
#1
Nov3013, 06:02 AM

P: 79

This picture is from http://www.ecircuitcenter.com/Circui...s/op_ibias.htm I am not very sure if I am getting the right idea of how the I_{B+} and I_{B} works, but when i solve it open up one of the I_{B} to solve the problem. for I_{B+}, (I_{B} is opened) V_{out} = V_{}( 1 + R_{2}/R_{1}), ..... V_{} = I_{B+}R_{3} V_{out} = I_{B+}R_{3}( 1 + R_{2}/R_{1}) for I_{B} (I_{B+} is opened) I_{B} = (V_{}  0) / R_{1} + (V_{}  V_{out}) / R_{2} V_{out }= V_{}(1+ R_{2}/R_{1}) + I_{B}....can i consider that when I_{B+} is opened, V_{+} = 0 = V_{} ? if I assume that I am right, i get what the page gives: V_{out} = R_{2}(I_{B}) one more thing, the I_{B+} is positive input biase current while the I_{B} is the negative input bias current? And the I_{input offset }= difference of I_{B} and I_{B+} >I_{B+}  I_{B} ? These two input bias current and input offset current are the parameters that always appear in the datasheet right? 



#2
Nov3013, 06:23 AM

P: 391

Try watch this video
http://www.youtube.com/watch?v=TxBJbZ0XFI Also how can you add current into resistor ? Vout = IB+R3( 1 + R2/R1) 



#3
Nov3013, 07:01 AM

P: 79

I consider that the I_{B+} current will pass throught the R3 and therefore the V_{+} is equivalent to (I_{B+})R_{3} . And i just apply the normal concept for finding the voltage amplification of operational amplifier circuit. 



#4
Nov3013, 07:31 AM

P: 391

Need help in understanding input offset curent of OpAmp
For sure we can use a superposition to find Vout.
First we assume IB+ > 0A and IB = 0A So have Vout'1 = I_{B+}*R3 * (1 + R2/R1) and this voltage is positive Next we open I_{B+} = 0A and we left with I_{B}. Vout'2 =  R2*I_{B} and for this IB current direction the output voltage is negative. And finally Vout = Vout'1 + Vout'2 = I_{B+}•R3•(R2/R1 + 1)  I_{B}•R2 



#5
Nov3013, 08:07 AM

P: 391

IB = (I_{B+} + I_{B})/2 and offset current Ios = I_{B+}  I_{B} so I_{B+} = IB + Ios/2 I_{B} = IB  Ios/2 



#6
Nov3013, 09:34 AM

P: 79

get it, thanks you very much !!



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