Electrical Engineering and Electronic Principles

ELECTRICAL ENGINEERING 0
ELECTRICAL AND ELECTRONIC PRINCIPLES

ELECRTICAL ENGINEERING 1
Question 1
Part A
As per the provided information, the relation among voltage values:
V0= 8V4+ 4V3+ 2V2+ V1....................... (1)
The below figure shows the suitable op-amp circuit as per the provided
inputs:
Now, applying Nodal at point x:
I= i1+i2+i3+i4
I= V1/R1+ V2/R2+ V3/R3+ V4/R4
Based on the output voltage relation, we can put:
R1= R, R2= R/2, R3= R/4 and R4= R/8
Now, I= V1/R+ V2/R/2+ V3/R/4+ V4/R/8
Or, I= 1/R[V1+ 2V2+ 4V3+ 8V4]
Here, output voltage (Vo1)= -R*I
So, Vo1= -( V1+ 2V2+ 4V3+ 8V4)
Now, the overall output can be reported as:

ELECRTICAL ENGINEERING 2
Vo= (-1)*Vo1
So, V0= 8V4+ 4V3+ 2V2+ V1
Part B
The significant application of the developed circuits it that it may be utilized
as a weighted resistor type DAC converter.
Question 2
Part A
Apply KCL at node V1:
-I+ V1-V/R1+ V1-0/R2= 0
V1*[1/R1+ 1/R2]= I+ V/R1........................ (1)
Using the concept of virtual ground:
Va= Vb= 0
I= Va-V1/Rf
Put, Va= 0
I= -(V1/Rf)
Or, V1= -Rf*I............................. (2)