Positive Feedback Issue

104

01:20:30

Why is it acting like negative feedback ?! .. Shouldn't it saturate the opamp at 15v ?

published 10 years ago

Striker27

10 years ago

This is acting like it should. Inverting terminal is grounded so the opamp is going push back untill the voltage at the positive terminal is also zero potential with respect to ground. Remember ideal opamp rules. V+=V- and current going into opamp input terminals are zero. Taking a KCL at the positive input terminal yields (Vs-0)/R1=(0-Vo)/R2. Solving for Voltage out in terms of Voltage source gives Vo= -(R2/R1)Vs. This setup is called a voltage inverter. To get a positive output. Try conecting the feedback to the inverting terminal. Hope this helps.

hurz

10 years ago

Its a schmitt trigger and only working as inverter here under clean simulator conditions. We had already some discussions about it, and if it would act the same wrong way in reality. Swap input terminals to get an inverter amp. Hope this helps.

BillyT

10 years ago

Helps me .

abobaker

10 years ago

@Striker27 well, thanks however the rule of V+=V- is valid only when it's negative feedback , as long as your out of saturation , and the open loop gain (ß*A) >>1 ..... Here in this case it's positive feedback making a schmitt trigger as hurz said

abobaker

10 years ago

@hurz Yeah this is great thanks .. So it's only because every thing is balanced here and is subject to the intial conditions, right ?

hurz

10 years ago

Right, i guess already at power on it wont find this mathematical stable single point of balance and directly goes into clipping as schmitt trigger.

MichaelFaraday

9 years ago

I agree that it's wrong. If you build this it will rail out positive or negative and stay there until you reverse the voltage on the positive input.

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