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Whats the point of the Op-Amp in this D/A converter

I've been tasked with building a digital to analogue converter for my class. Like so:

https://www.electronics-tutorials.ws/opamp/opamp11.gif

It needs to have 5 volts in on the inputs and 1 2 3 4 on the output corresponding to pressing 1 2 3 4. I understand how to do this, but I don't see why I can't just accomplish this with a voltage divider. If I eliminated the amp all together I think the circuit would achieve the same thing.

Thanks guys!
 
That is becasuse:
- 1. This is not a voltage divider. It doesn't work as a voltage divider at all.
- 2. Why: The negative input terminal at the opamp will always be at zeero, at least very close - probably less than your multimeter can measure.
- 3. And why is that: The negative terminal is keept close to zero because the opamp output will compensate so that the negative voltage will try to stabilize in a point where the current through Rf is excact the same as the current you provide via V1, V2, or V3.

Also: Try to do some calculations of voltage drop when different input voltage is applied - and remember to follow the rules of resistor value doubles for each.
Remember: the sum of currents into a node is always zeero.

Some important things to be aware of:
- The output function will generate a voltage representation of a binary input given that the resistor network is correctly set up.
- Opamps have limitations. It may not be able to deliver a voltage close to Vee/Vcc.
- Won't work unless you have a negative voltage supply also.
 
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