This is a very strange behaviour of a download site, isn't it?
I still can download it for free, as many times as I want. On both of my computers.
I think, I am not allowed to publish hardcopies, but I dont worry about distributing it to interested people, as it seems to be an academic paper.
Franz
I still can download it for free, as many times as I want. On both of my computers.
I think, I am not allowed to publish hardcopies, but I dont worry about distributing it to interested people, as it seems to be an academic paper.
Franz
So, at risk of getting out of my depth, I think this idea and that paper is a bit silly. My aim here is not to bash on anyone's viewpoint, just to examine the numbers. After all, most of the arguments in this thread are numbers-based arguments.
For one, the input resistor and positive input resistors contribute the most noise. You can never get output noise below the level of these resistors multiplied by gain. So if both resistors are 47k, you get 39nV/rtHz multiplied by your gain.
Now the feedback resistor increases proportionally to the input resistors as gain increases, but resistor noise does not increase proportionally with resistance. So at gain>>1, the feedback resistor will almost always contribute no significant noise.
So say your gain is 20. You need 47k input resistors, and a 940k feedback resistor. The input resistors contribute 39*20=780nV/rtHz (whoah!), and the feedback resistor contributes 123nV/rtHz. Adding those together, (780^2+123^2)^0.5=790nV/rtHz.
The contribution of the feedback resistor only increases output noise by 1.3%. You can gain no more than 1.3% no matter what trick you find to reduce the noise of the feedback resistor, because it's not where all the noise is coming from.
This brings us to the main conflict for anyone that wants to use an inverting amp the same way they would use a non-inverting amp. The inverting amp needs to use the same feedback network values as the non-inverting in order to have equivalent noise and performance in general. But for the inverting amp, the feedback shunt resistor must ALSO be the input resistor. And 680R (as a ballpark value) is usually considered too much of a load for standard sources. Here we must choose between low noise and low input impedance, or high noise and normal input impedance (unless we want to dramatically increase complexity).
Then there is the question of BW and distortion performance. Normally, low feedback impedance is better for distortion. For BJT inputs there is a point at which increased feedback resistance will proportionally increase output distortion (In my experience somewhere around 5k or less). There are also capacitive input currents which result in reduced BW and carry higher order harmonics in addition to this. For the inverting amp we increase feedback impedance by up to 100x. The aforementioned paper attempts to get around this by putting the feedback loop wide open at RF with a capacitor to ground in the T-network, completely ignoring the main cause of loss in BW which is the capacitive input currents. I think a better solution may be to just use a standard miller capacitor, which will increase BW by salvaging loop gain, rather than throwing it away. The result will be better output behavior and performance.
I think there is a widespread misconception that because the feedback resistor is "in the feedback path" that it is more important than the feedback shunt resistor which is "not". If you are looking at the engineering data, nothing could be further from the truth. My aim here is not to bash on anyone's viewpoint, just to point out that you can hardly claim any technical advantage to the T-network as has been implied. It's useful in some cases, just not in the ones I see here.
If you want to have the benefits of the inverting amp without the drawbacks, what makes the most sense to me is to just use an input buffer with as low an input resistor as it will handle well. Not unlike what was referenced in the original post:
Tube with Power IC Output Stage - JLTi
For one, the input resistor and positive input resistors contribute the most noise. You can never get output noise below the level of these resistors multiplied by gain. So if both resistors are 47k, you get 39nV/rtHz multiplied by your gain.
Now the feedback resistor increases proportionally to the input resistors as gain increases, but resistor noise does not increase proportionally with resistance. So at gain>>1, the feedback resistor will almost always contribute no significant noise.
So say your gain is 20. You need 47k input resistors, and a 940k feedback resistor. The input resistors contribute 39*20=780nV/rtHz (whoah!), and the feedback resistor contributes 123nV/rtHz. Adding those together, (780^2+123^2)^0.5=790nV/rtHz.
The contribution of the feedback resistor only increases output noise by 1.3%. You can gain no more than 1.3% no matter what trick you find to reduce the noise of the feedback resistor, because it's not where all the noise is coming from.
This brings us to the main conflict for anyone that wants to use an inverting amp the same way they would use a non-inverting amp. The inverting amp needs to use the same feedback network values as the non-inverting in order to have equivalent noise and performance in general. But for the inverting amp, the feedback shunt resistor must ALSO be the input resistor. And 680R (as a ballpark value) is usually considered too much of a load for standard sources. Here we must choose between low noise and low input impedance, or high noise and normal input impedance (unless we want to dramatically increase complexity).
Then there is the question of BW and distortion performance. Normally, low feedback impedance is better for distortion. For BJT inputs there is a point at which increased feedback resistance will proportionally increase output distortion (In my experience somewhere around 5k or less). There are also capacitive input currents which result in reduced BW and carry higher order harmonics in addition to this. For the inverting amp we increase feedback impedance by up to 100x. The aforementioned paper attempts to get around this by putting the feedback loop wide open at RF with a capacitor to ground in the T-network, completely ignoring the main cause of loss in BW which is the capacitive input currents. I think a better solution may be to just use a standard miller capacitor, which will increase BW by salvaging loop gain, rather than throwing it away. The result will be better output behavior and performance.
I think there is a widespread misconception that because the feedback resistor is "in the feedback path" that it is more important than the feedback shunt resistor which is "not". If you are looking at the engineering data, nothing could be further from the truth. My aim here is not to bash on anyone's viewpoint, just to point out that you can hardly claim any technical advantage to the T-network as has been implied. It's useful in some cases, just not in the ones I see here.
If you want to have the benefits of the inverting amp without the drawbacks, what makes the most sense to me is to just use an input buffer with as low an input resistor as it will handle well. Not unlike what was referenced in the original post:
Tube with Power IC Output Stage - JLTi
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