# Looking at feedback and some Opamp problems

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1. If the unity gain frequency is 5 MHz and the midband open-loop voltage gain is 200,000, what is the cutoff frequency?

2. An op amp has a voltage gain of 500,000. If the output voltage is 1V, what is the input voltage?

3. Name a few ICs other than op amps.

4. Above the cutoff frequency, what happens to the voltage gain of the 741C?

5. Describe some of the characteristics of the 741C.

Though in some applications the 741 is a good approximation to an ideal op-amp, there are some practical limitations to the device in exacting applications.

* The input bias current is about 80 nA.

* The input offset current is about 10 nA.

* The input impedance is about 2 Mega Ohms.

* The common mode voltage should be within +/-12V for +/-15V supply.

* The output impedance is about 75 Ohms.

* The voltage gain rolls off 6dB per octave starting at 100kHz.

* There is a finite input offset which must be zeroed by a resistor between pins 1 and 5. The input offset is typically 2mV to <6mV.

* The slew rate is 0.5V/microsecond.

* There is some temperature dependence.

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##### Solution Summary

This solution looks at some of the Opamp properties including determining closed loop gain of an OPAMP given the open loop gain, determining the properties of negative feedback, determining OPAMP Bandwidth using the Gain Bandwidth product expression. It also looks and tries to explain some of the non ideal, real OPAMP properites and explains what an OPAMP tries to do

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1. Gain Bandwidth product (GBP) applies to the performance of the Op-amp. This is expressed by (1) below.

GBP = frequency x voltage gain at that frequency in linear terms (1)

For the parameters given at unity gain we have

GBP = 5 MHz x 1 {from unity gain @ frequency of 5 MHz}

GBP = 5 x 10^6 x 1 = 5 x 10^6

At mid-band we have a voltage gain of A(mid) = 200,000 = 2 x 10^5 in linear terms

So we can determine the mid-band frequency {f(mid)} of this op-amp by equating the GBP at this frequency to the GBP we calculated earlier. So we can say

A(mid)*f(mid) = 5 x 10^6

{2 x 10^5}*f(mid) = 5 x 10^6

f(mid) = {5 x 10^6}/{2 x 10^5} = 25 Hz

Cutoff frequency {fc} is 2x the mid band frequency so

Cutoff frequency fc = 2 x 25 = 50 Hz ...

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