Amplifer clipping.....

[MiiB]

I am wondering....

Why is it that the most common configuration is a voltage amplification-stage with higher supply rails than the output current stage...eg 60 V on the front-end and 50 V to the output stage....???

The reason for the question is that i made some simulations on a circuit.. well on many circuits in fact..and most often the clipping performance of the front end is much nicer and cleaner than on the output-stage...
So the question is .. why not run the front en on lower supplies and the let the voltage clipping if any occur there...(offcouse you must consider the SOA of the output devices ect ect...and accept the extra dissipation of the like 5V higher rails)

But It will give you the advantage of not having an extra supply with higher voltage.. instead you can use the voltage already there and then chop it down with a high quality shunt or some other kind of regulation...

Shure mr Salas...can come up with a good circuit....exactly for that..

[AndrewT]

Hi,
start your amplifier proposal with the same voltage rails for all stages.

Can the performance of the amplifier be improved?

Would extra smoothing on the PSU help?
Would extra decoupling at the output devices help?
Would extra RC filtering on the drivers and/or pre-drivers help?
Would extra RC filtering on the voltage amp stage help?

The list of questions goes on and on and on .....

[bear]

The answer to your question, I think, is that in order to drive the outputs fully for most circuits (due to the drive requirements, losses, and the follower configuration) one needs more swing in the driver. However, your question is valid imo, and it is easy enough to build an amp of this type where the driver will clip before the output will by using a lower rail voltage for the driver section. It might be worth trying and seeing how it measures and sounds. I've often thought about doing just this myself.

_-_-bear

PS. keep in mind that the drive requirements are not static and depend in part on the load being driven - resistive loads NOT being totally indicative of an amps performance in the real world.

[Nico Ras]

Hi MiiB,

the reasons are as Bear described above, firstly to provide ample voltage swing if the output devices are lateral mosfets, but the second reason is the higher the applied rail voltage, the higher both second and third order intercepts, thus both harmonic and intermodulation distportion decreases.

Lowering the supply voltage would infact increase the distortion, both spurious and harmonically related for the same input signal (provided gain remains the same).

Nico

[bear]

Nico, the voltage rail differential is ~10% how much difference in the measured distortion?

_-_-bear

[djk]

An amplifier like a Leach can only come within about 7V of either rail. Using a higher voltage for the front-end helps, but can cause some 'sticking' in the slow output devices when coming out of clipping.

An easy fix is to add a Baker clamp like the APT model 1. A fast diode is connected from the Vas stage to the rail the outputs run from. Another fast diode is inserted in series with the Vas outputs to the bias network. The output stage is connected to the bias network.

The first diode pair conduct when the output voltage from the Vas exceeds the rail voltage for the output stage. The diodes in series with the bias drop the signal voltage applied to the driver transistors enough that the front-end clips before the drivers and outputs can saturate. The idea is that the faster low current devices in the front-end can recover from clipping (from the Baker clamp) in a better fashion than the slower high current outputs can.

[AndrewT]

the Leach I built and measured ran @ +-58.5Vdc when idling.
The PSU sagged to ~ +-56V when delivering full power to 8r0.
The peak of the sinewave voltage, when just short of clipping, was ~52.8Vpk, (174W into 8r0).
I wonder how much that peak voltage can rise, if the input stages are separately powered?

[djk]

About 5V when driving a 4R load.

[AndrewT]

Originally Posted by djk

An amplifier like a Leach can only come within about 7V of either rail.

Originally Posted by djk

About 5V when driving a 4R load.

the result of combining these two pieces of information is that the peak output into a 4r0 load is ~12V less than the supply rail voltage when common rails are used for the whole amplifier.

[djk]

"I wonder how much that peak voltage can rise, if the input stages are separately powered? "

The peak voltage when driving a 4R load rises about 5V with respect to the 7V~8V loss without a high voltage tier, so the peak output voltage is only down about 2V~3V, not 7V~8V (three pair of outputs).

On an original Leach with two pair of outputs running on ±57V I was able to get 52V peak at the point of clipping into a loudspeaker using dynamic program material. The speaker was nominally 8R, and over 20R through most of the midrange.