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### Ri and Ro of the Inverting Amplifier

Course: ITTC 412, Fall 2009
School: E. Kentucky
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Word Count: 511

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and 1/27/2003 Ri Ro of the Inverting Amplifier.doc 1/6 Inverting Amplifier We can use the concept of the virtual short to easily determine the input resistance of the inverting amplifier. Recall that the input resistance of an amplifier is: Ri and Ro of the Ri = vi ii For the inverting amplifier, it is evident that the input current ii is i1 : R2 i2 vi R1 v1 - ii = i1 v2 + ideal vo From Ohm's Law, we...

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and 1/27/2003 Ri Ro of the Inverting Amplifier.doc 1/6 Inverting Amplifier We can use the concept of the virtual short to easily determine the input resistance of the inverting amplifier. Recall that the input resistance of an amplifier is: Ri and Ro of the Ri = vi ii For the inverting amplifier, it is evident that the input current ii is i1 : R2 i2 vi R1 v1 - ii = i1 v2 + ideal vo From Ohm's Law, we know that this current is: ii = i1 = vi - v 1 R1 1/27/2003 Ri and Ro of the Inverting Amplifier.doc 2/6 The non-inverting terminal is "connected" to virtual ground: v1 = 0 and thus the input current is: ii = i1 = vi R1 We now can determine the input resistance: Ri = vi ii R = vi 1 vi = R1 The input resistance of this amplifier is therefore Ri = R1 . Now, let's attempt to determine the output resistance Ro. Recall that we need to determine two values: the short-circuit output current (ios ) and the open-circuit output voltage (voo ) . To accomplish this, we must replace the op-amp in the circuit with its linear circuit model: 1/27/2003 Ri and Ro of the Inverting Amplifier.doc 3/6 R2 i2 vi R1 ii = i1 + v 2 + - - v1 Roop ioop Aop (v2 - v1 ) vo = 0 ios From KCL, we find that: ios = where: i2 = and: v 1 - vo = R2 -Aop v1 i op o = -Aop v1 - vo Roop = Roop Therefore, the short-circuit output current is: ios v1 Aop v1 = - R2 Roop Roop - R2 Aop = R R op 2 o v 1 1/27/2003 Ri Ro and of the Inverting Amplifier.doc 4/6 The open-circuit output voltage can likewise be determined in terms of Aop and v1: R2 i2 vi R1 ii = i1 + v 2 + - - v1 Roop ioop Aop (v2 - v1 ) io =0 + voo - Here, it is evident that since io=0: i2 = -ioop where we find from Ohm's Law: i2 = v1 - ( - Aopv1 ) R2 + Roop v1 1 + Aop = op R2 + Ro 1/27/2003 Ri and Ro of the Inverting Amplifier.doc 5/6 and thus voo is: voo = v1 - R2 i2 1 + Aop = v1 - R2 v R2 + Roop 1 R + R op R2 1 + Aop o = 2 - R2 + Roop R2 + Roop Roop - R2 Aop v = R + R op 1 o 2 ( )v 1 Now, we can find the output resistance of this amplifier: Ro = voo ios Roop - R2 Aop R R op 2 o -1 Roop - R2 Aop = R + R op o 2 R2 Roop = R2 + Roop = R2 Roop In other words, the inverting amplifier output resistance is simply equal to the value of the feedback resistor R2...

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E. Kentucky - ITTC - 412
2/5/2003Superposition and OpAmp Circuits.doc1/4Superposition and Op-Amp CircuitsConsider this op-amp circuit:v1R1R2-v2R3ideal+voR4The easiest way to analyze this circuit is to apply superposition! Recall that op-amp circui
E. Kentucky - ITTC - 412
9/30/2003The Small Signal Equaton Matrix1/2The Small-Signal Equation MatrixWe can summarize our small-signal equations with the smallsignal equation matrix. Note this matrix relates the small-signal BJT parameters vbe , ib , ic , and ie . Colu
E. Kentucky - ITTC - 412
9/4/2003The Weighted Summer1/2The Weighted SummerConsider an inverting amplifier with multiple inputs!v1 v2 v3R1 R2 R3i1 i2 i3i-Rfideal+voFrom KCL, we can conclude that the currents are related as: i= and because of virtual g
E. Kentucky - ITTC - 412
1/31/2003SP 2.3-2.doc1/1Special Problem 2.3-2 An amplifier with an open-circuit voltage gain of Avo = 10 and an output resistance of Ro = 1K is connected in series to an inverting amplifier, as shown below. Find the value of R1 so that VO/VI =
E. Kentucky - ITTC - 412
1/31/2003SP 2.4-1.doc1/1Special Problem 2.4-1 The input VA() is a signal expressed as its Fourier Transform. You of course recall that the impedance of an inductor is Z = j L . Note both resistors have the same value. Determine VO() in terms o
E. Kentucky - ITTC - 412
1/31/2003SP 2.4-2.doc1/1Special Problem 2.4-2 Using two ideal op-amps, design a circuit which takes two inputs (VA(t) and VB (t) ) and produces at an opencircuit output the signal:d VA (t ) VO (t ) = - 2VB (t ) dtThe only capacitor that you
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
2/7/2003SP 2.6-1.doc1/1Special Problem 2.6-1 For the circuit below, find vO in terms of inputs v1 and v2. Find both the differential and common-mode gains of this amplifier. Determine CMRR. 9K 1K_ +vo4K2K2Kv1v2
E. Kentucky - ITTC - 412
2/7/2003SP 2.6-3.doc1/1Special Problem 2.6-3 Determine the open-circuit voltage gain of this amplifier.R4 =24K i2 vi R1 =1K R3 =4K-idealR2 =4K+vo
E. Kentucky - ITTC - 412
2/14/2003SP 2.7-2.doc1/1Special Problem 2.7-2 The amplifier below has a 3 dB bandwidth of 100 kHz.9K1K_vovi+1) 2) 3)What is the gain-bandwidth product of the op-amp used in this amplifier? At what frequency is the magnitude of th
E. Kentucky - ITTC - 412
2/14/2003SP 2.7-3.doc1/1Special Problem 2.7-3 Homer S. constructed the following circuit in his garage. He determined that the 3dB frequency (f3dB) of this circuit is 1 MHz - exactly the bandwidth he wanted. In order to generate more gain, Home
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
2/21/2003SP 4.6-5.doc1/1Special Problem 4.12 -5 It is known that the emitter voltage of the transistor in this circuit is at 2 volts with respect to ground (i.e., VE = 2.0 V). What then, is the value of resistor R?10.2 VR10 K 3.7 V = 100
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
2/21/2003SP 4.6-9.doc1/1Special Problem 4.6-9 For the circuit below, assume that the BJT is in the active region. Determine the base current and the collector voltage. Determine whether the active assumption was correct.10.7 V 1K 0.7 K = 78
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-10.doc1/1Special Problem 4.6-10 Assume the BJT is in active mode.1) Based on this assumption, find collector current iC, collector voltage VC and emitter voltage VE . 2) Determine if the original assumption was correct.15.7 V
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
4/4/2003SP 5.5-9.doc1/1Special Problem 5.5-8 The capacitors in this circuit are very large. Determine the small-signal voltage gainA = vo vi of this amplifier. vDetermine the small-signal input resistance of this amplifier.4.5 V 10 V10
E. Kentucky - ITTC - 412
4/21/2003SP 5.6-1.doc1/1Special Problem 5.6-1 In the circuit below, Q1 has K1=2 mA/V2 and Vt = 2.0 V. But, Q2 has K2 = 1 mA/V2 and Vt = 2.0 V. In other words Q1 and Q2 are not identical! 1) Determine R so that the drain current of Q1 is 8 mA. 2
E. Kentucky - ITTC - 412
11/10/2003SP 5.6-41/1 /1Special Problem 5.6-4 In the circuit below, Q1 has K=1.0 mA/V2 and Vt1 = 1.0 V. The transistor Q2 likewise has K = 1 mA/V2, but has a threshold voltage of Vt2 = 2.0 V. In other words Q1 and Q2 are not identical! The resi
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