AKSA said:
Yes, I'm aware of that.
Yes, I'm aware of that.
Nothing to prove wrong.
Then let me rephrase it so you perhaps can understand what I was asking.
If one directly feeds the opamp's output into its inverting input, creating a unity gain buffer, does that loop not contain the open loop gain of the opamp?
In other words, if you follow the loop around, the opamp's open loop gain is part of that loop. You have the input, the summing node, then the opamp's open loop gain, the output which goes into the feedback network which is then fed to the summing node.
Point I was getting at was that the opamp's open loop gain doesn't just disappear. If it did, then feedback wouldn't be able to do the things that it does.
So now let me get more to the point.
Let's say you've got an unknown opamp with an unknown open loop gain configured as a unity gain buffer.
You can determine that it's gain is 1. But you're not going to know what the open loop gain is and subsequently the feedback factor. So what do you do? You configure it as an open loop amplifier and measure its open loop gain.
If it's open loop gain is 100,000, you know that the feedback factor is also 100,000 and you've got 100dB of feedback.
Why is a transistor fundamentally any different?
If a transistor configured as a common-emitter amplifier with say 100 ohms for RC gives you a gain of say 100, is not the same transistor configured as a common-collector amplifier with 100 ohms for RE an amplifier which has a feedback factor of 100 and 40dB of feedback?
I don't see any fundamental difference here.
se
If an amplifier circuit has three stages: 1st stage gain = 15, 2nd stage gain of 75, and 3rd stage gain of 1, it has a open loop gain of 1125 (~67db). Feedback does not change the gain of each of the stages. The signal from the inputs is smaller with more feedback. The more feedback, the smaller the difference signal.
This is kind of interesting. So, with an op-amp with an OLG of 10,000. Configure it as a buffer. Apply 2VAC. Output is 2VAC. However, the difference signal at input of the first stage is 200uVAC. If the CLG is 10, then, of course, the signal is 2mVAC.
I imagine that something similar happens with single/compound active device circuits (meaning the gain of the devices remains the same OL v CL).
JF
This is kind of interesting. So, with an op-amp with an OLG of 10,000. Configure it as a buffer. Apply 2VAC. Output is 2VAC. However, the difference signal at input of the first stage is 200uVAC. If the CLG is 10, then, of course, the signal is 2mVAC.
I imagine that something similar happens with single/compound active device circuits (meaning the gain of the devices remains the same OL v CL).
JF
Charles Hansen said:
I assume you mean this:
Responding to an earlier comment, the gain device doesn't
know what mode (Common Drain, Source, or Gate) it is being
used in. It only sees the varying voltages and currents, and
these can easily be identical from the device's point of view.
Ok.
Essentially the same can be said of an opamp.
So what does this tell us?
se
Steve Eddy said:Hehehe. Yeah. I'm beginning to feel like I'm in a Kung Fu flashback.
Sorry to be so cryptic. It's a Procol Harum quote from one of my favorite albums, "Shine On Brightly". Words are here: http://www.procolharum.com/w/w0208.htm. The guy who transcribed it made a lot of run-on sentences out of it that probably weren't intended by the author.
I don't mean to be cryptic, so I'll say also that off the top, the
circuit driving a Sziklai pair sees the Vgs or Vbe variation of the
first device, and with the darlington it sees the variation of the
sum of both transistors, thus it is not the same, and the
darlington is seen to have less voltage gain, all other things
being equal.
circuit driving a Sziklai pair sees the Vgs or Vbe variation of the
first device, and with the darlington it sees the variation of the
sum of both transistors, thus it is not the same, and the
darlington is seen to have less voltage gain, all other things
being equal.
Nelson Pass said:
Hmmmm. How does the source driving the pair see just the Vgs/Vbe variation of the first device in the Sziklai but the Vgs/Vbe variation of both devices in the Darlington?
The source is referenced to ground and is in series with the source/emitter resistor. Since the current through the source/emitter resistor (I assume we're talking about an emitter follower configuration here) is the sum of the currents through both devices whether Sziklai or Darlington, then the source must be seeing the Vgs/Vbe variations of both devices whether Sziklai or Darlington, no?
se
Hello SY -
Well, you don't have any choice about this when it comes to the Darlington emitter follower. It will always be (slightly less than) unity gain.
Now the CFP is a different beast altogther. Instead of a common collector feeding a common collector (where there is no choice about what the gain is), we now have a common emitter feeding a common emitter. Each of these common emitter stages will have a gain that (before closing the feedback loop) will depend on what resistor values are chosen for that circuit.
To illustrate this, let's look at a circuit taken from Self's book on amplifiers. In one figure, the first transistor in a CFP has a 100 ohm collector resistor. There is no emitter resistor for the second transistor. The speaker load (nominally 8 ohms) forms both the emitter resistor for the first transistor and the collector resistor for the second transistor.
Now if we "break" the feedback loop in our minds eye, the first transistor will have a gain of 100/(re + 8) where re depends on the bias and is probably a couple of ohms. Let's just say that re = 2 for conveniece's sake, so that the gain of the first transistor is 10. The second transistor will have a gain of 8/re, where re is now around 0.5 (based on a likely bias current in the output stage). This gives a gain of around 16 for the second transistor.
So the composite gain with the loop broken is roughly 160 or 44 dB. Once the loop is closed to achieve unity gain, we now have 44 dB of feedback. This is just within the short 2-transistor loop created by the CFP and is in addition to any other (e.g., global) feedback loops that may exist in the amplifier.
The CFP will have lower measured steady-state distortion (into a resistive load) than the Darlington emitter follower because it has more feedback. Furthermore, it will be less stable.
Best regards,
Charles Hansen
Well, you don't have any choice about this when it comes to the Darlington emitter follower. It will always be (slightly less than) unity gain.
Now the CFP is a different beast altogther. Instead of a common collector feeding a common collector (where there is no choice about what the gain is), we now have a common emitter feeding a common emitter. Each of these common emitter stages will have a gain that (before closing the feedback loop) will depend on what resistor values are chosen for that circuit.
To illustrate this, let's look at a circuit taken from Self's book on amplifiers. In one figure, the first transistor in a CFP has a 100 ohm collector resistor. There is no emitter resistor for the second transistor. The speaker load (nominally 8 ohms) forms both the emitter resistor for the first transistor and the collector resistor for the second transistor.
Now if we "break" the feedback loop in our minds eye, the first transistor will have a gain of 100/(re + 8) where re depends on the bias and is probably a couple of ohms. Let's just say that re = 2 for conveniece's sake, so that the gain of the first transistor is 10. The second transistor will have a gain of 8/re, where re is now around 0.5 (based on a likely bias current in the output stage). This gives a gain of around 16 for the second transistor.
So the composite gain with the loop broken is roughly 160 or 44 dB. Once the loop is closed to achieve unity gain, we now have 44 dB of feedback. This is just within the short 2-transistor loop created by the CFP and is in addition to any other (e.g., global) feedback loops that may exist in the amplifier.
The CFP will have lower measured steady-state distortion (into a resistive load) than the Darlington emitter follower because it has more feedback. Furthermore, it will be less stable.
Best regards,
Charles Hansen
Steve Eddy said:
I did not assume such a resistor for either case, and adding one
creates a new ball game.
As far as the estimate of internal feedback of the Sziklai, I think
40 dB is way too high if you consider that there is invariably
a resistor Base-Emitter of the second transistor. Depending on
that value, a more typical figure is maybe 10 dB, very dependent
on the impedance of the source and the load.
This doesn't negate the central conclusion - is there more gain
with a Sziklai? Voltage yes, current not really. After that we
are arguing amounts.
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