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Why is the output of the IPS LTP considered a current output?
Why is the output of the IPS LTP considered a current output?
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Old 27th March 2013, 12:52 AM   #1
Dan Moos is offline Dan Moos  United States
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Default Why is the output of the IPS LTP considered a current output?

What defines whether the output of a stage is current or voltage? Self and Cordell's books both say that the output of the IPS LTP is in current form. What makes this so? Is it simply because of relatively low voltage gain?

What's more, when I simulated the simple amp in chapter 2 of Cordell's book, with a current mirror LTP load being the only improvement, the voltage output of the IPS looked horribly distorted while the current output was a pretty good looking sine wave. What's going on here?
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Old 28th March 2013, 12:59 AM   #2
Dan Moos is offline Dan Moos  United States
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'Bump'. Anyone?
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Old 28th March 2013, 03:54 AM   #3
Dave Zan is offline Dave Zan  Australia
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Quote:
Originally Posted by Dan Moos View Post
What defines whether the output of a stage is current or voltage? Self and Cordell's books both say that the output of the IPS LTP is in current form. What makes this so? Is it simply because of relatively low voltage gain?
When the output impedance of the LTP is much more than the load then the current is not much affected by variations in the load impedance, whereas the potential will vary to match. This is the case over much of the frequency spectrum when Miller compensation is used.
So the input potential difference between the two inputs determines the current out.
The effective definition is that if the output impedance is much more than the load then it is current output and if the output impedance is much less then ...well I am sure you can work that one out.

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What's more, when I simulated the simple amp in chapter 2 of Cordell's book, with a current mirror LTP load being the only improvement, the voltage output of the IPS looked horribly distorted while the current output was a pretty good looking sine wave. What's going on here?
Does the explanation above clarify this to you? IIRC Self comments on this too. It can look bad but if the so called VAS actually has it's output determined by the input current then it all works out.


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David

Last edited by Dave Zan; 28th March 2013 at 04:03 AM.
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Old 28th March 2013, 04:13 AM   #4
Dan Moos is offline Dan Moos  United States
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Ok, tell me if I'm on the right track. If I treat the ISP and VAS as a voltage divider, the VAS behaves (in this case anyway) like a resistor who's resistance is constantly changing in a non-linear way, so that a clean looking current creates a wacky voltage signal.

That part makes sense. Now what's confusing me is exactly how the VAS is able to use this signal. Wouldn't using a current as the input to the VAS be dependent on beta, where as using voltage would be dependent on transconductance, the latter which is more dependable, or at least my impression has been that that is the case?

It is also hard for me to understand how this wacky voltage signal doesn't affect the VAS output.

I assume that an improved VAS design makes for more linear loading of the ISP. Is that part of this?
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Old 28th March 2013, 01:11 PM   #5
Dave Zan is offline Dave Zan  Australia
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Originally Posted by Dan Moos View Post
Wouldn't using a current as the input to the VAS be dependent on beta, where as using voltage would be dependent on transconductance, the latter which is more dependable, or at least my impression has been that that is the case?
Yes, the beta is less predictable than the transconductance. But if there is sufficient gain then the current feedback thru the compensation (Miller) capacitor dominates and is very predictable.

More later if you want, bedtime here!

Best wishes
David
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Old 28th March 2013, 02:45 PM   #6
Dan Moos is offline Dan Moos  United States
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What about freqs below those where the miller cap is in play, which I thought basically included the bulk of the audio band? Or is the idea that any distortion harmonics will be in the miler caps band of influence.
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Old 28th March 2013, 05:35 PM   #7
DF96 is offline DF96  England
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The Miller cap dominates over all except the lowest part of the audio band, in most amp circuits.
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Old 29th March 2013, 01:19 AM   #8
Dave Zan is offline Dave Zan  Australia
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Quote:
Originally Posted by Dan Moos View Post
What about freqs below those where the miller cap is in play, which I thought basically included the bulk of the audio band? Or is the idea that any distortion harmonics will be in the miler caps band of influence.
With a two transistor (beta transistor added) VAS then the Miller capacitor is in play down pretty low, rather unpredictable because depends on beta, but below much of the audio band. And below that then the overall feedback takes care of the non-linearities. So we don't rely on the distortion harmonics to be in the Miller capacitors band of influence.

Best wishes
David
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Old 29th March 2013, 01:58 AM   #9
Dan Moos is offline Dan Moos  United States
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What does "in play" mean exactly. Remember, I'm pretty green at this stuff. I was under the impression that the miller cap is chosen so that unity gain is reached pretty high up in the audio band. Apparently I'm wrong.

In a vanilla miller compensation scheme, (just the one cap around the single VAS, or even a 2 transistor VAS as you mentioned), how many dB of cut is there at typically 20hz? How much dB of cut before the cap is no longer significantly relavent?
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Old 29th March 2013, 02:07 AM   #10
Bonsai is offline Bonsai  Europe
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The LTP is fed from a current source. The input signal 'steers' the the LTP current source into either one of the two LTP collector loads - be they resistor or mirror.

Anytime you connect a capacitor to a current - which is what you are doing in the VAS stage, you form an integrator, and a key property of an integrator is that it converts a current input into a voltage output (transimpeadance circuit).

As the other posters note, at very low frequencies and DC, its the VAS (better to call it by its proper term: Transimpedance stage or 'TIS') transistor beta that dominates the transfer function, and not the action if the input current and the Miller cap, Cdom.
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