MikeB said:
Now, i can't resist, it does exactly what an opamp is supposed to do, amplify the difference between + and - input. Your circuit will have identical collector currents for q2/q3 if both inputs have the same voltage... no unity gain, zero gain 🙄
Mike
The TRUTH at last!
Wait a minute, do we suddenly have to start discussing what unity gain and zero gain means? AFAIK, and also according to all uses of the terms I can find with google they mean exactly the same thing. Unity gain means that the input is multiplied by 1. Zero gain means that the output is amplified 0 dB. That is obviously the same thing.
If anybody disagrees with this, then please give your definitions of the terms.
If anybody disagrees with this, then please give your definitions of the terms.
With zero gain i meant no output. n * 0 = 0...
If you connect the - input with + input of an opamp, all it can output is what is left from the cmmr.
Of course amplifying with 0db is unity gain. I meant -INF dB.
Mike
If you connect the - input with + input of an opamp, all it can output is what is left from the cmmr.
Of course amplifying with 0db is unity gain. I meant -INF dB.
Mike
So, since some people seem never to read a datasheet 😉, let's take a look at an archetypical CFB op amp, the LM6181.
I really hope we can agree that this op amp contains a diamond buffer (ie. a unity gain buffer) s.t. the + input of the op amp is the input of this buffer and the - input of the op amp is the output of the buffer.
I really hope we can agree that this op amp contains a diamond buffer (ie. a unity gain buffer) s.t. the + input of the op amp is the input of this buffer and the - input of the op amp is the output of the buffer.
Attachments
Yes, that's right, but the subject was that an opamp will not output (at its output, not the - input) anything if the - input and the + input are connected together.
The schematic you showed has three pins, + input, - input and output.
Why should i use the - input as output ?
Mike
The schematic you showed has three pins, + input, - input and output.
Why should i use the - input as output ?
Mike
MikeB said:If you connect the - input with + input of an opamp, all it can output is what is left from the cmmr.
Nobody suggested there is a jumper between the inputs inside the op amp. There is a unity gain buffer between them, and the output voltage of the amp is a function of the output current of this buffer.
MikeB said:
Why should i use the - input as output ?
You shouldn't. You should use the buffer output as the - input of the amp.
This discussion was about the following sentence:
"...a unity gain buffer connects the non-inverting input to the inverting input.." and that is :bs:
unless that part was taken in the wrong subject and is only part of an unfinished sentence, in this case i get ....'d.
I have no longer any idea what we are discussing right now.
Of course i should use the buffer output as - input, that's why it is labeled "inv-input".
Obviously we did talk of completely different things...
Mike
"...a unity gain buffer connects the non-inverting input to the inverting input.." and that is :bs:
unless that part was taken in the wrong subject and is only part of an unfinished sentence, in this case i get ....'d.
I have no longer any idea what we are discussing right now.
Of course i should use the buffer output as - input, that's why it is labeled "inv-input".
Obviously we did talk of completely different things...
Mike
This phrase can be understood in 2 ways:
1, you get a unity gain buffer by connecting non-inverting input to the inverting input.
2, a unity gain buffer is between non-inverting input and the inverting input
Looks like mikeks and me took No1...
I am sorry for the trouble.
Mike
1, you get a unity gain buffer by connecting non-inverting input to the inverting input.
2, a unity gain buffer is between non-inverting input and the inverting input
Looks like mikeks and me took No1...
I am sorry for the trouble.
Mike
With the "a unity gain buffer is between non-inverting input and the inverting input" the concept of cfb gets more clear, if the outputimpedance of that buffer would be zero ohms, it would be in fact a pure cfb. With the limited outputimpedance it gets a mixture of cfb and vfb.
Mike
Mike
MikeB said:This phrase can be understood in 2 ways:
1, you get a unity gain buffer by connecting non-inverting input to the inverting input.
2, a unity gain buffer is between non-inverting input and the inverting input
No, in case 1 there is no buffer, just a connection. I think it is all quite clear if you read for instance:
http://focus.ti.com/lit/an/slva051/slva051.pdf
http://www.intersil.com/data/an/an9420.pdf
MikeB said:With the "a unity gain buffer is between non-inverting input and the inverting input" the concept of cfb gets more clear, if the outputimpedance of that buffer would be zero ohms, it would be in fact a pure cfb. With the limited outputimpedance it gets a mixture of cfb and vfb.
Mike
Exactly, and just as both traderbam and I have said, both VFB and CFB are theoretical models and real amps are a mix of these two models, but op amps are usually designed to approximate either model, that is having a very low or a very high impedance on the - input.
Christer said:So, since some people seem never to read a datasheet 😉, let's take a look at an archetypical CFB op amp, the LM6181.
I really hope we can agree that this op amp contains a diamond buffer (ie. a unity gain buffer) s.t. the + input of the op amp is the input of this buffer and the - input of the op amp is the output of the buffer.
LM6171 instead 🙂
http://www.diyaudio.com/forums/showthread.php?s=&threadid=2755&perpage=10&highlight=&pagenumber=1
Can I ask what is going on in the LM6171 internal schematic ? (It looks like they were lazy when drawing, and just simplified most of it)
The amp is referred to as VFA but I don't understand it

ash_dac said:
LM6171 instead 🙂
http://www.diyaudio.com/forums/showthread.php?s=&threadid=2755&perpage=10&highlight=&pagenumber=1
Can I ask what is going on in the LM6171 internal schematic ? (It looks like they were lazy when drawing, and just simplified most of it)
Now that's one funny op amp. 🙂
As I have understood it, and how it also seems from the simplified schematic, they have taken the 6181 CFB amp (or something very similar) and added an extra buffer on the - input to make this input high impedance, and we suddenly have a VFB version of 6181. We actually have two buffers here fighting with each other, but since the output voltage of the amp is a function of the current flowing between the outputs of these two buffers, it actually becomes a function of the voltage difference between the inputs, ie. it is a VFB amp but with some CFBish properties (I think).
Christer said:
Exactly, and just as both traderbam and I have said, both VFB and CFB are theoretical models and real amps are a mix of these two models, but op amps are usually designed to approximate either model, that is having a very low or a very high impedance on the - input.
Okay, now i perfectly understand the concept of cfb, for an amp with pure cfb, the voltage at - input would always be identical to the voltage at + input, the devices at the buffer output working as some "folded cascode", explaining the speed of cfb...
Only the current supplied to the - input is of interest.
Mike
MikeB said:
Okay, now i perfectly understand the concept of cfb, for an amp with pure cfb, the voltage at - input would always be identical to the voltage at + input, the devices at the buffer output working as some "folded cascode", explaining the speed of cfb...
Only the current supplied to the - input is of interest.
Mike
Exactly, though it is more common to use current mirrors than folded cascodes, but there are many variations on the concept. You can also just sense the current in the buffer output transistors with resistors to the rails and have an ordinare VAS, as is done for instance in the Zenquito power amp.
CMRR
Hi,
I found this formula.
CMRR = Av (diff) / Av (CM)
I have looked at the AD811 , and AD815 which IMO both have a terrible CMR . I assume this is due to the current feedback design but why is it that the CMR response is flat to 100K ? (this is different from the CMR response of the standard differential input e.g. AD826 which has a smaller 'flat region').
Hi,
I found this formula.
CMRR = Av (diff) / Av (CM)
I have looked at the AD811 , and AD815 which IMO both have a terrible CMR . I assume this is due to the current feedback design but why is it that the CMR response is flat to 100K ? (this is different from the CMR response of the standard differential input e.g. AD826 which has a smaller 'flat region').
if you really want to know the answer you should dig in here:
http://citeseer.ist.psu.edu/cache/p...zcas91.pdf/a-general-relationship-between.pdf
referenced here:
http://citeseer.ist.psu.edu/432448.html
http://citeseer.ist.psu.edu/cache/p...zcas91.pdf/a-general-relationship-between.pdf
referenced here:
http://citeseer.ist.psu.edu/432448.html
MikeB said:
Okay, now i perfectly understand the concept of cfb, for an amp with pure cfb, the voltage at - input would always be identical to the voltage at + input, the devices at the buffer output working as some "folded cascode", explaining the speed of cfb...
Only the current supplied to the - input is of interest.
Mike
Not really....See Cherry..
- Status
- Not open for further replies.
- Home
- Amplifiers
- Solid State
- Current feedback - not suitable for audio ?