For all problems below, assume that the op amp is almost ideal where Ri = ∞ , Ro = 0 , V+ =10V, and V- = -10V.
Watch for the + and – signs on the input terminals of the op amps as the noninverting and inverting input terminals of the
op amp are not always in the same orientation in each of the circuits.
1. Draw the voltage transfer characteristic, Vo vs. VIN, for the non-inverting amplifier circuit shown below. Note
that Vo is the node voltage at the output terminal of the op amp and is also the voltage drop across the dependent
voltage source that is part of the op amp model
a. Let VIN vary from -5V to +5V.
b. Label the following regions: positive saturation, linear region, and negative saturation.
VIN
VIN
VO
I=0 A
I=0 A
The op-amp configuration given above constitutes a non-inverting amplifier whose gain is:
This can be proved by applying KCL at the inverting (-) terminal of the opamp:
Rearranging gives:
(
)
(
)
Thus:
Because it is a nearly ideal operational amplifier, we have to be concerned about the values of V + and V-, the power
supply rails. These values are the largest magnitude that the output voltage of the op amp.
The values of VIN at which the output will reach V+ and V- and the op amp enters the positive and negative saturation
regions, respectively, can be calculated as follows.
The transition between linear and positive saturation occurs when Vo = V+:
The transition between linear and negative saturation occurs when Vo = V-:
Vo Vs. V IN
12
X: 1.667
Y: 10
8
Positive
Saturation
Linear
0
O
V (Volts)
4
-4
-8
-12
-5
2.
Negative
Saturation
-4
-3
X: -1.667
Y: -10
-2
-1
0
VIN (Volts)
1
2
3
4
5
Draw the voltage transfer characteristic for the inverting amplifier circuit shown below.
a. Let Vin vary from -3V to +3V.
b. Label the following regions: positive saturation, linear region, and negative saturation.
The op-amp configuration given above constitutes an inverting amplifier whose gain is:
This can be proved by applying KCL at the inverting (-) terminal of the opamp:
Rearranging gives:
(
)
(
)
(
)
Thus:
Because it is a nearly ideal operational amplifier, we have to be concerned about the values of V + and V-, the power
supply rails. These values are the largest magnitude that the output voltage of the op amp.
The values of VIN at which the output will reach V+ and V- and the op amp enters the positive and negative saturation
regions, respectively, can be calculated as follows.
The transition between linear and positive saturation occurs when Vo = V+:
The transition between linear and negative saturation occurs when Vo = V-:
Positive
Saturation
Linear
Negative
Saturation
3. A load resistor is attached to the output terminal of the inverting amplifier. The ideal op amp model has been
drawn within the symbol for the op amp, which is shown in red. Calculate the currents is, if, i, and iL when Vin =
1V.
I=0 A
V1
I=0 A
Since V1=V2=0 V,
Additionally, no current enters the opamp so that:
From inspection, the output voltage of the inverting amplifier is:
Finally, applying KCL at the output:
which simplifies to:
(
)
In summary:
𝑖𝑠
𝑅𝑓
𝑉
𝑅1 𝑅𝐿 𝑖𝑛
𝑉𝑖𝑛
,𝑖
𝑅1 𝐿
𝑖𝑓
𝑅𝑓 𝑉𝑖𝑛
𝑅𝐿 𝑅1
𝑖
For Vin=1 V, we have:
𝑖𝑠
𝑖𝑓
m , 𝑖𝐿
m
𝑖 =3.5 mA
4. Calculate the values for VL, is, if, i, and iL.
3V
Vo
3V
Since no current enters the opamp,
, and 3 V appears at the non-inverting input of the opamp as depicted above.
Since the inverting input (- terminal and also VL) follows the voltage of the non-inverting terminal, VL=3V. Similarly,
must also equal 0 A. Next, the configuration above is that of a non-inverting amplifier as in Problem 1(although drawn a
little differently). Thus, the voltage at the output of the opamp is:
(
)
Finally,
In summary:
𝑖𝑠
𝑖𝑓
, 𝑉𝐿
𝑖𝐿
m
𝑖
m
5. Determine the output voltage of the op amp, vo.
Iin
Vo
Iin
0V
V1
From inspection,
. Now, we can note that
is the voltage across the 50 kΩ resistor since the inverting terminal
of the opamp is at ground. Thus,
the output voltage, Vo.
. Next, we can apply KCL at V1 to determine
Substituting in the expression for Iin and rearranging the equation gives:
(
)
(
)
Since V- < Vo < V+, the solution is in the linear region of the op amp.
6. Determine the value for Is such that:
a. The amplifier has just entered the positive saturation region.
b. The amplifier has just entered the negative saturation region.
VOUT
VIN
The opamp is in the non-inverting configuration (in regards to VIN), so that we have:
(
Furthermore,
)
(
)
Saturation occurs when the output reaches the voltage rails of the opamp, ±10V. So, we can write:
m
a) Is = 1.333 mA at the transition between the linear and positive saturation regions.
b) Is = -1.333 mA at the transition between the linear and negative saturation regions
7. Calculate the values for Vs, Vo, if, i, and iL.
Because no voltage difference is allowed between the negative and positive input terminals of an op amp when there is a
feedback resistor present, then Vs must be equal to 3V.
Since no current flows into the negative (or positive) input terminals of the op amp, if must be equal to –1 mA.
Using KCL:
Applying Ohm’s Law:
(
)