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The integrated operational amplifier consists of a multi-stage direct-coupled amplifier with a high open-loop voltage amplification. In terms of operating principle, the integrated hunger wheel op amp can be divided into two aspects: linear application and non-limb linear application. ** In the working area, the linear amplification relationship between the output voltage UO and the input voltage UI is UO=AUO(U U-)=AUOUI, because the amplification AUO of the integrated operational amplifier is as high as 104 107, if the UO is a finite value, the depth negative feedback must be introduced to widen the linear region, and the linear operation circuit of the integrated operational amplifier is constituted.
In the case of engineering applications, the integrated op amp is regarded as the ideal op amp, that is, the technical indicators of the integrated op amp are rationally conceived, and the operational amplifier that meets the following conditions is called the ideal op amp.
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Summary. The efficiency of an integrated power amplifier circuit is generally between 50% and 90%, depending on the design and structure of the circuit. To improve the efficiency of the integrated power amplification circuit, the design and structure of the circuit should first be determined to ensure the stability and reliability of the circuit.
Secondly, a low-noise amplifier should be used to reduce the noise of the circuit and improve the efficiency of the circuit. In addition, low-resistance components are used to reduce the loss of the circuit and improve the efficiency of the circuit. Finally, an efficient power management system should be adopted to reduce power losses and improve the efficiency of the circuit.
In summary, to improve the efficiency of integrated power amplification circuits, it is necessary to determine the design and structure of the circuit, adopt low-noise amplifiers, low-resistance components, and efficient power management systems to reduce the noise, loss, and power loss of the circuit, thereby improving the efficiency of the circuit.
The efficiency of an integrated power amplifier circuit is generally between 50% and 90%, depending on the design and structure of the circuit. To improve the efficiency of the integrated power amplification circuit, the design and structure of the circuit should first be determined to ensure the stability and reliability of the circuit. Secondly, a low-noise amplifier should be used to reduce the noise of the circuit and improve the efficiency of the circuit.
In addition, low-resistance components are used to reduce the loss of the circuit and improve the efficiency of the circuit. Finally, it is necessary to adopt an efficient power management system to reduce power loss and improve the efficiency of the circuit. In short, to improve the efficiency of the integrated power amplification circuit, it is necessary to determine the design and structure of the circuit, and adopt a low-noise amplification plexus, low-resistance components, and an efficient power permeation management system to reduce the noise, loss, and power loss of the circuit, thereby improving the efficiency of the circuit.
Can you add, I don't quite understand it.
The efficiency of an integrated power amplification circuit depends on the ratio between its output power and input power, and in general, the higher the efficiency, the greater the simple volt output power of the amplifier. The efficiency of an integrated power amplifier circuit is typically between 50% and 90%, depending on the design of the amplifier and the components used. The efficiency of the integrated power amplifier circuit can be improved by changing the parameters of the components, such as changing the resistance, capacitance, feedback circuit, etc., to increase the output power of the amplifier.
In addition, the efficiency can also be improved by changing the structure of the amplifier, such as the use of a bipolar amplifier structure, which can improve the efficiency of the amplifier. In addition, it is also possible to improve the efficiency by changing the operating frequency of the amplifier, and generally speaking, the higher the operating frequency of the amplifier, the higher the efficiency of the amplifier. In addition, the efficiency can also be improved by varying the input voltage of the amplifier, generally speaking, the higher the input voltage of the amplifier, the higher the efficiency of the amplifier.
In short, the efficiency of the integrated power amplifier circuit can be improved by changing the parameters of the components, changing the structure of the amplifier, changing the operating frequency of the amplifier, and changing the input voltage of the amplifier.
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According to the requirements, you can write the operation expression: uo=10-500ui = - 10+500ui).
Therefore, an inverted summation adder is required (find the classical circuit in the analog textbook by yourself), with a gain of 500 times, plus a DC component.
Since the output voltage is up to 10V, it is recommended to use +-15V for power supply, and since the input is in the MV level, it is recommended to use the high-precision op amp OP07.
The resistance R at the input signal end is 200, and the feedback resistance RF is 100K to achieve an inverting gain of -500.
The DC additive signal is led from the -15V supply with a 200K adjustable resistor to the inverting terminal, and when the input voltage is zero, adjust the adjustable resistor (at about 150K) to make the output 10V.
If the gain requirement is accurate, you need to replace the R at the input with an adjustable resistor and adjust it so that the output voltage is zero at 20mV input.
By the way, your signal unit is written incorrectly, 20mv means 20 megavolts, which is 20 million volts! The millivolt "m" must be lowercase, it can't really be a 20000000v input, right?
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You just look for the book and rip off the circuit diagram with the op amp inverted magnification.
Gain = 10, according to the gain to determine RF and R1
over.-Only make moves, not do it for you--
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When talking about integrated op amps, it should be said that the chips with integrated op amps are used as functional modules, because each op amp has peripheral circuits to work, and a module is a small circuit made by adding these peripheral circuits.
The above figure shows a high-frequency amplification circuit module designed with an OPA695 integrated op amp.
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<> your picture is too small to see clearly, as shown in the picture above, the relationship is as follows.
uo = u1 - u2) *r1 + r3) /r2 + 1] *rf / rs)
This is a typical three-op amp differential amplification circuit, which can effectively suppress the common mode signal and amplify the differential mode signal, which is commonly used in precision instrumentation.
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1. Basic circuit: according to your requirements, the inverted phase amplification circuit can be realized;
2. The most troublesome problem: op amp (OP) selection.
1) The output voltage is plus or minus 30V, which means that the power supply voltage of the OP is generally above plus or minus 32V, which is a rare high-voltage op amp.
2) The bandwidth is 100kHz, and the gain is 20dB (10 times), and the two inherit each other to obtain the gain bandwidth product, which is at least 1m;
3) Under the condition of plus or minus 15V, the output of plus or minus 30V should be reached, and the output voltage of any op amp cannot be higher than the power supply voltage.
Conclusion: The conditions given are incorrect, please check for yourself.
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What is the use of design, I can provide it.
Not necessarily, but there are exceptions, generally speaking, there is more than half of the chance of success, and the case of failure is a small half, and then you will also face a very embarrassing situation, please don't try it if you are not determined!
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