building a non inductive load ?

[AndrewT]

Hi,
Hi want to test the output stage of a solid state amplifier and have assembled some dummy loads.

Can an RC pair in parallel to a slightly inductive 8r load be used to cancel the inductance?

How would one calculate the R & C values needed to cancel the inductance?

Can the effectiveness of the combination be tested to show that the inductance has been cancelled out?
Am I dreaming?

[Nordic]

Interesting question, but isn't an impendance basicaly the ratio of inductance to capacitance?

Hah, googled it

http://www.allaboutcircuits.com/vol_2/chpt_13/3.html

Impendance = SQRT (Inductance/capacitance)

[Piercarlo]

Originally posted by AndrewT


Can the effectiveness of the combination be tested to show that the inductance has been cancelled out?
Am I dreaming?

If you have an RL series circuit you can defeat the L components with an RC series circuit connected in parallel to the first and having the same time constant. R should be fo the same value for either branch but, supposedly that RL time constant lead to an high corner frequency (out of audio bandwidth), not of the same power.

Hi
Piercarlo

[AndrewT]

Originally posted by Piercarlo


If you have an RL series circuit you can defeat the L components with an RC series circuit connected in parallel to the first and having the same time constant. R should be fo the same value for either branch but, supposedly that RL time constant lead to an high corner frequency (out of audio bandwidth), not of the same power.

does this mean that the effective impedance of the zero inductance pair is R/2?
At what frequency does the impedance come down to R/2?
or
become significantly less than R?

[Geoff]

Originally posted by AndrewT
does this mean that the effective impedance of the zero inductance pair is R/2?
At what frequency does the impedance come down to R/2?
or
become significantly less than R?

Andrew

With a correctly sized capacitor, the resultant impedance becomes a pure resistance and is not frequency dependant. The resultant resistance remains as R.

I won't detail the mathematics here (email me if you want the details) but it can be shown that if you have a resistor R in series with an inductor L and the pair are in parallel with another resistor R in series with a capacitor C the resulting network impedance Z is L/RC (L in Henrys, C in Farads).

If we say that Z equals R (equals 8ohm) then the equation becomes R=L/RC. If L is known C can be deduced from C=L/R^2.

For example, if the two resistors are 8ohm and L is 1mH, C needs to be 15.6uF to give an overall network impedance of 8ohm (resistive).

However, this analysis is too simple for your application as the resistor in series with the capacitor will also have an inductance (as will the capacitor, along with some resistance of its own). How far one goes with the complexity of the calculation depends upon how close to a pure resistance you wish to achieve. Again, in view of the maths involve this is probably better discussed by email.

Perhaps a non-inductive power resistor would be an easier option.

Geoff

[Tim__x]

Carfully matched, the impedance will be flat, and equal to R.
Think about it at extreme frequencies, at dc the capacitor is an infinite impedance and is out of the circuit while the inductor is a short, so the impedance is equal to R. At inifinite frequency the inductor is an infinite impedance and the cap is a short, so again the impedance is equal to R.

[Piercarlo]

Originally posted by AndrewT
does this mean that the effective impedance of the zero inductance pair is R/2?
At what frequency does the impedance come down to R/2?
or
become significantly less than R?

Nothing of this happen. Simply the overall circuit appears to the outer world as a single resistor of value "R", although this is really composed of two series RL and RC branches connected in parallel. If time constants are set equal in the two branches with the same "R", the virtual "resonant circuit" composed by the parallel of L and C is completely damped and there is no any chance of resonance: L and C phase shift cancel each other and the only component appearing active to the outer world is just *2R* for each branch which, paralleled, give out the original *R*, theoretically for *all frequencies passing through*.

This nice solution was, some years ago, the base for a rather cruel quiz posed on the italian newsgroup of electronic "it.hobby.elettronica". The question was: if you have a so fashioned circuit well closed in box, how you can guess that the circuit is not a simple resistor but a complex network? Theoretically, whatever frequency you chose for "arousing" the real network is ininfluent: the circuit appear everly as a "innocent" resistor and as so effectively behave...
The solution of the quiz was not electrical but *physical* and related to the intrinsic noise which, if working temperature is changed by changing current passing through, reveal the real nature of circuit composed by *two*, not one resistor. At least until the circuit has reached a new thermal equilibrium.

Hi
Piercarlo

[jneutron]

Originally posted by AndrewT
Hi,
Hi want to test the output stage of a solid state amplifier and have assembled some dummy loads.

Can an RC pair in parallel to a slightly inductive 8r load be used to cancel the inductance?

How would one calculate the R & C values needed to cancel the inductance?

Can the effectiveness of the combination be tested to show that the inductance has been cancelled out?
Am I dreaming?

Yes, you are dreaming.

What power level?

What bandwidth.

What impedance.

What maximum inductance?

What b dot error max?

You can use a pair of Dale NI's wired to null the pickup error, or just use one of them.

Under 50 watts, I can post pics of a 250 picohenry load (it's actually 60 picohenries, I just can't measure it that low, only to 250 pico.

Why, actually, do you need to cancel the inductance?

Cheers, John

[john_ellis]

Hi Andrew T

The catch is that if you use a series RC network (or any other arrangement) which cancels the LR network, the "R" in the RC arm of the circuit also needs to be non-inductive...


I made a non-inductive resistor as follows:

150 mm length of aluminium rod about 12 mm dia
set of fins made from aluminium plates (three off) approx. 2x3 inches with slots cut into one end of the rod; and slots half-way along the plates so that the plates slotted into the rod to a depth of 1 inch forming a finned heat radiator

1 layer of high-temperature insulating tape (Kapton type) carefully wound so that no edges overlapped nor spaced from the previous turn

non-inductive wound (this means folding the wire into half so that the current flows up and back in opposite directions) resistance wire (insulated constantan or eureka) about 26 swg cut to the right length to give 8 ohms

this has to be tightly wound over the tube so that heat will flow from the wire into the rod

finish with another layer of high temperature insulating tape/

Did it work?

Was fine for 50W but wire burned on 100W ...

Geoff's right - buy a non-inductive resistor!

cheers
John

[jneutron]

Originally posted by john_ellis
I made a non-inductive resistor as follows:

What did it measure?

Originally posted by john_ellis

Was fine for 50W but wire burned on 100W ...

Had ta push it, dinna ya?

Cheers, John