Make Inductive Load Look Resistive

But it depends on the inductance being linear - which most examples it is not.

This calculator shows how it is commonly done with speaker drivers.

Not precise, as speaker inductance is not linear varies with frequency - since the core material is soft iron.

Transformer primary inductance is even worse cos the laminated core can saturate easily.

BTW:

See how leaving out the DETAILS of a question makes it near impossible to answer ??

Or was that your aim all along ??

.... Phil

Easy. You have R + L

Parallel it with R + C (in series)

Where C = L/R^2

Do the math (Laplace) Klaus, then you'll see how it works.

I think I have it right. Corrections may be necessary after wine night >:-} ...Jim Thompson

"Phil Allison"

.... Phil

Yup, I always remember it by equating the two times. RC=L/R. If you (the OP) have the voltage headroom, I sometimes pad the inductor with some more resistance. This helps to keep the needed capacitance lower. (It's also well worth going through the algebra for yourself.)

George H.

Capacitance works but then it only works at one frequency, F = 1 / (2*pi*sqrt(L*C)).

The only "frequency independent" way is to handle the reactive power by brute force (i.e., big enough transistors) in a switching converter, powered by DC. The converter tranforms load current into ripple current, so the DC supply will carry ripple corresponding to the reactive power of the load, which requires beefy filter caps.

The DC supply itself, finally, carries no reactive power because, hey, it's DC.

Another way to explain this is, the DC supply's filter caps are alternately switched to the load, transferring reactive power at the switching frequency without resonance.

Tim

You don't actually need algabra, or any math. It's obvious in the time domain:

in---------+----------------+ | | | | C L | | +--V1 +--V2 | | R R | | | | | | gnd gnd

Suppose the L/R and RC time constants are equal. Assume a 1 volt step at the input. Imagine the waveforms.

The voltage at V1 is a 1 volt step with an exponential decay. The voltage at V2 starts at 0 and increases to 1 volt with an exponential rise. The waveforms across the two resistors are complements so their sum is the input voltage. So the input current is Vin/R for all time.

The parts in either leg can of course be swapped.

This has been used in scope front ends to provide a 50 ohm resistive termination while driving a capacitive load, like the grid of a tube.

The R's (or padded R's, as George points out) have to be equal... easily seen by noting the "resistance" must be the same for very low or very high frequencies. ...Jim Thompson

You don't actually need algabra, or any math. It's obvious in the time domain:

in---------+----------------+ | | | | C L | | +--V1 +--V2 | | R R | | | | | | gnd gnd

Suppose the L/R and RC time constants are equal. Assume a 1 volt step at the input. Imagine the waveforms.

The voltage at V1 is a 1 volt step with an exponential decay. The voltage at V2 starts at 0 and increases to 1 volt with an exponential rise. The waveforms across the two resistors are complements so their sum is the input voltage. So the input current is Vin/R for all time.

The parts in either leg can of course be swapped.

This has been used in scope front ends to provide a 50 ohm resistive termination while driving a capacitive load, like the grid of a tube.

It's a two-pole diplexer--also dead useful in fast PD front ends.

Cheers

Phil Hobbs

Hi George,

The Laplace analysis is very informative to see _why_ it actually works.

Some >:-} can't do the math so they hand-wave time constant equivalencies, failing to note that the R's must be equal.

...Jim Thompson

The values are R and R. Sound equal to me. And it's obvious that they have to be equal.

Math and simulation are both crutches, useful when things are not obvious, but both subject to error. I've seen pages of beautiful math with some bad assumption or some small error halfway through, making the conclusion nonsense, with the author defending the nonsense result because the math says so. The author could (almost) push symbols around as he was taught, but didn't have any feel for what was actually happening.

There is a small window between when something is so simple that it's obvious, and when it's too complex for a closed-form solution. This particular case is obvious, if you look at it right.

To a sufficiently intelligent life form, everything would be obvious.

I note here that I posted an intuitive and correct explanation for why the input looks resistive. You alluded to a LaPlace "explanation" but didn't post it.

Been doing that since the mid-70's, when modems were acoustic, and used "muffs", and I found that magnetically driving the earpiece (outbound) had much lower distortion that driving the crappy carbon microphone.

All your stuff is "intuitive" and "obvious". How come so much of it doesn't work ?>:-}

Larkin, known as the PALwHUB, now with ED. ...Jim Thompson

Well, finish it.

Check my web site. I all works and about 98% of it worked and sold at rev A, first PCB etch, with no prototypes.

I do math when I have to, but filling out pages with algebra has no fundamental appeal.

Limp. Limp. Limp. ...Jim Thompson