anonymous
  • anonymous
Describe the operation of the circuit in the figure below. Assume that any non-zero output turns on the LED.
MIT 6.002 Circuits and Electronics, Spring 2007
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
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KenLJW
  • KenLJW
Assuming the output of OpAmp 2 not connected to OpaAmp1 OpAmp2 biased a +5 on + side V1<5 OpAMP2 output 10 biasing OpAmp1 at 50/16 on - side V1>5 OpAmp2 output 0 biasing OpAmp1 at 0 on - side 50/16 < V1 < 5 LED on 5 < V1 LED on Therefore LED on V1 > 50/16 LED off V1 < 50/16
anonymous
  • anonymous
Please just explain in details including how you get 50/16. I'm totally lost.

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KenLJW
  • KenLJW
OpAmp2 + bias 15 10/(5 +10) = 10 previous error When OpAmp2 - is < 10 output Vo2 = 10, is > 10 output Vo2 = 0 OpAmp1 - bias is Vo2 5/(5 + 5+ 6), 50/16 = 25/8 or 0 For OpAmp1 out of 10 requires Vin > Vo2 5/16 = 25/8, 0 OpAmp1 - = 0 when Vin > 10 therefore LED on OpAmp1 - = 25/8 when Vin < 10, therefore 25/8 < Vin < 10 LED on OpAmp1 - = 25/8, Vin < 25/8 LED off Therefore Vin > 25/8 LED on Vin < 25/8 LED off
KenLJW
  • KenLJW
All that is used in the analysis is the Voltage divider rule and the inverting principle of Op amps.

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