a short to AC. They also prevent noise and
help 'couple' the output of one stage of the circuit to the input of the next
stage to create a stable circuit. The values of the capacitor are not critical
because they are only for decoupling and do not affect VCC or VCM since they
are DC voltages.
The data sheet shows that this op-amp has an input bias
current that flows out of the op-amp's inputs of typically
20 nA. This current flows out of both the non-inverting
and inverting inputs through the resistors connected to these inputs. Show how
the operation of the circuit can be effected if, for example, R1 and R2, are
much larger. Explain what is going on.
The values of R1 and R2 can have a noticeable impact on the
circuit. The two resistors should ideally be the same value or close to the
same value to keep the output centered and allow for the largest output swing
possible. If the values of R1 and R2 are increased to much larger values, the
input bias current begins to impact the voltage divider circuit. Specifically,
the VCM terminal has two 10kΩ resistors in parallel with 20nA flowing
through to ground, resulting in a voltage drop of 0.1mV. Adding 0.1mV to a VCM
of 2.5V results in 2.5001V. Larger
resistors will result in a larger voltage added to VCM and a clipped output.
For example 100MΩ resistors results in 50MΩ*20nA and 1V added to
VCM. This is a large impact on circuit functionality.
What is the input offset current? What does this term
describe?
The input offset current is the difference between the input
bias currents of the op-amp terminals.
Thi input offset current of the LM324 at room temperature is typically
2nA.
Explain how the
following circuit can be used to measure the op-amp's offset voltage. Note that
if the output voltage is precisely the same as VCM then the op-amp has no
offset voltage (this is very possible).
To measure small offset voltages increase the gain by increasing RF to
100k or larger. Explain what is going on.
The circuit below in figure 9 applies the same input, VCM, to
both the inverting and non-inverting inputs of the op-amp. Assuming the op-amp
to be ideal, no current flows through Rf because the nature of the
op-amp causes Vm at the inverting terminal to be equivalent to VCM at
the non-inverting terminal. Thus, performing circuit analysis results in 
at the inverting
terminal. By Kirchhoff's Current Law, current entering a node equals
the sum of the currents exiting the same node and therefore, the current
through Rf would be zero. Consequently, if there is an offset
voltage, Vm will equal VCM plus the offset voltage. The inverting
amplifier will amplify this differential by a factor of 20. To measure the offset
voltage, measure Vout and VCM, calculate the difference between them
and divide by a factor of 20. The result will be the offset voltage.
Figure 9
Measure the offset
voltage of 4 different op-amps and compare them.
The four op-amps measured will be
the LM324, TL081, LM348, LM741.
Below in figure 10 is the
measurement of the LM324. Table 1 below
is the offset measurements Voffset is calcluated by taking the differece and
dividing by the gain of the op-amp in this case 20.
Figure 10
|
LM324 |
Volts |
|
VCM |
2.54 |
|
Vout |
2.50 |
|
Voffest |
0.002 |
Table 1
Below in figure 11 is the
measurement of the TL081. Table 2 below
is the offset measurements Voffset is calcluated by taking the differece and
dividing by the gain of the op-amp in this case 20.
Figure 11
|
TL081 |
Volts |
|
VCM |
2.70 |
|
Vout |
2.50 |
|
Voffest |
0.01 |
Table 2
Below in figure 12 is the
measurement of the LM348. Table 3 below
is the offset measurements Voffset is calcluated by taking the differece and
dividing by the gain of the op-amp in this case 20.
Figure 12
|
LM348 |
Volts |
|
VCM |
2.54 |
|
Vout |
2.50 |
|
Voffest |
0.002 |
Table 3
Below in figure 13 is the
measurement of the LM741. Table 4 below
is the offset measurements Voffset is calcluated by taking the differece and
dividing by the gain of the op-amp in this case 20.
Figure 13
|
LM741 |
Volts |
|
VCM |
2.70 |
|
Vout |
2.50 |
|
Voffest |
0.01 |
Table 4
Conclusion
Laboratory exercise three presented a quick overview of the
basic operational amplifier topologies, specifically inverting and
non-inverting topologies, as well as an opportunity to utilize the information
on an op-amp's datasheet and compare the values to the practical operation of
the amplifier in a circuit. This lab demonstrated the relationships of
parameters in operational-amplifier circuits, such as common mode voltage,
closed-loop gain, output swing and maximum allowable input signal which create
the foundation for learning to design with these multi-faceted devices.
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