OP AMP circuit applications

OP AMP circuit applications - Operational Amplifier Linear...

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Operational Amplifier Linear Circuit Applications Sattar Hussain sattar.hussain@ryerson.ca
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1- Basic Amplifier Concepts A signal amplifier is basically a two-port network whose function is to produce an output signal that is a larger replica of the input signal applied to its input port. where A is a constant referred to as amplifier gain . The above equation determines the amplifier voltage transfer characteristics, i.e., v o -v i relationship ) ( ) ( t Av t v i L = v L (t)
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2- Operational Amplifier Operational Amplifier (op amp) is a high gain active circuit element available in small-size, low-cost, and highly reliable package. It is an amplifier designed to perform mathematical operations of addition , subtraction , multiplication , division , differentiation and integration 3- The open-loop model Ideal op amp behaves as an ideal difference amplifier that amplifies the difference between two input voltages. v o =A V(OL) (v + - v - )=A V(OL) v d A V(OL) =open loop gain
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Op amp voltage supply limitation Amplifier circuit needs a dc power supply to provide almost all the output signal power and the power dissipated in the circuit. V S + and V S - determine what is called the upper saturation level and the lower saturation level of the amplifier, respectively, Amplifier v i v o input output V+ V- positive supply connection negative supply connection V S + =positive DC supply V S - =negative DC supply + < < S out S V t v V ) (
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An ideal op amp has the following characteristics: 1. Infinite open-loop gain, A V(OL) ≈ ∞ 2. Infinite input resistance, R i ≈ ∞ 3. Zero output resistance, R o ≈ 0 dynamic range ≈ 0 (disadvantage) A v(OL) (v + -v - ) V S + V S -
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8 Parameter Typical range Ideal values Open-loop gain, A v(OL) 10 5 to 10 8 Input resistance, R i 10 5 to 10 13 Output resistance, R o 10 to 100 0 Supply voltage, V S 5 to 24 V Typical ranges for op amp parameters
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Example Calculate the output voltage, v o, if the open-loop gain of the op amp, A v(OL) , is 100,000. Solution v 0 = A v(OL) v d = A v(OL) (v + - v - ) = A v(OL) (v 2 - v 1 ) = 10 5 (20-10) x 10 -6 = 1V Example Calculate the output voltage, v o ,if the open-loop gain of the op amp, A v(OL) , is 200,000. v 0 = A v(OL) v d = A v(OL) (v + - v - ) = A v(OL) (v 2 - v 1 ) = 2 x 10 5 (20 + 20) x 10 -6 = 8 V
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4- The closed-loop model feedback loop idal op amp R f is an external resistor that connected around the op amp from the output terminal back to the inverting terminal ( negative feedback) or to the non-inverting terminal (positive feed back) . The op amp is said to operate in closed-loop configuration To analyze the performance of an idealop amp circuit, an important assumption will be needed: that the current flowing into the input circuit of the amplifier is zero, or 0 = in i
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0 0 0 t v o (t) t v i (t) dynamic range distorted output L- V S + V S v S S v S A V v A V / / + < <
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4.1 The inverting amplifier i F =feedback current Applying Ohm’s law: , , (ideal op amp) thus, Note that, v + =0 ( connected directly to the ground) Since v out =A V(OL) (v + - v - ), This implies in F S i i i = + S S S R v v i = F out F R v v i = 0 = in i or i i i i F S F S = = + 0 + = = v A v v OL V out 0 ) ( Ideal op amp A V(OL) = S
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OP AMP circuit applications - Operational Amplifier Linear...

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