- All right, so we introduced operational amplifiers
in the last video.
And we talked about how using two simple principles,
they were quite easy to use
and to calculate all the stuff around them.
I realized that saying, "Hey, this is really easy,"
is not the same as actually showing you.
So in today's tutorial,
all I'm going to do
is go over three different operational amplifier circuits,
and show how you can use those two simple rules,
which I'll go over again in a second,
to solve these things in a very straightforward way.
So, as a reminder, the two rules that you need to remember
with op amps using this first version that I have
as an example,
is that these two voltages are always going to be the same,
and that no current goes into these inputs.
So first rule:
these voltages will always be the same on the inputs.
And the second rule
is that the inputs have no current going in or out of them.
All right.
So with that, let's jump into this,
realizing that I am horribly prone to making mistakes.
So I know how to do this.
I just hope I don't make any small math errors,
and throw it all into disarray.
And that is why we always do that sanity check afterwards.
So jumping into this first one,
we wanna take our time, look at it,
think about what are we doing, what are we looking for?
So we are trying to find V out.
And we can see that this input is tied to ground.
So we know that not only is this zero volts,
this right here is zero volts.
Now, once we know that,
this immediately simplifies a huge amount,
because we have this voltage, this resistance,
this resistance, and then this voltage,
and we know that no current goes in here.
So the current through this resistor
is the same as this resistor.
And since we know the current through this resistor,
because it's .5 volts minus zero over 10k,
then we know what this voltage is going to be.
So just did that at a very high level.
Let's actually work through it.
As always, I like to put Ohms law here in the corner.
I don't know why.
It just helps me visualize things better.
So let's find, I'll just call this i1.
And so i1 is just going to be
.5 volts minus zero volts over 10 kiloohms.
But we can see that this is about 50 micro amps,
or .05 milliamp.
So get 50 micro amps as the current through there.
So now we know the current through there,
because that 50 micro amps isn't going through here,
it is going all the way through.
So now, we have our current going this way,
so we want to make sure we don't mess up our signs.
This is still i1.
So we want to say that through that,
we have zero minus i1 over 25.
What am I doing?
Oops.
And immediately I see
that this is why I have that Ohm's law on the corner,
'cause that's not right.
I'm looking for the voltage.
So V0 is just going to equal i1 times 25,000.
So that'll just be, oh yeah, 50 micro amps,
times 25 kiloohms,
which, let's punch that into the calculator,
.00005 times 25,000, gives us 1.25 volts.
And that's it.
This first one, we now know what i1 is,
through these two resistors, which is 50 micro amps.
And now we know what this voltage is.
Wait a second, that's minus, ha!
Yes, yes, because of all of this
I know that this is zero volts.
And so for me to have a current flowing here,
this has to be less than zero
to keep that current flowing there.
So I should have had that, so it is minus 1.25 volts.
See, this is why we look at things,
and we have that sanity check.
We know that we can't have current flowing this way,
in current flowing this way,
because there's nowhere else for the current to go
in this node.
So the current goes here, current goes there,
which means that this has to be less than zero volts,
so that the current can flow out that way.
All right, so update.
The current is 50 micrograms,
and this voltage is negative 1.25 volts.
See, even on these easy ones, you can get messed up.
So with that, let's move on to the second one.
Hopefully I don't screw this one up again.
This is actually what they call
a current-to-voltage converter.
There's a couple of ways to do it,
but this is one of the simpler ways.
And so we're not going to get actual numbers outta this,
but this is supposed to show you
how you can take a current source,
and have a voltage output
that you can understand, and control.
So we're actually going to be looking
for a relationship here.
Now, like with the last one,
the previous demonstration we just did,
you can look at this, and you can say,
"Okay, this is zero volts,"
so you know that this is gonna be zero volts right there,
because those two are matched,
and you know that this current coming through here,
is not gonna go in there.
So you know that your current
through this resistor, is i1.
So that seems pretty straightforward.
So all we're doing now is,
we're trying to find a relationship between i1,
and our output voltage V0.
Okay, since we already know that this is i1,
now we're simply going to put it in regards to our voltage.
So using Ohm's law, we can say basically,
that our current is going to be
zero minus V0 over R equals i1.
All right?
And since we want want this to be in a ratio,
we will just move this stuff around.
But this is already the answer.
Immediately, the initial equation that we put together
is the answer we're looking for.
We're just gonna put it
in a way that's a little bit more convenient to work with.
And since we want to know it using basically,
R as our our multiplier,
let's try and get R on its own sides.
So let's multiply, multiply both sides by R.
That'll give us a negative V out. equals i1 R.
And then we can divide that i1 from both sides,
and multiply that over.
So we have V out over i1,
and that gives us V out over i1 equals negative R.
So if you ever have, somehow,
I can't even think of a possible situation,
in which me as an embedded-focused guy
would ever deal with it,
if I ever had a current source
that I wanted to convert into a voltage output,
I could use this,
and know that I can control the relationship,
simply by varying my value of R.
And then I would have a very linear response
between the amount of current going in,
and the amount of voltage going out.
And it'll be very nice.
But that's really it.
Again, this was made so simple
by the fact that we knew that that's gonna be zero volts,
and that that current
was just gonna be going through that resistor.
And we can disregard all of the real life issues of,
oh that's not gonna be exactly zero volts, and oh,
there might be a little bit of current in there.
We're not worrying about that at this point.
You can worry about that later.
Don't worry about it.
All right, so let's go to number three.
All right, so for number three,
this one might look a bit scary,
because on the last one, we only had one resistor.
Now we have four resistors, and a voltage source.
And like, "Oh my goodness is all going to heck!"
No
It's not a problem.
Still, we take our time and we look at it.
Now, unlike the other ones where it's very simple
to just say, "Hey, this one's tied to ground,"
and we can assume the other one's ground,
we have to do a little bit more calculation.
But the first thing we do is,
we realize that on our positive input,
there is no feedback from the output.
So that is basically gonna be our set point.
And so we have three volts here,
and we have what's obviously a very simple voltage divider.
So we have 4K and 8K, because again,
no currents going through there.
So we can just assume that this is a voltage divider.
Three volts, all of that we know by inspection,
very simple math,
but that's gonna be two volts,
'cause you're gonna drop one volt,
over the 4K, and two volts over the 8k.
So now once we've done that,
we know that this spot right here is two volts.
And suddenly things become much simpler.
So we are still looking for our V out here,
and now we know that's two volts.
So we can calculate our current
through this two kiloohm resistor.
And then once we know the current through that resistor,
we know the current through this resistor, because again,
there is no current into the op amp.
It is so simple when you're able to (inhaling)
take your time, think about it.
So much stuff simplifies when you're able to do that,
when you're able to just ignore those currents going in,
and assume that those are the same voltages.
So with that, let's do some really quick math,
and let's figure out what our current is through these.
I'll just call that i1.
Okay, so there we have i1 equals two volts
over 2000 ohms and one milliamp.
Whew.
Okay, so now,
since we know the currency there is one milliamp,
we know it's one milliamp through there.
So now we just have to find that voltage
in comparison to i1 right there.
So we know our V out minus two volts,
'cause don't forget that this is two volts,
this isn't ground.
So it's your voltage out, over two volts,
over five kiloohms, is going to be one milliamp.
All right, so we take that 5K and we multiply it
on both sides, and those units are gonna cancel out,
and give us V out minus 2V equals five volts.
There we go.
And then it's just simple math:
V out equals seven volts.
That's it.
That's the right answer.
It's a miracle.
All right.
Those were three pretty simple,
even though they might have looked
a little bit more complicated
they were pretty simple op amp circuits,
made simple because of all of those assumptions
that you can make with the op-amp.
So again, I hope you're not like me.
For some reason, op-amp,
it took me a while for them to sink in,
for it to really become intuitive.
And I'm not sure why, because looking at it now,
I'm like, oh, this is so straightforward.
You just make those assumptions,
and everything kind of falls into place, most of the time.
So I hope that this helped you
at least to get your feet wet,
in understanding how it can be simplified,
by making those assumptions.
And even though we used numbers for the first and the third,
and we just created an equation with the second one,
there are a lot of very common configurations,
which people have already come up with the equations
to describe it.
So even though this is really important to understand,
and to do, honestly,
a lot of the times when you use an op-amp,
you can just say,
"Oh, I would like to use it in this configuration.
"Here's the equation.
"I'm just gonna punch in these numbers,"
And that math is done for you.
But if you can't, you can see this is quite straightforward.
All right, I hope that helped.
I hope your understanding of op amps has increased.
I hope that you're a lot more comfortable than I was,
when I first started with first started with op amps,
and that you can see the value of using them in a circuit.
If you like this video, give it a like,
subscribe to our channel, all that good stuff.
(bell ringing)
And we will catch you in the next, next one.
Take care.