Discussion in 'Electronics Homework Help' started by cps13, Apr 22, 2017.

1. ### cps13

46
1
Feb 25, 2013
Hi,

I have a question in my HND about R/2R ladders, I have been given the diagram below and asked the following question:

The diagram of FIGURE 1 shows a 3 input R/2R digital to analogue converter circuit. By working back from the output (or otherwise) to find the equivalent input resistance in order to determine the input voltage, show that the output voltage values for the following digital input combinations are as shown below.

The voltages "as shown below" are:

1) When A= 1, B = 0 and C = 0 the output = 1/12 of Vin
2) When A= 0, B = 0 and C = 1 the output = 4/12 of Vin
3) When A = 1, B = 1 and C = 0 the output = 3/12 of Vin Source: https://www.dropbox.com/s/6k5utzcy41isudy/R2R.jpg?dl=0
(Sorry but I cannot seem to get the image to show up, I have shared it and made it public on dropbox but it always just comes up as a red X)

The trouble is that I cannot get these answers. Also doing research online there always seems to be another 2R resistor between point E and ground.

I was struggling trying to work it out by cancelling resistors out so I have simulated the circuit in multisim and get the following for each 3 combinations:

1) 1/5 of Vin
2) 2/5 of Vin
3) 3/8 of Vin

Can anybody tell me where I am going wrong?

thanks

Last edited by a moderator: Apr 22, 2017
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,499
2,839
Jan 21, 2010
As far as I can tell, the output voltage (expressed as a fraction of Vin) is independent of Vin and R.

This one can be confirmed by observation:

If you ground A and B and put some voltage Vin on C, you have a voltage divider with 2R above the junction and R below it. The output is 1/3 Vin (i.e. 4/12 Vin).

The answer to (1) is also possible by inspection, but I leave that up to you.

Once you have confirmed (1) and (2), I would try A=0, B=1, and C=0. Since none of the voltages (A, B, or C) are dependent, you can simply sum the result for outputs where a combination of them are 1. This will give you the answer to (3).

3. ### cps13

46
1
Feb 25, 2013
Ok so I think I understand what you are saying.

this is my attempt for explaining 1:

Working from point A, when A is high the voltage passes through a 2R/2R potential divider and becomes ½ of VIN at point F. The voltage then continues through a 2R/R potential divider and is further divided by 1/3 at point E, giving a voltage at point E of 1/6 of VIN (1/2 x 1/3). The voltage then continues through a potential divider of 2R/2R, so at point D it measures 1/12 of VIN (1/6 x 1/2).

This is my attempt for 2:

Working back from the output, when C is high it passes through a potential divider of 2R over R. Therefore producing an output voltage of 1/3 of VIN. So measuring at point D would produce 1/3 VIN.

Am I on the right tracks?

thanks

4. ### cps13

46
1
Feb 25, 2013
This is my attempt at 3

Firstly working out for A = 0, B = 1 and C = 0 working from point E, if B is high then the voltage passes through a 2R/R potential divider producing an output of 1/3 of VIN. This then in turn passes through a potential divider of 2R/2R and becomes 1/6 VIN (1/3 x 1/2).

It is known that when A = 1 this contributes a voltage value of 1/12 of VIN to the output.

Therefore 1/6 + 1/12 = 2/12 + 1/12 = 3/12 or 1/4

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,499
2,839
Jan 21, 2010
For parts 1 and 2, just calculate resistors in series and parallel to simplify the circuit.

My recommendation is to use superposition to solve part 3.

You have the correct answers, but feeding one potential divider into another does not give the result you describe.

Give the resistors numbers, then simplify the circuit by combining resistors that are in parallel or series with another resistor of an equivalent value. Doing this for A, B, and C individually high will give you the answer for the first 2 parts, and by superposition you can get the answer to the third part.

For example, for C only high, Raf is in parallel with Rfg. This is in series with Rfe. That combination is in parallel with Rbe. All that is in series with Red, and finally all of that is in parallel with Rdg. Now you can get the resistance between d and g (ground) and the result is a simple voltage division.

6. ### Ratch

1,094
334
Mar 10, 2013
Where or what is Vin?

Ratch

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,499
2,839
Jan 21, 2010
It is the voltage present on A, B, and/or C when they are in the "1" state.

8. ### Harald KappModeratorModerator

11,713
2,719
Nov 17, 2011
This is where you err. The potential divider is loaded by the resistors to the right which makes the voltage at F < 1/2*V(A).
I recommend you use the superposition theorem to solve this task. For this first calculation connect B and C to ground an work from right to left to evaluate the resulting parallel and series connection of resistors. Replace each pair of parallel or series resistors by a single resistor with the equivalent value (should be easy as only R and 2R are used ).
Your result should be a potential divider from A to ground 2R in the upper leg (A-F) and a smaller resistor in the lower leg (F-Gnd).

Repeat this for the other combinations.

This is effectively the same as Steve has explained using only a different wording.

9. ### Ratch

1,094
334
Mar 10, 2013
Ok, it looks like a node analysis is the way to go. Only three nodes.

Ratch  