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The R-2R ladder utilizes parallel resistances to create even ratios of currents. This is achieved by repeatedly connecting a pair of parallel resistors: one with resistance R and the other with resistance 2R (hence the name R-2R ladder), except for the last pair, where both resistors have the same resistance of 2R. The figure below illustrates the base of a 3-bit R-2R ladder.
Starting with the rightmost pair, notice that the parallel resistors have the same resistance. Because of this, the current going into that branch splits evenly. Let’s also observe the equivalent resistance at the node joining that pair with the previous one. Since the resistors are in parallel, their equivalent resistance would be (2R*2R)/(2R+2R) = 4R^2 / 4R = R.
The next pair provides the current that feeds into the aforementioned pair, but it is also fed by current coming from its preceding pair. Given the previous calculations, we know that the equivalent resistance of the rightmost pair is R. Since that pair was in series to the R resistor of this pair, the equivalent resistance of that branch is 2R, which matches the resistance in the other branch. So, once again, the current splits evenly. Additionally, the equivalent resistance of this pair is once again R since the resistances of the parallel branches are both 2R.
This pattern repeats itself for the leftmost pair. The current splits evenly once again. Now, this pair of resistors is fed by the reference current. As such, the current going through the leftmost 2R resistor is Iref/2, and the current going through the R resistor is also Iref/2. This current is then fed into the next pair of resistors, which splits it evenly again. Thus, the current through that 2R resistor is Iref/4. The current going into the next pair of resistors is also Iref/4, but then it is split evenly once more such that the current through both 2R resistors is Iref/8.
Notice that the source current has been divided three times, which matches the digital resolution that we had previously specified (n = 3). As such, we can attach switches to the ends of the 2R resistors (while the last one remains grounded), which in turn can either be flipped to ground or to an op-amp. The switches represent the bits, with the leftmost switch (controlling the most current) being representative of the MSB and the rightmost belonging to the LSB. When the switch is flipped to ground, the bit is a 0, whereas flipping the switch to the op-amp is treated as a 1.
Also notice that the equivalent resistance at the leftmost node is R. Thus, we can treat that entire network of resistors as just a series resistor of resistance R. If we attach a feedback resistor of resistance R to the op-amp, then by Ohm’s Law, R = Iref/Vref = Iout/Vout. Iout is the sum of all the currents flowing into the op-amp. In the case of the 3-bit R-2R ladder, if we flipped all the switches such that each branch connects to the op-amps, then Iout = I0 + I1 + I2 = Iref/2 + Iref/4 + Iref/8 = Iref * ⅞ = Vref / R * ⅞. Substituting Iout in the Ohm’s Law equation then gives us Vout = Iout * R = Vref / R * ⅞ * R = Vref * ⅞, representing the maximum output voltage that can be outputted by a 3-bit code.
Suppose we were given the 3-bit code 010. What would the output voltage be if the reference voltage was 16 V? Since bits 0 and 2 are zero, the switches corresponding to those bits must be flipped toward ground, whereas bit 1’s switch goes to the op-amp. Since I2 is the only current going through the op-amp, Iout = I2 = Iref / 4 = Vref / R * ¼. Thus, Vout = Vref / R * ¼ * R = Vref * ¼ = 16 * ¼ = 4 V. A shortcut to this method is by multiplying the reference voltage by the digital value (in this case, 010 = 2) and dividing by 2^n (n = 3, so 2^n = 2^3 = 8). So for this example, multiplying the reference voltage by 2/8 (or simply ¼) yields the output voltage.
Since we can limit our digital inputs to a finite amount of numbers, designing a DAC is much simpler than designing an ADC, which can have an infinite amount of rational number inputs even when operating only within a specific voltage range. ADCs also tend to be more expensive than DACs due to the circuitry needed. Nonetheless, ADCs are crucial to interfacing with technology by analog means.
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