Chapter 4 – Operational Amplifiers – Part 1 Negative Feedback Positive feedback Operational Feedback

Coupling of Amplifiers 1) Direct Coupling 2) RC Coupling RC-coupled transistor amplifier Direct-coupled transistor amplifiers. c b

Impedance-coupled transistor amplifier 4)Transformer Coupling 3) Impedance Coupling Impedance-coupled transistor amplifier 4)Transformer Coupling Transformer-coupled transistor b c

Operational Amplifiers v1 vi vo vo Feedback  Amplifier  Feedback Amplifier Note: small letters indicate ac voltage signals. vi – input signal voltage vo – output signal voltage  = amplifier gain = feedback factor, vo = Feedback voltage vi < v1 , positive feedback vi > v1 , negative feedback

Why Operational Amplifiers The performance of transistor amplifiers is enhanced by returning a fraction of the output signal to the input terminals. This process is called feedback. A particular feedback, known as operational feedback, is used in operational amplifiers. Advantage of feedback :- (i) frequency-response of transistor amplifier is improved. (ii) output waveform distortion is reduced. (iii) Amplifier performance is less dependent on the changes in transistor parameters due to aging or temperature effects.

Importance Operational amplifiers are important building blocks for a wide range of electronic circuits. Op-amps are widely used in a vast array of consumer, industrial, and scientific devices. Op-amps are widely used in measurements and control applications. Perform mathematical operations such as addition or integration of an input signal, differentiator, Schmitt Trigger .

Negative Feedback v1 vi vo vo Feedback  Amplifier  vi > v1 , negative feedback = amplifier gain  = feedback factor

Feedback Network: The feedback network is a simple combination of (i) resistors, or (ii) resistors and capacitors or, (iii) capacitors.

RF = Feedback Resistor ___ Feedback Network 1 ___ Feed back Network 2

The R2 sets (i) the bais on the first stage (i. e The R2 sets (i) the bais on the first stage (i.e. sets the base voltage of T1) and (ii) provides dc feedback. It stabilizes the direct-coupled amplifier. It is not part of the ac feedback circuit.

Effect of the feedback ratio on the frequency response Amplifier gain,  = 1000 (without RF). With increased feedback by using smaller RF, the amplifier gain reduces and the frequency response is extended. RF = 1200,  = 120/1200 = 0.1. So  = 1000 x 0.1 = 100. The gain ’ = 1/ = 1/0.1 = 10. The upper cutoff frequency = 15 mhz. Voltage feedback

Exercise – 1: An amplifier has an input of 10 mV and output of 2 V Exercise – 1: An amplifier has an input of 10 mV and output of 2 V. What is its voltage gain in dB? Solution : i) So, by convention, we define: gain = 20 log (Vout/Vin)       = 20 log (2V/10mV)       = 46 dB Exercise – 2: If the amplifier has a voltage gain of 50 dB, an input signal of 20 mV, what would be the output voltage? Solution: 50 dB = 20log(Vout/Vin) 50/20 = log(Vout/Vin) Vout/Vin = 10(50/20) Vout/Vin = 10(2.5) (or 10^(2.5)) Vout/Vin = 316.22 Vout = 316.22 x 20 mV = 6.324 V

Exercise – 3: A negative feedback of = 2 Exercise – 3: A negative feedback of = 2.5 x 10-3 is applied to an amplifier of open loop gain 1000. Calculate the change in overall gain of the feedback amplifier if the gain of the internal amplifier is reduce by 20%.

Current Feedback A feedback signal is proportional to the output current.

Operational Feedback Operational feedback is versatile negative-feedback connection. Capable of performing a number of mathematical operations on input signals. High-gain dc-coupled amplifiers are universally used in this application. Operational amplifiers exhibits high input impedance, low output impedance, and introduce a 180o phase shift between the input signal and the output voltage.

Operational negative Feedback Rf = Resistance for negative feedback connected between input and output. Feedback ratio () varies from unity for a high-impedance source to R/ (R + Rf) for a low impedance source.

The feedback signal is a portion of the output signal and , therefore, also 180o degree out of phase with the input signal. The voltage at point S (input of the amplifier) is so small that it is considered to be virtual ground. Virtual potential is a point in a circuit which is at ground potential (0V) but is not connected to the ground. Analyze the operational feedback circuit by applying Kirchoff’s Current Rule to the branch point S. Amplifier input impedance (ri) is large. The current in this branch (i.e. 2) is negligible. Therefore, the current in R is equal to the current in Rf.

Virtual ground circuit Example – 1 (not for examination) 1) The output voltage Vo = - V1 (Rf / R) = - (1/ 1) 10V = -10V. 2) The difference in potential in the circuit, VT = 10V – (-10V) = 20V. 3) The total resistance of the circuit, RT = 1 + 1 = 2 .

4) The total current in the circuit, IT = VT / RT = 20/ 2 = 10A. 5) The voltage drop across the resistance R, VR = R x IT = 1  x 10A = 10V. 6) The voltage at virtual point S = V1 – VR = +10V – (+10V) = 0V. 7) This is reconfirmed by computing the Vo. To compute the Voltage Vo, find the voltage drop across Rf and subtract this from the voltage at point S. The result should be the voltage Vo. 8) The voltage drop across Rf VRF = Rf x IT = 1  x 10A = 10V. and Vo = 0V – VRF = 0 V – 10 V = -10V.

Virtual ground circuit Example - 2 1) The output voltage Vo = - V1 (Rf / R) = - (10/ 1) 1V = -10V. 2) The difference in potential in the circuit, VT = 1V – (-10V) = 11V. 3) The total resistance of the circuit, RT = 1 + 10 = 11 .

4) The total current in the circuit, IT = VT / RT = 11V/ 11 = 1A. 5) The voltage drop across the resistance R, VR = R x IT = 1  x 1A = 1V. 6) The voltage drop across Rf VRF = Rf x IT = 10  x 1A = 10V. 7) The voltage at virtual point S = V1 – VR = +1V – (+1V) = 0V. So the point S is at virtual ground. 8) To reconfirm this result, compute the voltage Vo. 9) Vo = Voltage at point S – Voltage drop across Rf = 0V – VRF = 0 V – 10 V = -10V.

Virtual ground circuit Example - 3 Compute the output voltage Vo.