Hi,
this is a more or less theoretical question.
Imagine a differential pair fet input-stage. One gate is driven with a signal, the other gate is grounded via a resistor with 100R. The signal is taken from inverted output.
Does the 100R resistor contribute to the overall performance? Does its noise add to the circuit? It is said, every node of an active device serves as an input and an output: will it have signal voltage induced, and if so, at what magnitude?
I hope, it's not too silly
Rüdiger
this is a more or less theoretical question.
Imagine a differential pair fet input-stage. One gate is driven with a signal, the other gate is grounded via a resistor with 100R. The signal is taken from inverted output.
Does the 100R resistor contribute to the overall performance? Does its noise add to the circuit? It is said, every node of an active device serves as an input and an output: will it have signal voltage induced, and if so, at what magnitude?
I hope, it's not too silly
Rüdiger
But...
if I think of it as a basic op-amp, then it tries to hold its +/- pins at 0V.
If the noninverted input sees a NF-Signal, then the inverted input tries to mirror that signal, so a corresponding Voltage should be seen over that 100R resistor, which in turn should be amplified by a certain degree by Q2. -- Not so?
Wait a minute -- Do I need a feedback arrangement for this behaviour?
Rüdiger
if I think of it as a basic op-amp, then it tries to hold its +/- pins at 0V.
If the noninverted input sees a NF-Signal, then the inverted input tries to mirror that signal, so a corresponding Voltage should be seen over that 100R resistor, which in turn should be amplified by a certain degree by Q2. -- Not so?
Wait a minute -- Do I need a feedback arrangement for this behaviour?
Rüdiger
Yes, it will have an effect, though it may be too small to be of any consequence. The resistor will generate noise, as all resistive devices do, which will be amplified by the differential pair in the same way that the input signal is. The rms value of the noise at room temperature will be:
sqrt(1.6e-20*R*BW)
where R is the value of the resistor and BW is the bandwidth of the circuit (probably 20KHz in this case).
sqrt(1.6e-20*R*BW)
where R is the value of the resistor and BW is the bandwidth of the circuit (probably 20KHz in this case).
In principle, noise at the output would be the sum of noise voltages present at both inputs. Noise voltages sum in this way:
v=sqrt(v1^2+v2^2+v3^2...vn^2)
However, depending on the impedance seen by the other intput and the amount of noise produced by the own transistors, you may consider the noise from the 100 ohm resistor negligible.
v=sqrt(v1^2+v2^2+v3^2...vn^2)
However, depending on the impedance seen by the other intput and the amount of noise produced by the own transistors, you may consider the noise from the 100 ohm resistor negligible.
The 100R resistor would theoretically have no change on the circuit in terms of its gain because the transistor is a FET.
==> Look at the FET with its gate tied to ground through a 100R resistor. Ideal FETs draw no gate current, so there is no current going through that 100R. If there is no current through it, then there is no voltage drop across it, and the 100R resistor looks just like a short circuit (THEORETICALLY).
It would have an effect on the noise of the amp, but I'm not sure how to calculate this (see the other postings).
Also, to Onvinyl: yes you would need negative feedback to assume that both terminals are at the same potential. In an opamp, the virtual short principle only applies when there is negative feedback around the opamp, and there also must be more negative feedback than positive feedback around the opamp (if there is any positive feedback at all).
==> Look at the FET with its gate tied to ground through a 100R resistor. Ideal FETs draw no gate current, so there is no current going through that 100R. If there is no current through it, then there is no voltage drop across it, and the 100R resistor looks just like a short circuit (THEORETICALLY).
It would have an effect on the noise of the amp, but I'm not sure how to calculate this (see the other postings).
Also, to Onvinyl: yes you would need negative feedback to assume that both terminals are at the same potential. In an opamp, the virtual short principle only applies when there is negative feedback around the opamp, and there also must be more negative feedback than positive feedback around the opamp (if there is any positive feedback at all).
Um,..I'm pretty sure all you guys have it wrong. If the inverting gate was tied to ground it would thus be part of the NF network and provide little gain at all,..or conversely, there would be mucho grande feedback. This is on the premise that a much smaller output-to-input resistor was used.
Onvinyl, the 100 ohm resistor is in series with the input path and generates 100 ohms of noise which is a little over 1nV/rt Hz. Why is it sometimes there? It can be part of a feedback return path, an RF 'stopper' resistor, or used to match the 2 differential inputs with the input resistance in order to get better tracking of the two differential transistors by balancing the base current drops at each input.
Is 100 ohms a potential noise problem? Sometimes. For example, in one of my phono preamps, it would add 10dB of added noise, quite a lot. For a cheap IC op amp input, it might be barely measurable. It just depends.
Is 100 ohms a potential noise problem? Sometimes. For example, in one of my phono preamps, it would add 10dB of added noise, quite a lot. For a cheap IC op amp input, it might be barely measurable. It just depends.
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