Option C: Digital Technology

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.

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”.

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.

C.3.3: The gain of a non-inverting amplifier

Remember that

Vinput can be shown as

Vin = IR (the drop to 0 V)

Voutput can be shown as

Vout = I (RF + R)

If you assume current is equal:

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)

Gain VoutputVinput

VoutVin

RF RR

1RFR

= 10 kΩ

= 1 kΩ

0.5 V

0.5 V

C.3.3: The gain of an inverting amplifier

In an inverting amplifier

Vinput can be shown as

Vin = IR. (Voltage drops from Vin to 0 V)

As the voltage is already at zero, the voltage must drop to negative heading towards Voutput

Voutput = − IRF

If you assume current is equal:

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)

Gain VoutputVinput

VoutVin

RFR

= 2 V

5 kΩ

R0 V

0 V

0 V

1 kΩ

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)

Inverting Amplifier Voltages

Sketch the output obtained if the rising potential shown in the lower graph is applied to the inverting amplifier (with a supply of 10 V)

= 2 V

5 kΩ

R0 V

0 V

0 V

1 kΩ

Inverting Amplifier Voltages

Extending to negative voltages gives the image on the right

+ Supply Voltage

− Supply Voltage

Inverting Amplifier

+ Supply Voltage

− Supply Voltage

Non- Inverting Amplifier

− Supply Voltage

+ Supply Voltage

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.

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)

139.6 Ω

139.6 Ω 139.6 Ω

At 17 oC, find the potential at the + terminal (0.975 V)

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 once it crosses a low boundary? Why would you want to do that? (flip the diode, too cold)

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.

C.3.6: Solve problems involving circuits incorporating operational amplifiers

+ 9V

− 9V

C.3.6: Solve problems involving circuits incorporating operational amplifiers

If the supply is ± 13 V, use the circuit to the right to show that:

If Vin = 0, the circuit has negative saturation

The circuit has an upper threshold of 6.5 V

The circuit has a lower threshold of 0.8 V,

Set Vin = 0V, and set the output to negative saturation, and you will get negative voltages for V2 and V1

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.

Vout remains −13 V

R2R1

Vout = −13 V

I = 0.11 mAVx= −2.3V

Vin = 0 V

6.5 V

0.8 V

C.3.6: Solve problems involving circuits incorporating operational amplifiers

Upper ThresholdThe threshold would switch once Vx became greater than +3 V.

For Vx to equal +3, V2 must be +16V (to 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

R2R1

Vx= 3V

Vout = −13 V

V2= 16V

V1= 3.5V

Vin= 6.5V

If the supply is ± 13 V, use the circuit to the right to show that:

If Vin = 0, the circuit has negative saturation

The circuit has an upper threshold of 6.5 V

The circuit has a lower threshold of 0.8 V,

6.5 V

0.8 V

C.3.6: Solve problems involving circuits incorporating operational amplifiers

R2R1

Vx= 3V

Vout = +13 V

V2= 10V

V1= 2.2V

Vin= 0.8V

If the supply is ± 13 V, use the circuit to the right to show that:

If Vin = 0, the circuit has negative saturation

The circuit has an upper threshold of 6.5 V

The circuit has a lower threshold of 0.8 V,

Lower ThresholdWhen Vin > 6.5 V, Vout = +13 V

Now the drop from Vout to Vx is 10 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.

6.5 V

0.8 V