a.
If you look at all the resistors in parallel, at the right end, 2R in
parallel with 2R give a resistor of R, which is in series with another R
resistor, which gives a new 2R. This 2R is in parallel with another 2R and so
forth up to the left side. Thus at the end we have an R resistor for the
voltage Vr. Thus we have Vr = R*I so I = Vr/R, independent of the digital input
code.
b.
The output is such that Vo = - Rf*I- the maximum of I-
is by definition I = Vr/R.
The current I is divided
into different branches that have all a 2R resistor and thus the current is
divided by two each time there is a new branch.
Considering that the
switches can be on 1 or on 2 (and that is the value of the b’s) the current
will be each time divided by two and taken into account for the total I-
only if the value of the b is 1. We get the current I multiplied by the values
of the b’s and divided by two each time.
Thus we have the formula
asked for.
c.
We can realize that I+ + I- = I – In where In is
the current in the last resistor on the right.
This current’s value is I/2n
(because divided n times by two), which is 2-n*Vr/R
Thus we get I+ +
I- = I – In = Vr/R – 2-n*Vr/R = (1 – 2-n)*Vr/R
d.
We have b1, b2, b3, and b4.
We have V0 = -
Rf*Vr/R *(b1/2 + b2/4 + b3/8 + b4/16)
Rf = R = 10kΩ
and Vr = -16 V
Thus Vo =
16*(b1/2 + b2/4 + b3/8 + b4/16)
We have the
different combinations and the output
0000 gives V0 =
0V
0001 gives V0 =
1V
0010 gives V0 =
2V
0011 gives V0 =
3V
0100 gives V0 =
4V
0101 gives V0 =
5V
0110 gives V0 =
6V
0111 gives V0 =
7V
1000 gives V0 =
8V
1001 gives V0 =
9V
1010 gives V0 =
10V
1011 gives V0 =
11V
1100 gives V0 =
12V
1101 gives V0 =
13V
1110 gives V0 =
14V
1111 gives V0 =
15V
The output varies
continuously from 0V to 15V, it is thus a DAC which transforms the digital
variation (bits of 1 or 0) into analog (continuous).