The
lab is about understanding the operation of op-amps, as well as
being able to read the data sheet regarding the op-amp being used. It
is about basic topologies, finite gain and offset.
Lab Report should include:
Answers supported from the data sheet of the LM324.pdf op-amp.
Circuit simulation from LTspice, and hand calculation to support answers.
The first
experiment was about exploring the data sheet information for the LM324
op-amp, which is the op-amp that will be used to construct the circuits
in this lab.
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?
From the
schematic one can see that both Vp and Vm are at 2.5V each so that the
op-amp can work properly. Common mode voltage is the average of both
input voltages so here it is (2.5+2.5)/2=2.5V.
As for the
maximum and minimum common mode voltage for the op-amp to operate
properly, from the data sheet the minimum at room temperature is 0V and
the maximum is VCC-1.5. The VCC used in the experiment was +5V so
maximum common mode voltage is 5-1.5=3.5V.
it can be seen that the chosen common mode voltage (2.5V) is between the range of proper operation for the op-amp.
What is a good estimate for the op-amp's open-loop gain?
Open
loop gain is the gain of the op-amp without feedback. From the data
sheet one can know the specification of the open-loop gain associated
with the certain op-amp used.
In the data sheet, there were three ways one can find the open loop gain.
From
the large signal voltage gain the typical value of the open loop gain
is given as 100V/mV so that the number is not too large, so if we
change it to V/V we get 100/(10e-3)=100,000.
As for the voltage
gain graph, the wave depends on the power supply input. In our case +5V
were used so as highlighted that corresponds to ~100dB which is also
100k V/V.
Lastly the open loop frequency response is the gain depending on the frequency. The 100dB gain (100k V/V) falls under ~10Hz.
What is a good estimate for the offset voltage? For worst case design what value would you use?
Offset
voltage is the voltage that one needs to add to the input of the op-amp
so that the DC output of the op-amp is set to zero.
From
the data sheet the typical offset voltage is 2 mV, but a good estimate
for the offset voltage for the LM324 is 7 mV because it is not too high
as 9 mV and low as 5 mV so it is in between the maximum range. For
worst case design 9 mV would be used which is the maximum offset
voltage though not preferred.
In this experiment a inverting op-amp was built with a gain of -1V/V. The circuit has an DC offset of 2.5V, and decoupling capacitors. For the experiment a power supply was used to supply the Vcc of +5V, as well as power the op amp with V+
of 5V and V- of 0V. A function generator was used at the input with a
sine wave of 100mV amplitude and 1k Hz. The output was seen using a
oscilloscope.
AC coupling to show the wave swing.
DC
coupling showing offset
What is the common-mode voltage, VCM? Does VCM change? Why or why not?
The
common mode voltage is 2.5V, which is created by the two 10k ohms
resistors used and the 5V Vcc. When analyzing the circuit in DC the
capacitors act as an open, and from the circuit shown above, this
creates a voltage dividor that sets the VCM at a fixed 2.5V.
VCM={10k/(10k+10k)}*5=(1/2)*5=2.5V. The VCM does not change ideally
because the Vcc is a fixed DC source at 5V, and the two resistors as
well. But in reality the two resistors used are not exactly 10k ohms
and due to noise and other factors like temperature which can affect
the resistors can lead to a slight change in VCM.
What is the ideal closed-loop gain?
This
circuit represents an inverting op-amp configuration. The ideal
closed-loop gain of an inverting op-amp is -Rf/Ri, which in this case
is -5k/5k = -1.
What
is the output swing and what is it centered around? What happens if the
input isn't centered around around VCM, that is, 2.5 V? Provide a
detailed discussion illustrating that you understand what is going on.
The
positive output swing is 2.5V+100m = 2.5+0.1 = 2.6V, as for the
negative output swing it is 2.5V-100m = 2.5-0.1 = 2.4V. This can be
confirmed with the simulation done above.
The output swing is centered around the DC offset, which is 2.5V.
Op-amps
have an output offset, and without canceling the offset the output of
the op-amp can be saturated/clipped earlier than it should be. So as
mentioned earlier the offset voltage added to this certain op-amp was
2.5V so that the DC quiescent point is set to zero. This was done
because the sine wave input of the op-amp has a DC offset of 2.5V, so
when the VCM point is at the same potential the Ri resistor have a 0V
drop across it which results in the output being centered around 2.5V.
If the VCM was not set to be 2.5V, the output wave would be centered
around a different voltage, and since the op-amp has a V+ of 5V and a
V- of 0V, the output swing cannot exceed these values or it will clip
and not show a full wave. As seen in the hand calculation below, if the
voltage dividor was chosen so that VCM equals ~3.6V then the output
would be centered around 4.8V and with a swing of +/- 100mV the top
part of the signal will clip because it exceeds the limit of +5V.
Depending on the type of op-amp used this also matters because some
op-amps cannot go more than 90% of V+, and other op-amps like rail to
rail op-amps can get very close to 5V so without setting the right VCM
that can affect the output.
What is the maximum allowable input signal amplitude? Why?
The
maximum allowable input signal is limited by the V+ and V- which are
+5V and 0V. So to have a full wave signal output without clipping one
should stay in that range 0~5V. Ideally, since the offset is set to
2.5V this means that the maximum allowable signal is 5-2.5 = 2.5V and
0-2.5 = -2.5V. But in reality, as mentioned before not all op-amps can
get close to the V+ and V- values; therefore the swing can be saturated
though in calculation it could have given a result smaller than 5V or
larger than 0V. In the picture below, the positive peak is seen
to start saturating.
*Notice
the function generator has an offset of 1.19V instead of 2.5V. This is
due the impedence of the function generator that makes the offset twice
as much. So using the scope we decided on the offset that gave a very
close value to 2.5V.
What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
If
the gain is increased by 10 then the input signal is 1/10th of the
output. Thus, since ideally the maximum allowable signal amplitude is
+/-2.5V then 1/10th of that is +/-0.25V or +/-250mV. Doing this part of
the experiment caused the output positive peak of signal to start
to saturate as shown with the pictures below. The top is already
saturated and bottom peak is starting to saturate due to the gain
increase.
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 0.01 µF capacitors are used
as decoupling capacitors. The decoupling capacitors' purpose is to keep
the output signal stable and reduce the noise in the circuit. It is a
good habit to add decoupling capacitors from the node of the source to
ground in any circuit one makes so that the noise is reduced.
The values of the capacitors is not critical as they are only for decoupling and since the capacitors act as an open in DC and a short in AC it will not affect the circuit calculations.
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. What is the input
offset current? What does this term describe?
If the
resistors R1 and R2 are much larger than 10k ohms the input bias
current can change the VCM voltage which can change the DC quiescent
point and cause problems. This can be seen from doing a simple
calculation. If we ignore the Vcc for a minute and have both resistors
in parallel so that 10k || 10k = 5k ohms and since the 20nA is going
through these resistors to ground we get a voltage drop of 5k * 20n =
0.1 mV. This voltage drop is added to 2.5V so that VCM is now 2.6V and
we encounter the same problem disccused earlier about not having a VCM
of 2.5V. But with the 10k resistors the problem is not major. However,
having larger R1 & R2 in the MEG(10e6) or G(10e9) will give bigger
voltage that will be added to VCM and cause the output signal to
saturate since it will exceed the 0~5V limit.
Just as the
input offset voltage, there is an input offset current that is the
difference between the biasing currents of the op-amp terminals.
Biasing currents of the op-amp are the average of the inverting and
non-inverting terminal currents. In ideal cases since the op-amp should
have a very high input impedence there should be no current entering
the terminals and this problems is taken care of using the biasing
currents, and the biasing currents difference between the terminals is
the input offset current.To reduce the effects of the offset current one can match the impedences at the terminals so that the currents cancel and do not affect the output.
This experiment deals with the same circuit as in experiment #2 but this time the Ri and Rf are different as well as there is no AC component to the circuit, and instead of the AC input, there is VCM. Moreover, the gain is not -1 V/V now it is -20 V/V.
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.Measure the offset voltage of 4 different op-amps and compare them.
Since the inverting terminal is connected to VCM and not the AC source in series with a 1k ohms resistors, and the non-inverting terminal
is also connected to VCM. Assuming ideal op-amp, this will lead to no
current going through the 1k ohms resistor and will enable us to
measure the offset of the circuit. Due to the gain being 20V/V the
offset is amplified so when measuring the difference between the output
and VCM we divide by 20 to get the real offset value. This can be seen further in the calculation below.
When increasing Rf to 100k ohms this increases the gain to 100V/V which can help us see small offset values.
Four
different LM324 op-amps were used and the offset voltage of each was
measured. Each one had a different offset, and that is because each IC
chip is different than the other, and it is hard to manufacture the
exact same chip characteristics. The measurement was done using a
multimeter; the output was measured with one wire and the input from
the inverting side with the other wire, and the offset is shown in the
pictures below. Each offset measured was different from the other,
which shows the importance of conducting this test for each chip used
so that problems can be avoided.
*Notice to get the actual offset the value measured by the multimeter/100 gives the actual offset.
