The word, 'large' adds no meaning. The rest is pretty iffy, also.
This is good enough to be useful.
No. It uses negative feedback to make its two inputs more the same.
Yes. The output voltage is the result of some sort of voltage divider operation across the supply.
I wouldn't go that far. Besides, simple generalities are almost never 'right'. But sometimes they are close enough to right to be useful.
-- John Popelish
Without long explainations, is it fair to say :
1) A transistor uses change in voltage to control a large flow of current.2a) An op-amp amplifies small differences in voltages across its 2 inputs.
2b) An op-amp using negative feedback tries to make the amplified output signal more like the input signal.2c) The amplified voltage of an op-amp cannot be greater than the supply voltage. e.g. if supply voltage is 5V, it cannot be amplified to more than -2.5 to +2.5V.
Is all of the above right?
What if the feedback includes diode junctions, zeners or other nonlinear effects?
With some unstated assumptions, yes. I figured he would learn more if I nibbled at the edges of what he knows.
-- John Popelish
Yes.
Yes.
Yes.
Yes.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
A bit of a nit pick me thinks.
Depends on how one reads "more like the input signal". Your actual statement here is obviously correct, but I read this as negative feedback makes the "shape" of the output more like the shape of the input, i.e. feedback reduces distortion errors, which is indeed what feedback does. So, without further information on what the poster really means, it seems that his description, although crude is ok.
Well, I though he had a reasonable handle on it.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
Yes.
Yes, but that's not a sufficient definition for an "operational amplifier" specifically. The above definition would apply to any amplifier with differential inputs; to qualify as an "operational amplifier," there are generally the additional requirements of very high open-loop gain, very high input impedance, and very low output impedance.
No. The operational amplifier itself does not do this. Many useful circuits USING op-amps DO employ negative feedback, but not all.
This is true of any amplifier; more generically put, the output signal swing cannot exceed the limits of the "rails" (the positive and negative supplies, or the supply and the reference or "ground" point). In practice, it can only approach these limits, and the output waveform will begin to distort as the limits are closely approached.
Bob M.
Ahem. Methinks you both missed that fact that the op-amp ITSELF does not necessarily employ negative feedback in this manner AT ALL.
Bob M.
Maybe, but I major that's not what the poster cared about. One is trying to answer the question as one believes the answer should be. One has to make assumptions as to what the *main* point of the question is.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
Ahmmm.... now your gettig a tad complicated....imo...
I tried to take a somewhat different approach this time, you know, "Yes" instead of my usual...
Kevin Aylward snipped-for-privacy@anasoft.co.uk
Maybe, but if the original poster leaves thinking that this thing in the data book called an "op amp" behaves as described, I would submit that they are in for a rather rude awakening later on.
Bob M.
Again, I disagree. Sure, an op-amp can be used for many things, but the poster gave a pretty reasonable statement on what an op-amp *can* be used for. The fact that the op-amp has other uses is not really relevant. One has to start somewhere. What he stated was indeed, essentially correct.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
The point of op-amp circuits with negative feedback is that if the amp is able to bring the difference between its inputs back to zero by changing the output voltage, then it will do. Which means that the output voltage depends (almost) only on the inputs and the feedback network you're using, rather than the circuit that's been used to build the amp.
-- http://www.niftybits.ukfsn.org/ remove 'n-u-l-l' to email me. html mail or attachments will go in the spam bin unless notified with [html] or [attachment] in the subject line.
I think we're still disagreeing over the original poster's confusion between "op-amp" and "op-amp-based circuits". The statement that an op-amp, alone, with no further qualfication or description, uses "negative feedback" to achieve such-and
-such a result is incorrect on the face of it. From the original statements, I could easily see the original poster believing that an op-amp, all by itself, can act as a low-gain linear amplifier, and that's simply wrong.
Bob M.
Maybe, but I think its more of understanding what the poster thought he wrote, rather then what he actually wrote. Most beginners can't articulate what they mean correctly.
I still disagree. By itself, with a reasonable assumption, the statement is completly reasonable. There seems to be some implications here on the difference between necessary and sufficient, and other like terms. An op-amp circuit usually uses negative feedback to achieve such and such a result. This is an indisputable fact, so I simply don't have a major problem with that statement. The notion that there may be *additional* factors as well to *complete* the description is not necessarily relevant. Of course, the poster could believe how you interpret it. Maybe he/she will renter the discussion.
Maybe, if, so, the posters views would be incorrect, but I simply don't read that as what the poster is meaning, although that may be the case. I am filling in the blanks with the assumption that the poster has some idea that one connects up other bits and bobs to get actual op-amp circuit to work.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
No. You are wrong. The orginal claim was *100%* correct. Its not debatable.
Indeed it does as a side effect, however the transistor is absolutely and fundamentally a *voltage* controlled and operated device
Applying a *voltage* to the base emitter injects carriers into the base region. These carriers are then swept up by the collectors accelerating voltage. The fact that a few leak away through the base is just a nuisance. Ideally, there would be no base current at all.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
Depends how to interpret your questions. Litterally or trying to understand what you "really mean".
No. A transistor amplifies current (hFE). But by "overstearing" the transistor you can have it act as a current switch.
Yes.
In "normal" circuits you provide negative feedback to limit the amplification. Often a 100,000+ times amplification is to much.
Yes, but your conclusion is wrong. If you have 5V supply, the output can swing between 0V and 5V (allmost). If you have +/-5V supply the output can swing between -5V and 5V.
Sorry, no.
/Ingvar
The transistor base voltage varies very little over the range of collector current. Like you say, the base voltage relative to the emitter will usually be 600 to 700 mV, or maybe a bit more.
A general rule of thumb to calculate voltage gain is
40 times the difference of supply voltage to collector voltage. So, if the transistor were biased such that the collector is 3 volts and the supply is 6 volts, you get a voltage gain around (6-3)*40 = 120. So, for every millivolt change on the base, the collector should change about 120 millivolts.Doesn't much matter what the hFE is, other than it effects the input impedance.
-Bill
Well, as you state in your paper "Despite much literature that implies other wise..." I think it must be debatable, otherwise I can't see that "much literature" would imply so :)
From my point of view. If I feed 1mA into the base the Ic would be some hundreds (hFE) mA (if available), even though I know that hFE is not very accurate. On the other hand if I supplied 0.1V to the base, nothing much will happen at the collector. I can accept that it can be seen as a Voltage amplifier, but only around its working point (Vb ~0.6-0.7V).
Ingvar Esk
Its not debatable in the sense that it is accepted by any physicists that understands how transistors actually work. Its simply impossible to derive and sensible device equations on the assumption that base current controls emitter current. The most simple derived equation is:
Ie = Io.(exp(Vbe/Vt) - 1)
Its called the diode equation. It is derived/shown in *any* standard text book devoted to the device physics of the transistor. Its based on applying a potential to the junction. There is no iffs or butts about it.
Unfortunately there are rather a lot of non or semi technical books that use explanations equivalent to water down a pipe. The sooner novices get to grips with the fact that the ic=ib.hfe model, is a *grossly* simplified model, with limited use, the better. It causes never ending confusion.
You are looking at this from a way too naive point of view.
The transistor is a transconductance device. It outputs a current based on its input voltage. It can be accurately exponential over 6 decades. Sure, at low voltages, there is only a small current, but this doesn't change how the transistor operates.
Its not a mater of what you want to accept. Its how it is. You can't disagree with all the semiconductor physicista in the world, well not unless your name is Einstein anyway-)
What I am explaining is *THE* *standard* *accepted* physics of the situation. *Only* those *without* the academic background have transistor operation mistaken. Go and have a read of *any* standard text on semiconductor physics.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
Sorry - that's one model you can use, and it's often a useful one. But with respect to the actual physics of the device, it is more accurate to say that slight VOLTAGE variations across the base-emitter junction control the current "into" the collector. The standard "hybrid pi" model of the transistor is best for showing this in simple form.
Bob M.
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