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hi...

how common collector circuit work?
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Since the Emitter current = the Collector current, you are simply using the emitter current across a resistor to generate the output voltage. This circuit has little voltage gain, but Beta current gain and the output voltage is in phase with the base current.

Do you have a more specific question?
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Keith's explanation was technically quite correct, but you may have been looking for something a tad more basic and practical.  It just means that the collector of an output transistor is literally not connected to anything internal other than the pin itself.  Look at the schematic for an LM393 in the datasheet and examine the output pin circuitry.  It can only pull the pin to ground or allow it to float.  To get a logical high output, you use a pull-up resistor connected between the output pin and Vcc.

In contrast, look at the output of a rail to rail op-amp and note how the output pin is in a push pull fashion so that it can drive the output to the positive supply voltage.

The open collector allows the device to be easily connected to other circuitry that might be operating at a different voltage.
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Keith's explanation was technically quite correct, but you may have been looking for something a tad more basic and practical.  It just means that the collector of an output transistor is literally not connected to anything internal other than the pin itself.
No, thats "open collector".  We are talking about the common-collector configuration (I assume), almost always called
an "emitter follower"  BTW.

The voltage gain is about 1 (well very very slightly less than 1), current gain as explained above.  The normal reason
for this circuit is to take a high impedance signal and generate a low impedance output.  Its (small signal) output
impedance can be much lower than the emitter resistor, note.  The large signal behaviour is very asymmetric
which can be an issue.
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Sometimes a picture is helpful:

http://www.google.com/imgres?imgurl=http://www.tpub.com/neets/book7/0113.GIF&imgrefurl=http://www.tpub.com/neets/book7/25k.htm&h=409&w=200&sz=4&tbnid=wqNrCkF7sLFZYM:&tbnh=90&tbnw=44&prev=/search%3Fq%3Dcommon%2Bemitter%2Bcommon%2Bbase%2Bcommon%2Bcollector%26tbm%3Disch%26tbo%3Du&zoom=1&q=common+emitter+common+base+common+collector&usg=__-WL0cgGXst-IgfpTRK7V9yjbUsU=&docid=D3SObTRw49hXsM&hl=en&sa=X&ei=wsAiUauLIs_NqQGdl4GAAQ&sqi=2&ved=0CFIQ9QEwBQ&dur=567

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texas
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Keith's explanation was technically quite correct, but you may have been looking for something a tad more basic and practical.  It just means that the collector of an output transistor is literally not connected to anything internal other than the pin itself.
No, thats "open collector".  We are talking about the common-collector configuration (I assume), almost always called
an "emitter follower"  BTW.

The voltage gain is about 1 (well very very slightly less than 1), current gain as explained above.  The normal reason
for this circuit is to take a high impedance signal and generate a low impedance output.  Its (small signal) output
impedance can be much lower than the emitter resistor, note.  The large signal behaviour is very asymmetric
which can be an issue.

Oops, had open collector on the brain today.  Sorry again.
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http://www.allaboutcircuits.com/vol_3/chpt_4/6.html
If this would be helpful!
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