eeet-221/1-January/1-15_Notes.md

In this instance, assuming traditional current I_{R2}=2mA \ I_{R1}=1mA \I_{R3}=1mA But there’s nothing saying you can’t do it in a different direction and just say its negative.

Now that we’ve established that, back to the original problem.

Example 1

(note, the positive and negative are switched in the below diagram)

Givens: V_+=V_-\ I_+=0=I_- There’s three major nodes. There is the positive side of the op amp, the output, and the minus. The output is the only side that has current.

Ultimate goal: Find I_L

We know I_L={6V \over 2k\Omega}=3mA. But why

now, do the nodal in all locations, recommend starting with the top. 0={V_--0 \over 2k\Omega}+{V_--V_O \over 2k\Omega}\ V_O=2V_-\ V_+={V_O\over 2}=V_-\ 0={{V_O\over 2}-6V\over 2k\Omega}+{{V_O\over 2}-0V\over 1.2k\Omega}+{{V_O\over 2}-V_O\over 2k\Omega}\ 0=\cancel{V_O\over 2}-6V-\cancel{V_O\over 2}+{\cancel{2k\Omega} \over 1.2\cancel {k\Omega}}({V_O\over \cancel{2}})\ 6V*1.2=V_O=7.2V\ {V_+ \over 1.2k\Omega}=i_L\ {3.6V\over 1.2k\Omega}=i_L=3mA\ QED

Inverting Amplifier

Goal: Find {V_O\over V_{in}}

0={0-V_{in} \over 12k\Omega}+{0-V_o \over 180k\Omega}\ V_{O}=15V_{in}\ {V_O \over V_in}=15\ QED

Non-ideal op-amp

So, we can guess that$V_O=0$. Then you become the V_{CC}, but it doesn’t make sense.

So, guess V_O=0=2V. Then… uh… moving on.

Ulimately, the op-amp has finite gain. Ultimately, V_+\approx V_-. This leads to V_- to be some number close to 2V. The output will be pretty much 2V.

This is known as the ==Voltage Follower==.

The 2V input is the peak of a sine wave, and the output will get worse as frequency increases.