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?
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?
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)
Hah, googled it
http://www.allaboutcircuits.com/vol_2/chpt_13/3.html
Impendance = SQRT (Inductance/capacitance)
AndrewT said:
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
Re: Re: Re: building a non inductive load ?
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
AndrewT said:
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
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.
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.
Re: Re: Re: building a non inductive load ?
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
AndrewT said:
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
AndrewT said:
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
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
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
john_ellis said:
Sorry, meant inductance measurement.
Once you started dissipation, it was a foregone conclusion that the wire would expand and lose contact with the aluminum, you didn't have adequate transfer through either the kapton nor the wire insulation.
But it was a really neat construct..
My first low L resistor worked great, it was a coupla nanohenries. I was using a rmx 1450 (280 wrms) to fire it up for some tests, the amp being built into a mobile rig with an eq in the path..I normalized it at 1 Khz and 20 watts, then ran a sweep from a cd I burned..problem was, the eq was setup smileyface, and the lows toasted it..melted the solder holding the resistors in place.
We all learn from our mistakes...I must be a genius by now, regardless of what my kids say..
Cheers, John
ya killin me...john_ellis said:
What did it measure in inductance...
Ah, forgot to add...I'm makin a widgit that's 1 inch diameter and 8 inches long...I'm puttin a stainless heater on it in the exact same pattern you describe...but it's 1 mil thick, 28 milliohms per square, about 37 inches long.
I need to drop about a kilowatt into this for roughly 20 uSec and 5 hz rep rate, and was wondering if you could see the inductive drop due to eddy currents in the aluminum, as that'll slow the leading edge of the voltage form. Too much and it'll force me to integrate for power, rather than just voltage times current. I like it simple.
Cheers, John
Hi all,
everything happens when I go out for the night! didn't get in till 0345hrs, yawn.
I have paralleled up three different sets of power resistors, 4r 200W, 4r 50W and 8r 100W and intend making up some more.
I think power upto the low 100kHz, whether the lower harmonics should also be included might take this up to 1MHz.
Jneutron,
I can follow the first 5 questions although I don't have answers to all, but what does this mean?
everything happens when I go out for the night! didn't get in till 0345hrs, yawn.
I have paralleled up three different sets of power resistors, 4r 200W, 4r 50W and 8r 100W and intend making up some more.
I think power upto the low 100kHz, whether the lower harmonics should also be included might take this up to 1MHz.
Jneutron,
I can follow the first 5 questions although I don't have answers to all, but what does this mean?
when you asked max L, is that before cancellation or after?
AndrewT said:
When you take a meter or scope, and connect it across a resistor, you measure a voltage which represents the current through the resistor. If that current is changing, there will also be an error voltage due to the loop formed by the probe and the physical size of the resistor.
B is a measure of the flux, b dot is the rate of change of the flux. B dot error is the voltage caused by trapping the time varying flux, faraday's law of induction.
For low impedance current viewing resistors, sometimes this voltage error is very significant with respect to the voltage caused by the current across the resistor.
For a Dale NI type resistor, while the windings are bifilar, negating any solenoid based inductance, the size of the body and the loop it forces with the scope probes, high frequency currents will cause a non zero error, which will show up as phase shift. I've made some lower power resistors (50 watt) which have virtually no errors of this type.
So if you need absolute accuracy, low b dot error is required. If all you want is to remove the inductive component of the load, that is different.
Cheers, John
Hi Jneutron,
thanks for that explanation.
All I want is to remove a source of measurement error due to an unknown inductance when testing an audio amplifier at very high frequency. Rather than compensate test measurements I think it would be simpler to ensure each of my test loads had non resistive components at a low enough level that errors here can simply be ignored.
thanks for that explanation.
All I want is to remove a source of measurement error due to an unknown inductance when testing an audio amplifier at very high frequency. Rather than compensate test measurements I think it would be simpler to ensure each of my test loads had non resistive components at a low enough level that errors here can simply be ignored.
Hi Andrew T
Geoff explained how you can compensate the L of an inductive R, but you will need to know what inductance each of your resistors has in order to set the right compensation circuit. At this point it maybe that rather than compensate, which needs measurement and evaluation, it might be easier to buy?
(The output LR/CR filters on most amps are supposed to ensure that the load is resistive at high frequencies. Impedance matching throughout is I believe what Zobel intended; many LR/RC filters do not achieve that; and yet others are intended to ensure stability when capacitor loads are used, but don't actually use the right values for that either).
Arcol manufacture NI versions of their aluminium housed power resistors, but RS only sell the ordinary version.
Maybe if enough diy-ers are interested we could bulk buy some?
cheers
John
Geoff explained how you can compensate the L of an inductive R, but you will need to know what inductance each of your resistors has in order to set the right compensation circuit. At this point it maybe that rather than compensate, which needs measurement and evaluation, it might be easier to buy?
(The output LR/CR filters on most amps are supposed to ensure that the load is resistive at high frequencies. Impedance matching throughout is I believe what Zobel intended; many LR/RC filters do not achieve that; and yet others are intended to ensure stability when capacitor loads are used, but don't actually use the right values for that either).
Arcol manufacture NI versions of their aluminium housed power resistors, but RS only sell the ordinary version.
Maybe if enough diy-ers are interested we could bulk buy some?
cheers
John
AndrewT said:
Ok. If you are using sines, the methods being mentioned by others will work. But remember that those comp methods may need several cycles at the frequency to settle to that of a non reactive load. Great for sines, lousy for transients. My preference is to make the load pure resistive out to a ghz if possible, to remove all doubt...but that's me..
john_ellis said:
I previously mentioned the Dale NI type. They can be had in the 250 watt package, they are bifilar would to eliminate solenoidal inductance..
I recommend this option, but purchase two of them at double the desired impedance, and build the load with one on each side of an aluminum plate. This provides a magnetic null in the center of the resistors, there you can put the voltage measuring wire without worrying about field induced error voltages, it just requires machining a slot through the plate, or using two plates.
Cheers, John
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