Chapter 2 - Revision history

Cdxskier at 03:05, 18 January 2010

2010-01-18T03:05:59Z

← Older revision		Revision as of 20:05, 17 January 2010
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	*[[Ben Henry\|Henry,Ben]]		*[[Ben Henry\|Henry,Ben]]

			*[[Lau, Chris\|Christopher Garrison Lau I]]

			==Readers==

	*[[Lau, Chris\|Christopher Garrison Lau I]]		*[[Lau, Chris\|Christopher Garrison Lau I]]

Cdxskier: /* Reviewers */

2010-01-18T02:24:45Z

Reviewers

← Older revision		Revision as of 19:24, 17 January 2010
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	*[[Ben Henry\|Henry,Ben]]		*[[Ben Henry\|Henry,Ben]]

			*[[Lau, Chris\|Christopher Garrison Lau I]]

Fonggr: /* Problem 2.17 */

2010-01-13T19:13:50Z

Problem 2.17

← Older revision		Revision as of 12:13, 13 January 2010
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	[[Image:E2.17.PNG‎\|thumb\|300px\|Problem 2.17 after finishing step 4]]		[[Image:E2.17.PNG‎\|thumb\|300px\|Problem 2.17 after finishing step 4]]
	*Derive an expression for R in terms of the other resistor values, so that the output voltage due to the bias currents is zero.		*Derive an expression for R in terms of the other resistor values, so that the output voltage due to the bias currents is zero.
			*Writing two nodal equations for each terminal voltage: <math>\frac{-V_-}{R}-I_B=0</math>, <math>-I_B+\frac{-V_+}{R_1}+\frac{-V_+}{R_2}=0</math>
	*Solving the two nodal equations for the terminal voltage and then setting the two equations equal to each other gives: <math>-I_BR=-\frac{I_BR_2R_1}{R_1+R_2}</math>		*Solving the two nodal equations for the terminal voltage and then setting the two equations equal to each other gives: <math>-I_BR=-\frac{I_BR_2R_1}{R_1+R_2}</math>
	*Solving for R gives: <math>R=\frac{R_2R_1}{R_1+R_2}</math>		*Solving for R gives: <math>R=\frac{R_2R_1}{R_1+R_2}</math>

Fonggr: /* DC imperfections */

2010-01-13T18:55:56Z

DC imperfections

← Older revision		Revision as of 11:55, 13 January 2010
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	*Offset current is the difference between the bias currents. <math>I_{off} = I_{B+}-I_{B-}\,</math>		*Offset current is the difference between the bias currents. <math>I_{off} = I_{B+}-I_{B-}\,</math>
	*Offset voltage occurs when the output voltage is nonzero for zero input voltage.		*Offset voltage occurs when the output voltage is nonzero for zero input voltage.
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	==Canceling bias currents==		==Canceling bias currents==

Cdxskier: /* Canceling bias currents */

2010-01-13T07:03:50Z

Canceling bias currents

← Older revision		Revision as of 00:03, 13 January 2010
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	==Canceling bias currents==		==Canceling bias currents==
	*Because bias ~~currents flow~~ have equal magnitude and direction, it is possible to negate their effects.		*Because bias current flows have equal magnitude and direction, it is possible to negate their effects.
	**The orientation of the offset voltage and the direction of the offset current are unknown, thus it is not possible to correct for these parameters with a circuit design.		**The orientation of the offset voltage and the direction of the offset current are unknown, thus it is not possible to correct for these parameters with a circuit design.

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	==Reviewers==		==Reviewers==

	*[[Ben Henry\|Henry,Ben]]		*[[Ben Henry\|Henry,Ben]]

Cdxskier: /* Op Amp Nodal Analysis */

2010-01-13T07:02:37Z

Op Amp Nodal Analysis

← Older revision		Revision as of 00:02, 13 January 2010
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	*If negative feedback is present (and no positive feedback), then <math>V_+ = V_-\,</math>		*If negative feedback is present (and no positive feedback), then <math>V_+ = V_-\,</math>
	*Write nodal equations at <math>V_+\,</math> and <math>V_-\,</math>, but not at <math>V_o\,</math>		*Write nodal equations at <math>V_+\,</math> and <math>V_-\,</math>, but not at <math>V_o\,</math>
	**There is a voltage source inside the op amp. Writing a nodal equation at a voltage source adds an extra equation and an extra variable. ~~You gain no ground~~.		**There is a voltage source inside the op amp. Writing a nodal equation at a voltage source adds an extra equation and an extra variable. It doesn't make the problem any easier.

	=DC imperfections=		=DC imperfections=

Benjamin.henry: /* Problem 2.17 */

2010-01-13T02:17:43Z

Problem 2.17

← Older revision		Revision as of 19:17, 12 January 2010
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			==Reviewers==

			*[[Ben Henry\|Henry,Ben]]

Fonggr: /* DC imperfections */

2010-01-12T21:05:45Z

DC imperfections

← Older revision		Revision as of 14:05, 12 January 2010
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	#Ground the input		#Ground the input
	#Set the output to 0		#Set the output to 0
	~~:*Are we doing this so that we can make sure we don't have any other effects, yet the bias will still be there?~~
	#Add in the bias currents		#Add in the bias currents
	#Add in the new resistor to balance out the bias currents		#Add in the new resistor to balance out the bias currents
	#Set the two (input terminal) nodal equations equal to each other and solve for the new resistor		#Set the two (input terminal) nodal equations equal to each other and solve for the new resistor

			===Reasoning===
			*When the input is grounded, the bias currents will lead to an offset voltage. Arbitrarily setting the output to zero and adding in the bias resistor allows us to have current flowing through the bias resistor equal to even out the bias currents. Due to symmetry, the bias currents will cancel each other out.

	===Problem 2.17===		===Problem 2.17===

Fonggr: /* Ideal Op Amp Characteristics */

2010-01-12T20:57:55Z

Ideal Op Amp Characteristics

← Older revision		Revision as of 13:57, 12 January 2010
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	*[[Chapter_1#Amplifier_Models \| Zero output impedance]]		*[[Chapter_1#Amplifier_Models \| Zero output impedance]]
	*Infinite bandwidth		*Infinite bandwidth
	**~~To allow~~ for infinite gain regardless of ~~the~~ frequency~~? Instantaneous feedback?~~		**This means there are no inductors or capacitors in the model and allows for instantaneous changes (and infinite gain regardless of frequency).

	=Op Amp Nodal Analysis=		=Op Amp Nodal Analysis=

Fonggr: /* Amplifier Circuits */

2010-01-12T20:36:47Z

Amplifier Circuits

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	~~=Amplifier Circuits=~~
	~~==Buffer Amplifier==~~
	~~[[Image:BufferAmplifier.PNG\|thumb\|300px\|Buffer Amplifier ]]~~
	*Used to transfer voltage but not current to the following circuit. This amplifier can be used to negate the loading effects. No current flows through the amplifier, thus there is no voltage drop through the input resistor (going to the buffer amplifier).

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	~~==Inverting Amplifier==~~
	~~[[Image:InvertingAmplifier.png\|thumb\|300px\|Inverting Amplifier]]~~
	*Uses negative feedback to invert and amplify voltage. Using nodal analysis at the negative terminal, the gain is found to be <math>-\frac{R_2}{R_1}</math>
	*<math>R_{bias}=\frac{R_1R_2}{R_1+R_2}</math>
	*To get rid of unwanted DC components, a capacitor can be added inbetween <math>R_1\,</math> and <math>V_{in}\,</math>. In this case <math>R_{bias}=R_2\,</math>
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	~~==Summing Amplifier==~~
	~~[[Image:Summing_Amplifier.PNG‎\|thumb\|300px\|Summing Amplifier]]~~
	*<math>V_o=-R_f \left( \frac{V_3}{R_3}+\frac{V_2}{R_2}+\frac{V_1}{R_1}\right)</math>
	*If all resistances are equal, then the output voltage is the (negative) sum of the input voltages
	*<math>\frac{1}{R_{bias}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_f}</math>

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	~~==Noninverting Amplifier==~~
	~~[[Image:Noninverting_Amplifier.PNG‎\|thumb\|300px\|Noninverting Amplifier]]~~
	*<math>V_o=V_{in} \left(1+\frac{R_2}{R_1}\right)</math>
	*<math>R_{bias}=\frac{R_1R_2}{R_1+R_2}</math>
	*<math>R_{bias}\,</math> goes between the positive terminal and <math>V_{in}\,</math>
	*To get rid of unwanted DC components, a capacitor can be added inbetween the positive terminal and <math>V_{in}\,</math>. The bias resistor has the same value, and is placed inbetween the positive input terminal and ground.

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	~~==Differential Amplifier==~~
	~~[[Image:Differential_Amplifier_2.PNG‎\|thumb\|300px\|Differential Amplifier ]]~~
	*<math>V_o=V_2\frac{(R_1+R_f)R_g}{(R_2+R_g)R_1}-V_1\frac{R_f}{R_1}</math>
	*If you let <math>R_1=R_2\,</math> and <math>R_g=R_f\,</math> then the equation simplifies to <math>V_o=\frac{R_f}{R_1}(V_2-V_1)</math>