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Design Specifications - Electronics - Solved Assignment, Exercises of Computer Science

These are the Solved Assignment of Electronics which includes Simple Inverting Amplifier, Input Resistance, Current, Load Resistor, Input Voltage, Given Circuit, Voltage Equations, Node, Confirmed etc. Key important points are: Design Specifications, Simple Inverting Amplifier, Input Resistance, Current, Load Resistor, Input Voltage, Given Circuit, Voltage Equations, Node, Confirmed

Typology: Exercises

2012/2013

Uploaded on 03/22/2013

dhritiman
dhritiman 🇮🇳

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Assignment #1
#1-6 The design specifications for a simple inverting amplifier require an input resistance
of 10 k and a voltage gain, A = - 6.2
a) Prepare a design using an ideal OpAmp.
b) If a real OpAmp with the properties:
Ri = 2 M
A = 200 k
Ro = 75
is used, what error in the gain and input resistance will result due to the non-ideal
properties of the OpAmp?
Solution: Will be included this week
#1-11. Determine the current through the load resistor, RL, as a function of the input voltage,
vi, for the given circuit. Know that v2 = v1 and that iRL = v2/RL .
Solution: Write node voltage equations at v2 and v1:
@ v1 :
@ v2 :
, since v2 = v1
but (v1 - vo )/4.7k = (vi - v1 )/2.2k = (vi - v2 )/2.2k so:
.
Note also that iRL = v2 /RL .
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Assignment #

#1-6 The design specifications for a simple inverting amplifier require an input resistance of 10 kΩ and a voltage gain, A = - 6. a) Prepare a design using an ideal OpAmp. b) If a real OpAmp with the properties:

  • Ri = 2 MΩ
  • A = 200 k
  • Ro = 75 Ω is used, what error in the gain and input resistance will result due to the non-ideal properties of the OpAmp?

Solution: Will be included this week

#1-11. Determine the current through the load resistor, RL , as a function of the input voltage, v i , for the given circuit. Know that v 2 = v 1 and that iRL = v 2 / RL.

Solution: Write node voltage equations at v 2 and v 1 : @ v 1 :

@ v2 :

, since v 2 = v 1 but ( v 1 - vo )/4.7k = ( vi - v 1 )/2.2k = ( vi - v 2 )/2.2k so:

Note also that iRL = v 2 / RL.

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Simplifying yields the solution:

Confirmed using Pspice and vi = 1V and RL = 1k found in #1-11 Pspice folder. The result is the current through RL.

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