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I recognize that as a National Semiconductor schematic.Actually my question is not to find wrong in schematic. Question is please explain the working of analog multiplier provided schematic there.
Then use some proper software which doesn't have such restrictions, try LTSpice which can be found using Google.Chandu06450 said:Chandu06450]We use pspice student version and it didn't support for this diagram(no. of nodes are more than its limit).
The word "instrumentation" is misleading. An instrumentation amp won't work here. A modern quad op amp would.The op-amp used is ancient, use a modern instrumentation quad op-amp.
I recognize that as a National Semiconductor schematic.
In a BJT, Ic≈Ie=Is*(e^(Vbe/Vt)-1)), the well-known diode equation.
For Ic>>Is, Ic≈Is*e^(Vbe/Vt)
Ic/Is=e^(Vbe/Vt)
Taking the log of both sides,
Vbe/Vt=ln(Ic/Is)
In the attached schematic,
V(A)=Vbias-Vt*ln(I1/Is)
V(B)=Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is)
V(C)=Vbias-Vt*ln(I3/Is)
Vbe4=V(C)-V(B)
substituting,
Vbe4=Vbias-Vt*ln(I3/Is)-(Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is))
simplifying,
Vbe4=Vt*(ln(I1/Is)+ln(I2/Is)-ln(I3/Is))
Vbe4=Vt*ln(I1*I2/(I3*Is))
From the diode equation,
I4≈Is*e^(Vt*ln(I1*I2/(I3*Is))/Vt)
I4/Is=e^ln(I1*I2/(I3*Is)
Taking the log of both sides,
ln(I4/Is)=ln(I1*I2/(I3*Is))
I4=I1*I2/I3
Since R1=R4=R9=R11=10k,
Vout=V1*V2/V3
Q.E.D.
LOL!Could you restate that without using all those annoying numbers ??
=FB=