Personal Projects

Square Root Operation Circuit


  • Using an op-amp and BJTs, I achieved the square root operation between Vout and Vin to be:

Vout=0.1 √(Vin )

In Detail description and calculation:

Firstly, we have a voltage divider, getting a 4k/(36k+4k) Vin=Vin/10 at OA1, Op-amp with Q5 would act as a voltage to current converter: IC5=V+/1k=Vin/10k A, Q6 and Q7 section would be simply a current source of 100µA as many examples we did at the beginning of this course. Afterwards, looking at the BJT loop Q1, Q2, Q3, Q4 while ignoring base currents, we get the following relationship: IC1*IC2=IC3*IC4, and looking at Q1 and Q5, we know that IC1=-IC5=-Vin/10k A, while I_C2= -100µA fixed by our current source while IC3=-IC4. Therefore, Vin/10k*100µ=I_C3^2 and rearranging we get: I_C3= -√(Vin/(10k*100µ)) and simplifying we get, I_C3=-√(Vin )/10k .

Lastly, the 1kΩ resistor I added would simply act as a current to voltage converter and gives us the relationship we want: Vout= - IC3*1kΩ=√(Vin )/10k*1kΩ, and simplifying this expression:

Vout=0.1 √(Vin )