output impedance of op-amps

Hi there,

Which part of the internal circuitry of an op-amp helps achieve the requirement for an op-amp to have low output impedance, and how does it do that?

Thanks

Reply to
cruiser
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Agreed, and do add "negative feedback" to your searches.

Reply to
Charles Schuler

Do thy own homework.

Bob M.

Reply to
Bob Myers

"Charles Schuler"

** Not part of the "internal circuit " of an op-amp.

....... Phil

Reply to
Phil Allison

"cruiser"

** The use of a "voltage follower" output stage.

Low output impedance is due to inherent local negative feedback in the such voltage followers.

Most op-amps use complementary ( NPN + PNP ) transistors with output supplied from the two emitters.

....... Phil

Reply to
Phil Allison

I have done my research - I just don't understand it... What i've found is that the low output impedance is designed in the last stage (the output voltage amplifier stage) and it has something to do with a BJT in emitter follower configuration, which is loading a class AB amplifier. While I understand these two things separately, when they are put together, along with the other components in the op-amp, I can't follow the direction of the current flow and I don't know what the transistors are doing.

Reply to
cruiser
** Hey, Groper.

Alter your settings so the person and post you are replying to is quoted.

Posting " in mid air" is very bad etiquette.

** See "Emitter Follower Discussion".

formatting link

Then see "Feedback in the Emitter Follower".

Also, post a link ot the op-amp schematic you are looking at.

...... Phil

Reply to
Phil Allison

Emitter followers have the lowest possible output impedance of the three configurations. Check into that configuration ... emitter follower (or common collector). Google!

Class AB is simply a means to reduce crossover distortion. Don't let that confuse you as to output impedance ... they are separate issues.

Reply to
Charles Schuler

Look at the internals of a TL071. It should be very clear from that. And indeed how to set a specific figure.

Graham

Reply to
Eeyore

You need to learn something about circuit theory (and practice) then !

Graham

Reply to
Eeyore

Err.. actually...yes..very low, open loop, output impedance is often fundamentally due to internal negative feedback, to wit via the usual Miller compensation capacitor.

Typically, in the raw state, the main gain stage would be a very high impedance node indeed, often a cascade at top and bottom (ro=100sMeg?). Even driving double source followers, the transformation of this impedance to the emitters of the buffers, might still be relatively high. However, the Cbc capacitor at the high gain stage instead ensures that the driving impedance driving the buffer followers is of the order of 1/gm (say 100 ohms) of the gain stage transistor at mid band and above.

relevant formula:

zo = rs/hfe + re.

To wit, you don't get the re of the output buffer if the driving source is very high impedance, which it would be if no Miller cap. This one of the reasons its used it does several things at once, e.g. pole splitting.

--
Kevin Aylward
ka@anasoftEXTRACT.co.uk
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Reply to
Kevin Aylward

"Kevin Aylward"

** But that is not what the poster who used the phrase " negative feedback " was alluding to.

Or he would have said so.

....... Phil

Reply to
Phil Allison

I was referring to the original poster text.

"Which part of the internal circuitry of an op-amp helps achieve the requirement for an op-amp to have low output impedance, and how does it do that?"

It is the Millor capacitor that is part of the internal circuitry of an op-amp that helpes achieves an op amps low output impedance, in addition to emitter followers of course. In rail-to-rail output amps, the Millar cap at the output might be the only key feature that results in a low output impedance.

--
Kevin Aylward
ka@anasoftEXTRACT.co.uk
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Reply to
Kevin Aylward

output transistors, but low output impedance is usuall achived using external components.

Bye. Jasen

Reply to
jasen

"Kevin Aylward"

** But I clearly was not.

And you posted your remarks to me.

Your error.

....... Phil

Reply to
Phil Allison

Jane, you ignorant slut!

Reply to
Dr. Polemic

How exactly does a Miller cap help output resistance ?

Graham

Reply to
Eeyore

The output resistance of an emitter follower is:

ro = Zs/hfe + re., Zs is the driving impedance. If Zs is not low, ro is not = re

That is, the driving impedance gets reflected by hfe to the output.

Consider a high gain cascode stage. It might be say, Va/Ic.(gm.zx) =

100/1ma*100 = 10meg gm.zx being a gain increases of the raw 1/ho of the transistor, and left as an exercise for the reader to work out.

If there is a feedback impedance from collector to base, we have, approximately

io = vo . hie/(Zf +hie) . gm

That is, feeding a signal accross CE results in a potential divider driving the base, which turns on the collecter current.

Or

Zs = vo/io = (1 + Zf/(re.hfe)).re

Once Zf < hie, vo/io tends to 1/gm, i.e it looks like a diode impedance. Even before it gets that low, zo is way lower than, the 10meg raw output impedance.

--
Kevin Aylward
ka@anasoftEXTRACT.co.uk
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Reply to
Kevin Aylward

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