Electronics > Beginners
DAC circuit
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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