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1-16_Notes.md Executable file
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# Important info for lab
Input resistance
$R_{in}=R_d(1+A\beta)$
Gain
${V_o \over V_{in}}={A \over 1+A\beta}$
output resistance
$r_{out}={r_o\over 1+A\beta}$
$r_d$ is the resistance between the inputs.
$r_o$ is the resistance on the output after the source.
The source within is represented as $A_v(v_+-v_+)$
${R_i \over R_i + R_f}=\beta$ is the feedback factor. $\alpha={R_f\over R_i+R_f}$ is the control factor. A is the open-loop gain. (Closed-loop gain is when the output feeds back into the input)
$R_i$ is the input resistor
$R_f$ is the feedback resistor. It is ALWAYS back to the negative.
# Simple feedback op-amp for examples
![image-20200116163421345](image-20200116163421345.png)
## Gain
If A is finite, and $r_d$ and $r_o$ are neglected, then the gain = ${R_i+R_f \over R_i}={1\over \beta}$
## Input resistance
$$
goal\\
r_{in}={1V\over I_{in}}\\
.\\
0={v_-- 0\over R_i}+{v_-- V_{in}\over r_d}+{v_-- V_o\over R_f}\\
i_{in}={1- V_- \over r_d}\\
0=V_-({R_iR_f+r_dR_f+r_dR_i \over R_ir_dR_f})-{1\over r_d}-{A(1-v_-) \over R_f}\\
0=V_-({(R_iR_f+r_dR_f+r_dR_i(1-A)) \over R_ir_dR_f})-{1\over r_d}-{A\over R_f}\\
1-i_{in}r_d=V_-\\
0=(1-i_{in}\cancel {r_d})({(R_iR_f+r_dR_f+r_dR_i(1-A)) \over R_i\cancel {r_d}R_f})-{1\over r_d}-{A\over R_f}\\
0=(1-i_{in})(1+{r_d\over R_i}+{r_d(1-A)\over R_f})-{1\over r_d}-{A\over R_f}\\
0=1-i_{in}+{r_d\over R_i}+{r_d(1-A)\over R_f}-{r_di_{in}\over R_i}-{r_d(1-A)i_{in}\over R_f}-{1\over r_d}-{A\over R_f}\\
i_{in}(1+{r_d\over R_i}+{r_d(1-A)\over R_f})={r_d\over R_i}+{r_d(1-A)\over R_f}-{1\over r_d}-{A\over R_f}
$$

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EEET-222-02
Electronics II Lab
Professor: Richard Cliver
Semester: 2195 (2020 Spring)
Time Slot: Unknown
See EEET-221 for description of professor.
This course is very easy, with most labs having step by step instructions. The professor is extremely willing to give hints as to what to do if you and your lab partner are stuck, and we even managed to confuse him once or twice. Lab reports are very easy, no more than a few lines. Any lab supplies are already owned (should be purchased when taking DC Circuits if in the EET major) or supplied in the lab itself.

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![100hz](Untitled.assets/100hz.jpg)
$$
A={V_o\over V_i}={10.31V\over 98mV}\approx105
$$
![1khz](Untitled.assets/1khz.jpg)
$$
A={V_o\over V_i}={10.4V\over 116mV}\approx89.7
$$
![10khz](Untitled.assets/10khz.jpg)
$$
A={V_o\over V_i}={6.2V\over 116mV}\approx53.4
$$
![100khz](Untitled.assets/100khz.jpg)
$$
A={V_o\over V_i}={734mV\over 98mV}\approx7.49
$$

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1. Determine $R_{in}$ of a non-inverting amplifier when $A_O=200000$, $R_D=10M\Omega$, $R_f=10k\Omega$, and $R_i=1k\Omega$.
$$
R_{in}=R_d(1+A\beta);\ \beta={R_i\over R_i+R_f}\\
\beta={1k\Omega\over 11k\Omega}=0.0\overline{9}\\
R_{in}=10M\Omega(1+2_E6(0.0\overline{9}))\\
R_{in}=10M\Omega(181818)=1.\overline{81}T\Omega
$$
2. Determine $R_{out}$ of a non-inverting amplifier, given the above, where $R_O=75\Omega$.
$$
R_{out}={r_o\over 1+A\beta};\ \beta A=181818\\
R_{out}\approx{75\Omega\over181818}=412\mu\Omega
$$
3. Determine $A_{CL}$ of a non-inverting amplifier when $A_O=2000$, $R_f=10k\Omega$, and $R_i=1k\Omega$.
$$
A_{CL}\approx{1\over\beta};\beta=0.0\overline{9}\\
A_{CL}=11
$$
4. Repeat 3 where $A_O=200000$
$$
A_{CL}\approx{1\over\beta}\\
A_{CL}=11
$$

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In this lab, we build a series of operational amplifiers. We tested each one individually, to ensure that they worked. Then, we connected one to a sine wave, and another to a square wave, and connected the two together. This resulted in a final output waveform that looks like a square wave overlaid over a sine wave.
![](lab3.assets/sineWave.jpg)
$$
A_{CL_{theory}}={1\over \beta}={R_i+R_f\over R_i}={1k\Omega+2.2k\Omega\over2.2k\Omega}\\
A_{CL_{theory}}=1.45\\
A_{CL_{lab}}={V_o\over V_i}={6.8V\over2.13V}\\
A_{CL_{lab}}=3.19
$$
![](lab3.assets/squareWave.jpg)
$$
A_{CL_{theory}}={1\over \beta}={R_i+R_f\over R_i}={5.6k\Omega+1k\Omega\over5.6k\Omega}\\
A_{CL_{theory}}=1.17\\
A_{CL_{lab}}={V_o\over V_i}={5.3V\over2.13V}\\
A_{CL_{lab}}=2.49
$$
![](lab3.assets/vout3.jpg)

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# Skyler MacDougall
## Lab 3 : Intro to Op-Amps
---
### Circuit 1:
![image-20200220142619123](lab3.assets/image-20200220142619123.png)
### Circuit 2:
![image-20200220142643364](lab3.assets/image-20200220142643364.png)
### Circuit 3:
![image-20200220142751975](lab3.assets/image-20200220142751975.png)
### Circuit 4:
![](lab3.assets/circuit4.jpg)
$$
A_{CL_{theory}}={-R_f\over R_i}={-5.6k\Omega\over2.2k\Omega}=-2.\overline{54}\\
A_{CL}={V_o\over V_i}={50V\over1.99V}\approx25
$$
### Circuit 5
![](lab3.assets/circuit5.jpg)
$$
A_{CL_{theory}}={R_f+R_i\over R_i}={2.2k\Omega+ 5.6k\Omega \over 2.2k\Omega}=3.\overline{54}\\
A_{CL}={V_o\over V_i}={38V\over20.3V}\approx1.82
$$
---
Circuit 3 does not have distortion at the peaks of the waveform, making it better than Circuit 1. It also has no distortion in the middle of the waveform, making it better than Circuit 2. The inverting gain calculation does not appear to work in this instance (I may have grabbed the wrong resistor). The minus sign refers to the fact that the amplitude is inverted compared to the input signal. The non-inverting gain does not appear to work either. Finding the datasheets was rather difficult, although I would assume that in the real world, when buying devices, they come with some form of datasheet, either digital or physical.

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| | Typical | Maximum | Measured |
| -------------------------------- | ------- | ------- | --------- |
| Input Offset Voltage, V~io~ (mV) | $1.0$ | $5.0$ | $4.95$ |
| Input Offset Current, I~io~ (nA) | $20$ | $200$ | $4.464$ |
| Input Bias Current I~b~ (nA) | $80$ | $500$ | $33.4714$ |
The input offset Voltage refers to the voltage difference between the two inputs. The input offset current is the amount of current needed to force the opamps output to zero. The input bias current creates a voltage across an input impedance, reducing the losses from the opamp.

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Q: Does this system react the same to different forms of waves put into the mic?
A: The system has the same output as the input, so if the input is a sawtooth waveform, then the output will be similar. The square wave is not a perfect replication, but its very close.
Q: Why does the square wave not go through perfectly?
A: The square wave can't be replicated because the slew rate is not high enough.
Q: Why do you set the break frequency so much farther below the lowest signal frequency?
A: There may be some unexpected signals that dip slightly below your wanted lowest signal that you want to pick up. Moreover, it gives you a bit of wiggle room to make sure that all of the data you want will come through at 100% power.
Q: How did you find your values for your initial gain for -220?
A: Gain for an inverting amplifier is ${-R_f\over R_i}$. As we had to pick any value between $1k\Omega$ and $820k\Omega$, we picked easy numbers of $1k\Omega$ for $R_i$ and $220k\Omega$ for $R_f$.

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Initial Conditions:
| QD | QC | QB | QA |
| ---- | ---- | ---- | ---- |
| 1 | 0 | 1 | 1 |
Input 1:
| Clear | Load | D | C | B | A | EnableP | EnableT | Clock |
| ----- | ---- | ---- | ---- | ---- | ---- | ------- | ------- | ------ |
| 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | rising |
Output:
| QD | QC | QB | QA |
| ---- | ---- | ---- | ---- |
| 0 | 0 | 0 | 0 |
New
Input 2:
| Clear | Load | D | C | B | A | EnableP | EnableT | Clock |
| ----- | ---- | ---- | ---- | ---- | ---- | ------- | ------- | ------ |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | rising |
Output:
| QD | QC | QB | QA |
| ---- | ---- | ---- | ---- |
| 1 | 0 | 1 | 1 |
New
Input 3:
| Clear | Load | D | C | B | A | EnableP | EnableT | Clock |
| ----- | ---- | ---- | ---- | ---- | ---- | ------- | ------- | ------ |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | rising |
Output:
| QD | QC | QB | QA |
| ---- | ---- | ---- | ---- |
| 1 | 1 | 0 | 0 |

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