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rail to rail op-amp

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firstman

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I have problem about Rail-to-Rail op-amp.Rail to Rail op-amp use common emitter in it's output stage. Common emitter has high output resistance(collector).I do not understand why rail to rail op-amp(OP279) has low output resistance ,although it's output stage use common emitter.In datasheet (OP279) expressed output resistance is 22 ohms.It is very low output resistance.Please tell me,it is why? :confused:

http://www.datasheetcatalog.com/datasheets_pdf/O/P/2/7/OP279.shtml
 
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MikeMl

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When pulling low, the NPN transistor can turn on hard enough to be saturated, which means that the output impedance is low... Conversely, when the PNP is on, the output can pull high with a corresponding low output impedance. Therefore, "rail to rail" output.
 

kchriste

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The reason is negative feedback. Notice that they spec it at 1 Mhz and an Av (gain) of one. It'll be much lower at DC with Av = 1. See chart TPC 15 in datasheet.
 

audioguru

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The output impedance is 22 ohms only when it has a lot of negative feedback to make its gain only 1. The negative feedback is a high amount only at low frequencies and makes the output impedance much lower.

Figure 15 on the datasheet shows that at 1MHz the output impedance is 22 ohms when the gain is 1 but the output impedance is 100 ohms when the gain is 10 and 100 (because then the negative feedback is reduced).
The output impedance is close to zero ohms (when the gain is 1) at frequencies below 100kHz.
 

firstman

New Member
The output impedance is 22 ohms only when it has a lot of negative feedback to make its gain only 1. The negative feedback is a high amount only at low frequencies and makes the output impedance much lower.

F

thanks all for your replys. How can i know output resistance when negative feedback is not used.It is clear ,the negative feedback reduces output resistance.When negative feedback is not used,the output resistance must large,or not ,because of using common emitter in output stage?Please tell me ,how can i know value of ouput resistance when negative is not used.
 

audioguru

Well-Known Member
Most Helpful Member
The datasheet shows the gain dropping at high frequencies on a graph that shows the gain without negative feedback.
Simple arithmatic can calculate the output impedance without any negative feedback and uses the open-loop gain with the output impedance when negative feedback reduces the gain to 1.
 

mneary

New Member
An operational amplifier is useful when it has negative feedback. Without negative feedback, it becomes a comparator. Devices made as op amps often perform poorly as comparators because they are not designed to be used that way.
How can i know output resistance when negative feedback is not used.
It does not have one value. As you should have suspected, without feedback, its V/I curve (impedance) would be very high because of the common emitter output stage. When approaching one rail or the other the slope has to level off, making the impedance appear to be quite low.

All this is pointless without negative feedback. Then it becomes just a (poor) comparator with an op amp's part number.
 

firstman

New Member
Simple arithmatic can calculate the output impedance without any negative feedback and uses the open-loop gain with the output impedance when negative feedback reduces the gain to 1.

Thanks for your useful answer.But i am not very clear.If you can help me ,please calcuate output impedence without negative feedback(at open loop gain ).I think ,that the output impedence is very high.I am waiting for your reply.thanks all for useful answer.:)
 
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Roff

Well-Known Member
I think the confusion may result from an error in the datasheet. See attached.
The impedance is obviously closed loop.
 

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  • op279 output impedance typo.png
    op279 output impedance typo.png
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Roff

Well-Known Member
Thanks for your useful answer.But i am not very clear.If you can help me ,please calcuate output impedence without negative feedback(at open loop gain ).I think ,that the output impedence is very high.I am waiting for your reply.thanks all for useful answer.:)
It will be high, but impossible to calculate.
 

kchriste

New Member
Forum Supporter
I think the confusion may result from an error in the datasheet. See attached.
The impedance is obviously closed loop.
I agree. The bit that says, "Av = +1" is not open loop either.
 

mneary

New Member
I think ,that the output impedence is very high.
Correct. It is probably very high.

You can make up any number you want and it won't be wrong.

[edit] The schematic shown is only representative. It hides details in the construction that we don't care about. To calculate the actual output impedance you need to get these secrets from the manufacturer. Sometimes manufacturers put more characteristics in their SPICE models. These are also representative, and don't usually cover tolerances.[/edit]
 
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