Electronics > Beginners

DAC circuit

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nForce:
Sorry, but I still don't understand. What do you mean by D? DAC? And what is delta?

So DAC outputs a voltage between 0 and 3.3 V.  And because it's 12-bit DAC the resolution is 3.3/4095?
What about those op amps? And there is +10 V or -10 V, shouldn't be the range then 20 V?

Wimberleytech:

--- Quote from: nForce on January 26, 2019, 12:08:35 pm ---Sorry, but I still don't understand. What do you mean by D? DAC? And what is delta?

--- End quote ---

D is the digital input.  It has a value between 0 and 4095.  When D=0, the output of the DAC is 0, when D=4095, the output of the DAC is 3.3.

"delta" is a change.  0 to 3.3V is a delta of 3.3V.  1V to 2V is a delta of 1V.


--- Quote ---So DAC outputs a voltage between 0 and 3.3 V.  And because it's 12-bit DAC the resolution is 3.3/4095?

--- End quote ---
The size of an LSB is 3.3/4095.  The smallest unit it can resolve is 1 LSB.

--- Quote ---What about those op amps? And there is +10 V or -10 V, shouldn't be the range then 20 V?

--- End quote ---
The opamps do what opamps do.
Yes, -10V to +10V is a delta of 20V

Wimberleytech:
Here is the circuit with some example values.

Use your circuit analysis skills and analyze the signal path.

nForce:
So V1 = 1.65 V.

And your U1 op amp has a gain of 3. In my case since the resistors are all the same, it has a gain of 2.

So: 1.65 * 2 = 3.3, and because the delta is 3.3 then 3.3/3.3 = 1 which is A = 1.

Right? But why is delta important here?

Thank you for your effort.

Wimberleytech:
Delta is important because that is what the original "K" equation was calculating.  Calculating small signal gain.  Dont worry about the -1.65 V voltage source when calculating K.

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