Non Ohmic Constant Voltage Drop "Resistor"

Zener diode? ( over a limited current range.)

not sure but there are some very esoteric materials that exhibit a bit of constant voltage over a current range. I think one is what's used in the MOV protection devices. That's a passive device.

PS: to me, your subject line didn't match your question, almost didn't look at it. again, to me.

Battery ?>:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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I love to cook with wine.     Sometimes I even put it in the food.

slapping forehead, and yelling, now that's just embarrassing!

A battery might not be passive in the sense of a resistor or zener which never need charging.

I need to damp an oscillator down to zero current after one oscillation, critically damped. One oscillation is necessary for this to work. The voltage should damp down too, to a constant, maybe zero or at most a fraction of the maximum.

I may be able to use a conventional resister as well as or in place of the zener diode.

Bret Cahill

Try a regulator.. or maybe a zener diode..

Jamie

What kind of oscillator? If it's an LC, there is an optimum value resistor that, slapped across the tank, will kill the oscillation down to a very small voltage within one cycle time.

I make LC oscillators that start in nanoseconds and can be quenched in well under one cycle.

We had a thread in s.e.d. where I invented the fristor, the electrical equivalent to static friction. A fristor would kill an LC even faster than an ohmic resistor. I think.

--

John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

The 'after one oscillation' indicates a holdoff time; the 'down to zero current' indicates clamping. The "critically damped" phrase seems out of place here.

Don't you want a timed capacitor grounding switch? Or, a timed open-the-inductor-so-no-current-flows switch?

The damping can start well before the end of the first oscillation.

It's OK if it's somewhat underdamped.

Too complicated. Cost down to pennies is an issue.

I just don't want much current after 2 - 3 cycles.

Bret Cahill

Millihertz? Hertz? Megahertz? Gigahertz?

w.

0.6 Hz.

Any mechanical switch or relais can do that.

w.

r

Oh now that is interesting. (fristor) So I'm not sure how the analogy tran slates into electrical terms. But if you damp a mechanical oscillator with friction, such that it's "very hard"*, then you end up stopping the system in a state that still has some potential energy. (Unless you are lucky, o r smart.) Where the spring force is not enough to over come the frictional force. I'm not sure how that translates to the LC-fristor case. But mayb e it leaves some charge on the cap???

George H.

*very hard = frictional loss in ~one cycle is something like the energy s tored in the system.

Without knowing more about your overall system, like whether it already includes a microprocessor, consider just generating the exact waveform you want. At 0.6 Hz you won't be taxing any uP system.

Best regards,

Bob Masta DAQARTA v7.60 Data AcQuisition And Real-Time Analysis

Scope, Spectrum, Spectrogram, Sound Level Meter Frequency Counter, Pitch Track, Pitch-to-MIDI FREE Signal Generator, DaqMusiq generator Science with your sound card!

wow! I'm an analog guy, and *I* would be tempted to do this with a PIC-like chip! cheap, reliable, trouble free. and battery friendly. even the inards of an old Timex watch could do that. and those are cheap.

I know you know this, but still have to say it, an 'opposing' batery is the equivalent of static friction. So, yes, absolutely leaves energy in at some unknown state.

I did this once using an 'impulse' [think ball bat] The thing oscillates and just as it comes swinging near zero, whack it just right, like hitting it with a ball bat. It stops almost on zero. Or, probably better put, remove the energy from the system at the appropriate time.

The fristor is the equivalent to killing a mechanical resonator with sliding friction, not viscoscity. Note that the envelope decay is linear, not exponential.

It's not an exact analogy, because electronics can do cooler things than mechanics.

--

John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

For a normalized LC (L=1, C=1) you want about 0.57 ohms across it to kill it fast.

--

John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com