amplifier questions

[roofingboom]

so doing some tests i ran across a huge question that i cannnot answer yet
so in a class ab amp the input is usally a diff pair
so the diff amp amplifies the input and if its a npn diff pair the output of the diff pair so between GND and the postive rail
this output is usually fed to a drive which supplies enough current for the output stage
the output stage class B biased on so it is class ab
my question here is
where is the signal getting centered at GND so both the pnp and npn are used in the output stage
my ciruit right now the output from the diff pair is between GND and postive rail thorugh the driver to the class ab power output stage and only the top half is only on
where should the signal be centered at GND and how is this accomplished
?

[MikeB]

Hi !

A classic ClassAB-Design consist of three stages:

1, The Inputstage (using the diffpair)
2, The VAS (VoltageAmplificationStage)
3, The Outputstage (CurrentAmplification)

The answer to your question is the VAS.
The output from the diffpair is very low, normally < 1Volt
This signal is fed into the VAS, centering the voltage and
amplifying it to the full voltageswing. Just take a look
into existing schematics, you will find 2 "complementary" transistors
that are "connected" with their collectors, having their emitters
to the supplyvoltagerails with a small resistor (mostly ~150ohm)

Michael

[sreten]

Just to clarify, the output from the diff pair is current,
voltage swing is not needed and is detrimental, the
VAS converts this to full near rail to rail voltage swing.

So input is voltage to current, VAS current to voltage.

sreten.

[MikeB]

Okay, to be completely precise, the outputcurrent from the diffpair
is converted into voltageswing via a resistor, and then fed into
the base of the VAS-transistor...
Or did i miss something with the ohm's laws ?

[sreten]

Originally posted by MikeB
Okay, to be completely precise, the output current from the diffpair
is converted into voltage swing via a resistor, and then fed into
the base of the VAS-transistor...
Or did i miss something with the ohm's laws ?

The resistor sets the balance of the differential pair.

Input into the VAS is current.

The resistor cannot convert current to voltage as it has the
VAS base emitter diode effectively in parallel, this makes it
a virtual earth point, so linear voltage swing is not possible.

sreten.

[Tube_Dude]

Originally posted by sreten


The resistor cannot convert current to voltage as it has the
VAS base emitter diode effectively in parallel, this makes it
a virtual earth point, so linear voltage swing is not possible.

Not in the case where the VAS transistor has a emitter resistor (as it has , in many cases)...if it have a emitter resistor say 100 Ohms the input impedance of the VAS is 10 KOhms ( transistor with a gain of 100).
In that case the conditions that Michael talk about are correct..

[sreten]

Originally posted by Tube_Dude



Not in the case where the VAS transistor has a emitter resistor (as it has , in many cases)...if it have a emitter resistor say 100 Ohms the input impedance of the VAS is 10 KOhms ( transistor with a gain of 100).
In that case the conditions that Michael talk about are correct..

True there is some voltage swing, but its still current that counts.

sreten.

[Tube_Dude]

Originally posted by sreten



True there is some voltage swing, but its still current that counts.

As the input impedance of the VAS is 10k ...10 times greater than the driving impedance (the LTP load resistor are usually ~1k ) we must talk of voltage drive.

The some as the input of the LTP ...it need current to work but we don't say it is current drived ....

[sreten]

Originally posted by Tube_Dude



As the input impedance of the VAS is 10k ...10 times greater than the driving impedance (the LTP load resistor are usually ~1k ) we must talk of voltage drive.

The some as the input of the LTP ...it need current to work but we don't say it is current drived ....

Jesus,

This is hardly the thread to get technical and argue simple
details with someone who knows what they talking about.

Most of the frequency range current drive is into Cdom.

And trying to arrange voltage swing is pointless.

sreten.

[RetroAudio]

roofingboom - If I understand your real question correctly, I'm not sure it's been answered yet. The signal is "centered" with reference to the output stage by means of the collector of the VAS stage in the top half using everything it "sees" electrically as it's load. That means that the complimentary npn trannie on the static bottom half as well as the whole output stage AND it's speaker load is the load to that upper collector. This load for most practical purposes is quite high in impedance. It is this load that the current from the upper pnp VAS transistor collector sees, and what makes "floating" voltages occur, keeping the levels betwixt the + and - power supplies. Further, if no speaker load were connected, voltage swings would still occur, but no current would be drawn from the output stage(emitter follower). If a load was connected, then the swings would be referenced to ground, thus "centering" it. (???) Now, some amp topologies have a resistor connected from that collector/output node to ground, in which case it's meant to be the primary load that the upper pnp's collector sees,..in which case most of the current will go through it forming the "centered" voltage. So, the first case describes a floating voltage (at least within that node), and the resistor case describes a voltage referenced to ground. Hope that helps, starting out in discrete amp design is a bit overwhelming I'll have to admit. If I misunderstood you,..then,...never mind.......