Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

subtractor circuit using OP AMP

Status
Not open for further replies.

Arrowspace

Banned
Full Member level 3
Joined
Jan 23, 2015
Messages
186
Helped
3
Reputation
6
Reaction score
3
Trophy points
18
Activity points
0
Can some one explain how it is working

- - - Updated - - -

According to me it should be -4.5 Vdc
 

Attachments

  • Subtractor.JPG
    Subtractor.JPG
    41.5 KB · Views: 89

Your math is off. :wink: You apparently forgot about how R7 and R8 affect the non-inverting gain.

The voltage at the op amp (+) input is 2.5V, multiplied by the gain of the op amp, which is +2 (due to R7 and R8), giving an output of +5V for that input

The +7V at the minus input is multiplied by the op amp gain of -1, giving an output of -7V for that input.

Using superposition, the op amp output is thus -2V, exactly as you simulated.
 

Normally all values are equal for a unity gain differential amp ( 10K) to minimize offset from input bias current.

inverting gain= -R7/R8 *7V = (-1) * 7V
Non Inv gain = 1+ |inverting gain| * V+in= (1+1) * 5V*R5/(R5+R6)= (1+1) 5v * 1/2= 5v

Using superposition rules the output-7+5 =2 the difference.

As a check, the Vin-=Vin+ whenever the output is in the linear range ( not saturated)
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top