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Kindly help me in solving this.. Op amp is ideal, calculate Vo. Your hhelp will be really appreciated

What type is this Op amp? What is formula for calculating Vo.

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    \$\begingroup\$ I'm voting to close this question as off-topic because it looks like homework with no attempt to a solution. \$\endgroup\$
    – Rev
    May 17 '16 at 6:03
  • \$\begingroup\$ Yup... positive feedback is creating all confusion... otherwise this is really simple.. \$\endgroup\$ May 17 '16 at 6:13
  • \$\begingroup\$ What type of opamp? The question tells you : an ideal one. \$\endgroup\$ May 17 '16 at 10:27
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When dealing with operational amplifiers you have to consider two things : First there is no current coming in or out of op amp (not true in real life applications). Second , voltages at the positive and negative inputs of the opamp is always equal. When we look at the positive input rail, you can see that output voltage and the ground form a voltage divider circuit. if we call the voltage at the input, Vin. Vin = (Vo x 200000)/ 300000, Vin = 2Vo/3 . Remember what i said before voltages at the positive and negative inputs of the opamp is always equal.That means voltage of the inverting input rail is also 2Vo/3.Now remember rule number one : there is zero current in the input rail , which means current coming from the 400mV source is going straight to output without visiting the input.İf we calculate the equation for that : (400-2Vo/3) /5000 = (2Vo/3-Vo)/50000 . If you solve the equation, you can see that Vo equals to 631.57 mV.

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  • \$\begingroup\$ Maybe i'm misreading things but afaict he has more positive feedback than negative feedback which means the amplifier wont be acting as an amplifier at all. \$\endgroup\$ May 17 '16 at 5:54
  • \$\begingroup\$ If your analysis gives +632mV, how can you then flip it to -632mV? If the true answer is -632mV there must be something wrong with your analysis. \$\endgroup\$
    – Chu
    May 17 '16 at 12:22
  • \$\begingroup\$ While technically correct, this answer is extremely misleading because what you have calculated (+631.57 mV) is a metastable state -- a state that cannot persist if there's the slightest perturbation (noise) in any of the signal levels. Because of the strong positive feedback, the only stable states for the output voltage are +10V and -10V, the supply rails. Also, your final statement is just wrong. \$\endgroup\$
    – Dave Tweed
    May 17 '16 at 12:38
  • \$\begingroup\$ Yes - the answer from Dogus Ural is wrong. As soon as the power is switched-on the opamp goes into saturation due to the dominating positive feedback. \$\endgroup\$
    – LvW
    May 17 '16 at 20:14

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