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# Oa05 - R f R a V i i a i f V O R L OA 4 V p = V n = V i and...

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OA 1 OPERATIONAL AMPLIFIERS An operational amplifier (or op amp) may consist of more than two dozen transistors, one dozen resistors, and a capacitor. However, it may be as small as a resistor. Because of its small size and simple external operation, an op amp is usually considered as a single circuit element. Eight Lead DIP Package: Chip Numbers represent pins Circuit Symbol: V S = DC power supply input, typically ± 15 V. Amplification = 100,000 or greater. For an ideal op amp, A = . Limitation of V O : V O max V S ; V O max INVERTING AMPLIFIER n o p _ + V s - R f V O R i R L V i i i i f + _ - + + - V n V p Offset Null Non- inverting Input 8 7 6 5 1 2 3 4 NC V S + Offset Null Invertin g Input Output V S - Negative Power Supply Positive Power Supply

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OA 2 V p is connected to ground, therefore V p = 0. Since we assumed V p = V n , then V n = 0 We know by KCL that i i = i n + i f , but i n = 0 so i i = i f V V R V V R i n i n o f - = - V n = 0 V R V R i i o f = - The voltage gain G = (V O / V i ) : G R R f i = - SUMMING AMPLIFIER R f V O R a R L V c + - V b V a R b R c + _ i f
OA 3 V p = 0 = V n i f = i a + i b + i c V V R V V R V V R V V R V R V R V R V R n o f a n a b n b c n c o f a a b b c c - = - + - + - = - + + NONINVERTING VOLTAGE AMPLIFIER

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Unformatted text preview: _ R f R a V i i a i f + + _ V O R L _ + OA 4 V p = V n = V i and V O = i f R f + i a R a i f = i a , and V O = i a (R f + R a ) Since i a = (V n / R a ) = (V i / R a ) V O = V i (R f + R a ) / R a = V i (R f / R a + 1) EXAMPLE 1 Find the output voltage V O in the following op amp circuit. Saturation voltage is ± 12 V. V O = -(3.5 V) x 10k Ω / 5 k Ω = -7.0 V EXAMPLE 2 Find the output voltage V O in the following op amp circuit.-+ V O 5 k Ω 15 k Ω V i = 3.5 V + _ 10 k Ω-+ 6 k Ω 10 V 4 k Ω 6 V 12 k Ω 10 k Ω V O OA 5 V o = - (-10 V / 4 k Ω + 6 V / 6 k Ω ) 12 k Ω = 18.0 V EXAMPLE 3 Find the current thru the 6k Ω resistor in the following op amp circuit. Saturation voltage is ± 15 V. V O = i f (5 k Ω + 10 k Ω ) = (3.5 / 10 k Ω ) (15 k Ω ) = 5.25 V _ 5 k Ω 10 k Ω 3.5 V + + _ V O 6 k Ω...
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Oa05 - R f R a V i i a i f V O R L OA 4 V p = V n = V i and...

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