R-2R Ladder Circuits
An explanation of how they work.
First Steps
Here are some basic concepts that will help to understand.
Circuit Diagrams
These are the components we will be using:
| Rail | Ground | Resistor |
|---|---|---|
| Represents the high voltage source | Represents ground (0V) | Represents a resistor with resistance $R$ |
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Voltage Drop Across a Resistor
In the circuit below, a single resistor is placed between the high voltage source and ground.
The voltage at the top is $V_{dd}$, and the voltage at the bottom is $0$. This means that as the current travels down the length of the resistor, the voltage is dropping along the way. By the time we get to the bottom, the voltage is all gone and it’s 0V.
Resistors in Series (Voltage Divider)
If the voltage is dropping as we go along the length of the resistor, what happens if we look at the voltage halfway along the resistor? We can break that resistor into two resistors like this:
Depending on the values of $R_1$ and $R_2$, the voltage at the mid point will be different. Lucky for us, it is very intuitive. If $R_1$ and $R_2$ are equal, that means half the resistance is above the midpoint and half the resistance is below the midpoint. In that case, by the time we travel halfway along the full resistance (all the way to the midpoint), half the voltage will be gone. So the voltage at the midpoint would be $\frac{V_{dd}}{2}$. If instead $R_1$ was 1% of $R$, then the voltage at the midpoint would be 99% of $V_{dd}$ because only 1% of the voltage drop has occurred by the time we travel 1% of the way across the total resistance.
This is what is typically called a “Voltage Divider” because it is dividing the voltage drop into 2 parts. The equation for the voltage at the midpoint can be written like this: $V_{midpoint} = \frac{R_2}{R_1 + R_2} * V_{dd}$. Usually since the top is fixed at $V_{dd}$ and the bottom is fixed at ground, we call the midpoint the “output” of the circuit, since that’s the voltage we care about (the voltage the circuit is “producing”).
Furthermore, we can also follow this process of splitting the resistors backwards by combining two resistors. Resistors in series can be simply added together. In this case, $R_1 + R_2 = R$. We can use this recursively to reduce circuits with many resistors in series down to a single resistor (if you have three resistors in series, you can combine the first two and now you’re left with just two in series, which you can combine again).
Resistors in Parallel
If we have two resistors in parallel, we can’t add the resistance values. Instead, consider the following circuit:
Here, an amount of current will flow through each of the resistors according to the size of the resistor. The fact that there are two paths for the current to flow through means that more current will flow overall. The formula for parallel resistors is as follows:
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$
Another way to write it:
$R = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$
This section doesn’t have as good of an explanation, but there’s the formula.
R-2R Circuit Concepts
An R-2R circuit is meant to convert a number to a voltage. What that means is if we have a number between 1 and N, we want to turn that into a voltage between 0V and $V_{dd}$. To do this, we use a binary representation of the number and a set of input nodes in the circuit.
Binary Numbers
For computer systems, we usually only use 2 voltages and we call them “high” and “low.” Digitally, they correspond to 1 and 0, respectively. In these examples, high is $V_{dd}$ and low is 0V.
A number can be represented in binary by a number of bits. For example, a 1-bit number only has 1 bit, whereas a 10-bit number has 10 bits. The number of different values that an N-bit number can have is $2^N$. So a 1-bit number can have 2 values, while a 10-bit number can have 1024 different values.
Mapping Binary Numbers to Voltages
To map numbers into voltages, we want to divide the output voltage range into equal parts. For example, if we had 100 possible different numbers, we would want the number to scale the output voltage by 1% of the maximum each time. If we had 10,000 possible values for our number, we would want steps of 0.01% of the maximum voltage. In other words, the higher the resolution of our number (the more possible values we have), the more possible values for the output of the circuit there are.
There is a slight problem with this system that you might have already noticed. If we want to go from 0 to $V_{dd}$ by 1% each step, there will actually be 101 steps because we include 0 as one of the values, and we also include $V_{dd}$ as one of the values. For this reason, R-2R circuits leave the maximum value out. So in reality, this kind of circuit with 100 steps would go from 0V to $\frac{99}{100}V_{dd}$.
However, that creates a nice little formula we can use. The output voltage is equal to the value of the number divided by the total number of numbers times $V_{dd}$. Take a look at the following table that is based on a 1-bit number. A 1-bit number can only be either 0 or 1, so there are only 2 rows in the table.
| Value | Binary | Voltage |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | $\frac{1}{2}V_{dd}$ |
You will notice that the 2 possible output values are 0 and $\frac{1}{2}V_{dd}$, but that we can’t get $V_{dd}$. Now take a look at a similar table for a 2-bit number.
| Value | Binary | Voltage |
|---|---|---|
| 0 | 00 | 0 |
| 1 | 01 | $\frac{1}{4}V_{dd}$ |
| 2 | 10 | $\frac{2}{4}V_{dd}$ |
| 3 | 11 | $\frac{3}{4}V_{dd}$ |
We have 4 possible value with a 2-bit number. Here is one more table for a 3-bit number, with 8 possible values.
| Value | Binary | Voltage |
|---|---|---|
| 0 | 000 | 0 |
| 1 | 001 | $\frac{1}{8}V_{dd}$ |
| 2 | 010 | $\frac{2}{8}V_{dd}$ |
| 3 | 011 | $\frac{3}{8}V_{dd}$ |
| 4 | 100 | $\frac{4}{8}V_{dd}$ |
| 5 | 101 | $\frac{5}{8}V_{dd}$ |
| 6 | 110 | $\frac{6}{8}V_{dd}$ |
| 7 | 111 | $\frac{7}{8}V_{dd}$ |
You should see the pattern where the output is based on the value of the number divided by the total number of possible values.
Let’s take a look at how we can build a circuit that provides this function.
R-2R Circuits
For an R-2R circuit to work, we need 1 output and N inputs, where N is the quantity of bits in our number. Each input will correspond to a single bit of the number.
1-Bit R-2R Circuit
Here is a 1-bit R-2R circuit. The output node is on the bottom-right, between the two resistors. At the top, we have an input called “bit 1” that can either be set to 0V or $V_{dd}$.
If we set the input node to 0V, then we have created a situation where no current will flow. There is no voltage at the input and there is no voltage at the ground, so there will be no voltage at the midpoint (output). This matches the first row of the 1-bit table above.
If we set the input node to $V_{dd}$, then we have created a voltage divider! The top is $V_{dd}$ and the bottom is ground. We can redraw it like this:
Looking at the Voltage Divider section above, we can use the formula:
$V_{output} = \frac{2R}{2R + 2R} * V_{dd} = \frac{1}{2}V_{dd}$
So now you can see how if we set the input to 0V, we get 0V out, and if we set the output to $V_{dd}$, we get $\frac{1}{2}V_{dd}$. This matches with the 1-bit table above. Now let’s look at a 2-bit R-2R circuit.
2-Bit R-2R Circuit
Here is a 2-bit R-2R circuit. This time there are 2 input nodes called “bit 1” and “bit 2.” There are 4 possible states for the circuit (bits 1 and 2 are both low; bit 1 is high and bit 2 is low; bit 1 is low and bit 2 is high; or bits 1 and 2 are both high). You can use the 2-bit table above to visualize all the possibilities.
Let’s look at each case one at a time.
First Case: Bits 1 and 2 Are Both Low
Both bits are low, so we ground the inputs.
Here, both inputs are at 0V and the ground is at 0V. Similarly with the 1-bit R-2R circuit, no current can flow, so the output is just 0V.
Second Case: Bit 1 is High and Bit 2 is Low
Bit 1 is high, so we connect it to $V_{dd}$. Bit 2 is low, so we ground it.
This case is a little more complicated. First, we need to determine the voltage at the very center node (at the “T” junction). It is labelled with a question mark. To do that, let’s first rearrange the circuit a little so that it looks more familiar.
Now let’s combine the 2 resistors on the left-hand fork by using the series resistor formula. $2R + R = 3R$.
Now we have 2 resistors in parallel that we can combine using the formula:
$R = \frac{1}{\frac{1}{3R} + \frac{1}{2R}} = \frac{6}{5}R$
Hey, what do you know, we’re left with a very familiar circuit again. This time we calculate the voltage in the middle as
$\frac{\frac{6}{5}R}{2R + \frac{6}{5}R} * V_{dd} = \frac{3}{8}V_{dd}$
So now we have our original circuit, but we know that the voltage at the “T” junction is $\frac{3}{8}V_{dd}$.
The voltage at that node will remain constant, so from here we can look at the circuit like this
Hmm… this should look familiar too. Yep, it’s an other voltage divider! I won’t redraw it vertically this time; let’s skip to the formula:
$\frac{2R}{2R + R} * \frac{3}{8}V_{dd} = \frac{1}{4}V_{dd}$
So for this case (bit 1 is high and bit 2 is low), we found that the output voltage of the circuit is $\frac{1}{4}V_{dd}$.
Third Case: Bit 1 is Low and Bit 2 is High
Bit 1 is low, so we ground it. Bit 2 is high, so we connect it to $V_{dd}$.
This one is a little easier. First let’s redraw the circuit a little.
We can reduce the bottom 2 parallel resistors to $\frac{1}{\frac{1}{2R} + \frac{1}{2R}} = R$.
Now let’s reduce those bottom 2 series resistors to $R + R = 2R$.
What do you know, another voltage divider. This time the voltage is $\frac{2R}{2R + 2R} * V_{dd} = \frac{1}{2}V_{dd}$, and that is the output of the circuit for this case.
Fourth Case: Bits 1 and 2 are Both High
Both bits are high, so we connect them to $V_{dd}$.
Once again, we need to do an intermediate calculation to find the voltage at the “T” junction (labelled with a question mark). Let’s rearrange the circuit.
Now we can combine the 2 series resistors into $R + 2R = 3R$.
Now we have 2 parallel resistors, so we can combine them into $R = \frac{1}{\frac{1}{3R} + \frac{1}{2R}} = \frac{6}{5}R$.
This is a voltage divider again, and the middle voltage is $\frac{2R}{\frac{6}{5}R + 2R} * V_{dd} = \frac{5}{8}V_{dd}$. Let’s go back to the original circuit and solve for the output now.
Once again, the voltage at that node is constant, so we can look at the circuit like this:
Finally, we can calculate the output voltage of the circuit using this final voltage divider. But be careful! All the voltage dividers we have seen so far have been connected to ground. This one goes between $V_{dd}$ and $\frac{5}{8}V_{dd}$ instead. This means that we have to modify our formula a little.
Remember how we came up with the formula? We said that as you travel along the resistor, you are dropping your voltage from $V_{dd}$ down to ground. Well, in this case we are not dropping all the way to ground. So instead of taking the fraction of the drop and multiplying by $V_{dd}$, let’s multiply by the total voltage drop. In this case, it’s $V_{dd} - \frac{5}{8}V_{dd} = \frac{3}{8}V_{dd}$. That gives us the amount of voltage the output is above the lower-voltage node. So in order to get the actual output voltage, we need to add the voltage of the lower node as well.
So our overall voltage divider formula for this circuit will be $\frac{5}{8}V_{dd} + \frac{R}{2R + R} * \frac{3}{8}V_{dd} = \frac{3}{4}V_{dd}$.
2-Bit R-2R Circuit Review
We found that for the first case, the output was 0V; for the second case, the output was $\frac{1}{4}V_{dd}$; for the third case, the output was $\frac{1}{2}V_{dd}$; and for the fourth case, the output was $\frac{3}{4}V_{dd}$. If we look back to the 2-bit table above, we see that these are the exact values we wanted to get. So now we can see that constructing the 2-bit R-2R circuit this way will allow us to convert numbers to voltages just like how we planned.
N-Bit R-2R Circuit
We can extend the circuit as far as we want to. Here is an 8-bit version.
Doing all the math to compute what the voltages are is a pain, but we can use what we learned earlier and just stick to the table.