This is section C3 for the IB Physics SL program. Enjoy!

1. Option C: Digital Technology

2. C.3.1: State the properties of an ideal operational
amplifier
1. Infinite gain
2. Infinite input impedance
1. Impedance = the resistance between the input
terminals.
3. Zero output impedance
C.3.2: Draw circuit diagrams for both inverting and non-
inverting amplifiers (with single input) incorporating
operational amplifiers.

3. C.3.3: Derive an expression for the gain of an inverting
amplifier and for a non-inverting amplifier
The image to the right
shows an operational
amplifier, known as an op-
amp. Technically this is an
inverting amplifier.
The op amp has 2 inputs
and 1 output, as shown.
15V is known to be the
“supply”.

4. 1. The output voltage cannot exceed the supply, so the max and min output voltage can
be 15 V
2. The gain of an ideal op amp is infinite. for our purposes we will assume it is 1.0 x 106.
3. If the gain is 1.0 x 106 and the Input difference is 5 V, the Output voltage would be 5
V.
4. If the gain is 1.0 x 106 and the Input difference is 5 mV, the Output voltage would be
5000 V.
1. As the maximum output voltage is set at 15 V, the voltage reverts to 15 V. This
is called saturated.
5. Considering #3, if the difference between the inputs is larger than 15 V, the amp is
saturated. This is very small, so we could say that the input voltages are the same.

5. C.3.3: The gain of a non-inverting amplifier
Voutput
Remember that Gain
Vinput
Vinput can be shown as
0.5 V
Vin = IR (the drop to 0 V)
Voutput can be shown as
Vout = I (RF + R)
= 10 kΩ
If you assume current is equal: 0.5 V
Vout RF R RF
1 = 1 kΩ
Vin R R
Questions:
What is the gain of the amplifier? (11)
What is the output voltage? (5.5 V)
What is the current through RF? (0.5
mA)
What is the pd across RF? (remember it
is a potential difference drop from output
to Vin) (5 V)

6. C.3.3: The gain of an inverting amplifier
Voutput
In an inverting amplifier Gain
Vinput
Vinput can be shown as
Vin = IR. (Voltage drops from Vin to 0 V)
As the voltage is already at zero, the voltage 5 kΩ
must drop to negative heading towards Voutput 1 kΩ
Voutput = − IRF R
0V
If you assume current is equal: 0V
=2V
Vout RF 0V
Vin R
Questions:
What is the gain of the amplifier? (−5)
What is the output voltage? (10 V)
What is the current through R? (2 mA)
What is the pd across R (remember it is a
potential difference drop from Vin to 0 V) (2
V)

7. We say that the inverting input is a virtual earthpoint. That
is, its potential is zero but it is not physically connected to
earth.
There are two very important rules to remember about
Inverting Amplifiers or any operational amplifier for that
matter and these are.
No current flows into the input terminals
The differential input voltage is zero as V− = V+ = 0 (Virtual
Earth)

8. Inverting Amplifier Voltages
Sketch the output
obtained if the rising 5 kΩ
1 kΩ
potential shown in the R
lower graph is applied 0V
0V
to the inverting =2V
amplifier (with a supply 0V
of 10 V)

9. Inverting Amplifier Voltages
+ Supply Voltage
Extending to negative
voltages gives the
image on the right
− Supply Voltage

10. + Supply Voltage
Inverting
Amplifier
− Supply Voltage
+ Supply Voltage
Non- Inverting
Amplifier
− Supply Voltage

11. C.3.4: Describe the use of an operational amplifier circuit as a
comparator
A comparator is a circuit set up so that the input
potentials are different, and creates a binary
circuit dependent on the difference. The output
will be saturated intentionally.
If the + terminal is greater, the output will be +.
If the − terminal is greater, the output will be −.
By varying one of the potentials using a
temperature dependent resistor (thermistor) or a
light dependent resistor (LDR), you can create a
situation dependent circuit. A diode is also helpful,
a device which allows current to flow in only one
direction.

12. C.3.4: Describe the use of an operational amplifier circuit as a
comparator
The circuit is designed to go off if the
temperature rises above18 oC.
R1 = R3 = R4 = 139.6 Ω. Anything
above that, and the diode will stop
the current.
For R1 and R3, find voltage and
current at 17 oC . (1V, 7.2 mA)
At 17 oC, find the
potential at the +
terminal (0.975 V)
139.6 Ω
If the supply is 9 V
and the gain is
1,000,000, what
would the output be?
( 9V)
How could we
change this circuit
so that the bell rings 139.6 Ω 139.6 Ω
once it crosses a low
boundary? Why
would you want to

13. C.3.5: Describe the use of a Schmitt trigger for the reshaping of
digital pulses
A Schmitt trigger is a type of comparator
The output of a Schmitt trigger has two possible
values, HIGH and LOW. If the thresholds are
+1 V and -1V, then if the input is above 1V the
output will be HIGH. It will only switch to low
when the voltage drops below -1V.

14. C.3.6: Solve problems involving circuits incorporating
operational amplifiers
+ 9V
− 9V

15. C.3.6: Solve problems involving circuits incorporating
operational amplifiers
Vout = −13 V
If the supply is ± 13 V, use the
circuit to the right to show
that: R1 R2
If Vin = 0, the circuit has
negative saturation I = 0.11 mA
The circuit has an upper Vx= −2.3V
threshold of 6.5 V
The circuit has a lower
threshold of 0.8 V,
Vin = 0 V
Set Vin = 0V, and set the output to
negative saturation, and you will get 6.5 V
negative voltages for V2 and V1
0.8 V
If Vout = −13 V, I = 0.11 mA, Vx =
−2.3 V
Vx = −2.3 V is obviously less than +3,
so you would have negative
saturation.

16. C.3.6: Solve problems involving circuits incorporating
operational amplifiers
Vout = −13 V
If the supply is ± 13 V, use the V1= 3.5V
circuit to the right to show
that: R1 R2
If Vin = 0, the circuit has
negative saturation Vx= 3V V2= 16V
The circuit has an upper
threshold of 6.5 V
The circuit has a lower Vin= 6.5V
threshold of 0.8 V,
Upper Threshold
The threshold would switch once Vx 6.5 V
became greater than +3 V.
For Vx to equal +3, V2 must be +16V (to 0.8 V
counter the – 13V saturation)
If V2 = 16V, R2 = 100 kΩ, then I = .16
mA. If I = .16 mA, V1 = 3.5 V.
For Vx to be greater than 3V, Vin must
accommodate for V1 and still have 3 V
left over, so Vin must equal 6.5V

17. C.3.6: Solve problems involving circuits incorporating
operational amplifiers
Vout = +13 V
If the supply is ± 13 V, use the V1= 2.2V
circuit to the right to show
that: R1 R2
If Vin = 0, the circuit has
negative saturation Vx= 3V V2= 10V
The circuit has an upper
threshold of 6.5 V
The circuit has a lower Vin= 0.8V
threshold of 0.8 V,
Lower Threshold
When Vin > 6.5 V, Vout = +13 V 6.5 V
Now the drop from Vout to Vx is 10 V 0.8 V
If V2 = 10 V, R2 = 100 kΩ, then I = .1
mA. If I = .1 mA, V1 = 2.2 V.
For Vx to be less than 3 V, 2.2 V will
already be used up by V1, so Vin only
needs to be 0.8 V.