hi, for an opamp internal circuit, we connerct a ce-cb config in the output stage rite,
a ce cb config has high output impedence. but an opamp has a very low output impedence.
can anyone explain me y this s so or have i understood wrong.

 

Hi,
have a look at page 2 of this datasheet (LM324)
http://www.national.com/ds/LM/LM124.pdf

The output devices are arranged as emitter followers (common collector) which gives low output impedance.


The other important point is that Op-Amps are often used in closed-loop configurations with feedback. This makes the output impedance appear lower, as the op-amp is actively driving the output pin to the correct voltage regardless of internal losses, as long as the currents are within the device limits.

As a demonstration of this, in a typical closed-loop amp, you can connect a moderate value resistor between the chip output and the circuit output, and as long as the feedback is taken from after the resistor, the circuit output will work just the same for small signals.

This is actually common practice when driving high-capacitance loads with some types of op-amp, the resistor stops the amp oscillating.

See figure three here, note 'Riso' at the amp output:
http://www.planetanalog.com/features/showArticle.jhtml?articleID=206501946

 

oh ok..
consider this mc1435 ckt. it has cb ce config at its output stage, so it has to have high output impd rite??

this ckt is given as the general circuit for all opamps in the text books..

 

I would say that must be a 'simplified' representation of the actual chip design.

The MC1435 Data sheet also shows that same emitter follower output stage with a 2K resistor from output to negative supply, but the data characteristics do not fit that configuration.

Datasheet here: http://www.datasheetarchive.com/pdf-datasheets/Datasheets-110/DSAP0010025.pdf


Example: With a +/- 6V supply, the short-circuit output current is given at typically 17mA.

Although the emitter follower stage could source significant current, the 2K resistor in the schematic would only sink 6mA with the output shorted to V+ (with a total 12V between V+ & V-).

The opamp must actually have an active output circuit to allow it to sink that much current.

However, the data also says it can only guarantee +/- 2.5V into a 10K load..

I think it is a rather poor example for educational purposes.
(The datasheet specifies the output impedance is 1.7K)

 

ok...
thanks a lot, sir..
i'll let u know if i get any more doubts..

 

oh ok..
consider this mc1435 ckt. it has cb ce config at its output stage, so it has to have high output impd rite??

this ckt is given as the general circuit for all opamps in the text books..

The output stage in your attachment is not CB-CE, nor is it similar to the output stage in the MC1435 datasheet.

 

I would say that must be a 'simplified' representation of the actual chip design.

The MC1435 Data sheet also shows that same emitter follower output stage with a 2K resistor from output to negative supply, but the data characteristics do not fit that configuration.

Datasheet here: http://www.datasheetarchive.com/pdf-datasheets/Datasheets-110/DSAP0010025.pdf


Example: With a +/- 6V supply, the short-circuit output current is given at typically 17mA.

Although the emitter follower stage could source significant current, the 2K resistor in the schematic would only sink 6mA with the output shorted to V+ (with a total 12V between V+ & V-).

The opamp must actually have an active output circuit to allow it to sink that much current.

However, the data also says it can only guarantee +/- 2.5V into a 10K load..

I think it is a rather poor example for educational purposes.
(The datasheet specifies the output impedance is 1.7K)

The output stage in the MC1435 datasheet is not an emitter follower. It is a Sziklai pair (compound transistor), and the output is from the collector, which explains the 1.7k output impedance spec.

 

I was not clear on the 'CB-CE' bit, but opamp schematic is the same as one section of the MC1425 as per the datasheet.

(Note the datasheet shows both sections, the lower drawn inverted).

 

hi, for an opamp internal circuit, we connerct a ce-cb config in the output stage rite,
a ce cb config has high output impedence. but an opamp has a very low output impedence.
can anyone explain me y this s so or have i understood wrong.

Most op-amps have emitter follower output stages, which are inherently low impedance even without feedback.

 

I was not clear on the 'CB-CE' bit, but opamp schematic is the same as one section of the MC1425 as per the datasheet.

(Note the datasheet shows both sections, the lower drawn inverted).

Look closely. They are not the same.

 

Yep, there are a few details different, I was only looking specifically at the output transistor & load resistor.

The changes *could* be transcription errors? The mid section is missing a load resistor, without which it could not work.

There again, the links from the input stage to the mid stage are reversed, which would account for the inverted output between the two designs.

 

Yep, there are a few details different, I was only looking specifically at the output transistor & load resistor.

The changes *could* be transcription errors? The mid section is missing a load resistor, without which it could not work.

There again, the links from the input stage to the mid stage are reversed, which would account for the inverted output between the two designs.

I don't see the inverted output between the two designs. See the annotated attachment.

 

Why are you looking at the antique MC1435 opamp? It has nothing that is common with a modern opamp and is no longer available:
Its max supply voltage is very low.
Its max gain is very low.
Its input impedance is low.
Its frequency response is very poor.
Its supply current is high.
It needs external compensation capacitors.
Its output is class-A instead of class-AB.

 

Why are you looking at the antique MC1435 opamp? It has nothing that is common with a modern opamp and is no longer available:
Its max supply voltage is very low.
Its max gain is very low.
Its input impedance is low.
Its frequency response is very poor.
Its supply current is high.
It needs external compensation capacitors.
Its output is class-A instead of class-AB.

Only because the OP brought it up.

 

now i am even more confused..
can i have a detailed conceptual analysis of the outout impedence..
including tat ce cb confirmatory part pls....

 

now i am even more confused..
can i have a detailed conceptual analysis of the outout impedence..
including tat ce cb confirmatory part pls....

I think the word you want is "complementary". It is not "ce cb". There are no common base transistors of either of the output stages that we looked at.
This will probably just confuse you more, but a cascode pair is made up of a CE stage, followed by a CB stage. It has nothing to do with the discussion here.
I won't attempt to post an analysis of a Sziklai pair, other than to say it is a feedback circuit, and has current gain ≈β1*β2. You can get a little more info by reading the Wikipedia article that I referenced.

 

The school teacher should be fired for having students analyse such an old garbage opamp.

 

The school teacher should be fired for having students analyse such an old garbage opamp.

He's probably been out of industry so long (maybe never), that that's all he knows. If this is the case, it's not good.