Industrial Electronics DAE 32003 Industrial Electronics Jabatan Kejuruteraan Elektrik, PPD

Chapter 5: Operational Amplifier (Op-Amp) 2.1 Op Amp structure and properties 2.2 Op Amp characteristics 2.3 Inverting Op Amp 2.4 Non-inverting Op Amp 2.5 Op Amp circuits 2.6 Op Amp as interface device to input signal 2.7 Function of voltage follower

OP AMP structure & properties

2.1 OP AMP structure & properties It’s a five-terminal device: 2 input terminals 1 output terminal 2 power supply terminals The input terminals are identified as + and - , which does not indicate polarity but how the output responds if an input terminal of the same polarity is applied. The + input terminal is called the “non-inverting” input, which means if a positive signal is sent to this input, the output signal will be positive. Conversely, a negative input yields a negative output. The – input terminal is called the “inverting” input, which means if a positive signal is sent to this input, the output signal will be negative while a negative input yields a positive output.

2.1 OP AMP structure & properties And for the power supply terminals, the positive supply is connected to the +ve terminal and the negative to the –ve terminal. The two power supply terminals allow the output voltage to swing to either a positive or negative voltage with respect to ground.

2.2 OP AMP characteristics There are 3 important characteristics of op amps that make them ideal amplifiers : High input impedance High voltage gain Low output impedance

2.2 OP AMP characteristics Assume open loop voltage gain, AVOL is very large approximately ~ 200,000. Vo = AVOL(Vni-Vi) Vo = AVOLVdiff Vdiff = Vo/AVOL = 15 / 200000 = ~0.07mV (quite small) Since Rin is considered infinity (~1MOhm), Ii = Vdiff / Rin = ~0V / 1MOhm = 0 A

2.3 Inverting OP AMP The inverting amplifier, will amplify the input, Vs and invert the value which is going to be negative value. A technique called feedback is used to control the gain of this device, and it is accomplished by connecting a resistor from the output terminal to an input lead. A negative feedback circuit is shown below.

Both input terminals have high impedances, therefore they do not allow current to flow into or out of them. The potential at the negative input lead is called 0-volt virtual ground (that is, it acts like a 0-volt ground). The positive input lead is connected to an actual 0-volt ground potential. Because point VG is 0V, there’s a voltage drop of 2V across the 2kΩ resistor RIN. 1mA flows through it. The 1mA can’t flow into the op amp. Therefore it flows up through the 10kΩ feedback resistor, RF , developing a 10V across it. Because VOUT is measured with respect to the virtual ground, its voltage is -10V.

The voltage gain, AV of the op amp is determined by: AV = VOUT / VIN = 1 + (RF / RIN) The gain is influenced by the resistance ratio of RF compared to RIN. The larger RF becomes compared to RIN, the larger the gain. The gain will always be greater than 1. VOUT = [1 + (RF / RIN)] x VIN

IR1=VR1 / R1 = (Vs-Vi) / R1 = (Vs-~0) / R1 = Vs / R1 = 0.5 / 1kOhm = 0.5mA IRF= (Vi -VOUT) / RF = (0-VOUT) / RF = -VOUT / RF So, VOUT= -IRF x RF = -IR1 x RF *(IRF=IR1) = - (0.5mA) (20kOhm) = -10V (Since it is inverting, the output will be –ve) Closed loop voltage gain, AVCL= VOUT / Vs = -10V / 0.5V = -20 AVCL= VOUT / Vs = -Vs (RF/R1) / Vs AVCL= -RF / R1=-20

2.4 Non-inverting OP AMP The value, Vs will be amplify but not going to be invert. V-terminal = Vs IR1 = Vs / R1 So, Vs = IR1 x R1 IRF= (VOUT - Vs) / RF But IR1 = IRF Vs / R1= (VOUT - Vs) / RF VOUT – Vs = (RF / R1) Vs VOUT = (RF / R1)Vs + Vs = VS[(RF/R1)+1] VOUT / Vs = (RF / R1) +1 = 1 + (RF / R1) = AV So, AV = 1 + (RF / R1) Thus, for R1 = 1kOhm and RF = 20kOhm, Vs = 0.5V as before, the non inverting amplifier provide voltage gain of AV= 1 + (20k / 1k) = 21 VOUT = 21 Vs = 21 (0.5) = 10.5V

2.5 OP AMP circuits – Comparator amplifier This device compares the voltage applied to one input to the voltage applied at the other input. Any difference between the voltages drives the op amp output into either a positive or negative volt. Inverting input voltage < non-inverting input voltage = positive output voltage Inverting input voltage > non-inverting input voltage = negative input voltage Inverting input voltage = non-inverting input voltage = zero output voltage

2.5 OP AMP circuits – Comparator amplifier

2.5 OP AMP circuits – Comparator amplifier Comparators are used in many industrial applications as sensor monitors. For example, the comparator can monitor analog sensor such as temperature and vibration sensors in environmental testing equipment. When the temperature, VIN falls below a preset temperature, VREF, the comparator activates its load, the heating relay and the heat is turned on. When the vibration sensor VIN exceeds a preset number VREF, the comparator activates its load, a relay and the vibration stopped.

2.5 OP AMP circuits – Summing amplifier When 2 or more inputs are tied together and the applied to an input lead of an op amp, a summing amplifier is developed. This type of amplifier is capable of adding the algebraic of sum of the AC or DC signals. The current for each input is calculated and then summed to obtain the resulting current flow through RF. The VOUT is determined by multiplying IRF and RF

2.5 OP AMP circuits – Summing amplifier I1 = V1 / R1 I2 = V2 / R2 IT = (0-VOUT)/RF but IT= I1+ I2 -VOUT / RF = (V1 / R1)+(V2 / R2) -VOUT = [V1/R1+V2/R2] RF So, VOUT = -[V1/R1+V2/R2] RF Let R1 = 10kOhm, R2 = 20kOhm, RF = 40kOhm, V1=1.2V, V2= -1.9V So, -VOUT = [V1/R1+V2/R2] RF -VOUT = [1.2/10k+(-1.9)/20k] 40k = 1V VOUT = -1V

2.5 OP AMP circuits – Differential amplifier The differential amplifier finds the algebraic difference between two input voltages. Neither the inverting nor the non-inverting input is grounded. Instead, the signals are applied to both inputs at the same time, and the difference between them is amplified. If the signals are the same, the output voltage is zero. The difference between this two signal is considered an error The error is going to be VOUT. The gain is 1 since it only comparing. VOUT = (V2 - V1) x (RF / R1) Example: V1 = 2.2V , V2 = 1.5V RD = RF = 86kohm, VOUT = ? R1 = R2 = 10kohm VOUT = (1.5 - 2.2) x (86k / 10k) = (0.7) X 8.6 = -6V

2.5 OP AMP circuits – Integrator amplifier An integrator is an amplifier circuit that continuously increases its gain over a period of time. The magnitude of the output is proportional to the period of time that the input signal is present.

2.5 OP AMP circuits – Integrator amplifier In standard amplifier, which uses two resistors, has output which is nearly instantaneous. But output of an integrator is not instantaneous, it is function of time, which is controlled by the resistor capacitor network, R1 and C1. When input signal change, the capacitor changes to a new value and the time it takes to charge to a new value is T, the integral time.

2.5 OP AMP circuits – Integrator amplifier VOUT = - (1 / R1C1) X ∫VIN dt Cut off frequency, fc=1 / (2πRFC1) Stop act as integrator when f < fc and act as inverting amplifier.

EXAMPLE Given the integrator amplifier, sketch and label the values of the output signal Vout for Vin that is 1kHz square wave with a peak voltage of + 1 V (2Vp-p).

EXAMPLE - solution The input signal is specified as square wave varying between +1V and -1V at 1 kHz rate, T= 1/f = 1/1kHz = 1ms which means that the input will be +1V for half time, or 0.5ms and at -1V at 0.5ms.

EXAMPLE - solution For t = 0 to 0.5ms, Vi=1V Vo1= -(1/R1C1) X ∫VIN dt + Vo(0) =-1/(10kΩ x 0.01µF) x + 0 = -10000 x = -10000 x 0.5ms = -5V For t =0.5ms to 1.0ms Vo2= -(1/R1C1) X ∫VIN dt + Vo1 =-1/(10kΩ x 0.01µF) x + (-5V) = [-10000 x ] - 5V = [-10000 x (-0.5ms)] -5V = 0 V

2.5 OP AMP circuits – Differentiator amplifier The output of differentiator is proportional to the rate of change of input; VOUT = - (RFC1) x dVIN/dt Cut off frequency, fc = 1/(2πR1C1) If f > fc , it stops acting as differentiator and act as inverting amplifier.

EXAMPLE Calculate VOUT in figure above where RF = 2.2k Ohm and C= 0.001uF and where VIN is ramp input that goes from +5V to -5V in time given from figure shown below.

EXAMPLE - solution For t = 0 to 50us VOUT = - (RFC1) x dVIN/dt = -(2.2k)(0.001u) x 10/(50u) = -0.44V For t=50us to 100us VOUT = -(2.2k)(0.001u) x [-10/(50u)] = 0.44V For t=100us to 150us VOUT =-(2.2k)(0.001u) x 10/(50u) For t= 150us to 200us

2.7 Function of voltage follower A unique non-inverting amplifier is the voltage follower, which has its output connected directly to its inverting input, thus producing an output that is equal to the non-inverting input voltage both amplitude and polarity. Because the output is equal to the input, the gain is always 1, and so the voltage follower is sometimes referred to as a “unity gain amplifier”.