Read the write-up seen below before coming to lab.
Lab
description
Understand how to read data sheets for op-amps
Understand how basic op-amp circuits work including max input signal and max output signal
Understand how and why different op-amps will behave differently
Lab Requirements
using the LM 324 op-amp build and answer the set of questions
Gathering information from datasheet Common mode voltages
From the data sheet, it can be seen that the
minimum VCM will be 0 Volts while the maximum VCM can be Vcc-1.5 which
is 3.5 Volts when a 5 Volt Vcc is used.
Estimate Open Loop Gain
Looking at the large signal voltage gain from the data sheet, we can
estimate the open loop gain for the op amp. The data sheet says that
the large signal voltage gain is typically 100 V/mV. This converts to about 100K
V/V.
The open circuit gain will depend on the frequency
of the circuit
beucase as frequency increases the gain of the op amp (accoridng
to the data sheet). Therefore it can be seen that the typical values
for large signal voltage gain in the chart is assuming a frequency of
around 10 Hz becuase 10^(100/20) ~ 100K V/V
Estimate for offset voltage A
good estimate for the offset voltage would be 2 mV becuase that is the
typical value as listed on the data sheet. For worst case design, one
would use the largest offset voltage possible which in this case would
be 9mV. The reason behind this is beucase the larger the offset
voltage, the larger difference between the 2 input terminals than
what is expected in calculations.
Experiment with scope set to AC
coupling removes offset
Scope set to
DC coupling, able to see offset
The circuit above is a
op-amp connected in the inverting typology with a gain of 1. Notice
that the purpose of VCM purposefully set to 2.5 volts so that there is
no current going through the resistors and the gain of DC remains 1
regardless of the values of the resistors while the AC part can
be amplified. This can be seen by changing the RF value to 10K. The DC
portion will still have a gain of 1 becuase there is no current flowing
through the resistors meaning no voltage drop. The AC portion on the
other hand will have a gain of -10k/5k =2. VCM is needed for this
circuit to work beucase if VCM is not there, the DC part will be
amplified and the circuit will easily be saturated (becuase VCC is 0
and 5 volts, this sets the absolute max swing of the output voltage
assuming ideal conditions). One of the purposes of this circuit is
amplify a signal that has an offset. The VCM value will have to be the
same value of the input offset to ensure that the DC offset will just
be passed through the circuit becuase of no DC current flow.
Questions
What is the common-mode voltage, VCM? Does VCM change? Why or why not?
The
value of VCM is physically set to 2.5 Volts beucase of the DC voltage
source and voltage divider. This value cannot change beucase the
voltage source is held to a constant 5 volts. To ensure that the value
of VCM will not change due to noice, the 2 .01 u capacitors will charge
based on how noise effects the circuit. Since the capacitors are so
small, they will charge and discharge quickly resulting in the 2.5 Volt
DC line to be stable regardless of noise.
What is the ideal closed-loop gain?
The
ideal closed-loop gain is 1 for DC voltages and -1 for AC voltages. The
reason for this is becuase when considering DC, there is no current
flowing resulting in a gain of 1 from Vin to Vout. When considering AC,
the formula for gain in an inverting op-amp is -Rf/Ri which in this
case turns out to be -1. Since the input is 100mv and the output is
-100mv, this causes a 180 degree phase shift and the output to lag the
input.
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
output voltage swing is still 100 mV since the AC gain is -1 and since
the DC gain is +1, the offset voltage is still at 2.5 volts. This means
that the signal is centered at 2.5 Volts with a 100mV swing. Since the
purpose of VCM is to match the input offset and make the DC gain 1, if
the value of VCM is changed, then the DC gain will be some other number
than 1 and this will depend on the resistances of Rf and Ri. Let us
assume that VCM is now 0 Volts and Rf is changed to 10k. Now when
calculating DC gain, the gain is -2 and the AC gain is also -2. This
would mean that the output signal will have an offset of -5 Volts with
an amplitude of -200mV. Since the -Vcc terminal of the op amp is
grounded, this will cause the whole signal to sauturate. Now if VCM was
set to the same value of the offset of the input (purposelly), the DC
gain will remain 1 while now the AC gain is -2. This will result in the
output having a 2.5 Volt offset and an amplitude of -200mV and none of
it will saturate. As menchioned above the purpose of VCM centered at
2.5 is so that the the op amp circuit will only amplify the AC part
since any significant amplification of the DC portion will lead to
saturation (especially since this is an inverting op-amp and the -Vcc
is tied to ground).
What is the maximum allowable input signal amplitude? Why?
Since
Vcc and -Vcc are tied to 5 Volts and 0 Volt respectively, the maximum
output swing of Vout can only be from 0-5 Volts. Therefore since the
offset of the output signal will always be 2.5 volts because of VCM,
the amplitude should can be maxed out to be from -2.5 to 2.5 volts
(negative voltage in AC just means a phase shift). Althought this
calculated voltage swing can only occur in ideal conditions. In the lab
when we tested our circuit, when applying an amplitude of 1.9 Volts,
the peaks start to clip. This is again beucase in a real circuit, the
output swing cannot physically reach Vcc, it can only get close.
Output signal (blue) clipping
Settings on function generator
What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
Again
since Vcc and -Vcc are set to 5 Volts and 0 Volts respectively, the
maximum output swing of Vout can only be from 0-5 volts. Since this
circuit has a gain of 10 and the offset is still 2.5 Volts, the
theoritical maximum input in ideal conditions would be 2.5(maximum area
to swing)/10(gain) which is 250mV. In the lab when increasing the input
amplitude to 210 mV, the output peaks started to clip.
Output signal (blue clipping)
Setting on function generator
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
purpose of the .01 uF capacitors from Vcc and VCM to ground is to act
as a coupling capacitor which means it acts as an AC short. Since it is
preferable for VCM to be as stright as possible, the coupling capacitor
(AC short) will short all the higher frequencies to ground (any changes
in the signal from noise). Another way to think of this is to think
that as every little change in voltage in the source occurs, the
capacitors will compenstae for that change and stirghetn out the
signal.
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?
With
a bias current of 20nA, if R1 and R2 are increased significantly, this
will change the value of VCM and therefore the circuit will no longer
be set to have no DC gain. If there is 20nA flowing into VCM, we can
apply superposition on just that portion of the circuit and see the
increase in VCM. With R1 and R2 set as 10k, the voltage contribution
from the bias current will be 20nA*(10K || 10K) which is .1 mv which is
relatively small. if R1 and R2 are much larger, for example 100MEG, the
contribution from the bias current will be 20nA*(10MEG || 10 MEG)
which is .1 Volt. This voltage contribution from the bias source is
added to the voltage contribution from VCC which is 2.5 V for both
cases. When R1 and R2 are 10MEG, this will make VCM 2.6 and that will
change the output offset voltage to 2.7 V (see bleow). This will become
a problem if the gain of the circuit is not set to -1 because in that
case, instead of just a change in offset, the circuit will mostlikly
saturate.
Output voltage
when VCM is 2.6V instead of 2.5
Using our example from
above the offset current is 0 becuae each the positive and negative
inputs have 20nA flowing out and offset current is defined as the
difference between the bias currents. In practical applications, the
current flowing out of the positive and negative input are not the same
and therfore it will introcduce an offset current. The bias current is
defined as the average of the 2 currents flowing out of each input.
Experiment 2
Circuit 2
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.
To
offset voltage of an op-amp is the measure of the difference between
the positive input voltage and the negative input voltage. Therefore,
this can be done by measuring the change in voltage from VCM to Vout
beucase these 2 values are expected to be the same. The reason why that
this circuit would be better to measrue offset voltage is becuase the
offset voltage may be so small that most multimeters cannot pick it up.
There this circuit has a gain of 20, this will amplify the offset
voltage so that the multimeter can pick it up. Dividing the offset
current measured form the circuit above will give the ture value of the
offset of the op amp. For our experiement, we set the value of Rf
to 100k to have a gain of 100 to gurantee that we will be able to read
a value off the multimeter. In that case, we had to divide the reading
off the multimeter by 100 to get the offset value of the op-amp.
Below are the measurement of 4 different op amp's offset current.
36.5mV/100=365uV
51.5mV/100=515uV
38.5mV/100=385uV
66.2mV/100=662uV
Notice that the values of the offset
voltages in different op-amps are slightly different. This is cuased by
the internal components of the op-amp such as matching of transistors.
This will not have much effect in circuits becuase usually, the offset
voltage is accounted for in calculations.
