Hi Max
I'm not sure what you need translated, you could have been a bit clearer about that.
If you look at the page you are liking to you find the text that you have copied and two diagrams.
"As expected, the CFA is here 5 time faster than the VFA. And this is not due to the miller cap, but the input stage topology: the way the feedback signals are substracted from the input ones.
This is due to the two input stage topologies transconductances: The Long-tail Pair is "compressive" while the CFA is "expansive" (© Richard Marsh).
In other words, think about the "current on demand" behavior of the CFAs, too well described in numerous papers to be pontificated here ;-)"
I have attached the diagrams.
The diagram on the left is showing the I/V transfere function of a LTP inputstage. It shows how much current you will get out depending on the input voltage, on the right diagram you see the transfer function of a Diamond inputstage.
You can see that for a LTP you will not get more current out even if you increase the input voltage, when you have passed a certain level, but for the diamond the current will increase with increasing input voltage and at a certain input voltage the current will will increase extremly. COD
As you know an H-bridge inputstage in this context are two diamonds and will have a sinh transfer function.
I hope this was understandable.
Attachments
Syn08 is right a single transistor input stage will have COD, but only in one direction. The one transistor inputstage will have an exponential transfer function.
If you take two single transistor input stages one NPN and one PNP connect them together and biasing them with two diodes between the bases and use two current mirrors you will get COD in both directions.
sinh(x)=(e^(x) - e^(-x))/2🙂
The amplifier topologies are asymptotic to a value depending on resistors not infinity. In the case of re = 50 Ohms and a 1K feedback R the sinh-ness is pretty weak.
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Before answering your questions, if needed, I would like you specify your position which is not very clear to me.
COD requires symetric topology but is a native property of CFA which can be asymmetric ?
Re=50 is emitter resistor, R = 1k is Rg or Rf // Rg, nothing indicates closed loop gain is 21.
Well, for a non inverting configuration, about 21 😀, depending on the open loop gain, if it's large enough. Loading the input with Rg||Rf~47 ohm affects a little the open loop gain Tz/(Ro+Rg||Rf) instead of Tz/(Ro+Rg) but not much the closed loop gain.
Could you please formulate, when applies, your replies as questions like "why is the closed loop gain 21" rather than statements "nothing indicates closed loop gain is 21". It would certainly stimulate the discussion.
Did I ever say "COD requires symmetric topology"? Where did you get this from? Do I detect a language barrier here?
There is no Rg, g = 1 follower, was that not obvious? Same is true of an H-bridge VFA with 1k degeneration (very little expansive action).
Thank you Hans,
I have questions/comment on your setup :
- a VFA with a resistor connected to the 2 inputs is detected as a CFA.
- how to conclude on the feedback operation mode while the DUT is in open loop ?
Hi Rodor, thanks for the clarification,I haven't followed this thread closely.....this has become very long thread with much arguing over fine points.
One point that is not widely discussed is the subjective differences between the topologies, and sound reproduction qualities is of course the final goal.
What are the subjective differences, anybody ?.
Dan.
Thank you.
Of course R=1k linearize the transfer at the expense of the gain
As you can see, nothing is evident for me .....but I'm not alone.
We must certainly live in parallel universes. Draw the schematic after breaking the loop and see what you get. Read any application note about the CFA closed loop gain, for example http://www.ti.com/lit/an/sloa021a/sloa021a.pdf and check out the closed loop gain at pp. 3-4
To add insult to injury, if Rg||Rf=1k and Rf=1k then arithmetic says this is nonsense, something like 0=1, or Rf/Rg=oo
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