Symmetric Amplification

David Eather schrieb:

Hello,

gravity is everywhere on the long way between the sun and pluto, otherwise all big and small planets between both would not keep their orbits as they do. Of course gravity is still present outside the orbit of pluto, because there are a lot of other Kuiper belt objects. You need an infinite distance to any mass to get zero gravity.

Bye

A symmetric amplifier, of course.

John

Depends on your frame of reference and your definitions doesn't it?

have

Because you love to argue definitions and minutiae endlessly, and argue that nothing is ever perfect. Go for it.

John

Nonsense. Light travels straight in free space. OTOH, the universe curves around gravitational wells.

Firstly, it's not exactly an amplifier (it's a transducer, volts-in/impedance-out). Second, it's not active at DC inputs (cold filament sheds no light). Third, the response is roughly V**2 for high enough input voltages to make the filament glow. So, it has a U characteristic with the only symmetry point being in the middle of a deadzone. And it rectifies an input signal, more than amplifies it ("U" characteristic and not '/").

Not an ideal exzmple, if I read the question right.

have

up

volts-in/impedance-out).

It can be used as a gain element, just like the devices the OP named.

Zero in, zero out. A logical requirement for a polarity-independent device.

Seems perfect to me.

Got anything better?

John

volts-in/impedance-out).

It also means the small-signal response is nil at the only operating point that allows it to be called symmetric. Missing output response is a pretty harsh criticism for an alleged amplifier

The Hall effect. A Hall sensor with some suitable bias generates an output voltage proportional to the transverse magnetic field. Positive field, positive voltage; negative field, negative voltage, and no asymmetry between the magnetic polarities. The input has to be a current into a field coil, or a magnetic field input of some kind, and the output is some piddly matter of millivolts. At least, the characteristic curve is "/", i.e. symmetric and everywhere-sensitive-to-input near zero input (at the operating point).

have

LINEARITY is broken, not symmetry.

The OP didn't ask that the transfer function be linear (tubes, transistors, fets, all amplifiers are nonlinear) but asked that it be polarity symmetric. He mentioned asymmetric electron flow; an LDR is polarity insensitive, as is an incandescent lamp. Read his post.

Suggest a symmetric amplifier topology or device yourself.

Or check this convenient box:

---- / / NO THANKS

have

up

volts-in/impedance-out).

That's a good one. Beats whining.

John