FET saturation and Power amplifier saturation

[[email protected]]

Can anyone please help me understand the relationship between a FET's
saturation region and the saturation
point of a power amplifier? From what I have read, the FET has three
regions of operation: ohmic, saturation and cut-off. The description
of the cut-off region seems the most straightforward:
Vgs < Vthreshold, Ids = 0, i.e. no current flow, so the Power
amplifier is like an open switch.

In the ohmic region:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)*Vds. Ids is dependent on
Vds, the Power amplifer is like a voltage controlled resistor.

I don't understand the purpose of the ohmic region from a power
amplification perspective, but it is the saturation region that really
has me confused. In saturation:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)^2. Ids does not depend on
Vds, but increases as Vgs increases.

As I understand it, this is called the 'saturation region' of the FET
because for any particular value of Vgs, Ids is constant for all
values of Vds > Vgs - Vthreshold. So Ids is saturated.

What I can't figure out is, at what point does a power amplifier
biased in the FET saturation region become saturated i.e. when the
power amplifier reaches its maximum output power and begins to
compress? If you plot Ids against Vgs you get like a rising
exponential. My thinking is that the part of the exponential where the
slope is zero corresponds to the power amplifier saturation. Is this
logical?

Any suggestions welcome.

Thanks
mees

 

[Joerg]

Can anyone please help me understand the relationship between a FET's
saturation region and the saturation
point of a power amplifier? From what I have read, the FET has three
regions of operation: ohmic, saturation and cut-off. The description
of the cut-off region seems the most straightforward:
Vgs < Vthreshold, Ids = 0, i.e. no current flow, so the Power
amplifier is like an open switch.

In the ohmic region:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)*Vds. Ids is dependent on
Vds, the Power amplifer is like a voltage controlled resistor.

I don't understand the purpose of the ohmic region from a power
amplification perspective, but it is the saturation region that really
has me confused. In saturation:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)^2. Ids does not depend on
Vds, but increases as Vgs increases.

As I understand it, this is called the 'saturation region' of the FET
because for any particular value of Vgs, Ids is constant for all
values of Vds > Vgs - Vthreshold. So Ids is saturated.

Think of a power amplifier as a voltage regulator. It puts out a voltage
half way between V+ and V-, in other words zero volts on a symmetrical
supply, when there is no input. When you strike a cord on a guitar it
attempts to steer the output proportionately to the input. The scale
factor depends on where you have set the volume.

In this mode the FETs are neither off nor saturated.

What I can't figure out is, at what point does a power amplifier
biased in the FET saturation region become saturated i.e. when the
power amplifier reaches its maximum output power and begins to
compress? If you plot Ids against Vgs you get like a rising
exponential. My thinking is that the part of the exponential where the
slope is zero corresponds to the power amplifier saturation. Is this
logical?

Saturation sets in when you have the volume at 10 and play "Stairway to
Heaven" with full gusto and the neighbors hollering at you :-D

Seriously, when the gate is driven more above Vth but there is no more
voltage headromm left to increase current then the current into the load
remains flat, which causes clipping. The FET will reside in saturation
but it doesn't help. On the other cycle it's the opposite FET going to
the same extreme.

Then there are class D amplifiers where the FETs are switched between
fully off and saturated but that's a whole 'nother ballgame.

 

[Tim Williams]

You are confused by lingo, and possibly characteristics.

In the FET world, since at low Vds, the characteristic looks ohmic, and the
resistance varies with gate voltage, it was named the "linear region".
Likewise, as Vds rises, Ids "saturates" to some value, thereby naming that
region as well.

This is in contrast to BJTs, which because they act more like diodes, the
voltage-saturation region, where Vce is small, was named the saturation
region. The linear region is where linear signal amplification takes place,
which is out in the middle of flatsville (where Ice varies little for a
change in Vce).

And nowadays, MOSFETs are widely used as "saturated" switches, which
horrendously means the voltage is saturated = as low as it can get.

Confused enough yet?

Now, to actually address your question. MOSFETs, BJTs, tubes,
burnt-galvanized-steel-negative-resistance-junctions, whatever -- all
operate in the linear region, where a change in input causes a larger change
in the output. The output can be current or voltage, it only matters that
the circuit around it is capable of turning that into a useful output.

On the curve of input vs. output, there is always SOMETHING which limits the
maximum output. Usually, it's that a transistor is bumped up right against
the power-supply rail and therefore simply cannot supply a higher voltage.
It can be said that the amplifier is saturated (cannot go any closer) to
that rail. This is analogy to switching circuits, which being designed for
switching, are usually better at it. Rail-to-rail amplifiers, typically
built with MOSFETs, are capable of switching-grade coverage, which seems to
be what your question was about.

Tim

 

[Phil Allison]

Can anyone please help me understand the relationship between a FET's
saturation region and the saturation point of a power amplifier?

** First you need to get the meaning of the term " saturation " right -
beware, it changes depending on the context.

" Saturation: a condition in which a quantity no longer responds to some
external influence " is the definition that applies to electronic devices
and circuits best.

Note, the term is not locked specifically to current, but an expected
response to an input.

With a single MOSFET , the input is gate voltage and the response is the
resulting drain current.

However, with most amplifiers, the input may be voltage or current and the
output is typically voltage.

So, amplifiers are said to " saturate " when some output VOLTAGE limit is
reached.





........ Phil

 

[D from BC]

Can anyone please help me understand the relationship between a FET's
saturation region and the saturation
point of a power amplifier? From what I have read, the FET has three
regions of operation: ohmic, saturation and cut-off. The description
of the cut-off region seems the most straightforward:
Vgs < Vthreshold, Ids = 0, i.e. no current flow, so the Power
amplifier is like an open switch.

In the ohmic region:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)*Vds. Ids is dependent on
Vds, the Power amplifer is like a voltage controlled resistor.

I don't understand the purpose of the ohmic region from a power
amplification perspective, but it is the saturation region that really
has me confused. In saturation:
Vgs > Vthreshold, Ids ~ (Vgs - Vthreshold)^2. Ids does not depend on
Vds, but increases as Vgs increases.

As I understand it, this is called the 'saturation region' of the FET
because for any particular value of Vgs, Ids is constant for all
values of Vds > Vgs - Vthreshold. So Ids is saturated.

What I can't figure out is, at what point does a power amplifier
biased in the FET saturation region become saturated i.e. when the
power amplifier reaches its maximum output power and begins to
compress? If you plot Ids against Vgs you get like a rising
exponential. My thinking is that the part of the exponential where the
slope is zero corresponds to the power amplifier saturation. Is this
logical?

Any suggestions welcome.

Thanks
mees

FET's set up with no feedback clip different than amplifiers with
feedback.


D from BC

 

[default]

The description
of the cut-off region seems the most straightforward:
Vgs < Vthreshold, Ids = 0, i.e. no current flow, so the Power
amplifier is like an open switch.

Saturation is the opposite - like a shorted (closed) switch. Device
is turned on completely and its resistance is as low as it ever gets -
irrespective of power supply or signal fluctuations for all practical
purposes.

BJT also exhibit the same sort of thing - but enough current in the
base and the collector emitter voltage is as low as it can get.

Clipping in an amp is usually the result of running out of power
supply - not necessarily because the mosfet is saturated, in a well
designed amp it will still have some headroom left or it wouldn't be
linear in the region it is used in.

 

[Joerg]

FET's set up with no feedback clip different than amplifiers with
feedback.

Yep, and when you place a weak FET amp such as a CD4049UBE hanging on a
lowish VCC up front you can almost make the whole thing sound like a
tube amp ;-)

 

[Tim Williams]

Clipping in an amp is ... or it wouldn't be
linear in the region it is used in.

So what is clipping but a...nonlinearity?

If you mean keeping certain parts of the circuit in their linear range while
others crash, well no, that still doesn't count as what I think you said,
because something went out of bounds. I mean, you could intentionally add
clamps to limit voltage and current swing and call it a feature, but it's
still nonlinear. The best thing you can do is design and build your circuit
to handle such things gracefully. I'm listening to an amp right now that
comes out of saturation within microseconds, with a perfectly smooth
recovery -- no transient top or bottom.

Tim

 

[MooseFET]

Yep, and when you place a weak FET amp such as a CD4049UBE hanging on a
lowish VCC up front you can almost make the whole thing sound like a
tube amp ;-)

The RCA ones sound better.

 

[[email protected]]

So what is clipping but a...nonlinearity?

If you mean keeping certain parts of the circuit in their linear range while
others crash, well no, that still doesn't count as what I think you said,
because something went out of bounds. I mean, you could intentionally add
clamps to limit voltage and current swing and call it a feature, but it's
still nonlinear. The best thing you can do is design and build your circuit
to handle such things gracefully. I'm listening to an amp right now that
comes out of saturation within microseconds, with a perfectly smooth
recovery -- no transient top or bottom.

Tim

The amplifier "sticking" issue (as Tim alludes to) is a big deal.
Everything has a breaking point, but failing gracefully is a selling
feature. One of the best ways of seeing this in real life is to look
at the "residual" output of a distortion analyzer. This also shows the
onset of clipping. I'm not sure if anyone ever did a write up on using
a distortion analyzer in this manner, but it is common knowledge. By
comparing input versus output, you can determine if the source of
distortion is at the zero crossings, slewing related, soft clipping,
hard clipping, and sticking.

 

[Joerg]

So what is clipping but a...nonlinearity?

If you mean keeping certain parts of the circuit in their linear range while
others crash, well no, that still doesn't count as what I think you said,
because something went out of bounds. I mean, you could intentionally add
clamps to limit voltage and current swing and call it a feature, but it's
still nonlinear. The best thing you can do is design and build your circuit
to handle such things gracefully. I'm listening to an amp right now that
comes out of saturation within microseconds, with a perfectly smooth
recovery -- no transient top or bottom.

As long as it has a really mean sound when you strum the twang box real
hard ;-)

 

[default]

As long as it has a really mean sound when you strum the twang box real
hard ;-)

Yeah well . . . have to define "amplifier" and "signal processor," if
that's the case. Guitar amp is a special function - part of the
instrument.

 

[default]

So what is clipping but a...nonlinearity?

If you mean keeping certain parts of the circuit in their linear range while
others crash, well no, that still doesn't count as what I think you said,
because something went out of bounds. I mean, you could intentionally add
clamps to limit voltage and current swing and call it a feature, but it's
still nonlinear. The best thing you can do is design and build your circuit
to handle such things gracefully. I'm listening to an amp right now that
comes out of saturation within microseconds, with a perfectly smooth
recovery -- no transient top or bottom.

Tim

The best thing you can do is to design your system so that it remains
linear up to the "threshold of pain." Anything less is a compromise.

 

[Fred Bartoli]

Le Tue, 14 Aug 2007 19:16:24 -0700, MooseFET a écrit:

The RCA ones sound better.

Why didn't RCA made a 3 inverters version in octal socket?
That would've been a real killer...

 

[Fred Bartoli]

Le Wed, 15 Aug 2007 07:45:53 -0400, default a écrit:

The best thing you can do is to design your system so that it remains
linear up to the "threshold of pain." Anything less is a compromise.

To me, with some 'modern music' anything above a mW is above the
threshold of pain.

 

[default]

Le Wed, 15 Aug 2007 07:45:53 -0400, default a écrit:



To me, with some 'modern music' anything above a mW is above the
threshold of pain.

I have to agree with you there. The cannons in the 1812 Overture take
a lot of amplifier and big speakers . . . Ride of the Valkyrie,
Bolero, Totentaz, etc..

When all is said and done, speakers are the weak link followed by
listening rooms. Headphones.

 

[D from BC]

Le Wed, 15 Aug 2007 07:45:53 -0400, default a écrit:



To me, with some 'modern music' anything above a mW is above the
threshold of pain.

In 2006 and 2007, will there be any songs that will become classics
recognized for decades?

D from BC

 

[default]

In 2006 and 2007, will there be any songs that will become classics
recognized for decades?

D from BC

Even a blind squirrel gets a nut or two.

I think the pop music industry has taken all the inventiveness, spirit
and soul out of music. Too much emphasis on marketing to the
detriment of the music.

If the Internet can keep the open free-for-all spirit that it
currently enjoys - it is just possible that music will see a
resurgence that could put the renaissance to shame.

But I don't expect that will happen. Too much money Too much greed
Too many corporations and politicians

 

[Joerg]

Even a blind squirrel gets a nut or two.

I think the pop music industry has taken all the inventiveness, spirit
and soul out of music. Too much emphasis on marketing to the
detriment of the music.

If the Internet can keep the open free-for-all spirit that it
currently enjoys - it is just possible that music will see a
resurgence that could put the renaissance to shame.

But I don't expect that will happen. Too much money Too much greed
Too many corporations and politicians

I think it can happen but not on the radio. They seem to mostly play
what the big labels want them to, over and over. Or, ahem, what the
labels pay for. However, the web allows solo artists to thrive. Most of
them seem to still follow the old routes though, by local distribution
of their products. I know a guy who writes his own songs, lyrics, then
sings and plays the guitar. A real one-man show.

 

[Michael A. Terrell]

In 2006 and 2007, will there be any songs that will become classics
recognized for decades?

Sure. In the 'Country Music' and 'Bluegrass' genres.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida