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- Thread starter Samtheguy
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In summary: The gain bandwidth product is most accurate for the non-inverting configuration, so make sure you review my hint in reply #4 above.

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uart

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Samtheguy

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Okay, that seems more sensible.

I have attemped Question 1. Am I correct in saying that you would draw the Inverting Op-amp circuit diagram, remove the op-amp component since the inputs have a very large input impedance (and thus no current flows into them), and label the virtual Earth point between the feedback and input resistors?

Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10

I have attemped Question 1. Am I correct in saying that you would draw the Inverting Op-amp circuit diagram, remove the op-amp component since the inputs have a very large input impedance (and thus no current flows into them), and label the virtual Earth point between the feedback and input resistors?

Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10

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uart

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Samtheguy said:Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10

Yes that is correct, however this question is more about frequency response and gain bandwidth product.

From the question see if you can figure out the following.

- What is the opamp DC gain (without feedback)

- What is the opamp 3dB bandwidth (dominant pole)

- What is the opamp gain bandwidth product.

* BTW. There is a bit of a trick with this question in that the relevant closed loop gain (for calculating gain-BW product) for the inverting opamp is actually 1+|A_cl|.

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Samtheguy

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A_d = 2x10^5 / (1+jf/7) and set f=0 since it is DC. Therefore, I get A_d = 2x10^5. In dB, this is a gain of 106.02 dB. To find the -3dB point, do 106.02 - 3 = 103.02dB and convert back to magnitude using, 10^(103.02/20) = 141579.

Am I on the right path to the solution?

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uart

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Samtheguy said:

A_d = 2x10^5 / (1+jf/7) and set f=0 since it is DC. Therefore, I get A_d = 2x10^5. In dB, this is a gain of 106.02 dB. To find the -3dB point, do 106.02 - 3 = 103.02dB and convert back to magnitude using, 10^(103.02/20) = 141579.

Am I on the right path to the solution?

Yep A_0 = 2x10^5 is good. :)

The -3dB point does occur at approx 141400 (gain) but that's a

More importantly however you don't actually need to find that -3dB gain (141400) in order to find the -3dB frequency. Just look at the transfer function

Hint : What is the magnitude of (1+j) ?

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Samtheguy

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The magnitude of (1+j) is sqrt(2) or 1.414. So therefore f would be 7Hz at the -3dB point.

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uart

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Samtheguy said:The magnitude of (1+j) is sqrt(2) or 1.414. So therefore f would be 7Hz at the -3dB point.

Yep that's correct. So therefore the gain bandwidth product is 7x2x10^5 = 1.4x10^6 Hz.

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Samtheguy

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Okay. So Gain-bandwidth product = Open-loop gain X cut-off frequency. Is this the answer to the small signal bandwidth or is there a lot more computation to do? (I'm guessing there is more to do)

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- #10

uart

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Samtheguy said:

No there's not much more to it, that's the simplicity and beauty of the "gain bandwidth product" method.

The open loop gain times the open loop -3dB freq is approx equal to the closed loop gain times the closed loop frequency. That is, the gain bandwidth product is approximately a constant.

BTW. The gain bandwidth product is most accurate for the non-inverting configuration, so make sure you review my hint in reply #4 above.

An operational amplifier is an electronic device that amplifies the input voltage signal and outputs a larger version of the signal. It consists of a high-gain differential amplifier and additional circuitry to provide desired characteristics such as high input impedance, low output impedance, and a wide range of frequency response. The input voltages are compared to each other and the difference is amplified, which results in a larger output voltage.

There are two main types of op-amps: inverting and non-inverting. In an inverting op-amp, the input voltage is applied to the inverting input terminal and the output voltage is the inverted version of the input voltage. In a non-inverting op-amp, the input voltage is applied to the non-inverting input terminal and the output voltage is the same polarity as the input voltage. Other types include summing, differential, and instrumentation op-amps, which have different configurations and purposes.

To analyze an op-amp circuit, you can use the golden rules of op-amps. These rules state that the input currents are zero, the input voltages are equal, and the output voltage is such that the difference between the input voltages is amplified by the gain of the op-amp. To solve problems, you can use circuit analysis techniques such as Kirchhoff's laws and Ohm's law to determine the values of the unknown components in the circuit.

Op-amps have a wide range of applications, including signal amplification, filtering, and signal conditioning. They are commonly used in audio and video equipment, sensors, and control systems. They are also used in mathematical operations such as addition, subtraction, differentiation, and integration. Additionally, op-amps are used in feedback circuits to stabilize and control the gain of the circuit.

When choosing an op-amp, you should consider factors such as the required input and output voltage ranges, the desired gain and bandwidth, and the power supply requirements. Other important considerations include the input and output impedance, noise level, and temperature range. It is important to carefully read the datasheet of the op-amp to ensure that it meets your specific needs for your application.

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