Hey Guys,

Quick question. The book I'm reading states that with an inverting amplifier setup, the greater the negative feedback, the higher the closed-loop cutoff frequency. In other words, the smaller Av(cl) is, the higher f2(cl) is.

Could someone please explain to me why this is? I'm really trying to understand how the ration of Rf/R1 impacts how much voltage is fed back to the input.

Thanks,

Austin

 

Pic here: http://forum.allaboutcircuits.com/showpost.php?p=752476&postcount=14

My explanation here: http://forum.allaboutcircuits.com/showpost.php?p=752480&postcount=15

 

Hey Guys,

Quick question. The book I'm reading states that with an inverting amplifier setup, the greater the negative feedback, the higher the closed-loop cutoff frequency. In other words, the smaller Av(cl) is, the higher f2(cl) is.

Could someone please explain to me why this is? I'm really trying to understand how the ration of Rf/R1 impacts how much voltage is fed back to the input.

Thanks,

Austin

It actually REDUCES the gain everywhere else, so the "cutoff" frequency pretty much ceases to exist.

 

The most understandable definition anyone ever gave me of this phenomenon was this: the op-amp's open-loop bandwidth is pretty bad. Negative feedback always has the effect of "fixing" the output so it more closely resembles the input (this is why the amplitude goes down...it's moving closer to the original amplitude at the input.) But NFB doesn't distinguish what *way* the two signals are different...just the more of it you apply, the closer the output signal gets to the original input. Bandwidth doesn't really increase, in one sense, it's just that the bandwidth errors get corrected.

 

In a functioning op-amp circuit, the current through the feedback resistor is equal and opposite of the current in the input resistor. The input and the feedback currents sum together at the inverting input and resolve to zero, relative to the non-inverting input.

In other words, the output drives the inverting input to match the non-inverting input through the feedback resistor. The lower the feedback resistor is (relative to the input resistor), the faster the output can drive the inverting input to match the non-inverting input, thus yielding a wider bandwidth.

Output slew rate is another player in gain product bandwidth. The farther the output has to go, the longer it will take.

 

In a functioning op-amp circuit, the current through the feedback resistor is equal and opposite of the current in the input resistor. The input and the feedback currents sum together at the inverting input and resolve to zero, relative to the non-inverting input.

In the case of an inverting amplifier the current (less the bias current into the inverting input terminal) flows in both input and feedback resistors in the same direction between input & output terminals. Is this what you mean?

In the case of a non-inverting unity gain buffer one can argue that no feedback current flows - other than the input bias current(s).

 

The gain-bandwidth product of a standard voltage-feedback op amp is mainly determined by the internal single-pole compensation required for stable closed-loop gain. This compensation pole adds a 20dB/decade rolloff to the open loop bandwidth. This has the effect of making the bandwidth inversely proportional to the closed-loop gain.

Here's a typical open loop response of such an amp:

The small-signal bandwidth (ignoring any slew-rate limits) can be determined by drawing a straight line from the closed loop equivalent non-inverting (noise) gain in dB on the left axis to where it intersects the open-loop response. Thus, for example, if the NI closed-loop gain was 30dB, the bandwidth would be about 100kHz.

A current-feedback op amp does not require this single-pole compensation for stable closed-loop gain so its bandwidth is only mildly affected by the closed-loop gain value. That's why it's often used for high gain amps needing high bandwidth.

 

In the case of an inverting amplifier the current (less the bias current into the inverting input terminal) flows in both input and feedback resistors in the same direction between input & output terminals. Is this what you mean?

In the attached inverting op-amp,
- Input voltage = +1V
- Output is -10V
- The inverting input is held at Gnd.

The current through the input resistor is the same as the current through the feedback resistor. The two currents sum to zero (volts) at the inverting input.

I have found that viewing the components in the feedback path as being in a constant current circuit aids with circuit analysis. One of the best circuits that demonstrates this is an op-amp integrator. A constant current applied to a capacitor yields a linear ramp.