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Operational Amplifier Part 2 (Electronics 2) — Transcript

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0:01hi hello welcome back okay so this is
0:03our
0:04second video for um operational
0:07amplifier and
0:08from this for this videos we're going to
0:10discuss the
0:11two basic op amp circuit which are the
0:15inverting amplifier and the
0:16non-inverting amplifier
0:18okay so to start with we will be
0:21discussing the
0:22inverting amplifier okay so um
0:25the inverting amplifier is we can say
0:28the most
0:28um basic op amp circuit okay so
0:32um we have here this
0:35type of amplifier as you can see this
0:39uses a um negative feedback to stabilize
0:44the overall voltage gain so where is the
0:46negative feedback here so
0:48the negative feedback is our rf here
0:51okay so why do we
0:52call this a negative feedback because
0:54from the output
0:56it it goes back to our input and
0:58particularly the
1:00inverting input okay so we call this as
1:04your
1:04negative feedback okay so why do we need
1:08why is there a need for a negative
1:10feedback
1:11okay so as we discussed from the
1:14previous um
1:15video uh the open loop voltage gain for
1:18an
1:19operational amplifier is too high okay
1:22and since this is it is too high it is
1:25unstable
1:26to be used of uh to be of any use
1:30without the form of
1:32feedback okay so usually it is
1:35uh between the range of twenty thousand
1:37to two hundred thousand
1:39okay so this is a very unpredictable
1:42voltage gain
1:43and this voltage gain will be useless
1:46okay so um
1:49therefore we place a
1:53feedback just to somewhat stabilize our
1:56overall
1:57voltage gain okay so as you can see we
2:00we kept the figure simple
2:04by eliminating first or by not showing
2:07first the power
2:08supply but always remember that we
2:10always have a positive
2:11and a negative voltage supply in our
2:15operational amplifier okay so to start
2:18with
2:18uh we have an input vi so this is our vi
2:22this is your v in which is um
2:25directly connected to the inverting
2:28terminal
2:29of our operational amplifier that's why
2:33we have here an inverting amplifier
2:36so your vi drives the inverting input
2:39through our resistor r1 this is our
2:41resistor r1
2:43so the inverting input v2 okay
2:46so the inverting input v2 this one plus
2:50or minus v2 here is amplified by the
2:52open loop voltage gain
2:54to produce an inverted output voltage
2:57so if we have here our input voltage
3:01here as you can see you have an
3:03amplified voltage with a 180 degrees
3:07phase reversal so we have an inverting
3:09amplifier
3:10okay so our rf here
3:14so output voltage is fed back to the
3:16input through our feedback
3:18resistor rf okay so rf results in a
3:22negative feedback
3:24so because the output is 180 degrees
3:28out of phase of your input okay so how
3:31can we say
3:34how can we say that uh the
3:37feedback resistor is somewhat stabilizes
3:40our overall voltage gain okay so
3:43if our uh none if our open loop voltage
3:46gain increases for any reason so it
3:50it uh increases significantly
3:53so the output will increase okay so once
3:56the output is increased
3:59a feedback it will feedback more voltage
4:02to the inverting input
4:04okay so which reduces v2 since
4:07since this is negative it will be
4:09opposing so it will be
4:11uh it will cause in a decrease in
4:14v2 okay so let's repeat once our
4:19once our open loop voltage gain
4:22significantly increases so
4:24it produce a an
4:27okay uh so an increase in
4:31output okay so that output will be will
4:33be
4:34uh feedback to your circuit
4:37which makes your v2 smaller because it
4:40is negative it will be
4:42opposing so that means since v2 becomes
4:45small even though our
4:46uh open loop voltage gain increases your
4:49output
4:50will uh will
4:53not or will just very slightly increase
4:57okay so small that it is hardly
4:59noticeable
5:00okay so therefore if we have an increase
5:03without the feedback and your over
5:05your open voltage gain increases your
5:08output will be increasing significantly
5:10but because of the feedback here okay so
5:13your
5:14v2 here will be adjusting to whatever
5:17increase
5:18in our open loop belt voltage gain okay
5:21so because when the output increases the
5:23feedback
5:24also increases since this is negative
5:26okay so it will cause a decrease in our
5:29v2
5:29so if this is um quite smaller even
5:32though the voltage gain
5:33increases so your output will not be or
5:36will not
5:37increase significantly so that is how
5:41it stabilize your overall voltage gain
5:44okay so the output voltage is fed back
5:47to the input
5:48through feedback resistor rf any changes
5:51in v2
5:52produced by the input voltage are
5:54opposed by the
5:56output signal okay so that is the effect
5:59of our
6:00negative feedback okay
6:03so we have here what we call the
6:07virtual ground okay so when we connect a
6:10piece of wire
6:11between some point in the circuit and
6:13the ground we all know that the
6:15uh that the resulting voltage there
6:20will be equal to zero at the same time
6:23that that wire from some point in the
6:26circuit and the ground
6:27uh will provide a path for for current
6:30to flow to the ground
6:31so that specific wire okay that
6:34that particular wire we call that a
6:37mechanical ground
6:38but that is very different from the
6:41concept of
6:42virtual ground so what is the concept of
6:44virtual ground
6:45virtual ground is widely used shortcut
6:49for analyzing and inverting amplifier
6:52so very virtual ground actually is very
6:55unusual
6:56why so it is just like a half ground
7:00why because it is a short for voltage
7:04yet it is open for current okay so we
7:07all know that
7:08when we have a short circuit your
7:10voltage is
7:11zero but your current is maximum okay
7:14but
7:14this virtual ground here is very unusual
7:17because
7:17it is short for voltage at the same time
7:21since it is not there it is just
7:23virtually there
7:24um it is open for for current
7:27so the virtual ground is uh
7:32the virtual ground is provided for us to
7:35have a very simple approach
7:37in our analysis of of an operational
7:40amplifier
7:41so the virtual ground is based on the
7:44ideal of um
7:45characteristics so what are those ideal
7:47op um characteristics so this is our
7:50virtual ground as you can see it is
7:53in a form of a broken lines okay so what
7:57do we mean by the
7:58broken lines there so this is just for
8:00our assumption it is not a mechanical
8:02ground in such a way that
8:04there is really a ah real wire that is
8:07connected from this point
8:09going to the ground so we are just
8:11assuming that there is a virtual ground
8:13there and we are basing that on your
8:16ideal operational amplifier
8:19so what do we mean by that okay so since
8:22our in is infinite so what do we mean by
8:24r in is infinite
8:26um we have a very very high input
8:29impedance and if we have a very very
8:31high input impedance
8:32that makes our i2 the current entering
8:36your
8:37operational amplifier equal to zero
8:40okay and another um
8:43another ideal characteristic of our op
8:46amp
8:47is our open loop voltage gain is
8:49infinite or very very high so if we have
8:52a very very high voltage
8:54gain so voltage gain is equivalent to
8:57output voltage divided by ven
9:00okay so how are you going to make your
9:03voltage
9:05over open loop voltage gain infinite so
9:07if our v in is equal to zero
9:09so therefore we are going to assume that
9:12this is equivalent to
9:13zero this is our basis for our
9:17virtual ground here okay so we're going
9:19to assume that no current will enter
9:21your operational amplifier
9:23and at the same time our voltage
9:27okay the input voltage to your
9:29operational amplifier will also be equal
9:32to
9:32zero okay so
9:36um the voltage gain
9:42so the voltage gain for our for our
9:45inverting amplifier is given by this one
9:48so this is
9:49what we call av so for voltage gain and
9:52this time we have a closed loop
9:54so it is no longer open loop so our our
9:57usual value for a 741 which is 100
10:00thousand
10:01okay so this is no longer equivalent to
10:04100 000 because there is already a loop
10:06okay in the form of our rf okay so
10:10the closed loop voltage gain
10:13this is the voltage gain when there is a
10:15feedback path between the output and the
10:17input so how did we arrived with
10:19negative rf
10:20all over r1 here so since we have here a
10:24virtual ground
10:25that means no input will enter our
10:29operational amplifier
10:30at the same time no current will enter
10:33our operational amplifier
10:35that only means that all of the current
10:37coming from our b
10:38in will directly flow across your
10:42feedback resistance here okay so if we
10:45are going to
10:46look at work if you're going to look at
10:49our circuit there
10:56what is our
11:00okay so what is our input there our
11:03input there
11:03is equivalent to your v in so your input
11:07voltage there
11:08is equal to v in and that is equivalent
11:11to
11:12i in multiplied by
11:16r1 so this is your input
11:19voltage okay now what is our output if
11:23you're going to look at your output
11:25okay so your output is directly
11:27connected
11:28across your rf but this is the negative
11:32sign
11:33okay the negative terminal for your
11:35output
11:36meaning the negative terminal for our rf
11:39because
11:40your current will enter here this is
11:41positive and your current will
11:43will leave here which is negative giving
11:46us our v
11:46out which is connected at the negative
11:48terminal
11:49so that will become
11:54our v out is equivalent to
11:57negative okay i in multiplied by
12:01your rf okay so what is your
12:04closed loop so this time your cl there
12:07stands for your
12:08closed loop so what will be your closed
12:10loop voltage gain
12:12so your closed loop voltage gain is
12:14simply equivalent to your
12:16output voltage divided by your
12:19input voltage so what is our output
12:22voltage our output voltage is minus
12:24i input multiplied by rf divided by our
12:28input
12:29which is equivalent to i input
12:31multiplied multiplied by
12:32our r1 since we are talking about the
12:35same input current because no current
12:37will enter your operational amplifier
12:40so we're just going to cancel this out
12:43leaving us
12:44our voltage gain equivalent to minus
12:47rf or the feedback resistance divided by
12:50r1 so this is the closed loop voltage
12:54gain for
12:54any inverting amplifier okay so the
12:58negative sign there
12:59signifies that there is a 180 degrees
13:02phase shift between the input and the
13:05output
13:06signals okay so as we can see
13:10we have a very very simple design
13:13here okay so we will have what for
13:16example
13:16uh you are the designer and you are
13:19asked
13:20to to build okay an amplifier
13:23wherein your your output will be 50
13:26times of your
13:27your input okay so all you have to do is
13:30to get an
13:30op amp and then connect a 50 kilo ohms
13:34at the feedback resistance and then a
13:35one
13:36a one kilo ohm at your r1 so 50 kilo
13:40ohms divided by
13:41one that will automatically give you a
13:44negative 50
13:46voltage gain so the point that you place
13:49your input at the inverting terminal
13:51that will already give you an automatic
13:54inverting
13:55output so that is how
13:58uh how your operational amplifiers
14:01amplifier who works as an inverting
14:05amplifier
14:05so what is the counter part of the
14:08inverting up
14:10in our previous discussions so this is
14:12just like
14:13a common emitter configuration because
14:16if you have a common emitter
14:18configuration
14:19you will have an output that is 180
14:21degrees out of phase
14:22with your input however if you are going
14:26to
14:27be asked to design a common ammeter
14:29amplifier
14:31with a gain equivalent to 50 you cannot
14:34just give
14:35the design right away because you still
14:37have to consider your circuit what will
14:39be your r1 what will be your r2 what
14:41will be your rc what will be your
14:44your re so you will be starting with the
14:47dc analysis
14:48you will be going to the ac analysis
14:50because before you can even solve for
14:52your
14:53again which is equivalent to 250
14:56but this is a very very straightforward
15:00approach
15:00okay i want to change my my gain i want
15:03it to become 100 so what are you going
15:06to do so you just change your
15:08your rf to 100 kilo ohms and then our r1
15:11is
15:12still equivalent to one kilo ohm so that
15:14is already a change in your
15:16voltage gain okay so as you can see
15:21there is no characteristic of op amp
15:24anymore so when we are talking about
15:26inverting amplifier
15:28what what uh
15:32tells you okay what what what
15:35uh what parameter gives you your voltage
15:38gain it is not anymore what of amp you
15:41are using
15:42so your gain will be dictated by what
15:45value of rf and r1
15:47that you are using okay so it says here
15:52so it says here that um
15:57the ratio of the feedback resistance to
16:00the input resistance
16:01will tell you how much is your voltage
16:04gain
16:05okay and automatically we have less than
16:09your uh open loop voltage gain because
16:13if this is 741
16:15the open voltage gain of this is
16:17somewhat near 100 thousand
16:19but because we make use of a feedback we
16:22somewhat
16:23uh reduce that into a more usable and a
16:26more practical value of
16:28a voltage gain so this is how
16:32your inverting amplifier will look like
16:36compared to a common emitter
16:38configuration
16:39when we are going to make use of an op
16:41amp all you need
16:43is the value of your feedback resistance
16:45and your
16:46r1 here okay
16:50so talking about the input impedance for
16:53an inverting amplifier so the input
16:55impedance with closed loop
16:57um circuit is just equivalent to
17:00r1 so for example our r1 is equivalent
17:03to one kilo ohm so that is already your
17:06input impedance and like for common
17:08emitter configuration in solving for the
17:10input impedance you still have to solve
17:13for the emitter current
17:14and then solve for the small re and then
17:17multiply it by beta
17:19so that you can get your input impedance
17:21so in
17:22um an operational amplifier you're
17:24working as an inverting amplifier
17:27all you have to do is to look for the
17:28value of z
17:30of of r1 and therefore that is already
17:33your
17:33input impedance closed loop
17:38okay so for the bandwidth okay so the
17:41greater the negative feedback
17:43the higher the closed loop cut off
17:47frequency okay so the smaller your
17:51voltage gain close loop voltage gain
17:55the higher your f2 so what is f2
18:00f2 is your cut off frequency
18:04okay or that is where your frequency
18:08will start decreasing okay so as you can
18:11see
18:11if your if your voltage gain is very
18:14very high
18:15you have a lower cut-off frequency
18:19but as your um gain decreases
18:23your f2 likewise
18:27they increase somewhat they increase so
18:30technically it is a trade-off what do
18:32you want in a circuit do you want a
18:34circuit having a higher gain
18:36so therefore our frequency will just be
18:38limited
18:39or if you want a somewhat lower gain
18:42then our frequency will not be
18:44limited remember the frequency dictates
18:47what application are you
18:50are you going to use your op amp
18:56okay so we have here your f2
18:59or your cut off frequency so every time
19:02uh all will start with zero because this
19:04is a
19:05an operational amplifier this is
19:07directly coupled
19:09okay so our question is what will be
19:11your uppercut of frequency
19:13so that is equivalent to f unity
19:16divided by the closed loop voltage gain
19:19plus one
19:20however since your closed loop voltage
19:23gain is somewhat greater than one
19:25okay we can neglect the value of one
19:27okay so
19:28that will simplify our formula to f
19:30unity which is equivalent to the
19:33closed loop voltage gain so in solving
19:36for your f unity
19:38now that is equivalent to the
19:41open uh closed loop voltage gain
19:43multiplied by your
19:45f2 cl okay or the
19:48upper cut off frequency so this f unity
19:51now is
19:52no longer equivalent to 1 megahertz
19:56okay so that will be equivalent to 1
19:58megahertz if our
20:02if our gain is equivalent to 1 but if we
20:05have a higher gain it will no longer be
20:07equivalent to
20:081 megahertz and f unity here is what we
20:11call the gain
20:12bandwidth product okay so why gain
20:15bandwidth product because this is a
20:17product of your
20:18gain which is your close loop voltage
20:20gain and your f2cl which is your
20:23bandwidth
20:23because since we start with zero any
20:26value of f2 here
20:27will give you the value of your
20:29bandwidth okay so this is called the
20:32gbp or the gain bandwidth product
20:37for the bias and offset so this is just
20:40the same with our differential
20:43amplifiers and with the values that we
20:45have
20:46discussed in our previous um
20:50video in our video number one okay so
20:53bias and
20:54offsets the negative feedback also
20:56reduces
20:57the output error or caused by the input
20:59bias current
21:00input offset current and the input
21:03offset voltage
21:04why because the although this one will
21:07be constant okay so your v1 error your
21:10v2 error with your v3 error we have
21:12already discussed this in our
21:14previous video our gain here
21:17will technically
21:20go down okay so unlike before if this is
21:23open loop this is equivalent to 100
21:26thousand
21:26but if we decrease your closed loop
21:29voltage
21:30gain therefore we are also decreasing
21:32our
21:33error voltages okay so we have a
21:37resistor compensation and offset nulling
21:40so usually this is equivalent to our rb2
21:44in our formula will be equivalent to r1
21:47parallel
21:47rf okay so to nullify or to
21:52to help uh offsetting this bias and
21:56offset currents and voltages
21:58we usually place an rb1 so instead of
22:02grounding it we usually place an rb1
22:05which is instead of grounding your
22:08second input
22:09we place an rb1 which is always
22:11equivalent
22:12to your rb2 so later we will be
22:15discussing that
22:17okay so we have here an example it says
22:20here
22:25so what are the closed loop voltage gain
22:28and bandwidth
22:29and what is the output voltage at one
22:31kilohertz
22:32and what is the output voltage at one
22:35megahertz okay so
22:42we have this solution for that
22:48so the voltage gain is very simple okay
22:52so how do we solve for that
22:54for the voltage gain so the voltage gain
23:05is equivalent to rf
23:08okay which is your feedback resistance
23:11which is equivalent to
23:12negative 70 of uh minus so the minus is
23:15because it is an inverting amplifier
23:17so that is 75 kilo ohms divided by 1.5
23:21we have a gain which is equivalent to
23:2350. so the output will be increased 50
23:26times and the negative sign
23:28is sigma will signify that there is a
23:30180 degrees phase difference between
23:33your
23:33input and your output there okay so to
23:36solve for the bandwidth
23:38so what do we mean by the bandwidth so
23:40we just have to solve for
23:48okay so we just have to solve for your
23:51f2 okay because um our f1 is
23:55always equal to zero so since this is
23:57equal to zero
23:58automatically what value the value of
24:00your f2 will always be your
24:02bandwidth so your bandwidth is your f to
24:05close
24:06loop okay so we have to specify their
24:08closed loop because
24:10we are we already have a feedback and
24:12that is equivalent to f unity
24:15divided by your closed loop voltage gain
24:18so since we have a 7
24:194 1 our f unity is constant
24:22so it is a standard value which is
24:24equivalent to 1 megahertz
24:26multi divided by our closed loop voltage
24:28gain which is equivalent to 50.
24:30no need to place the negative sign
24:33anymore because we don't have
24:34such a way that such as a negative
24:37bandwidth
24:38so the negative sign is very important
24:40in the gain
24:41because the negative sign will signify
24:44whether you are inverting or you are
24:47using an
24:47inverting or a non-inverting amplifier
24:51okay so we have here 1 megahertz divided
24:54by 50
24:55so that will give you 20 kilohertz
24:58so that means if we're going to get the
25:01the value of negative 50 in terms of
25:05decibel
25:06that will give us 34 db okay
25:09so from zero up to 20 kilohertz
25:13that is equivalent to 34 db so this
25:17is our f2 and from your f2
25:20that will be your negative 3 db point so
25:23that will
25:24start decreasing at a value of 20 db
25:27per decade until it reaches
25:31our one megahertz range
25:34okay so that is what we mean by the
25:37bandwidth
25:38now we are asked to solve for the output
25:41voltage
25:43if our frequency is one kilohertz
25:46so obviously one kilohertz is below
25:5120 kilohertz since it is below 20
25:54kilohertz
25:55what does that mean that means from 0 to
25:5820 kilohertz
25:59your gain is constantly equivalent to
26:02negative 50. so if we have an input
26:05voltage which is equivalent to 10 volts
26:08all we have to do is to multiply it by
26:11negative 50
26:12so we can get the output voltage which
26:14is equivalent to
26:16negative 500 volts peak to peak
26:19okay so from zero up to 20 kilohertz
26:23our gain is equivalent to
26:26negative 50 okay or it will be
26:29increasing your input by
26:3150 times however okay however
26:35once it reaches 20 kilohertz it will be
26:38decreasing
26:41by 20 decibels per decade
26:44and our next question is what will be
26:47the output
26:48at one megahertz so at one megahertz
26:52our v out is equivalent to negative 10
26:55millivolts it's not zero okay it's not
26:58zero
26:58because one megahertz is the unity
27:00frequency
27:01for an operational for a 741
27:04operational amplifier so what does that
27:07mean
27:08at this point we just have a unity gain
27:10so what do we mean by a unity gain
27:12so your input is just equivalent to your
27:16output but with a 180 degrees
27:20phase reversal why because we placed
27:23your
27:23input at the inverting terminal
27:27so that is how we solve a problem
27:30using an an inverting amplifier
27:34okay so this is very simple class our
27:37computation
27:39our computations are very very simple so
27:42it just has something to do with how you
27:45compute your
27:46voltage gain and you just have to uh
27:49put in your mind that uh
27:53until when okay until when your
27:55operating uh
27:56operational amplifier is working and
27:59until when
28:00or until what frequency will it not
28:02start to
28:03work anymore okay
28:10so we have here another example so
28:13actually this is
28:14this is an example of how your error
28:17voltage
28:18will uh will of affect will affect your
28:22your circuit okay so what is the output
28:25voltage when your v in
28:26is equal to zero you use typical values
28:29so what do you mean by typical values if
28:32you're not very familiar
28:33with operational amplifier or your 741
28:37it's just like saying that we have here
28:41an incomplete circuit
28:43or a circuit with an incomplete given
28:49or we can just say that um
28:53maybe the output is also equal to to
28:55zero okay so what does that mean
28:59and what are the typical values okay so
29:01what are the typical values
29:09so from our given circuit here
29:14so 4741 our input bias current
29:18from our table from our first video
29:21regarding
29:22operational amplifier
29:25your input current input bias
29:29is equivalent to 80 nano amperes your
29:32input offset current is equivalent to 20
29:35nano amperes
29:36and your input offset voltage is 2
29:39millivolts
29:40okay so first we need to solve for your
29:42rb2
29:43okay so first our rb2 is equivalent to
29:48r1 parallel our rf okay so that is 1.5
29:53kilo ohms parallel 75 kilo ohms
29:55and that is equivalent to 1.47 kilo ohms
29:59okay so all we have to do is to
30:02compute this using our given formulas
30:05but since in our original um circuit we
30:08don't have rb1 because um
30:10our uh our rb1 or the
30:14or the non-inverting input is
30:18grounded so therefore we don't have an
30:21rb1 there okay so what we have is your
30:24rb2
30:25because it is uh the resistance
30:28in uh in the base of our inverting
30:31amplifier so if we're going to imagine a
30:33differential
30:34amplifier so therefore
30:38for our formula here rb1 this is equal
30:40to 0 and this is negat
30:431.47 so this is equivalent to negative
30:451.47
30:46anyway these are just direct
30:48substitution okay
30:50uh the same is true with your v2 error
30:54and your v3 error so v3 error is just
30:57equivalent to your
30:58input offset voltage okay so you can see
31:01what is the effect of the
31:05what is the effect of your closed loop
31:06voltage gain so the closed loop voltage
31:09gain will
31:10minimize your volt error your input
31:12error voltage
31:13so why how will it minimize your input
31:16voltage
31:16error because instead of multiplying
31:19this by 100
31:20000 for an open loop voltage gain
31:23we will just make use of the closed loop
31:25voltage gain which is roughly equivalent
31:27to 50 which is 75
31:29divided by 1.5 okay having uh
31:33giving us an error equivalent to plus or
31:35minus 107
31:37millivolts okay so what do we mean by
31:40compensating resistance
31:42so the compensating resistance class is
31:45that
31:46from our original um circuit we can
31:49place
31:50instead of instead of um
31:54instead of placing okay instead of
31:57placing a
31:58uh this
32:02uh inverting in non-inverting input to
32:04the ground
32:05we place here an rc and this rc
32:08is equivalent to your rb2
32:12so usually this is equivalent to rb2 or
32:14the nearest
32:15possible um resistor
32:18available nearer to your value of rb2
32:22okay so so that we can
32:26we can nullify okay so we can uh what
32:29you call this we can
32:30um since they are for example
32:35rb1 so if we place an rb1 which is
32:37equivalent to our b2 so
32:39if we subtract this this will be
32:40equivalent to zero so therefore
32:43we can at least nullify or we can
32:45invalidate one
32:46error okay and also one way of
32:50uh one way of decreasing your your
32:53offset voltages and
32:55currents is by placing your nulling
32:58resistance so if you can still remember
33:00so you have a nulling circuit in um
33:03pin number one and in pin number five so
33:06we place here uh
33:08uh potentiometer wherein we can adjust
33:11the value of the potentiometer
33:13so that we can um decrease likewise the
33:16value of your
33:17offset voltages and your current so
33:20those are just
33:21um the different ways on how you're
33:23going to improve those
33:25error voltages of course those are error
33:28voltages and since it is error we don't
33:29like that
33:30why because you don't have any input
33:33there so
33:33supposedly you don't have any output
33:36okay but
33:37since it is uh since it
33:40it is uh non-ideal once your operational
33:44amplifier was manufactured they're
33:46already
33:47okay um offset voltages and
33:51currents so all we have to do is to at
33:54least
33:54minimize them by using this different
33:57techniques
33:58okay okay
34:01so next is um another
34:04basic of amsterdam okay so another basic
34:08pop-up circuit class is the
34:09non-inverting amplifier
34:11okay so again just like the inverting
34:15amplifier this also uses a negative
34:17feedback
34:18to stabilize the overall voltage gain
34:21okay
34:21so um this is the equivalent circuit for
34:24a
34:25non-inverting amplifier as you can see
34:27we still have a negative feedback
34:29because the feedback is
34:30still placed on the inverting input
34:33however
34:34we use or we place your input at your
34:38non-inverting terminal okay so
34:42part of the output voltage is fed back
34:45to the input
34:46through this voltage divider here but
34:49our
34:49input is placed at the non-inverting
34:52terminal therefore making this a
34:54non-inverting amplifier
34:55if we have here your v in we have an
34:58amplified version here of your input
35:00signal
35:01which is which is in phase with your
35:04input voltage
35:09so we have here the input the input
35:11voltage
35:12ven drives the non-inverting input okay
35:15so the input voltage is amplified to
35:17produce the in
35:18phase output voltage so the voltage
35:20across r1
35:22is the feedback voltage applied to the
35:24inverting
35:25input okay so
35:29um the difference between v v
35:32one here and v two here must be very
35:35very small
35:36okay so why because if you can still
35:38remember
35:39we also have such a thing as virtual
35:42short
35:42in our non-inverting amplifier
35:46okay and this is how we uh
35:49this is how we are going to explain the
35:52virtual short in a non-inverting
35:54amplifier
35:55so since r in is infinite okay both
35:58input currents both current at the
36:01positive
36:02and both current at the negative must be
36:05equal to zero
36:06and also since your avol is infinite
36:10v1 the difference between v1 and v2 here
36:13is also equal to zero so what happens
36:16here
36:18the value okay or the voltage being
36:21brought back against v2 here
36:23will be um equivalent to our
36:27v in here so if v in is increasing
36:30okay we have a feedback here which
36:33is equivalent to your um v in but
36:36in opposite polarity so therefore
36:40if we subtract them the answer is always
36:42equal to
36:43zero so if this decreases the feedback
36:47here will also decrease to the same
36:49amount of our input here
36:51so that is what we call bootstrapping
36:54okay
36:54so the non-inverting input pulls the
36:57inverting input
36:58up or down to an equal value so that
37:01when we subtract them with v1 minus v2
37:05the answer is always equivalent to zero
37:08so that is our
37:09virtual short here okay so how do we
37:13solve for the voltage gain
37:14so the voltage gain here is almost the
37:18same
37:18with the inverting amplifier except that
37:20we add one to the ratio of the
37:22resistance okay and also
37:24the output is in phase with the input
37:27therefore there is no negative sign
37:29used okay so how did we arrived
37:33with that value okay so as you can see
37:37whatever your uh voltage across your r1
37:41here
37:41that is equivalent or that is equal to v
37:44in because
37:45v1 minus v2 is always equal to
37:48zero okay so and since
37:51because of the virtual ground okay we
37:54know that no
37:55current must enter your inverting input
37:58so your current i y i one here is just
38:01equivalent to our current
38:02i1 here okay so they are just equal
38:05because of that
38:07because of the virtual short or the
38:09virtual ground okay so
38:11if we're going to
38:14elaborate this one what do we have
38:18okay so our input there by looking at
38:21the figure
38:22that is equivalent to your voltage
38:25across
38:25r1 and the voltage across r1 is
38:28equivalent to your i
38:29input multiplied by your
38:32r1 here okay how about your output
38:35so our output is placed between
38:39of course the the output terminal
38:43so that is the whole voltage
38:46from rf to the ground which is
38:48equivalent to your
38:50r1 and since the same current flows
38:53across rf
38:55and your r1 that it does that
38:58just shows that rf and r1 are in
39:02series because i in there is just equal
39:05so what is your output voltage so your
39:07output voltage is just the current
39:09multiplied by your rf plus your
39:11r in based on the circuit okay and
39:15it's positive right because the positive
39:17terminal is connected
39:19at the output unlike in the inverting
39:21amplifier
39:22so your v out there is equivalent to i
39:25in
39:26multiplied by the sum
39:29of your r1 and your rf so going now
39:33getting now the voltage gain so we place
39:36here a closed loop voltage gain because
39:38we have already a feedback
39:40so that will give you v out all over
39:44v in okay so our v out is equivalent to
39:47i
39:47in r1 plus rf
39:53divide by you have here i in
39:57multiplied by r1 since we have the same
40:00r in
40:00this will just be cancelled out and that
40:03will give us
40:04r1 plus rf all over
40:07r1 okay simplifying this this will give
40:11us one plus rf all over
40:14r1 which is the voltage gain
40:17for a non-inverting amplifier okay
40:21so if we wanted for example uh
40:24again equivalent to 50. so we're going
40:27to get a combination of rf
40:29all over r1 that will give us 49 okay
40:32because um we still need to add one
40:36so that we can have a gain which is
40:38equivalent to
40:40250 okay so the only difference between
40:44the
40:44between the inverting and the
40:46non-inverting is
40:48of course we don't have a phase reversal
40:50and at the same time we have
40:52a plus one and of course you will be
40:54placing the input voltage at the
40:57non-inverting terminal okay so
41:00the the circuit is very simple all you
41:03have to do is to place r1 and your
41:06rf okay so that is uh the non-inverting
41:10amplifier so what is the counter part of
41:12the non-inverting amplifier in our
41:16previous topics okay so the
41:18non-inverting amplifier
41:20is like the common base configuration
41:22because the common base configuration
41:25is capable of producing high voltage
41:27gain and it has no
41:29phase reversal so the counter part of
41:31the inverting amplifier is the common
41:34emitter configuration and the counter
41:36part of the common base is the
41:38non-inverting amplifier
41:41okay
41:45so for the other quantities it is the
41:47same
41:48so it is the same with with what we have
41:51discussed in the non-inverting in the
41:52inverting amplifier so the voltage gain
41:55can be a trade-off with
41:56bandwidth okay so do you want a higher
42:00voltage gain then we have a lesser
42:02bandwidth or
42:03if are we going to are we going to
42:06accept
42:07lower voltage gain and have a wider
42:10bandwidth instead
42:11okay so the negative feedback increases
42:14the input impedance so the input error
42:16voltages
42:17caused by the input bias current input
42:20offset current and input offset voltages
42:22are analyzed the same way as
42:24a sweda inverting amplifier and at the
42:27same time
42:28your rb2 is also equal to r1
42:31parallel rf
42:34okay so what is the effect
42:37of the output error voltage so your
42:40output error voltage reduces your peak
42:42to peak maximum peak to peak voltage
42:45okay so the greater the output error
42:47voltage the smaller your
42:48peak to peak value so for example if we
42:51have here a
42:54uh plus or minus 15 volts so we are
42:57expecting
42:58roughly uh uh 30 volts peak to peak
43:01but in reality we can only have at least
43:03one to two volts um
43:05the uh difference so that is given by a
43:08plus
43:08and minus 14 or 28 volts peak to peak
43:11but
43:12if we have an error voltage we can just
43:15have a 20 volts peak to peak
43:17output voltage so what causes the the
43:20lowering down of your output voltage so
43:23this is due to the
43:24error voltages so um if we can
43:28okay we wanted to eliminate those errors
43:31actually if we cannot eliminate it
43:33totally at least we lower it down so
43:35that we can still have what we wanted
43:38output voltage uh that we
43:41designed for okay so
43:44that's it for our
43:48that's it for our second um video
43:51regarding our operational amplifier so
43:54i just um i just discussed the inverting
43:58and the non-inverting
43:59because when we know the inverting and
44:01the non-inverting it is
44:02already very simple or easier to
44:05understand
44:06the other applications for an
44:08operational amplifier
44:10okay so um for our third video
44:13we will be discussing more applications
44:17for
44:17our operational amplifier okay so again
44:21thank you for watching
44:22and don't forget to hit the like button
44:24and after watching okay so thank you
44:26very much and
44:27god bless

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