Download Special Problem 2-3

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Special Problem 2.3-3
Determine the output voltage vout (t) of the circuit below:
vin (t)
R2 =3K
R1 = 1K
ideal
I=2 mA
vout (t)
+
R3 =1K
The first step in EVERY circuit analysis problem is to lable all currents
and voltages:
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vin
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i1
R1 = 1K
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R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
i
v

v3

ideal
vout
+
R3 =1K
i3
Q: I looked and looked at the notes, and I even looked at the
book, but I can’t seem to find the right equation!
A: That’s because the “right equation” for this circuit does not
exist—at least yet.
It’s up to you to use your knowledge and your skills to
determine the “right equation” for the output voltage vout !
Q: OK, let’s see; the output voltage is:
vout 
????
I’m stuck. Just how do I determine the output voltage?
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A: Open up your circuit analysis tool box. Note it consists of
three tools and three tools only:
Tool 1: KCL
Tool 2: KVL
Tool 3: Device equations (e.g., Ohm’s Law and the virtual
short).
Let’s use these tools to determine the “right” equation!
First, let’s apply KCL (I’m quite partial to KCL).
Note there are two nodes in this circuit. The KCL for the first
node is:
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i1  i2  i  I
i1
vin
i2
R1 = 1K
R2 =3K
 v2 
 v1 
i
I=2 mA
v
i
v

v3

ideal
vout
+
R3 =1K
i3
Note the potential of this node (with respect to ground) is that
of the inverting op-amp terminal (i.e., v  ).
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The KCL of the second node is:
I  i  i3
vin
i1
R1 = 1K
R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
i
v

v3

ideal
vout
+
R3 =1K
i3
Note the potential of this node (with respect to ground) is that
of the non-inverting op-amp terminal (i.e., v  ).
Now for our second tool—KVL.
We can conclude:
vin  v1  v 

v1  vin  v 
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vin
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i1
R1 = 1K
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R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
i
v

v3

ideal
+
R3 =1K
i3
vout
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And also:
v   v2  vout
vin
i1
v2  v   vout

R1 = 1K
R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
i
v

v3

vout
ideal
+
R3 =1K
i3
And likewise:
v   v3  0

v3  v 
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i1
R1 = 1K
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R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
i
v

v3

vout
ideal
+
R3 =1K
i3
Finally, we add in the device equations. Note in this circuit
there are three resistors, a current source, and an op-amp
From Ohm’s Law we know:
i1 
v1
R1
i2 
v2
R2
And from the current source:
I 2
And from the op-amp, three equations!
i3 
v3
R3
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i  0
i  0
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v  v
Q: Yikes! Two KCL equations, three KVL equations, and seven
device equations—together we have twelve equations. Do we
really need all these?
A: Absolutely! These 12 equations completely describe the
circuit. There are each independent; without any one of them,
we could not determine vout !
To prove this, just count up the number of variables in these
equations:
We have six currents:
i1 , i2 , i3 , i , i , I
And six voltages:
v1 ,v2 ,v3,v  ,v  ,vout
Together we have 12 unknowns—which works out well, since we
have 12 equations!
Thus, the only task remaining is to solve this algebra problem!
Q: OK, here’s where I take out my trusty programmable
calculator, type in the equations, and let it tell me the answer!
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A: Nope. I will not be at all impressed with such results (and
your grade will reflect this!).
Instead, put together the equations in a way that makes
complete physical sense—just one step at a time.
First, we take the two device equations:
I 2
and
i  0
And from the second KCL equation:
I  i  i3
vin
i1
2  0  i3

R1 = 1K

i3  2
R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
0
v

v3

ideal
+
R3 =1K
2
vout
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Now that we know the current through R3 , we can determine
the voltage across it (um, using Ohm’s law…).
v3  i3R3  2 1   2
vin
i1
R1 = 1K
R2 =3K
i2
 v2 
 v1 
i
v
I=2 mA
0
v
ideal
vout
+

2

R3 =1K
2
Thus, we can now determine both v  (from a KVL equation) and
v  (from a device equation):
v   v3  2
and
v  v  2
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i1
R1 = 1K
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R2 =3K
i2
 v2 
 v1 
i
v  2
I=2 mA
0
v  2
vout
ideal
+

2

R3 =1K
2
Now, inserting a device equation:
i  0
into the first KCL equation:
i1  i2  i  I

i1  i2  0  2

i1  i2  2
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i2  2 R = 1K
1
i2
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R2 =3K
 v2 
 v1 
0
v  2
I=2 mA
0
v  2
vout
ideal
+

2

R3 =1K
2
From KVL we find:
v1  vin  v 

v1  vin  0

and:
v2  v   vout

v2  2  vout
v1  vin
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vin
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i2  2 R = 1K
1
i2
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R2 =3K
 vin 2  
 2vout  
0
v  2
I=2 mA
0
v  2
vout
ideal
+

2

R3 =1K
2
So, from Ohm’s law (one of those device equations!), we find:
i1 
v1
R1
i2  2 

and:
i2 
v2
R2
vin  2

1
i2 

i2  vin  4
2  vout
3
Equating these last two results:
vin  4 
2  vout
3

vout  14  3vin
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Thus, we have at last arrived at the result:
vout  14  3vin
vin
vin  2 R = 1K
1
vin  4
 vin 2  
R2 =3K
 3vin  12  
0
v  2
I=2 mA
0
v  2
ideal
vout  14  3vin
+

2

R3 =1K
2
Note an alternative method for determining this result is the
application of superposition.
First we turn off the current source (e.g., I  0 )—note that
this is an open circuit!!!!!!!!
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R2 =3K
R1 = 1K
ideal
I=0
vout
+
R3 =1K
Note this is the same configuration as that of an inverting
amplifier!
Thus, we can quickly determine (since we already know!) that:
vout  
R2
3
vin   vin  3vin
R1
1
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R2 =3K
R1 = 1K
ideal
vout  3vin
+
R3 =1K
Likewise, if we instead set the input source to zero (vin  0 —
ground potential!), we will find that the output voltage is 14
volts (with respect to ground):
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R2 =3K
R1 = 1K
ideal
I=2 mA
vout  14
+
R3 =1K
From superposition, we conclude that the output voltage is the
sum of these two results:
vout  14  3vin
The same result as before!