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1-January/1-13_Notes.md

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+“I really enjoy torture-teaching people.” - Professor Cliver
+
+This course is… Op-amps. And its about imperfections with op-amps.
+
+For exams:
+
+1 sheet per exam. (for the final (3rd test) you get 3 sheets)
+
+# Ideal op-amp
+
+Stands for operational amplifier.
+
+Amplifies current, amplifies voltage. Can be used for signal conditioning.
+
+What are some operations it can do?
+
+- Sum
+- subtract
+- integrate
+- differentiate
+
+It has 5 pins. The diagram below is an ideal, and thus is missing the power and ground pins. There is text in their respective locations. The high and lower power pins are dependant upon the datasheet.
+
+![image-20200113161920562](image-20200113161920562.png)
+
+## Goals in life
+
+- $V_+ =V_-$
+- $I_+=I_-=0$
+
+Ideal op-amps complete these goals. We will play with this a little, then we will move on to more complex cases.
+
+The output is ultimately somewhere between the high power and the low power.
+
+## Example diagram
+
+![image-20200113163253802](image-20200113163253802.png)
+
+***Results using an ideal op-amp***
+
+$i_L={V_{in} \over R}={6V\over 2k\Omega}=3mA$
+
+$v_o=7.2V$
+
+^ These are found through nodal.
+
+Nodal is about finding a voltage at any point in the circuit.
+
+Nodal is currentIn = currentOut
+
+Alternatively:
+${V_{from} - V_{to} \over R}={V_{from} - V_{to} \over R}$
+
+Next class, we’ll discuss why $i_L\ne{V_{in}\over R+R_L}$
+
+
+

1-January/1-15_Notes.md

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+![image-20200115160404149](image-20200115160404149.png)
+
+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)
+
+
+
+![image-20200113163253802](image-20200113163253802.png)
+
+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 ${V_O\over V_{in}}$
+
+![image-20200115163502894](image-20200115163502894.png)
+$$
+0={0-V_{in} \over 12k\Omega}+{0-V_o \over 180k\Omega}\\
+V_{O}=15V_{in}\\
+{V_O \over V_in}=15\\
+QED
+$$
+
+# Non-ideal op-amp
+
+![image-20200115164231523](image-20200115164231523.png)
+
+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.

1-January/1-22_Notes.md

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+# Finite gain op-amps
+
+$V_O=A(V_+-V_-)$, which implies that $V_+\ne V_-$
+
+![finiteGain](finiteGain.png)
+$$
+0={V_--0\over R_i}+{V_--V_O\over R_f}\\
+{V_O\over R_f}=V_-({1\over R_f}+{1\over R_i})\\.\\.\\.\\
+{V_O \over V_{in}}={A\over 1+A\beta}\approx{1\over \beta}\\\therefore\\
+{V_O\over V_{in}}=1+{R_f\over R_i}
+$$
+
+## Contrasted with infinite gain op-amps
+
+$$
+{V_{in}-0\over R_i}={V_O-V_{in}\over R_f}\\.\\.\\.\\
+{V_O\over V_{in}}=1+{R_f\over R_i}
+$$
+
+You might be saying, well, whats the difference?
+
+The finite op-amps have the $A$ values in there. We then assume $A$ is giant, so it goes back to the infinite.
+
+# Inverting Amplifier
+
+![invertingOpAmp](invertingOpAmp.png)
+
+Givens:
+$V_O=A(0-V_-)$
+$$
+0={V_--V_{in}\over R_i}+{V_-V_O\over R_f}\\
+V_-({1\over R_f}+{1\over R_i})= V_{in}({1\over R_i})+V_O({R_f})\\.\\.\\.\\
+V_-=V_{in}\alpha+V_O\beta\\.\\.\\.\\
+{V_O\over V_{in}}=-{R_f\over R_{in}}
+$$

1-January/1-29_Notes.md

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+# Example 1
+
+![example1](1-29_Notes.assets/1_29NotesEx1.png)
+
+## Ideal
+
+$$
+{-v_i\over R_i}+{-V\over R_1}=0\\
+V={-R_1v_i\over R_i}\\
+\\[30pt]
+{V\over R_2}+{V-v_o\over R_3}+{V\over R_1}=0\\
+{-R_1v_i\over R_2R_1}+{-R_1v_i\over R_3R_1}+{v_i\over R_i}={v_o\over R_3}\\
+{v_o\over v_i}={R_1\over R_i}+{R_3\over R_i}(1+{R_1\over R_2})=A
+$$
+
+# Differential Amplifier
+
+![differentialOpAmp](1-29_Notes.assets/differentialOpAmp.png)
+$$
+{V-V_1\over R}+{V\over AR}=0\\
+{AV\over AR}+{V\over AR}={V_1\over R}\\
+{V_1A\over A+1}=V\\[25pt]
+{V-V_2\over \cancel R1}+{V-V_o\over A\cancel R}=0\\
+V-V_2+{V_1\over 1+A}={V_o\over A}\\
+{V_1\cancel{(A+1)}\over \cancel{(A+1)}}-V_2={V_o\over A}\\
+\underline{\overline{|V_o=A(V_1-V_2)|}}
+$$
+
+# Finite Gain
+
+![nonInvertingOpAmp](1-29_Notes.assets/nonInvertingOpAmp.png)
+$$
+A_{CL}={1+{R_f\over R_i}\over 1+{1\over A\beta}}\hspace{15pt}\beta={R_i\over R_i+R_f}\\
+$$
+![invertingOpAmp](1-29_Notes.assets/invertingOpAmp.png)
+$$
+A_{CL}={1+{R_f\over R_i}\over 1+{1\over A\beta}}\hspace{15pt}\beta={R_i\over R_i+R_f}\\
+$$

1-January/hw1.md

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+# ***Skyler MacDougall***
+
+## ***Homework 1: Due 1/20/2020***
+
+### Chapter 2
+
+1. Consider the circuit of Figure P2-1 with $R_i=10k\Omega$ and $R_f=50k\Omega$
+
+    ![hw1Q1](hw1Q1.png)
+
+    1. Determine the closed-loop voltage gain.
+
+    $$
+    A_{CL}={V_o\over V_i}={R_f\over R_i}=-5
+    $$
+
+    2. Determine the input impedance of the circuit.
+
+    $$
+    r_i=R_i=10k\Omega
+    $$
+
+    3. Determine the ideal output impedance of the circuit.
+
+    $$
+    r_o=0\Omega
+    $$
+
+    4. Determine the peak input voltage $v_i$ (peak) for which linear operation is possible.
+
+    $$
+    V_{ipk}={V_{sat}\over |A_{CL}|}=({13\over 5})or(2{3\over 5})
+    $$
+
+    5. Determine the output voltage for each of the following values for the following input voltages.
+
+    $$
+    V_o=A_{CL}v_i
+    $$
+
+    | $v_i$ (V) | $v_o$ (V) |
+    | --------- | --------- |
+    | 0         | 0         |
+    | -1        | 5         |
+    | 2         | -10       |
+    | -3        | 15        |
+    | 4         | -20       |
+
+3. Consider the circuit of Figure P2-3 with $R_i=10k\Omega$ and $R_f=50k\Omega$
+
+![hw1Q3](hw1Q3.png)
+
+1. Determine the closed-loop voltage gain.
+
+$$
+A_{CL}={V_o\over V_i}={R_f+R_i\over R_i}={6\over 5}
+$$
+
+2. Determine the input impedance of the circuit.
+
+$$
+r_i=R_i=10k\Omega
+$$
+
+3. Determine the ideal output impedance of the circuit.
+
+$$
+r_o=0\Omega
+$$
+
+4. Determine the peak input voltage $v_i$ (peak) for which linear operation is possible.
+
+$$
+V_{ipk}={V_{sat}\over |A_{CL}|}=({13(5)\over 6})or(10{5\over 6})
+$$
+
+5. Determine the output voltage for each of the following values for the following input voltages.
+
+$$
+V_o=A_{CL}v_i
+$$
+
+| $v_i$ (V) | $v_o$ (V)    |
+| --------- | ------------ |
+| 0         | 0            |
+| -1        | $-6\over5$   |
+| 2         | $12\over5$   |
+| -3        | $-18\over 5$ |
+| 4         | $24\over 5$  |
+
+5. For the circuit of Problem 2-1 with $v_i=-2V$, assume an external load of $R_L =2k\Omega$ is connected to the output. Determine the total op-amp output current.
+
+$$
+A{CL}=-10\\
+v_i=2V\\
+v_0=-20V\\
+{V\over R}=I\\
+{20V\over 2k\Omega}=I={1\over100}A=10mA
+$$

2-February/2-10_Notes.md

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+# Non inverting op amp
+
+![](2-10_Notes.assets/nonInvertingOpAmp.png)
+$$
+A_{CL}={A_{ideal}\over1+({1\over A\beta})}= {A_{ideal}\over1+({k_n\over A})}\\
+{V_o\over V_i}=k_n
+$$
+
+$$
+A\ne A_0\ne A_{CL}
+$$
+
+# Inverting op amp
+
+![](Untitled.assets/invertingOpAmp.png)
+$$
+A_{CL}={A_{ideal}\over1+({1\over A\beta})}= {A_{ideal}\over1+({K_n\over A})}\\
+{V_o\over V_i}={-R_f\over R_i}
+$$
+
+# Open loop gain
+
+$$
+A={A_0\over 1+j({f\over f_1})}\\
+A_0=given\\
+f=freq\\
+f_1=cutoff\ freq
+$$
+
+## Throw back time
+
+$$
+i=Ae^{j\omega t}\\
+e^{j\theta}=cos\theta+jsin\theta
+$$
+
+Given the above two, we can say:
+$$
+A={A_0\measuredangle0\over\sqrt{1-({f\over f1})^2}\measuredangle tan^1({f\over f_1(1)})}\\
+\downarrow\\
+|A|_{dB}=20logA_0-20log(\sqrt{1+({f\over f_1})^2})\\
+\measuredangle A=-tan^{-1}({f\over f_1})
+$$

2-February/2-17_Notes.md

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+# Gain
+
+Let $f_1=10Hz,\ A_0=1_E6$
+$$
+A={A_0\over1+j{f\over f_1}}\\
+|A|={A_0\over\sqrt{1+({f\over f_1})^2}}\\
+|A|_{dB}=20(logA_0-log(\sqrt{1+({f\over f_1})^2}))
+$$
+
+| $f$    | gain (dB)      |
+| ------ | -------------- |
+| $0.1$  | $\approx120$   |
+| $1$    | $\approx120$   |
+| $10$   | $\approx120-3$ |
+| $100$  | $100$          |
+| $1_E4$ | $80$           |
+
+$B=A_0f_1$, where $B$ is the unity gain frequency
+
+
+$$
+A_{OL}={A_0\over 1+j({f\over f_1})}\\
+A_{CL}={A_{IDEAL}\over 1+{k_n\over A_{OL}}}
+$$
+==$A_{CL}={{A_{IDEAL}\over1+{k_n\over A_0}}\over 1+j({fk_n\over B(1+{k_n\over A_0})})}$==
+
+^ This is INCREDIBLY important. It also leads to the following:
+
+==$Break\ frequency\approx {B\over k_n}$==, where  $B$ = unity gain frequency
+

2-February/2-19_Notes.md

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+Unity gain frequency = $12MHz$
+
+non-inverting amplifier
+
+
+
+Calculate closed loop 3-dB bandwidth for the amplifier if the ideal gain is $A_{IDEAL}=10$ (assume $A_0=1_E6$)
+$$
+A_{CL}={{A_{IDEAL}\over1+{k_n\over A_0}}\over 1+j({fk_n\over B(1+{k_n\over A_0})})}\\
+closedloop\ 3dB\ bandwidth={B\over k_n}\\
+noninverting\ \therefore\ A_0=k_n\\
+BW=f_1={B\over k_n}={12MHz\over10}=1.2MHz
+$$
+Do it again, with an inverting amp with an ideal gain of -100.
+$$
+A_{IDEAL}={-R_f\over R_i};\ k_n=1+{R_f\over R_i}\\
+k_n=A_{IDEAL} + 1\\
+f_1={12MHz\over101}
+$$
+
+
+![](2-19_Notes.assets/opAmpCapConversion.png)
+
+Given $X_C={1\over jwc}$
+
+Break frequency = $f_1={1\over2\pi RC}={1\over 2\pi \tau}$
+
+This is important. This determines what scope to buy.
+
+Looking at a square wave, you want to see a $33MHz$ signal undisturbed. Given a specific rise time, $BW={0.35\over risetime}$. Rise time is typically $1\over10$ of your frequency. This is your absolute minimum. If you want something that probably won’t effect the system, get a 5x bandwidth from the above calculation. If you want something overkill, get something 10x bandwidth above. 
+
+Rise time is the time from 10% to 90%.
+$$
+0.1V_{in}=V_{in}(1-e^{-t_1\over\tau})\\
+0.9V_{in}=V_{in}(1-e^{-t_2\over\tau})\\
+t_1=-\tau ln(0.9);\ t_2=-\tau ln(0.1)\\
+risetime=t_2-t_1=-\tau(ln(0.1)-ln(0.9))\\
+risetime={-ln(0.1)-ln(0.9)\over2\pi BW}\approx{0.35\over BW}
+$$
+
+
+
+# Example 1
+
+![](2-19_Notes.assets/example1.png)
+
+Unity gain = $B=1MHz$. Find 3dB BW ($B\over k_n$) and risetime ($0.35\over BW$)
+$$
+{V_o\over V_{in}}=(1+{1k\over2k\|3k})\approx1.833\\
+{B\over k_n}=545kHz;\ risetime = 640ns
+$$
+
+# Slew Rate
+
+Slew rate is pedal to the metal. $V_{out}$ grows at a constant speed.
+
+For the 741, the slew rate is $0.5V/\mu s$ max. 
+
+If you try to pull more current than it can output, it will just not do it. There will be distortion on an AC waveform. This distortion looks like a sawtooth function. 
+
+$f_{max}<{slewrate\over2\pi V_o}$

2-February/2-24_Notes.md

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+# Finite Bandwidth Effect
+
+This is a curve from low to high
+
+$T_R$= rise time
+
+$BW_{CL}$= closed loop bandwidth
+
+$B$=unity gain frequency
+$$
+BW_{CL}=\frac{B}{k_n}\\
+$$
+
+# Finite Slew Rate Effect
+
+This is what produces the sawtooth effect
+
+$T_{SR}$=rise time
+
+$S$=slew rate, measured in $V/\mu s$
+$$
+f_{sr}=\frac{S}{2\pi V_o}=\frac{1}{T_{sr}}
+$$
+
+# Example Problem
+
+A non-inverting amplifier with a ==gain of 50== is used to amplify a signal whose ==peak input is 250mV== and ==frequency range is from DC to 15kHz==. 
+
+Find the unity gain frequency given $BW_{CL}=60kHz$. (Also could be worded as “at a bandwidth of four times the frequency range”)
+$$
+A_{CL}=M(w)=\frac{A_{CL_{DC}}}{\sqrt{1+{f\over BW_{CL}^2}}}\\
+A_{CL}=\frac{50}{\sqrt{1+{15kHz\over (60kHz)^2}}}\\
+A_{CL}=48.5=k_n\\
+B=BW_{CL}k_n=60kHz(50)=\\
+\overline{\underline{|B=300kHz|}}
+$$
+
+Find the slew rate.
+$$
+S=2\pi V_0f_{sr};\ V_0=A_{CL}V_{io}=50(250mV)\\
+S=2\pi (50)(250mV)(20kHz)\\
+\overline{\underline{|S=1.57V/\mu s|}}
+$$

2-February/2-3_Notes.md

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+Think of an opamp as actually two nested op amps.
+
+The inner is ideal. The outer… not.
+
+on the inside of the + input, there is a battery ($V_{io}$) in series into the ideal. There is a current source going toward the circuit away from ground in parallel with both the negative and positive inputs
+
+
+
+![biasAndOffset](2-3_Notes.assets/biasAndOffset.png)
+
+
+$$
+V_o=V_{io}({1\over\beta})\\
+V_o=I_{b+}({1\over\beta})\\
+V_o=I_B(R_C{1\over\beta}-R_f)\\
+R_C=R_i||R_f
+$$

3-March/3-23_Notes.md

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+# Non-inverting Schmitt trigger
+
+Amplifiers have negative feedback. The shmitt trigger has positive feedback
+
+$R_i$ and $R_f$ are in the same spot, and may be called the same thing, but they are VERY different.
+
+![image-20200323161120819](3-23_Notes.assets/schmittTrigger.png)
+
+The trigger points are at ${-R_i\over R_f}V_{o_{sat}}$
+
+In this instance:
+$$
+{10k\Omega\over100k\Omega}(13V)=-1.3V
+$$
+This means that any value under -1.3V will have an output value of -13V. Once it is saturated at 13V, it takes a voltage less than +1.3V to bring it back down. This is according to the graph seen below.
+
+![image-20200323163440794](3-23_Notes.assets/image-20200323163440794.png)
+
+
+
+![image-20200323165148330](3-23_Notes.assets/image-20200323165148330.png)
+

3-March/3-2_Notes.md

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+![](3-2_Notes.assets/comparator.png)
+
+This is a comparator.
+
+It is not to be confused with an amplifier.
+
+
+
+Comparators don’t have feedback. As such, they work as such:
+$$
+V_o=A_{OL}(V_+-V_-)
+$$
+
+
+“Roughly 13, yeah! Thats almost 3,000,000!”
+
+
+
+![](3-2_Notes.assets/comparator2.png)
+
+You can draw this as a graph, with 3 lines. One at the high saturation level, one at the low saturation level, and a vertical line connecting the two, determined by the grounded/static value.
+
+If $V_i$ is on the minus, the +x portion of the graph is negative. If $V_i$ is on the plus, this is switched.
+
+Alright. Lets make it more complex
+
+![](3-2_Notes.assets/comparator3.png)
+
+![](3-2_Notes.assets/comparator4.png)
+
+These two images are equivalent.
+
+When $I_B>>>0$, the $V_o = 0$. When $I_B=0$, $V_o=5V$
+
+In the above example, $V_-= {15k\Omega(-15)\over 10k\Omega+15k\Omega} = 9V$
+

3-March/3-30_Notes.md

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+# Integrators and Differentiators
+
+![](3-30_Notes.assets/integrator.png)
+
+capacitors will be in very large pico, high frequency goes down to small pico
+$$
+{V_o\over V_{in}}={-Z_f\over R_i}\\
+Z_f={R_f\over1+j\omega R_fC}\\
+{V_o\over V_{in}}={{-R_f\over R_i}\over 1+j\omega R_fC}
+$$
+Integrators are, in essence, a low band pass filter. Differentiators are High-pass filter.
+
+Cutoff is as follows:
+$$
+f_c={1\over2\pi R_f C}
+$$
+These are better than passive circuits in lower frequency ranges, mainly due to their smaller cost.
+
+As frequency rises, the capacitor eats more of the feedback power, dropping the output power.
+
+at low frequency, the resistor is the only one.  use ${-R_f\over R_i}$ for gain in these cases
+
+At high frequencies, the capacitor is the only one. 
+
+## Differentiator
+
+![image-20200330164145326](3-30_Notes.assets/image-20200330164145326.png)
+
+RiRf does not exist, its a strait shot to ground
+
+at low frequencies, the capacitor is the only one. At low frequencies the resistor is the only one
+
+break frequency is 
+$$
+f_c={1\over 2\pi R_iC}
+$$

3-March/3-4_Notes.md

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+So, the problem with these, is the hair trigger. So, we’re gonna add positive feedback. There’s two ways of doing this:
+
+![](3-4_Notes.assets/positiveFeedback.png)
+
+This is the easy way. Its called an Inverting Schmitt Trigger. 
+
+It looks like an amp, but the output goes to the plus. 
+
+Given $V_o=13V$, $V_+=6.5V$. Given $V_o=-13V$, $V_+=-6.5V$. 
+
+Changing $V_i$ from 0 doesn’t change anything, until hitting 6.5. This drops the output back to -13V. $V_i$ changing to -6.5 makes the $V_o$ +13 again.
+
+It essentially adds a buffer between the triggers. 

4-April/4-15_Notes.md

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+Gain = 1+R~f~
+
+R~f~ base values are defined by the table in the handout
+
+Start Here
+
+![image-20200415164612530](Untitled.assets/image-20200415164612530.png)
+
+Use a C of 0.039uF
+
+Fc = 200Hz
+
+divide all resistors by2 pi 200Hz fc
+
+resistors come out to 20.4k ohms

4-April/4-1_Notes.md

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+# Filters
+
+low pass filters, high pass filters, band pass filters, and band stop filters
+
+![image-20200401162522106](4-1_Notes.assets/image-20200401162522106.png)
+
+## Two pole low pass filter
+
+$W_c=1n/s$
+
+![image-20200401163404384](4-1_Notes.assets/image-20200401163404384.png)
+
+increasing the resistances by a factor of 1k and reducing the capacitor value by 1k will leave you with the same value.
+
+To set the break frequency, divide the resistance and the capacitor by the break frequency

4-April/4-20_Notes.md

@@ -0,0 +1,50 @@
+# Two Pole Band Pass Filter Response Equations:
+$$
+M_{dB}=20log({M_o\over\sqrt{1+Q^2({f\over f_0}-{f_0\over f})^2}})\\
+f_0=\sqrt{f_1f_2}\\
+where\\
+f_1\ and\ f_2\ are\ cutoff\ frequencies\ and\ BW=f_2-f_1\\
+and\\
+M_o=max\ gain\\
+f_{1,2}=f_0\sqrt{1+({1\over2Q})^2}\mp {BW\over2}\\
+Q={reactive\over real}={\omega L\over R}=will\ be\ given\\
+Q={f_0\over BW}
+$$
+
+![image-20200420160648576](4-20_Notes.assets/image-20200420160648576.png)
+
+## Example:
+
+two pole band pass filter
+
+center frequency of 800Hz
+
+Q=20
+
+
+
+Find 3dB Bandwidth
+$$
+Q={f_0\over BW}\\
+BW={f_0\over Q}={800Hz\over20}\\
+\overline{\underline{|BW=40Hz|}}
+$$
+find $f_1$
+$$
+f_{1}=f_0\sqrt{1+({1\over2Q})^2}-{BW\over2}\\
+f_{1}=800Hz\sqrt{1+({1\over2(20)})^2}-{40Hz\over2}\\
+$$
+find $f_2$
+$$
+f_2=f_1+BW=780Hz+40Hz=820Hz
+$$
+Find response at frequency
+
+`use chart given on mycourses`
+
+can also use $M_{dB}$ equation above, but thats gross
+
+![image-20200420163801196](4-20_Notes.assets/image-20200420163801196.png)
+
+![image-20200420164431655](4-20_Notes.assets/image-20200420164431655.png)
+

4-April/4-2_Notes.md

@@ -0,0 +1,29 @@
+![image-20200406161225167](4-2_Notes.assets/image-20200406161225167.png)
+
+Start here.
+
+Multiply the resistor by a nice value. Divide the capacitor by the same value.
+
+Divide the capacitor by $2\pi f$
+
+This sets the break frequency and makes the parts reasonable values
+
+Another way to put it:
+$$
+{V_{out}\over V_{in}}={1\over1+j(2\pi fRC)}\\
+|{V_{out}\over V_{in}}|={1\over \sqrt{1+({f\over f_C})^2}}
+$$
+adding another cap adds another $j(2\pi fRC)$
+
+# Two pole filter
+
+![image-20200401163404384](4-2_Notes.assets/image-20200401163404384.png)
+$$
+{V_o\over V_i}={1\over\sqrt{1+({f\over f_C})^4}}
+$$
+($f\over f_C$) is to the 2n, where n is the number of poles (read as: the number of unique resistors)
+
+With more poles, it becomes more complex, to the point where 5 or 6 pole filters aren't worth calculating
+
+Adding additional poles is as simple as... look it up, becuase it sucks to come up with it manually
+

4-April/4-8_Notes.md

@@ -0,0 +1,32 @@
+How to determine the order filter I want
+
+We want a 7 pole filter with a 3dB cutoff frequency at 8kHz
+
+What is $M_{dB}\ @f=12kHz$
+$$
+M_{dB}=20log({1\over\sqrt{1+({f\over f_{3dB}})^{2n}}})\\
+M_{dB}=20log({1\over\sqrt{1+({12kHz\over 8kHz})^{14}}})=-24.67dB\\
+$$
+This was a low-pass filter
+
+Now for a high pass filter
+$$
+M_{dB}=20log({1\over\sqrt{1+({ f_{3dB}\over f})^{2n}}})\\
+M_{dB}=20log({1\over\sqrt{1+({12kHz\over 8kHz})^{14}}})=-24.67dB\\
+$$
+We can do this stupid cacluation thats rough on the head, or we can just use the graph
+
+(image forthcoming)
+
+
+$$
+10^{M_{dB}\over10}-1=({f\over f_c})^{2n}
+$$
+Ok, cool, but how do you build it
+
+![image-20200401163404384](4-8_Notes.assets/image-20200401163404384.png)
+
+Use two of these
+
+There is a chart (incoming) that can be used to calculate cap values
+

_Homework/hw10.assets/test2.sch

@@ -0,0 +1,38 @@
+<Qucs Schematic 0.0.19>
+<Properties>
+  <View=0,0,800,800,1,0,0>
+  <Grid=10,10,1>
+  <DataSet=test2.dat>
+  <DataDisplay=test2.dpl>
+  <OpenDisplay=1>
+  <Script=test2.m>
+  <RunScript=0>
+  <showFrame=0>
+  <FrameText0=Title>
+  <FrameText1=Drawn By:>
+  <FrameText2=Date:>
+  <FrameText3=Revision:>
+</Properties>
+<Symbol>
+  <.PortSym 40 20 2 0>
+</Symbol>
+<Components>
+  <OpAmp OP1 5 430 270 -26 -42 1 0 "1e6" 0 "15 V" 0>
+  <Port Vout 1 470 270 4 -32 0 2 "2" 0 "analog" 0>
+  <GND * 1 400 380 0 0 0 0>
+  <Vdc V1 5 400 340 18 -26 0 1 "7.5V" 1>
+  <Vdc V2 1 250 290 18 -26 0 1 "1 V" 1>
+  <.SW SW1 1 590 60 0 63 0 0 "" 1 "lin" 1 "V2" 1 "-13 V" 1 "13 V" 1 "521" 1>
+</Components>
+<Wires>
+  <400 370 400 380 "" 0 0 0 "">
+  <400 290 400 310 "" 0 0 0 "">
+  <250 250 400 250 "" 0 0 0 "">
+  <250 250 250 260 "" 0 0 0 "">
+  <250 320 250 380 "" 0 0 0 "">
+  <250 380 400 380 "" 0 0 0 "">
+</Wires>
+<Diagrams>
+</Diagrams>
+<Paintings>
+</Paintings>

_Homework/hw10.md

@@ -0,0 +1,70 @@
+# Skyler MacDougall
+
+##  Homework 10: Due 3/30/2020
+
+15. Consider the circuit below, where the $100k\Omega$ resistor is a $100\Omega$:
+    ![](hw10.assets/hw10q15.png)
+    
+    1. Calculate the threshold voltage.
+        $$
+        V_T=V_{sat}*{R_2\over R_1+R_2}=\pm13V*{100\Omega\over 20.1k\Omega}\\
+        V_T=\pm0.0646V
+        $$
+        
+    2. Sketch the input-output characteristic curve.
+        ![image-20200401174147982](hw10.assets/image-20200401174147982.png)
+    
+    3. Sketch the output if the following waveform is connected to the input.
+        ![](hw9.assets/figure1.png)
+
+![image-20200401175636700](hw10.assets/image-20200401175636700.png)
+
+
+
+17. Consider the circuit in question 15 with the following modifications:
+    $$
+    R_1=\cancel{20k\Omega}27k\Omega\\
+    R_2=\cancel{100\Omega}68k\Omega
+    $$
+
+    1. Calculate the threshold voltage.
+        $$
+        V_T=V_{sat}*{R_2\over R_1+R_2}=\pm13V*{68k\Omega\over 68k\Omega+27k\Omega}\\
+        V_T=\pm9.305V
+        $$
+        
+    2. Sketch the input-output characteristic curve.
+        ![image-20200401180613400](hw10.assets/image-20200401180613400.png)
+    
+    3. Sketch the output if the following waveform is connected to the input.
+
+![image-20200401180628999](hw10.assets/image-20200401180628999.png)
+
+25. Using a general-purpose op-amp, design an open-loop zero crossing detector based on the following logic:
+    $$
+    V_o\approx13V\ for\ v_i>0\\
+    V_o\approx-13V\ for\ v_i<0\\
+    $$
+    
+
+![image-20200401172005749](hw10.assets/image-20200401172005749.png)
+
+29. An open-loop comparator is desired to provide the following logic:
+    $$
+    V_o\approx13V\ for\ v_i>7.5V\\
+    V_o\approx-13V\ for\ v_i<7.5V\\
+    $$
+
+![](hw10.assets/image-20200401172118446.png)
+
+31. An inverting Schmitt trigger circuit is designed for use as a zero-crossing detector. Based on saturation voltages of $\pm13V$, the hysteresis transition voltages desired are about $\pm25mV$. Determine an acceptable design using 5% resistors.
+
+![image-20200401181423413](hw10.assets/image-20200401181423413.png)
+
+35. A circuit is to be designed to convert a 6V sine wave to square wave oscillating between -13V and 13V. A control is to be provided for possible delay adjustment as follows:
+
+    1. At one extreme setting , the output square wave will be $180^\circ$ out of phase with the sine wave.
+    2. At the other extreme, the output wave will undergo its transition at positive and negative peaks of the sine wave. 
+
+    Specify the value of any fixed resistance and the range of any variable resistance used.
+

_Homework/hw11.md

@@ -0,0 +1,18 @@
+# Skyler MacDougall
+
+##  Homework 11: Due 4/13/2020
+
+3. 
+
+
+
+17. 
+
+
+
+19. 
+
+
+
+25. 
+

_Homework/hw12.md

@@ -0,0 +1,18 @@
+# Skyler MacDougall
+
+##  Homework 12: Due 4/20/2020
+
+5. 
+
+
+
+7. 
+
+
+
+9. 
+
+
+
+13. 
+

_Homework/hw2.md

@@ -0,0 +1,129 @@
+
+
+# Skyler MacDougall
+
+##  Homework 2: Due 1/29/2020
+
+7. Consider the circuit below, with $R=1k\Omega$.
+
+    ![hw2q7](hw2.assets/hw2q7.png)
+
+    1. Determine the transconductance.
+        $$
+        g={1\over R}={1\over 1k\Omega}=100\mu S
+        $$
+
+    2. For $V_1=10V$ determine $i_L$ for $R_L=500\Omega$.
+        $$
+        i_L=i_i={V_i\over R}={10V\over1k\Omega}\\
+        i_L=1mA
+        $$
+
+    3. Repeat (2.) for $R_L=1k\Omega$.
+
+        $i_i=i_L \therefore i_L = 1mA$ 
+
+    4. Determine the maximum value for $R_L$ for linear operation.
+        $$
+        i_iR_L<V_{sat};i_i=1mA;V_{sat}=13V\\
+        R_L=1.3k\Omega
+        $$
+
+9. Consider the circuit below.
+
+    ![hw2q9](hw2.assets/hw2q9.png)
+
+    1. Determine the transconductance.
+        $$
+        g={1\over R}={1\over 2k\Omega}=500\mu S
+        $$
+
+    2. For $V_i=6V$ and $R_L=1.2k\Omega$, determine $i_L$.
+        $$
+        i_L={V_i\over R}={6V\over 2k\Omega}\\
+        i_L=3mA
+        $$
+
+    3. Determine the maximum value for $R_L$ for linear operation.
+        $$
+        R_Li_L<{V_{sat}\over2}
+        $$
+        
+
+9. Consider the circuit below.
+
+    ![hw2q13](hw2.assets/hw2q13.png)
+
+    1. Determine the value of $\beta$.
+        $$
+        \beta=1+{R_2\over R_1}=1+{8k\Omega\over2k\Omega}\\
+        \beta=5
+        $$
+        
+    2. For $i_i=0.6mA$ and $R_L=1k\Omega$, verify linear operation.
+        $$
+        (R_2+\beta R_L)(i_i)<V_{sat}\\
+        (8k\Omega+5(1k\Omega))(0.6mA)<13V\\
+        13\cancel{k}\Omega*0.6\cancel{m}A<13V \\
+        7.8V<13V
+        $$
+
+    3. Determine $i_L$ for the conditions in (2.)
+        $$
+        i_L=\beta i_i\\
+        i_L=3mA
+        $$
+        
+
+
+15. Consider the circuit below.
+
+    ![hw2q15](hw2.assets/hw2q15.png)
+
+    1. Write an equation for $V_O$ in terms of the three input voltages.
+        $$
+        V_O=-(V_1+2V_2+4V_3)
+        $$
+
+    2. Determine $V_O$ given $V_1=10V;V_2=3V;V_3=-7V$.
+        $$
+        V_O=-(10V+2(3V)+4(-7V))=28V-16V\\
+        V_O=12V
+        $$
+
+    3. Determine $V_O$ given $V_1=8V;V_2=-4V;V_3=5V$.
+        $$
+        V_O=-(\cancel{8V}+\cancel{2(4V)}+4(5V))\\
+        V_O=-20V
+        $$
+        
+
+35. Design a current controlled voltage source to have a transresistance of $2k\Omega$. Then determine the peak value $i_{i_{pk}}$ permitted for the input current for linear operation.
+
+![hw2q35](hw2.assets/hw2q35.png)
+$$
+V_O=-Ri_i<V_{sat}\\
+2k\Omega*i_i<13V\\
+i_i=6.5mA
+$$
+
+39. Design a linear combination circuit to combine two signals as follows:
+
+$$
+v_o=-4v_1-8v_2
+$$
+Using the following specifications:
+
+1. $R_{in}\ge10k\Omega$ at both inputs
+2. All resistance values $\le100k\Omega$
+
+![hw2q39](hw2.assets/hw2q39.png)
+
+41. Design a balanced closed-loop differential circuit to combine two signals as follows:
+    $$
+    v_o=3(v_1-v_2)
+    $$
+    Use resistances in the range of $10k\Omega-100k\Omega$.
+
+![hw2q41](hw2.assets/hw2q41.png)
+

_Homework/hw3.md

@@ -0,0 +1,165 @@
+
+
+# Skyler MacDougall
+
+##  Homework 2: Due 1/29/2020
+
+1. Given an op amp with the following characteristics:
+    $$
+    A_O=80k\\
+    r_d=400k\Omega\\
+    r_o=50\Omega\\
+    R_i=3k\Omega\\
+    R_f=47k\Omega
+    $$
+
+    1. Find the feedback factor
+        $$
+        \beta={R_i\over R_i+R_f}={3\over50}=0.06
+        $$
+
+    2. loop gain ($A_{CL}$)
+        $$
+        A_{CL}\approx{1\over \beta}=16.\overline6
+        $$
+
+    3. exact closed loop gain ($A_{CL}$)
+        $$
+        A_{CL}={A_O\over 1+A_O\beta}={80k\over 1+(80k)(0.06)}\\
+        \overline{\underline{|A_{CL}=16.663|}}
+        $$
+
+    4. Closed loop input resistance ($Z_i$)
+        $$
+        Z_i=(1+A_O\beta)r_d=(1+(80k)(0.06))(400k\Omega)\\
+        \overline{\underline{|Z_i=2G\Omega|}}
+        $$
+
+    5. output resistance ($Z_o$)
+        $$
+        Z_o={r_o\over 1+A_O\beta}={50\over 1+(80k)(0.06)}\\
+        \overline{\underline{|Z_o=10m\Omega|}}
+        $$
+
+    6. Compare with question 3-8
+
+        The gain changes are comparable.
+
+
+
+
+
+3. Repeat question 1(a-e) with the following changes:
+    $$
+    R_i=\cancel{3k\Omega}=30k\Omega\\
+    R_f=\cancel{47k\Omega}=270k\Omega
+    $$
+
+    1. Find the feedback factor
+        $$
+        \beta={R_i\over R_i+R_f}={1\over10}=0.1
+        $$
+
+    2. loop gain ($A_{CL}$)
+        $$
+        A_{CL}\approx{1\over \beta}=10
+        $$
+
+    3. exact closed loop gain ($A_{CL}$)
+        $$
+        A_{CL}={A_O\over 1+A_O\beta}={80k\over 1+(80k)(0.1)}\\
+        \overline{\underline{|A_{CL}=9.99875|}}
+        $$
+
+    4. Closed loop input resistance ($Z_i$)
+        $$
+        Z_i=(1+A_O\beta)r_d=(1+(80k)(0.1))(400k\Omega)\\
+        \overline{\underline{|Z_i\approx3G\Omega|}}
+        $$
+
+    5. output resistance ($Z_o$)
+        $$
+        Z_o={r_o\over 1+A_O\beta}={50\over 1+(80k)(0.1)}\\
+        \overline{\underline{|Z_o=6m\Omega|}}
+        $$
+
+
+
+
+5. Find $A_{CL}$ given
+    $$
+    R_i=1k\Omega\\
+    R_f=99k\Omega
+    $$
+
+    1. $A_O=\infin$
+        $$
+        \beta={1\over100}\\
+        A_{CL}={A_O\over 1+A_O\beta};\ A_O>>0\\
+        A_{CL}={1\over \beta}\\
+        \overline{\underline{|A_{CL}=100|}}
+        $$
+
+    2. $A_O=10^5$
+        $$
+        A_{CL}={10^5\over1+(10^5)({1\over100})}\\
+        \underline{\overline{|A_{CL}=99.\overline{900099}|}}
+        $$
+
+    3. $A_O=10^4$
+        $$
+        A_{CL}={10^4\over1+(10^4)({1\over100})}\\
+        \underline{\overline{|A_{CL}=99.\overline{0099}|}}
+        $$
+
+    4. $A_O=10^3$
+        $$
+        A_{CL}={10^3\over1+(10^3)({1\over100})}\\
+        \underline{\overline{|A_{CL}=90.\overline{90}|}}
+        $$
+
+    5. $A_O=10^2$
+        $$
+        A_{CL}={10^2\over1+(10^2)({1\over100})}\\
+        \underline{\overline{|A_{CL}=50|}}
+        $$
+
+
+
+
+
+7. Find $A_{CL}$ given
+    $$
+    R_i=10k\Omega\\R_f=510k\Omega
+    $$
+
+    1. $A_O=\infin$
+        $$
+        \beta={1\over52}\\A_{CL}={A_O\over 1+A_O\beta};\ A_O>>0\\A_{CL}={1\over \beta}\\\overline{\underline{|A_{CL}=52|}}
+        $$
+
+    2. $A_O=10^5$
+        $$
+        A_{CL}={10^5\over1+(10^5)({1\over52})}\\ 
+        \underline{\overline{|A_{CL}=51.97|}}
+        $$
+
+    3. $A_O=10^4$
+        $$
+        A_{CL}={10^4\over1+(10^4)({1\over52})}\\ \underline{\overline{|A_{CL}=51.73|}}
+        $$
+
+    4. $A_O=10^3$
+        $$
+        A_{CL}={10^3\over1+(10^3)({1\over52})}\\
+        \underline{\overline{|A_{CL}=49.50|}}
+        $$
+
+    5. $A_O=10^2$
+        $$
+        A_{CL}={10^2\over1+(10^2)({1\over52})}\\ 
+        \underline{\overline{|A_{CL}=34.11|}}
+        $$
+
+
+

_Homework/hw4.md

@@ -0,0 +1,98 @@
+# Skyler MacDougall
+
+##  Homework 4: Due 2/10/2020
+
+9. Given the circuit below:
+    ![hw4q9](hw4.assets/hw4q9.png)
+    
+    1. Determine the noise gain ($K_n$) for the circuit.
+        $$
+        K_n={1\over \beta};\ \beta={R_i\over R_i+R_f}\\
+        K_n={68k\Omega+2k\Omega\over2k\Omega}\\
+        \overline{\underline{|K_n=35|}}
+        $$
+        
+    2. Use the result to calculate the exact signal gain at DC and low frequencies if $A_o=10^5$.
+        $$
+        K_n={1\over \beta};\ \beta={1\over 35}\\
+        A_{CL}={A_o\over1+A_o\beta}={10^5\over1+(10^5)({1\over35})}\\
+        \overline{\underline{|A_{CL}=3.9998\approx4|}}
+        $$
+        
+
+
+
+11. Given the circuit below:
+     ![](hw4.assets/hw4q11.png)
+    
+    1. Determine the noise gain ($K_n$) for the circuit.
+        $$
+        K_n={1\over\beta};\ \beta={R_i\over R_i+R_f};\ R_i=12k\Omega||24k\Omega=8k\Omega\\
+        K_n={8k\Omega+120k\Omega\over8k\Omega}\\
+        \overline{\underline{|K_n=16|}}
+        $$
+        
+    2. Use the result to calculate the exact gain factors for the two signals if $A_o=5\times 10^4$.
+        $$
+        K_n={1\over \beta};\ \beta={1\over 16}\\
+        A_{CL}={A_o\over1+A_o\beta}={5\times10^4\over1+(5\times10^4)({1\over16})}\\
+        \overline{\underline{|A_{CL}=15.9949\approx16|}}
+        $$
+        
+
+
+
+13. For the circuit shown in problem 9, assume the following:
+    $$
+    V_{io}=1.2mV\\
+    I_b=60nA\\
+    I_{io}=8nA
+    $$
+
+    1. Determine the magnitude of the output DC voltage $|V_{o1}|$ produced by the input offset voltage.
+        $$
+        V_{o1}=V_{io}(\alpha);\ \alpha={R_f\over R_i+R_f}={34\over35}\\
+        V_{o1}=1.2mV({34\over35})\\
+        \overline{\underline{|V_{o1}=1.6mV|}}
+        $$
+        
+    2. With $R_c=0$ determine the magnitude of the output dc voltage $|V_{o2}|$ produced by the input bias currents.
+        $$
+        V_{o2}=R_c(\alpha)i_b^+-R_f(I_b^2);\ R_c=0\\
+        i_{io}=i_b^+-i_b^-;\ i_b={i_b^++i_b^-\over2}\\
+        8nA=i_b^+-i_b^-;\ 120nA=i_b^++i_b^-\\
+        i_b^+=64nA;\ i_b^-=56nA\\[16pt]
+        V_{o2}=0-68k\Omega(56nA)\\
+        \overline{\underline{|V_{o2}=-3.808mV|}}
+        $$
+        
+    3. Determine the optimum value of $R_c$.
+        $$
+        R_{c_{ideal}}=2k\Omega||68k\Omega\\
+        \overline{\underline{|R_{c_{ideal}}=1.94k\Omega|}}
+        $$
+        
+    4. Given your new value for $R_c$, find $|V_{o2}|$.
+        $$
+        V_{o2}=R_c(\alpha)i_b^+-R_f(I_b^2);\ R_c=0;\ i_b^+=64nA;\ i_b^-=56nA\\
+        V_{o2}=1.94k\Omega({34\over35})(64nA)-3.808mV\\
+        \overline{\underline{|V_{o2}=-3.688V|}}
+        $$
+        
+
+
+
+25. An op-amp is used at DC and very low frequencies. A closed loop gain of 200 is required. Specifications indicate that the error due to finite open loop gain cannot exceed 0.1%. Determine the minimum value of the DC open loop gain required.
+
+I am unsure how to do this problem. It feels like there is not enough information do this problem, but I can’t seem to wrap my head around it.
+
+27. Assume the design of problem 25, with the following additional parameters:
+    $$
+    DC\ output\ due\ to\ input\ offset\ voltage \le100mV\\
+    DC\ output\ due\ to\ input\ offset\ current \le5mV\\
+    $$
+
+    1. Determine the maximum value of input offset voltage allowed for the op-amp.
+    2. When an op-amp is selected to meet the requirements for the above, assume that $I_{io}=12\mu A$. Calculate the maximum value of $R_f$ permitted, assuming that a compensating resistors will be used.
+
+Due to this question being directly related to question 25, I cannot do this question either.