Microcomputer Circuits and Processes

this R is in series with the next horizontal R, and so on. For a ladder of
any length, the input resistance is just R.
The second feature is to do with currents in the vertical resistors.
Assume 16mA flows in through P and out of Q. Look at the resistors
around point A, shown in figure 4.19.
point A
16mA 8 mA R
2R
R
Qo---
......• --------'
Figure 4.19
Showing how currents split at nodes. The value of the rightmost resistor R is the
equivalent resistance ofthe circuit to the right of point A in figure 4.18.
The vertical resistance, R, at A is of course the equivalent resistance
of all resistors to the right of A, according to the first feature mentioned.
You can see at once that the 16 rnA must divide into two 8 rnA currents
at point P, since the resistance of both branches from P to Q is the same.
The point is that the current is halved at each of the nodes, where
vertical and horizontal resistors join together. So, as shown in figure
4.20, the currents flowing in the vertical resistors form a binary
sequence.
16
2R
R
Qo---E:---4"'---~--"----E---
16 8 4 2
--*""--'--~--""'"
Figure 4.20
All currents shown in this '4-bit' R-2R ladder are in rnA. The ladder has the property of
dividing the input current of 16mA into a binary sequence.
This resistor ladder can be used very easily to make a precision
digital-to-analogue converter. The circuit is shown in figure 4.21.
The operational amplifier in this circuit is connected as a current-tovoltage
converter. (If you have not studied op-amps in detail, if the
current into the' -' or 'inverting' input is 1 rnA, then the output voltage
is -1 rnA x 1 kQ = -1 V.) In figure 4.21, the switches So to S3 lead their
currents to the I -to- V converter, which receives 8 + 4 + 2 + 1= 15 rnA,
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