Op amps explained

Op amps explained

• 1.
  Option C: DigitalTechnology
• 2.
  C.3.1: State theproperties 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 anexpression 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 gainof 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 gainof 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 thatthe 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 Sketchthe 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)
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  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 theuse 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 theuse 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 theuse 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 problemsinvolving circuits incorporating operational amplifiers + 9V − 9V
• 15.
  C.3.6: Solve problemsinvolving 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 problemsinvolving 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 problemsinvolving 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.