An operational amplifier (often op amp or opamp) is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. In this configuration, an op amp produces an output potential (relative to circuit ground) that is typically 100,000 times larger than the potential difference between its input terminals. Operational amplifiers had their origins in analog computers, where they were used to perform mathematical operations in linear, non-linear, and frequency-dependent circuits.
The popularity of the op amp as a building block in analog circuits is due to its versatility. By using negative feedback, the characteristics of an op-amp circuit, its gain, input and output impedance, bandwidth etc. are determined by external components and have little dependence on temperature coefficients or engineering tolerance in the op amp itself.
Op amps are used widely in electronic devices today, including a vast array of consumer, industrial, and scientific devices. Many standard IC op amps cost only a few cents; however, some integrated or hybrid operational amplifiers with special performance specifications may cost over US$100 in small quantities. Op amps may be packaged as components or used as elements of more complex integrated circuits.
The op amp is one type of differential amplifier. Other types of differential amplifier include the fully differential amplifier (similar to the op amp, but with two outputs), the instrumentation amplifier (usually built from three op amps), the isolation amplifier (similar to the instrumentation amplifier, but with tolerance to common-mode voltages that would destroy an ordinary op amp), and negative-feedback amplifier (usually built from one or more op amps and a resistive feedback network).
So basically I am trying to give an output of Vo = 10(V2-V1)
From Figure 9 Example Gain of first Op Amp = Rf / R1, if R1 & R2 are equal.
What's throwing me off is using 5 resistors to create a circuit rather than 6 or just 3. My initial thoughts were the following:
To use the first loop...
Hi,
I am using an Apex PA443DF operational amplifier to drive a sinusoidal signal at 100 V amplitude across a very large resistor with resistance of about 5000 Mega ohms. The amplifiers are set up as noninverting with a gain of 22 and function very well for small resistive loads. The sinusoids...
1. Using the frequency equation I know the time period is 1 millisecond.
2. The duty cycle (50%) equation tells me that the pulse width and the amplitude is 1 V.
3. Thus the input waveform looks like this : (?)
4. As the slew rate is 0.5 volts per microsecond, the output voltage would take...
Here I let ##R_2 = 2 ## kohms and ##R_1 = 1## kohms. Using this suggests that ##R_3 = 0## and ## R_4 = 1 ## kohms (?!)
The resulting circuit is
Is this the correct way to solve this?
PS : I just realised that I have interchanged ##V_1## and ## V_2## in the diagram!
The open loop voltage gain is given as :
$$ u(s) = \frac{u_o}{1+\frac{s}{w_o}} = \frac{100}{1 + \frac{s}{40}}$$
Where u_o is the d.c. voltage gain and w_o is the pole.
The op amp that is given is:
And I am told to use the non ideal op amp model as follows:
Well my guess is that I can...
Homework Statement
The op amp has a near ideal level 1 model with G = 5000V/V, ri = inf, ro = 0
How would I obtain the feedback function?
Homework Equations
I know I have to find the feedback function which is:
f = - (ΔV/Vin)
The Attempt at a Solution
I will first drive using the...
Homework Statement
Derive the expressions for the voltage gain (Gv) of the following op amp:
Homework Equations
In = Ip = 0
Vp =Vn
The Attempt at a Solution
I can use KCL, and the fact that In and Ip are both 0, to derive the two equations, one from the top node and the other from...
Homework Statement
The opAmp network shown in figure 68 is a transimpedance amplifer that employs both negative and positive feedback. Assuming that the op-amp is ideal, calculate the value for the output voltage for Jg is 1mA.
The circuit and question are this...
Homework Statement
Homework Equations
CMRR = Av/Acm
Acm= Δ/R , Δ = (2 x Tolerance of Resistor).R
The Attempt at a Solution
I have to admit I am a bit confused by the premise of the problem.
My understanding is Vout is used to vary Vcharge in some way, so that Icharge is always 1.3 A. Then...
Homework Statement
Homework Equations
[/B]
CMRR = Av/Acm
Acm= Δ/R , Δ = (2 x Tolerance of Resistor).R
The Attempt at a Solution
I have an issue with part e) and f) but here are all my workings
c) For this part, Acm would be:
10 x10^-3
This makes 2.5mV -> 0.25 uV (which is the same...
Homework Statement
2. Homework Equations
3. The Attempt at a Solution
For the first set of questions:
I've worked through to part 6), at which is I encountered my first problem. I'm not entirely sure what the question is asking. Is it as if there would be a capacitor between...
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