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Komok [63]
3 years ago
11

1. What is an op-amp? List the characteristics of an ideal op-amp

Engineering
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

An opamp is an operation amplifier. It takes an input signal and amplifies it on the output side.

An ideal opamp should have infinite impedance at its input, infinite gain on the output, and zero impedance on the output

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Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

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Which band has an average of $3.58 per hour of parking?
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C it would be c because that has more and the others have less
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What is the probability that Tina will NOT wear a white t-shirt on the first day of her trip?
katrin2010 [14]

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4/5

Explanation:

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3 years ago
List six possible valve defects that should be included in the inspection of a used valve?
olchik [2.2K]

Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

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3 years ago
If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit
mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}

\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}

choose R2, R1 such that it will maintain required  ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0
3 years ago
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