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 impedanceC.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 V2. 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 VoutputRemember that Gain VinputVinput can be shown as 0.5 V Vin = IR (the drop to 0 V)Voutput can be shown as Vout = I (RF + R) = 10 kΩIf you assume current is equal: 0.5 V Vout RF R RF 1 = 1 kΩ Vin R RQuestions: 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)
C.3.3: The gain of an inverting amplifier VoutputIn an inverting amplifier Gain VinputVinput can be shown as Vin = IR. (Voltage drops from Vin to 0 V)As the voltage is already at zero, the voltage 5 kΩmust drop to negative heading towards Voutput 1 kΩ Voutput = − IRF R 0VIf you assume current is equal: 0V =2V Vout RF 0V Vin RQuestions: 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)
We say that the inverting input is a virtual earthpoint. Thatis, its potential is zero but it is not physically connected toearth.There are two very important rules to remember aboutInverting Amplifiers or any operational amplifier for thatmatter and these are. No current flows into the input terminals The differential input voltage is zero as V− = V+ = 0 (Virtual Earth)
Inverting Amplifier VoltagesSketch the outputobtained if the rising 5 kΩ 1 kΩpotential shown in the Rlower graph is applied 0V 0Vto the inverting =2Vamplifier (with a supply 0Vof 10 V)
Inverting Amplifier Voltages + Supply VoltageExtending to negativevoltages gives theimage on the right − Supply Voltage
+ Supply Voltage Inverting Amplifier − Supply Voltage + Supply VoltageNon- Inverting Amplifier − 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) 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
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 Vout = −13 VIf the supply is ± 13 V, use thecircuit to the right to showthat: 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 VSet Vin = 0V, and set the output tonegative saturation, and you will get 6.5 Vnegative voltages for V2 and V1 0.8 VIf Vout = −13 V, I = 0.11 mA, Vx =−2.3 VVx = −2.3 V is obviously less than +3,so you would have negativesaturation.
C.3.6: Solve problems involving circuits incorporating operational amplifiers Vout = −13 VIf the supply is ± 13 V, use the V1= 3.5Vcircuit to the right to showthat: 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
C.3.6: Solve problems involving circuits incorporating operational amplifiers Vout = +13 VIf the supply is ± 13 V, use the V1= 2.2Vcircuit to the right to showthat: 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.