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OP AMP Application Part 2 (Electronics 2) — Transcript

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0:00hi hello um for this video we're going
0:03to continue on the practical
0:05applications for
0:06operational amplifiers okay so from the
0:09previous videos
0:10we have discussed several applications
0:12such as um
0:14inverting non-inverting um unity gain
0:17and then we had our summing amplifier
0:20and then we had our differential
0:22amplifier
0:23okay so for this video we're going to
0:25discuss three more practical
0:27applications for operational amplifier
0:30okay so the first application
0:33is what we call an integrator okay so
0:36what does an integrator do
0:38so an integrator is used to provide an
0:42output voltage
0:43which is proportional to the integral of
0:46the input voltage
0:47okay so for an uh integrator circuit
0:50usually our input is a dc okay so
0:54our input is a dc level and its integral
0:58is a linearly increasing or decreasing
1:01ramp output
1:02okay so depending on the on our input
1:07so our input is a dc level and its
1:10integral is a linearly increasing
1:12ramp output okay so an integrator
1:15is similar to a scalar okay except that
1:18the feedback component
1:20is a capacitor instead of a resistor
1:23okay so how does uh how do we connect
1:27uh how do we connect our components
1:30in an operational amplifier for it to
1:33operate as an
1:34integrator okay so this is how
1:38a true integrator looks like okay so
1:41instead of
1:42a uh feedback capacit
1:46uh resistor here or rf we will be
1:49placing a capacitor
1:51instead okay so we still have here your
1:53r
1:54and we have here our input which is
1:56usually in
1:58dc okay so how do we uh
2:01get the output which is equivalent to 1
2:04all over
2:04rc negative 1 all over rc
2:08integral of v i or the integral of your
2:11input dt
2:12okay so usually um we have here a
2:16negative sign
2:17because if you're going to look at your
2:19uh
2:20input it is directly connected or
2:23applied
2:24at the inverting terminal of our
2:26operational amplifier thus we have there
2:28your
2:29negative sign okay so how do we
2:34how do we
2:37derive the equation for our integrator
2:41circuit
2:42okay so as you can see
2:47our input there okay or our output our
2:51input current is equivalent to your i in
2:54okay so that is equivalent to your
2:58i in or
3:01your i input there is called your ict
3:05okay so that is your ict
3:09okay so that ict
3:12uh current here okay so our input
3:15current here is the same as the output
3:18that will flow across our capacitor okay
3:21why so if you can still remember our
3:23um uh virtual grounding okay
3:27so we said uh that there is no current
3:30that will enter our operational
3:32amplifier so
3:33to do that that will be your i your
3:36current entering or flowing across your
3:38r here is the same current that will
3:40flow across your capacitor there okay so
3:43therefore those two currents are just
3:47equal okay so we have to recall
3:51okay from your um circuit analysis
3:54lecture okay so what is the voltage
3:58relationship
4:02for a capacitor so what
4:05is the voltage relationship for a
4:10capacitor okay so the voltage
4:12relationship for a capacitor
4:14or um the current okay so the current
4:18how do we solve for the current
4:20in a capacitor so that is equivalent to
4:23the capacitance
4:24okay multiplied by the
4:28okay so the derivative of your vc
4:32or your voltage with respect to time
4:35okay so for for current to flow across
4:38your capacitor
4:39there should be a change in voltage with
4:42respect to
4:43change in time okay so you can you can
4:46uh
4:47you can remember this from your notes in
4:49your circuit analysis uh
4:51lecture okay so getting now the values
4:54for your input
4:56and our output um current which happens
4:59to be
5:00equal okay so by nodal analysis we know
5:04that your v i here the input voltage
5:07okay minus from our 0.2 let us call that
5:100.2 equivalent to v2
5:13divided by our r is equal to
5:16okay so the capacitance multiplied by
5:21the derivative of our uh so
5:24plus or minus so that this is our v2
5:27okay minus it is connected at our output
5:30so that is equivalent to vot
5:34and then we have here your dt okay
5:37so uh recalling again
5:40our um
5:42[Music]
5:43our virtual grounding okay so we also
5:47said that your v1 is equal to v2 or
5:50uh there is no voltage that will uh
5:53that will enter your operational
5:55amplifier
5:56so therefore in this formula okay our v2
5:59is equal to zero so our v2 here is
6:03equivalent to
6:04zero okay so we have here your vi
6:08multi uh your vi with respect to time
6:12divided by our r is equal to c okay so
6:15we have here a negative vot so let us
6:18just place the negative sign here
6:20dvo or dv out all over
6:23dt okay so to to remove the derivative
6:27there okay so to remove our d there we
6:30have to integrate
6:31both sides of our equations okay so
6:35we need to integrate both sides of our
6:38equations
6:42okay so from zero to t
6:46so from zero to t so we have here your
6:48dt we have here your
6:50dt okay so what happens now is
6:54this will be cancelled out okay so
6:57that will give us your output voltage
7:02so that will give us your output voltage
7:04since r
7:05is a constant okay so since r is a
7:08constant we
7:09we bring your uh capacitance here so we
7:12have
7:12minus one all over rc okay
7:16and then we have the integral of um
7:20vi the integral of your input
7:23okay dt from zero to
7:26t so this is the output voltage
7:30if we have an integrator so if we
7:33place a capacitor in our feedback
7:37instead of a feedback resistor
7:40okay so that is how we derived
7:44our equation
7:47so we have here what do we mean by true
7:49integrator okay so what's the problem
7:52with a
7:52true integrator circuit
7:55so we have here um difficulties with
7:59true integrator circuit number one
8:01operational amplifiers used in true
8:03integrator circuits
8:05must have extremely low dc offset
8:07voltage and bias currents
8:10okay however general purpose op amps are
8:13virtually unusable
8:14for true integration okay so but we can
8:18have
8:19one special type of op amp it is called
8:21a chopper stabilized amplifier
8:23which uses a mechanical switching
8:26process to correct continually
8:28for offset and bias effects okay so
8:31to to eliminate these difficulties
8:35okay what we do is we have an ac
8:39integrator instead okay so uh
8:42instead of just having so it's like the
8:44true integrator
8:45is like a theoretical integrator circuit
8:49but in practice in
8:50reality what we are using is an ac
8:53integrator
8:54circuit so what's the difference the
8:57only difference is that
8:58we are placing an rf or a feedback
9:01resistor
9:02okay uh which is in parallel with your
9:07capacitor
9:08okay because actually up if we don't
9:11have
9:11uh on rf here and you're really
9:14operating in dc condition
9:17okay so at a very very low frequency
9:20what happens to your capacitor at a very
9:22very low
9:23frequency so what happens to your
9:25capacitor is it becomes an open circuit
9:27so this is
9:28just like an infinite circuit so without
9:31this
9:31rf here what happens is that you're
9:34you're just having a
9:36an open loop okay an open loop
9:39connection
9:40okay so that open loop connection is no
9:42more an integrator circuit
9:44so what we do is we place an rf here so
9:47that
9:48when you see here opens okay
9:51during dc we have a
9:55finite value instead of an infinite
9:58which is equivalent to that open
10:00circuit okay so we use an rf in parallel
10:04with the capacitor
10:06to drop the gain to a finite value at
10:10at dc because at dc and without this rf
10:13here
10:14we just have an open loop
10:18connection okay
10:21so uh we have here the brake frequency
10:24so
10:25we need the brake frequency so that we
10:27know when is the
10:28what frequency is your capacitor starts
10:32to operate okay so we know in your
10:34previous discussions that
10:36your capacitor has something to do with
10:38the frequency of your
10:39circuit okay so at a certain frequency
10:42your capacitor will start to short out
10:46uh
10:46to to become a short circuit so without
10:49reaching that specific frequency
10:51your capacitor is just acting like an
10:53open circuit
10:54so what frequency is that so we call
10:56that as your
10:58break frequency okay so uh your break
11:01frequency is equivalent to 1 all over 2
11:04pi
11:04rf multiplied by your capacitance
11:07or that is just equivalent to the the
11:10reciprocal of your time constant
11:14or your period which is your tf there
11:16okay
11:17so when your brake frequency is much
11:19much
11:20greater than the input frequency of your
11:24input signal you have a uh
11:28gain okay so you have a gain which is uh
11:32uh approximately equal to negative rf
11:35all over ri so that means
11:38uh our capacitor or our
11:42integrator is just functioning as a
11:44normal
11:45okay so it is functioning as a normal
11:50uh normal inverting amplifier okay so
11:55however when you're
11:56when you're break frequent when your
11:58input frequency is greater than your
12:00break frequency so that is the time
12:03when your um operational amplifier is
12:06functioning as an
12:08integrator circuit okay and for
12:10simplicity
12:11we we use a other books other references
12:14they use a prime here okay so when you
12:17see a prime here so that means we are
12:20using an ac integrator instead of a true
12:24integrator circuit okay
12:27so this uh are this is the simple
12:31design procedure for a for an ac
12:34integrator circuit okay so
12:36um number one we're going to assume that
12:39the desired processing function is
12:41so vot prime okay so that means we have
12:45a feedback resistor in parallel with
12:47your capacitance okay
12:49it's equivalent to negative one all over
12:51r i c so your r here
12:53is the input resistance okay
12:57um integral of your v input dt
13:00okay so the first step is you're going
13:02to select a fixed value of capacitor
13:05okay so uh we are going to make use of
13:09high quality stable capacitors made of
13:12polyester in or
13:13mylar so these are the capacitors with
13:16no polarities okay so we are going to
13:19make use here of a non-electrolytic
13:22capacitors okay so uh the first
13:25uh thing to do is you have to choose
13:27okay so as a designer
13:29uh you can choose any value of capacitor
13:32that you want
13:33okay so um after choosing the value of
13:36your capacitor
13:37we're going to solve for the for the
13:40resistor
13:41resistor ri okay using this uh
13:44scalar factor here so
13:48since k is equivalent to uh r1
13:51oh multiplied by c so ri is equivalent
13:54to 1 all over
13:56the scale factor k multiplied by your
14:00chosen capacitor okay so um
14:04you can also choose ri so you can also
14:07choose ri and then
14:08finally you can solve for c however um
14:12in practical circuit components
14:15availability
14:17there are wider ranges of available
14:19resistors compared to capacitors so
14:22there are only few
14:23available um
14:26commercially used values for capacitors
14:29so
14:30if you choose the value of r and our
14:33resistors have tolerances you can always
14:35choose
14:36those values that are that is closest to
14:39your
14:40actual um
14:43actual solved value here however for
14:45capacitor if we choose
14:47the exact resistance we might have some
14:50difficulties in choosing the
14:52the nearest value for the capacitance so
14:55in practice
14:55we choose the capacitor and then we just
14:58solve for r
14:59and then we choose the the the
15:03closest value for an available
15:07commercially available value for a
15:09resistor
15:10okay so we now have the two important
15:13components for your um
15:15ac integrator circuit you have now your
15:17ri and we have your c so what is lacking
15:20is your feedback
15:21resistor okay so there are two ways for
15:24you to solve for ri
15:26so first is using the time constant
15:29which is equivalent to
15:30rf their time constant tf divided by
15:33your chosen c
15:36where in your tf is equivalent to 10
15:38times your uh
15:40period or time cons at uh your period
15:43okay so
15:44lowest period so that is your tl so
15:47since we are
15:49uh afraid that uh the at a lower
15:52frequency range your capacitor is not
15:54yet
15:55turned on or it is not yet um shorted
15:58so we use the lowest frequency here okay
16:01so
16:01we're in your tl is equivalent to one
16:04all over
16:04fl where in fl is your lowest frequency
16:08the lowest input frequency of your input
16:11signal
16:12okay so you can use the time constant or
16:15also you can use the formula for the
16:17break frequency
16:18and the formula for break frequency is
16:20equivalent to 1 all over 2 pi
16:23fb multiplied by c where in your fb
16:26or break frequency is equal to 0.01 of
16:29your lowest frequency okay so since we
16:33have here two formulas okay
16:35um do not expect that they will yield
16:38the same result okay so actually
16:40they will have a different value
16:43okay so uh but you can use any of those
16:47two values or
16:48if one value is an exact available
16:50resistor and the other one is not yet
16:53exact so you may choose the
16:54exact one or if the if one value is very
16:57very high or very very low
16:59so you have to choose the more um
17:02the more practical value among the two
17:06okay so if either or both of the
17:09resistance values are either excessively
17:11large or small
17:12then we we choose another value for
17:14their capacitor and then we start again
17:17with the with the procedure
17:18okay so if uh if your resistor
17:22uh computed resistance is too large then
17:25you choose
17:26or increase the value of your capacitor
17:28and vice versa so if it is too low
17:30so you decrease the value of your chosen
17:33capacitor okay so
17:35um to summarize all you have to do
17:38is to select or to design or to
17:42get the value for your three components
17:45so your three components are the
17:47the ri your c and your um
17:50and your rf actually you just have to
17:53solve for the two resistors because you
17:55can choose the value for your
17:57capacitor okay so a practical
18:00application
18:02for um practical application
18:06for integrator is it is some it is used
18:09for
18:10wave shaping applications okay so
18:13we have here an example we're in we get
18:16the value for the
18:17output peak to peak voltage and the
18:19output peak to peak voltage is
18:20equivalent to k
18:22actually your k is your one all over rc
18:25okay so that is your scalar okay so um
18:30before so the scalar for
18:35for the inverting and then inverting is
18:37just your rf all over ri or rf all over
18:40ri plus one so
18:41our scalar here for an integrator is
18:43your negative one all over rc although
18:45we we don't need to place the negative
18:47sign there okay
18:49so we need to compute okay so for the
18:52net area so we have your
18:54absolute value sign here so that means
18:57we we do not place the negative value we
18:59just get the amplitude
19:01of the of the net area having the same
19:04signal okay so what do we mean by this
19:07formula
19:08okay so we have here an example so it
19:11says here
19:12um the square wave below is applied to
19:15the input of an ac
19:16integrator with v o t okay with v
19:19output voltage equivalent to negative
19:22one thousand so this is this negative
19:241000 is your k this is your scalar
19:27okay so after steady state conditions
19:30are established determine the peak to
19:32peak
19:32output voltage and we have to plot it
19:35okay so what we need to remember here
19:43is your output voltage okay so what is
19:46our peak to peak output voltage
19:48so your peak to peak output voltage here
20:00so your peak to peak output voltage here
20:02is equivalent to k
20:04multiplied by the absolute value of the
20:07area
20:08okay net area okay so if you're going to
20:11look at the problem
20:13it's like during positive half you have
20:16your
20:17your voltage as your so it's it's like a
20:21rectangle
20:22okay so where is your amplitude is your
20:25um height or length
20:29and uh and the time
20:32okay or the period is your weed okay
20:36so having um having that shape it's like
20:40our area
20:41is in the form of a rectangle and the
20:42rectum and the area for our rectangle is
20:45just
20:45length times sweet so in this case it is
20:48the amplitude
20:49which is 10 volts and the weed which is
20:52your
20:54uh and the weed which is equivalent to
20:57your time or your period
21:03that is equivalent to 0.5 millisecond
21:07okay so in this case class it is k
21:10so your k is 1 000 no need to place your
21:13your negative sign there and then your
21:16area
21:17is like 10 volts multiplied by 0.5
21:21milli second okay so
21:25if you're going to recall what is our
21:27scalar
21:28okay so what is our scalar for our
21:32um integrator this is equivalent to one
21:35all over rsc
21:36and we know that rc is time constant
21:39okay so our unit there is what is a time
21:42constant so that is equivalent to
21:44seconds
21:45so therefore your millisecond here will
21:48be cancelled with your
21:50scalar here so the output is just
21:53equivalent to
21:54volts okay so your output voltage is of
21:57course in terms of
21:58voltage so computing for this
22:01our output peak to peak voltage is
22:04equivalent to
22:081 10 multiplied by
22:110.5 millisecond okay
22:15multiplied by the scalar which is
22:181000 and that is equivalent to 5
22:21volts okay so that is your peak to peak
22:24value
22:25so how are we going to sketch our
22:28graph or how are we going to plot our
22:31graph so we said
22:32from our first slide that for an
22:34integrator
22:35with a dc input you have a ramp as your
22:39output
22:40okay so if our if our
22:43input for example is like this so this
22:45is a negative this is 10 volts this is a
22:48negative going ramp
22:50so therefore our ramp is going down
22:53so for example our input is like that so
22:56this is your
22:57input here this is your 10 volts like a
22:59dc
23:00so this is up or up going okay or
23:03positive going
23:04your ramp will be like this one okay so
23:08combining now the two so that we can
23:10plot okay so an integrator
23:12dc as your input and ramp as your output
23:15so this
23:15is just like a dc because for the entire
23:19uh for the entire um duration
23:22of your cycle it is what it is
23:25equivalent to 10 volts
23:27for the entire duration of your cycle
23:29this is equivalent to negative 10 volts
23:31okay so if you're going to plot that now
23:34your
23:35output will look like this okay so this
23:38is your
23:39output looks like so for positive
23:4310 volts so this is up going down so we
23:46have a negative going around here and
23:48this is
23:48up down going up so we have a positive
23:51going
23:51ramp here and our output is 5 volts
23:55peak to peak so we have here positive
23:572.5
23:58to negat from 2.5 to negative 2.5 so
24:02that we can have a peak to peak value of
24:055 volts okay so that is one
24:08uh practical application of an
24:11integrator so why is it called a wave
24:13shaper
24:14because we change the the the
24:17wave or the shape of your waveform okay
24:20so from a square wave
24:22or a rectangular wave as our input
24:25we have a uh triangular waveform in our
24:30output so we reshape your input signal
24:33okay so if we want some applications
24:36that that looks like this
24:37therefore we can use an integrator
24:40circuit
24:41okay so aside from an integrator circuit
24:45your up can't always can also operate as
24:48a differentiator
24:49okay now what does a differentiator do
24:52okay so your differentiator
24:55um is used to provide an output voltage
24:57which is proportional to the rate of the
25:00change of the input voltage
25:01so actually your differentiator is just
25:04like the
25:04the reverse of your um integrator
25:07circuit okay so
25:08this is the inverse mathematical
25:10operation to that of an integrator
25:12so in this case we have your ramp as an
25:15input
25:16okay or a chain you have a changing
25:18input and then upon passing
25:21and then upon passing through a
25:22differentiator circuit you will have an
25:25uh a dc at the output okay
25:28so it's the the inverse mathematical
25:31operation for
25:32an integrator so the output voltage is
25:35proportional
25:36to the re the read to the derivative of
25:39the input voltage
25:40the constant of proportionality or your
25:43scale factor which is equivalent to
25:45negative rc
25:46okay so how does this look like okay so
25:49actually for a differentiator we're just
25:51going to change
25:53the we're just going to change the
25:58we're just going to change the
26:03position of the capacitor so instead of
26:06placing the capacitor
26:08in the feedback path between the input
26:10and the output
26:11we will be placing your capacitor at the
26:13input terminal so again the input is
26:16still applied at the inverting terminal
26:18therefore we still have a
26:20negative sign in our output formula
26:23okay so how do we derive
26:28how we how do we derive for the output
26:30voltage for
26:31a true differentiator circuit okay so
26:34we have here
26:38again so from the circuit um
26:42your your ict again is the same
26:45current that will pass through across
26:47your resistor
26:49okay so for our
26:52input voltage okay so for our input
26:55voltage
26:56and that is a capacitor so your input
26:59voltage here
27:04is equivalent to what or sorry your
27:07current
27:08your input current here okay is
27:11equivalent to c
27:13multiplied by the change in your input
27:15voltage
27:16divided by your change in time okay
27:20and that is equal so that is also equal
27:22to the output current which is also your
27:25ict
27:26and since that is just a resistor okay
27:29so this is just iris
27:30store but your resistor is connected um
27:34in a reverse polarity so this is plus
27:36minus while this is
27:38a plus minus so we have your plus and
27:40minus so we have a
27:41negative at our input so they don't have
27:44the same polarity
27:45what is your current there so your
27:47current there is your
27:49v o t or v multiplied by your
27:53um v divided by
27:58r okay so solving now for your output
28:03okay so solving now for your output we
28:05can um place your
28:06r here okay so we can um
28:10cross multiply your i there so we have
28:12your minus rc
28:13multiplied by dvi all over
28:16dt so this is our formula this is our
28:20output formula for our true
28:23differentiator circuit so it says here
28:27that
28:27you should have a change in your input
28:30so that we can have your
28:31output here because when we don't have
28:34change in our
28:35input our output is automatically equal
28:38to
28:39zero okay so there must be a
28:43change or it your input must not be
28:46equivalent to a dc source so the
28:49amplitude response is very small so the
28:52problem with true differentiator is that
28:54the ampli the amplitude response is very
28:57small at low frequencies
28:58okay but it increases linearly as the
29:02frequency
29:03increases okay so we have here an
29:06example we have here a true
29:08differentiator um
29:09circuit produces an output voltage
29:11equivalent to
29:13dvi dt so that means our k there is just
29:16equal to one
29:17so our rc there is just equal to one
29:19okay
29:20so where vi is the input voltage so this
29:23is our input voltage
29:25so the signal voltage at the left is
29:27applied as the input
29:29to ah to an uh a differentiator circuit
29:33here so let us sketch the output
29:35and label significant up levels okay so
29:38um how are we going to compute for this
29:41so how are we going to analyze this
29:43actually this is just very simple okay
29:47um we have here we will be
29:50we will be analyzing each uh each input
29:54okay in each inputs that we have so for
29:56example
29:58let us analyze our input from
30:02so it represents change in change in
30:05voltage which
30:07with respect to change in time um
30:10somewhat
30:12is like what so it is just like a slope
30:17okay so for example we have from
30:32so we have for example during the time
30:40during the time from 0 to
30:430.5 seconds okay so what happens during
30:48the time
30:48um from 0 to 0.5 seconds so we have your
30:52change in voltage
30:53which with respect to change in time
30:56okay so just like a slope
30:58okay so you have your y and x so your
31:01slope is
31:02from from y2 minus y1
31:05and then your x is x2 minus x1 so that
31:09will give you
31:10six minus zero okay so from point
31:13from from zero to point five that is
31:15equivalent to point five
31:16minus zero so this is your
31:20output voltage okay so change in vi with
31:24respect to change in time
31:25so six divided by 0.5 that will give you
31:2912 volts okay now how about from
31:340.5 okay to 4.5
31:39okay so your output voltage from 0.5 to
31:424.5 is as you can see
31:44there isn't any change from our
31:47output from our voltage okay so from
31:50this point
31:55so from this point this is 6 and from
31:58this point this is also 6
32:00so that will give us 6 minus 6 divided
32:03by we have here 4.5
32:05minus 0.5 but this is equal to zero so
32:08automatically
32:10from from um from 4.5 seconds
32:13from 0.5 second to 4.5 seconds we don't
32:16have
32:16an output voltage y because we don't
32:19have a change in our input
32:21so for a differentiator circuit you only
32:24have an
32:25output once you have and a change in
32:27your
32:28input voltage okay so let's continue
32:33from time okay so from time
32:38equivalent to 5 seconds from
32:414.5 seconds to five
32:45seconds okay so what is your voltage
32:48there so your voltage there is from
32:50zero so your output there is your change
32:52in voltage is from zero to six
32:55okay so 0 minus 6 divided by 5 minus 4.5
33:00so in this case we have negative 6 all
33:02over 0.5
33:04so this is equivalent to
33:08um
33:11negative 12 volts per second so we have
33:14here your
33:15um so the unit is um since this is
33:19change in
33:20voltage with respect to change in time
33:22so that is voltage divided by second
33:24so these are the scenarios for our three
33:28different
33:28inputs so this is your first input that
33:31is a
33:32ramp okay that is a positive going ramp
33:36our second um case is that we have a
33:40dc as our input so therefore there is no
33:42change in our input and finally
33:44we have a negative going ramp as in our
33:47output
33:48so we have a negative here and we have a
33:50positive there okay per second
33:52so what do we have there how do we draw
33:55this okay how do we sketch this
33:56so this is now your answer so from this
34:04okay so from from from zero to six okay
34:07or from zero to point five we have here
34:09your 12
34:11volts a sure output and then we don't
34:13have your
34:14output from 0.5 to 4.5 and then finally
34:17we have a negative 12
34:19output for a uh from 4.5 to 5 seconds
34:23so again we have a an
34:26in a differentiator circuit okay so when
34:29you have a ramp as
34:30your input you have a dc as azure output
34:34but if your input is um but if your
34:37input is
34:38uh dc there is no change in input with
34:41respect to time so your output is always
34:43equal to
34:44zero so that is the application for a
34:47differentiator
34:48circuit okay so the differentiator
34:52circuit can be used to detect sudden
34:54changes in the signal and to generate
34:57trigger voltage based on these signals
34:59okay as the transition becomes sharper
35:02the magnitude of the output voltage
35:04also increases however class the problem
35:07with a differentiator circuit
35:09is that um true differentiator circuits
35:13are not used often
35:14why because um it is uh the noise
35:18tends to have a sudden abrupt changes so
35:21what is the
35:22how do we how do we represent noise
35:25so noise is represented as spikes
35:28changing spikes like this one
35:30okay so that means since noise is
35:34changing
35:35uh your true differentiator circuit
35:37might
35:38might be confused that your noise is
35:42your
35:42actual input signal because the input
35:45signal for
35:46the signal for noise is changing
35:49so for a true differentiator circuit
35:52noise is accentuated
35:54okay so thus a true differentiator
35:56circuit is very noisy
35:58okay so we don't want that type of
36:01amplifier circuit so what do we do
36:03so this difficulty may be alleviated by
36:06using a low
36:07frequency differentiator instead so how
36:10do we
36:11how do we uh how do we
36:15how does a low frequency differentiate
36:18or differs
36:19from a uh difference from
36:23a true differentiator circuit so we just
36:26have to
36:27add okay so earlier for the integrator
36:30circuit we add a parallel resistor here
36:33so no use
36:34so we just place a uh capacitor here
36:38okay so this is our solution for a low
36:41frequency differentiator circuit
36:43so that our noise will not be
36:46accentuated okay so
36:51the break frequency is equivalent to 1
36:54all over 2 pi ri
36:55multiplied by c so this is now your your
36:58scalar ri times c
37:00okay so when your input frequency is
37:03greater than your
37:04your break frequency your capacitor is
37:07just like acting
37:09or is acting just like a uh
37:12a short circuit no so you your formula
37:16is just
37:16or your your gain is just like a normal
37:20inverting amplifier but when you're when
37:23your frequency is much
37:24much greater than your uh um break
37:27frequency
37:28your gain becomes um 2 pi so
37:31omega is 2 pi 2 pi f okay
37:35multiplied by rf multiplied by c
37:38okay so the design is just like what we
37:41had in our integrator circuit
37:44so we're going to assume that the
37:45desired processing function is specified
37:48as vot
37:49so this time our k is rc previously our
37:52k is 1 all over rc
37:54okay change in vit dt so
37:57let h this time we are talking about the
37:59higher
38:00or the highest time period or the
38:04highest frequency so for integrator we
38:06are considering the lowest frequency
38:08so for a differentiator we are
38:09considering the highest frequency
38:12okay so this is the reciprocal which
38:14represents the
38:15the period of the highest frequency
38:17component so
38:18first again is we select the value for a
38:21capacitor so for the same reason as what
38:23we did in our integrator
38:25because there are more available
38:27resistors
38:28compared to capacitors so we just choose
38:31the value of your
38:32capacitor okay so you have to select a
38:35fixed value of c
38:37um do not use an electrolytic capacitor
38:39so electrolytic capacitors these are the
38:42capacitors which has
38:44polarity so we we use those with no
38:47polarities or the non-electrolytic
38:49capacitors okay unless the voltage
38:51across the capacitor
38:53always has one polarity so from the
38:55desired value of k so this time we can
38:58compute for rf okay
39:00rf is equivalent to k all over c
39:03so after solving for rf we have to
39:06determine your ri so this time
39:09our ri is i can be computed in two types
39:15so in time constant criterion so ri is
39:18equivalent to 0.01 of your th all over c
39:21or that is equivalent to 0.01 of f h all
39:24over c
39:26or the break frequency criterion which
39:28is equal
39:29equivalent to 1 all over 2 pi fbc or
39:32so since this is higher so that is 1
39:34over 2 pi
39:3610 f h multiplied by c okay so if
39:39either or both of the resistance values
39:42are excessively large or small
39:44we change the value of capacitor and
39:46then repeat the entire design okay so if
39:49the rs are too large
39:50increase the value of your c and vice
39:53versa
39:55so this is an example okay so a low
39:58frequency differentiator is desired for
40:00a particular application to perform the
40:02operation
40:03so when you say design okay um
40:06how do we decide so we designed this in
40:08such a way that you have
40:10to supply the values for our three
40:13components so what are those three
40:15components so you have your
40:16capacitor of course you have your rf of
40:19course and then you have your ri
40:21okay so the first thing to do okay so
40:27the first thing to do is we choose okay
40:29the value
40:30for our capacitor
40:33okay so um you have to
40:41you have to uh
40:44there are several um
40:48sites where you can have your
40:52commercially available values for
40:54capacitors or
40:55you can even look at your multi-seam or
40:57even
40:58uh on in those values that we have used
41:01in our
41:02laboratory classes previously okay so
41:05for example
41:06i will be choosing the value for my
41:08capacitor so this is just your
41:10this is just your choice okay so you you
41:13can choose any
41:14as long as it is a commercially uh of a
41:17commercially available
41:18value for your capacitor okay so for
41:21example i will be choosing
41:23the value of a capacitor electrolytic
41:25capacitor okay
41:27so that is for example i will be
41:28choosing one point one micro farad okay
41:31so there is
41:32point zero one micro farad there is
41:34point zero zero one micro farad
41:36or there is a point uh zero two micro
41:38farad point zero zero two micro farad so
41:41those are the different um widely used
41:45values for capacitors
41:46okay so for example i will be choosing
41:480.1 micro farad
41:50so the first thing to do is i will be
41:52solving rf
41:53okay so rf is equivalent to
41:56um my capacitor divided by
42:00ah sorry um this
42:04the scale factor k divided by my chosen
42:07value for
42:08capacitor okay so it says in the problem
42:11that your k
42:12is equivalent to 0.001 so that is your
42:16scalar
42:16um no need to place the negative sign so
42:20the negative sign there just signifies
42:22that your input is applied at the
42:24inverting
42:25terminal okay divided by 0.1 microfarads
42:29where this the where does the 1.1
42:31microfarad came from so
42:32i just chose that okay so this will give
42:36me a value
42:37okay solving for that so that is .001
42:43by 0.1 micro
42:48farad so that will give you 10 kilo ohms
42:51okay so um is there a commercially
42:56available resistor having a value of 10k
42:58so of course there is so
43:00therefore we can use rf be equal to
43:03um 10 kilo ohms okay then we choose ri
43:06so we have two criterions in choosing
43:09your ri
43:10so the first is using the time constant
43:13criterion
43:15so our time constant criterion so th
43:18your highest period is equivalent to 1
43:20all over fh
43:22so it says in the problem that um
43:25uh the highest frequency there is one
43:28kilohertz
43:29okay so therefore this is one all over
43:32one k
43:33so that will give you one milli second
43:36okay so in this case
43:39solving for ri that will give you 0.01
43:43multiplied by 1 millisecond so 0.01
43:49multiplied by 1 millisecond based on our
43:52formula
43:53divided by our capacitance which is 0.1
43:57micro farad
43:58okay so solving for this you have
44:03.01 multiplied by
44:06one melee divide by
44:100.1 micro so that will give you
44:14100 ohms okay so s 100 ohms available
44:18of course so if 10k is available so as
44:21100 ohms but
44:22we we still have our second criterion so
44:25our second criterion is based on our
44:27break frequency
44:29okay and your ri based on your break
44:32frequency is equivalent to 1 all over 2
44:34pi
44:36multiplied by 10 times your highest
44:38frequency
44:39and that is equivalent to 1 kilohertz
44:42okay
44:44multiplied by our capacitor capacitance
44:47which is 1.1 micro farad okay so solving
44:50for this
44:51we have
44:542 times pi times 10
44:58times 1k times 0.1
45:01micro okay then we get
45:04the inverse of that that will give you
45:08159.15 ohms
45:09actually the two values are not very
45:11very far
45:12from each other but since we're not yet
45:15um
45:16sure if there is an exact value of 159
45:20although there is a close value because
45:21there is a tolerance
45:23when it comes to resistance we we might
45:26as well choose the safer one which is
45:28100
45:29ohms okay so therefore
45:32your your ri is 100 ohms our capacitor
45:36is c
45:37is 0.1 micro farad and finally our rf is
45:4010 kilo ohms so
45:42when you say design all you have to do
45:43is to uh
45:45draw your differentiator circuit
45:48using this values okay so we have here
45:51your answer
45:53so this is your answer
45:58okay so we have there your
46:01differentiator circuit okay then lastly
46:05your operational amplifiers can also
46:08uh be used as a
46:12comparator okay so when we say
46:14comparator
46:15this from the from the term itself it
46:18can compare
46:19so it compares a signal voltage
46:24okay it compares a signal voltage on um
46:27one input of an operational amplifier
46:31with a known voltage called a reference
46:33voltage on the other
46:34input okay so it is nothing more than an
46:37open loop op amp with two analog
46:39inputs and a digital output okay so you
46:42will be inputting an
46:44analog uh changing value for your
46:46voltage input
46:48and then you have a digital for your
46:50output okay so
46:51the output may be positive or negative
46:54saturation voltage
46:55depending on which input is larger okay
46:58so
46:58how does uh
47:03how does a how does a comparator
47:07looks like so it says in our definition
47:10it is just
47:11like a what it is just like
47:14a um
47:18open loop okay so how does an open loop
47:21circuit looks like so it is the one
47:24having no
47:25feedback okay so when you have an input
47:28you don't have
47:29an an output so we have here for example
47:32minus n plus
47:33so this is the typical value we usually
47:36place for example a reference voltage
47:39here
47:39okay so let's say we have a reference
47:42voltage which is equivalent to 2 volts
47:45and then we place your input here
47:49so this is your input voltage okay so we
47:52have here for example
47:53positive 10 volts and we have here
47:55negative 10 volts and we have here your
47:57output
47:58voltage so uh you can have uh
48:01an input which is analog or changing and
48:04at the output
48:05you just have two values it's either 10
48:07volts or just
48:08a negative 10 volts so when your input
48:12is greater than 2 volts okay so you have
48:15an
48:15output which is equivalent to positive
48:1710 volts
48:18but when your input is lesser than 2
48:20volts you have an
48:21output which is equal to negative 10
48:24volts so you just have two
48:26values for your output so you can have
48:28several input
48:30but it will compare uh the comparator
48:33circuit will compare it with your
48:34reference voltage which is 2 volts
48:36and then we'll give you an output just
48:39the 10 volts if your input is greater
48:42than or negative 10 volts if your grid
48:44input
48:44is less than okay so you can also make
48:48use of
48:49zero as your reference okay so when when
48:52we
48:52use zero as your reference it's just
48:55like you have your minus and plus
48:57so we place your uh your negative uh
49:00your inverting terminal grounded so that
49:03means what do we mean by zero
49:05by ground then so it is zero okay and
49:07then we place your
49:08input here we place your input here
49:12and then you have your positive v here
49:14and your negative
49:15v here so what does that mean
49:18all values or all input that is greater
49:22than zero
49:23okay so uh all inputs that is greater
49:26than zero
49:27we have an output that is equivalent to
49:29positive 10 and all inputs that is
49:32uh less than zero you have an output
49:35which is equal to negative volts
49:37so you have an ac
49:42you have an ac as your input so for
49:45example this
49:46are the values which are greater than
49:47zero so your output is equivalent to
49:51nega positive v so these are the values
49:54which are less than zero
49:56so therefore you output is negative
49:59v okay so you have an analog signal for
50:01your input and you have a digital signal
50:04for your
50:06for your output okay so that is how
50:11an operator do okay or functions
50:15so can can you use uh
50:18these two inputs both uh these these two
50:22terminals
50:23just for inputs okay so can you use
50:25these two terminals just as an input
50:27so yes we also have this
50:31type of connection okay so we have
50:34here um if we place uh
50:37here your negative input and when we
50:40place here your positive input okay so
50:43we still have your positive v here
50:46and you still have your negative v here
50:48so what will be your
50:49output oh hey so um
50:52it depends okay so when your
50:56negative input here is uh has a greater
50:59value or a greater amplitude compared to
51:01this one
51:02so we have an output which is equivalent
51:04to negative
51:05but if it is of this uh of
51:08uh but if your non-inverting input is of
51:11higher magnitude compared to your
51:13inverting input
51:15therefore you have a positive assure
51:17output
51:18okay so that is how a comparator
51:21does okay so usually these are this is
51:23being used in circuits where in
51:26you have a reference voltage and then
51:28when your input reaches this
51:30specific voltage you have an actuator
51:33you have a switch to be turned on to be
51:35turned off
51:36and so on okay so that is how your
51:39comparator functions okay
51:40so to summarize we have from the
51:43negative scalar to a positive scalar to
51:46a unity gain amplifier
51:48you have a summing amplifier you have
51:50different uh
51:51uh subtractor okay so you have
51:55integrator you have differentiator and
51:56then you have your comparator so what do
51:58we what did you notice
52:00so your operational amplifiers can
52:03practically perform mathematical
52:06operations
52:07okay so that is where the name
52:09operational amplifier came from
52:11okay so before they really use it for
52:14mathematical
52:15operations that's why it is called
52:17operational
52:18amplifiers okay so that's it
52:22thank you for listening and um don't
52:25forget to hit the like button
52:28okay so thank you very much and god
52:31bless okay
52:34bye

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