Determining Bandwidth & Impedance of an Amplifier

[dave12]

Ok I have this question that I am finding hard to answer. I understand that you guys won't answer homework questions but all I ask is a little point in the right direction. Heres the question anyway.

An amplifier has an open-loop, gain of 10,000 and a bandwith of 2MHz. Its input and output impedances without feedback are 50Kohm and 500ohm respectively. If the gain is subsequently reduced to 50, by application of negative feedback determine
a) The new bandwidth
b)The feedback fraction Beta
c)input impedance
d) Output impedance.

I think I can work out the Feedback fraction I make it to be, 0.0199.

But I am not sure on the rest, my notes don't tell me any way of working out the new bandwith or impedances.

Please can someone help me? Any formulas or hints apprecited.

Thanks in advanced Everything helps!

 

[mmwave]

Originally posted by dave12
Ok I have this question that I am finding hard to answer. I understand that you guys won't answer homework questions but all I ask is a little point in the right direction. Heres the question anyway.

An amplifier has an open-loop, gain of 10,000 and a bandwith of 2MHz. Its input and output impedances without feedback are 50Kohm and 500ohm respectively. If the gain is subsequently reduced to 50, by application of negative feedback determine
a) The new bandwidth
b)The feedback fraction Beta
c)input impedance
d) Output impedance.

I think I can work out the Feedback fraction I make it to be, 0.0199.

But I am not sure on the rest, my notes don't tell me any way of working out the new bandwith or impedances.

Please can someone help me? Any formulas or hints apprecited.

Thanks in advanced Everything helps!

It would be ridiculous to give you this problem without covering feedback amplifiers first. Any decent circuit design textbook will have a chapter on feedback (See Gray & Meyer Ch 8 for e.g). You can't determine the input & output impedances with the information you've given because the result depends on whether it is series-series, shunt-shunt, series-shunt or shunt-series feedback. That is you have to know how the 'input' signal is sensed & how the output signal is applied by the feedback circuit.

For a start, gain with feedback is a/(1+af) = 50 and a = 10000. Solve for f. For the rest, you need to read a chapter on feedback & then ask questions. I don't think anyone can explain it all to you here.

 

[ravenprp]

Hi there,

Determining the bandwidth and impedance of an amplifier can be a tricky task, but with the right formulas and understanding of the concepts, it can be easily solved.

First, let's define some terms:
- Bandwidth: This refers to the range of frequencies that an amplifier can amplify without significant loss of gain.
- Impedance: This refers to the resistance to the flow of electric current in a circuit.

To calculate the new bandwidth, we can use the formula:
New Bandwidth = Original Bandwidth / (1 + Feedback fraction)

In this case, the original bandwidth is 2MHz and the feedback fraction is 0.0199. Plugging these values into the formula, we get:
New Bandwidth = 2MHz / (1 + 0.0199) = 1.98MHz

To calculate the feedback fraction, we can use the formula:
Feedback fraction = (1 / Gain with feedback) - (1 / Open-loop gain)

In this case, the gain with feedback is 50 and the open-loop gain is 10,000. Plugging these values into the formula, we get:
Feedback fraction = (1 / 50) - (1 / 10,000) = 0.0198

Next, to calculate the input impedance, we can use the formula:
Input Impedance = (Open-loop input impedance) / (1 + Feedback fraction)

In this case, the open-loop input impedance is 50Kohm and the feedback fraction is 0.0199. Plugging these values into the formula, we get:
Input Impedance = 50Kohm / (1 + 0.0199) = 49.75Kohm

Finally, to calculate the output impedance, we can use the formula:
Output Impedance = (Open-loop output impedance) * (1 + Feedback fraction)

In this case, the open-loop output impedance is 500ohm and the feedback fraction is 0.0199. Plugging these values into the formula, we get:
Output Impedance = 500ohm * (1 + 0.0199) = 509.5ohm

I hope this helps guide you in the right direction. Remember to always double check your calculations and plug in the correct values for each formula. Good luck!