Hi, i need to understand the required calculations to complete the DAC table

Hi, i need to understand the required calculations to complete the DAC table 

If you need to work it all out, then go back to your inverting amplifier feedback theory. You have nine volts available  to selectively supply each of the input resistors Rin. If a switch is on, then the current in that resistor is 9/Rin. Calculate the current flowing in each resistor where the switch is on, and add all of the resulting currents together to give the total current into the summing junction of the amplifier.  All of the total current is similarly flowing into the summing junction via the feedback resistor, this must be the case to give 0 volts at the summing junction, which is a virtual earth. Hence the output voltage is (-the sum of the input currents*Rfeedback).You only have to do one calculation, preferably with only one resistor turned on(chose the one that gives a change in the first output value supplied in your table) from there on in, the rest of the table may be solved by inspection, for if you look at the input resistor values you will notice that they follow a significant sequence, or weighting. Cheers,Richard

Richard's explanation is a good way to look at it but if you have trouble understanding that way, here is another way to do the same calculation.All the resistors in the paths with the switches are in parallel IF the switch is closed. The resistors with open switches can be ignored. Calculate the equivalent resistance of the parallel resistors. From that you should be able to find the output voltage by using the formula for an inverting op-amp circuit. (Vout = -Vin *(Rin/Rf))Richard's method is actually easier to calculate in the long run. You should try both ways and try to understand why they are equivalent.

Hi Richard thanks.my main problem in that DAC circuit is how to complete the 8 table Coulomb, when i have 4bits only ? 

I will try to explain again...Using the example of the fourth line which has the switches for the 10K and 20K on. There are two ways you can calculate this.Using Richard's method, you know the current for the 10K and 20K resistors since one side is at 9V and the other side is ground. This current adds together and is the current going to the minus pin of the op-amp. This current goes through the 450 Ohm and generates the output voltage. This works the same as with only one switch on only the currents are combined.The other way to look at it is that with the two switches on, the 10K and 20K are in parallel so you can calculate an equivalent resistance then use the equivalent resistance to calculate the output as you did on the previous lines.Both methods will give the same result. The key to understanding this is to see that the resistors will combine when you have more than one switch on.

Hi EphraimLet's start from basics.   Forget the resistors and the switches for a minute.  What you have here is an inverting amplifier configuration:
Now from your op-amp theory you should know that VOUT = -(VIN x R1/R2)Now we know VIN (it's connected to +9V), and we know R1 = 450 Ohms.  R2 will depend on the position of the switches. So again to keep it simple let's take the case shown in your diagram.  Only the switch on the 20K resistor is closed, the rest are open.  so plug all those values int our formula above:VOUT  = - (9 x 450 / 20000)  = - 0.2025 V = -202.5 mVWhich is almost exactly what your simulator gets!So now lets try it with only the 40K resistor in circuit;  Put 40000 into the above formula instead of 20000 and we get -101.25 mV.    Use 10K and we get -405 mV, and with 5K we get -810 mV.Do you notice a pattern here?  Starting from 40K, every time we halve R2's value we get double the output voltage.  Lets try now with the next value in your tabel, with the 10K AND 20k resistors in circuit.  10K in parallel with 20K gives us 6.667 K (I'm assuming you know how to calculate resistors in parallel so I won't do it here).  SO plug that into the formula:VOUT  = - (9 x 450 / 6.667)  = - 0.6075 V = -607.5 mV
Notice here that this is the SUM of what we got with the 10K and 20K resistors on their own!Now here I must take issue with whoever is giving you these problems.   Instead of labelling your resistor columns 0 1 2 3 I would label them 8 4  2  1 as shown here - I have also relabelled the columns and shown VOUT directly in mV:
Now we can show how a DAC works.  You see the 20K column is labelled 2.  We can say that the output voltage is going to be 100 mV x 2 (disregard the odd few extra mV here).  The 10K column is labelled 4, and the output is 400 mV.  The 1,2,4,8 that I have put in the colums are the WEIGHTINGS of each resistor that is in circuit, that Richard was talking about above.  So you can say that the output is (Sum of weightings) x 100 mV.  Lets try it with the other values we calculated:
For the 40K resistor only we get 1 x 100 mV - correctFor the 10K and 20K resistors in circuit we get (2 + 4) = 6 x 100 mV = 600 mV - correctSo you can now very easily calculate your output voltage for any combination of the switches, just by adding together the weightings for those switches / resistors.Again I must take issue with the originator of these problems.  He has not show a 1 in the 40K column for any of the values shown.You said in your reply to Richard:My main problem in that DAC circuit is how to complete the 8 table Column, when i have 4bits only ?    (FYI it's column, not coulomb, and it's actually 8 ROWS and 4 Columns- 5 if you include the output.)Well - this is again a fault of your question setter.  You have 4 bits, and your binary theory tells you that 24 is 16.  So the question has only given you half the possible values for this DAC.  I'd suggest you redo the table with all 16 values and see what you get.Does this make it any clearer?Cheers / David