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LCR meter to measure inductance of power transformer?

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picotube

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power transformer inductance

Hello,

Do you think it's appropriate to use LCR meter to measure inductance of power transformer, or soft-iron type of inductors? Do you think this method will add some eddy-current loss during the measurement? Pls advise.

Thanks
 

Mr.Cool

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transformer inductance measurement

absulutely not. instead, connect your inductor/transformer in series with IGBT and hall-effect current sensor. then hit your IGBT with a pulse (say 80uS) and measure the current sensed on an oscilliscope.

now you can calculate your inductance via:

V = L * dI/dT

where V is the bus voltage and dT is your pulse width. dI is found from the current sensor output. solve for L.

that is the only way to get accurate results, good to within 5 - 10%. also, for transformers, you must ensure that you have no magnetizing current remonance (this is a B-H curve thing). if you have remonance you can not trust your results. to determine this, hit your IGBT with many pulses and calculate the L value. each time it should be the same (say, within 2%). if you see that the inductance is growing then you can not trust your results. if this is the case, then you have poorly designed transformer, or your voltage/pulse width selected does not represent the application to which the transformer will be used. therefore adjust as necessary. (you can zero out your magnetizing current by applying AC current across your inductor.. use a variactor for this).

goodluck
Mr.Cool
 

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inductance of a power transformer

Do you think the turn-on resistance of IGBT will give some errors in inductance measurement?
 

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measure inductance lcr circuit oscilloscope

no problem here. typically your transformer inductance is orders of magnitude higher inductance than the internal impedance of the IGBT. especially given that you are testing a POWER transformer and not some little nH inductor.

Mr.Cool
 

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how to measure inductance of a transformer

by the way, the IGBT circuit i described should be a half bridge (two igbts) only because the upper igbt has a body diode that can use to disipate the stored energy in the inductor during pulse OFF period. and you should use a large cap across the half bridge because only caps can discharge that kind of energy in short period of time. a power supply can't do that.. i hope these things go without say cause there is a whole list of "gotchas" like this.

Mr.Cool
 

picotube

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how to measure transformer inductance

What's about ferrite core type of transformer?
Is it OK to measure its inductance using LCR meter?
 

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inductance measuring circuit

ummm... i think u should not consider making/measuring this transformer of yours.

the inductance is created by the windings, not the core. the core stores the energy (flux). therefore it really wouldn't work if you stuck an LCR meter on two ends of a ferrite core. i think you would read nothing, or atleast nothing useful.

Mr.Cool
 

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picotube

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measuring output transformer inductance

Hi Mr Cool,

I'm aware that the inductance is one of the properties shown by current carrying conductor, e.g. inductor/transformer winding. The objective of winding the conductor around a soft-iron or ferrite core is to increase the inductance, and therefore, the energy storage.

I noted that some LCR meter provides several frequency settings, e.g. 60Hz, 120Hz, 1kHz, 10kHz, 100kHz, etc. Some LCR meter provides frequency setting lower than 60Hz (e.g. Agilent 4284A, frm 20Hz ~ 1MHz). Do you think it's appropriate to measure the inductance of a transformer (or motor phase winding) using Agilent 4284A at 20Hz setting? If NOT, what are the possible ERRORs of such measurement (using Agilent 4284A, set as 20Hz, to measure the inductance of power transformer)?

Thanks
 

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measuring inductance of a motor

perhaps i misunderstand your question. yes it is possible to measure inductance with LCR meter of transformer or motor phase, the later probably measured line-to-line meaning divide result by 2 for armature inductanse. my only concern is your transformer impedance must be of acceptable value as compared to internal to LCR measuring bridge. how do u know ahead of time? this is difficult to answer, but it is helpful if you have an idea of what your inductance is going to be. recall impedance is calculated: X_L = 2*pi*f*L ; f = frequency & L = what you want to know

so, now you see why LCR meter has adjustable frequency. you can set it such that resultant impedance is acceptable value as that required by LCR meter.. sort of a what came first, chicken or egg?

this is why i recommended pulsed method. you just set the circuit up, hit transformer with a pulse and measure result. but it requires some components and a current probe and oscilliscope. to be quite honest, measuring inductance of transformer or motor phase is difficult to get right (considering its such a simple task). but once you have done it you will never forget & it will be forever useful.

Mr.Cool
 

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how to calculate the power in lcr as heat

Mr.Cool said:
by the way, the IGBT circuit i described should be a half bridge (two igbts) only because the upper igbt has a body diode that can use to disipate the stored energy in the inductor during pulse OFF period. and you should use a large cap across the half bridge because only caps can discharge that kind of energy in short period of time. a power supply can't do that..
Mr.Cool
Does the circuit configuration (to measure inductance using 'pulse method') you meant is the one shown in figure below?

Mr.Cool said:
now you can calculate your inductance via:

V = L * dI/dT

where V is the bus voltage and dT is your pulse width. dI is found from the current sensor output. solve for L.
I'm not sure about the way to get dI. Do you mean find the dI = Ipeak from the scope display, or need some integration to find dI from the signal captured by the scope?
Thanks
 

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characterizing a transformer using an lcr meter

yes you have it exactly. and i like your use of hall sensor :)

V = L * di/dt

to measure the value V a differential probe works well. put the diff prob across the inductor AT THE TERMINALS and twist the diff prob leads together. since voltage changes instantly in an inductor you will have your "V" measurement.

the output of the hall effect sensor is isolated due to the interal transformer, so you can use a regular probe. read the hall-effect sensor datasheet for transformation ratio and suitable measuring resistor. measure the voltage generated across the measuring resistor on the oscilloscope. this represents your dI current, and since it ramps up (current does not change instantly in an inductor) you will also get your dT value from the oscilloscope. actually you are measuring the voltage across the resistor and will have to calculate back to discorver what the primary side current is and use *that* values as the dI in the formula.

now you have all the variables you need to solve for L. what is even better about this method is that you can test for inductor saturation limit. at what current does your inductor saturate?? or transformer... well, raise the current up and pulse the unit. the current you are measuring at the output of the hallefect sensor will ramp up as usual, but then all of a sudden will ramp up at a MUCH higher rate. at this junction represents the peak current your inductor can handle before it saturates.

the nice part about using the pulses is that if you have a catastrophic failure, or your inductor saturates much lower value than you were expecting.. your equipment and inductor are not likely to be damaged, since the pulse will end after a few microseconds anyways :)

what is a good pulse width to use?? this is a good question. i'll let you think about it for a while..

Mr.Cool
 

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inductance measurement using lcr meter

I built a Matlab/Simulink simulation model (see figure) of the proposed inductance measurement circuit. The sample inductor used in the simulation is 10uH/15mohm. The period of the pulse is 5ms (50% dutycycle). Vbus in the simulation is 5Vdc. The current and voltage waveforms across the inductor are shown in the figure.

Based on calculation:
V = 5V
di = 260A (approximately, derived from the current plot)
dt = 4ms / 2 = 2.5ms

V = L * di/dt

=> L = dt/(V*di) = 2.5ms / (5V * 260A) = 2.4uH
CORRECTION => L = V*dt/di = 4V * 2.5ms / 260A = 38uH.

The result is not equal to 10uH. Would you pls advise where I did wrongly?

Thank you very much.
 

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measuring inductance with oscilloscope

nice work :)

when determining di/dt you should calculate only in the region that is linear. using 2.5ms you are including the saturation area of your inductor, so this confuses the results.

also, 5 ms pulse is a very LOOOOOooooong pulse. try again using 5uS and measure the rise time of the di/dt in the linear area. if you can, use matlab to give you exact coordinates to make it easier for you to calculate di/dt --> (Y2 - Y1) / (X2 - X1)

Looking forward to seeing your results.

Mr.Cool
 

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measuring inductance

The pulse width is 50us.
Therefore,
L = V*dt/di = 4V * 50us / 19A = 10.5uH (close to 10uH)

Mr.Cool said:
what is a good pulse width to use?? this is a good question. i'll let you think about it for a while..
The pulse width should be choosen in such a way that the di/dt is linear. Pls correct me if I'm wrong.
 

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how to use measure inductor lcr

picotube said:
The pulse width is 50us.
Therefore,
L = V*dt/di = 4V * 50us / 19A = 10.5uH (close to 10uH)

Mr.Cool said:
what is a good pulse width to use?? this is a good question. i'll let you think about it for a while..
The pulse width should be choosen in such a way that the di/dt is linear. Pls correct me if I'm wrong.



When doing this experiment in a real circuit, won't do it whit a duty cycle so high. Or the core wil saturate and burn the circuit!

So, use 50us to ON, but let's OFF for a longer time.
 

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the uses of lcr meter

i do not think you should run a continuous pulsing circuit though. one single pulse is sufficient. controlled by a push button by the user. use a pot resistor to set the pulse width. if you run a continuous pulsing circuit (duty cycle) then you run the risk of over heating due to inductor/transformer saturation.

when the core saturates the flux linkage = 0 and your current has rises very fast as it is limited only by the wire resistance. you will see this on your pulsing circuit because the slope of the hall-effect current sensor output signal will increase dramatically. but your circuit is protected because of SINGLE pulse circuit.

you are right about the pulse width. it should remain linear. but there is more to it than that. it should remain linear in the NON-saturated area of the B-H curve. the pulse width should be the maximum time that your application circuit would see. for example, if you were switching the transformer at 15kHz, then your period is 66.6 uS. Yes a duty cycle may change the ON time, but the maximum on time is if the dutcy cycle is at 100%, which would mean a maximum pulse of 66.6 uS. use this as your period in your circuit.

if your transformer is line frequency, 60 Hz, then your pulse should be set to 16mS. hit the pulse, make sure the slope is linear and non-saturated and you're good.

enjoy
Mr.Cool
 

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calculating inductor saturation point

afonso said:
When doing this experiment in a real circuit, won't do it whit a duty cycle so high. Or the core wil saturate and burn the circuit!
Mr.Cool said:
if you run a continuous pulsing circuit (duty cycle) then you run the risk of over heating due to inductor/transformer saturation.
As I don't really understand, sorry to ask the following questions:
1) How a 'high duty cycle' or a 'continuous pulsing' will cause the core/inductor/transformer to be saturated ?
2) Why when the core/inductor/transformer is saturated, it will cause 'overheating' or 'burn the circuit' ?


Mr.Cool said:
when the core saturates the flux linkage = 0 and your current has rises very fast as it is limited only by the wire resistance.
3) Why when the core saturates, the flux linkage = 0 ? If I'm not mistaken, I learned before that when the operation of an inductor or a transformer reaches the saturation region, the flux linkage is no more increasing linearly with current anymore. Or, in other words, the flux linkage does not increase as the same rate as the current increases. But the flux linkage is not zero, right? Probably the rate of change of flux linkage when the core/inductor/transformer is saturated is 'small', so that the induced voltage ( Lens Law, or V = d(flux-linkage)/dt ) is not large enough and therefore the current increases rapidly. I studied this few years ago so some of my memory cells might be damaged already and therefore remember wrongly. I'm very confusing now. Pls advise and pls correct me if I'm wrong.
 

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power transformer sensor

this is also from memory.. which has not been kind to me lately.

when the core saturates the permiability is reduced to very small number. since inductance is proportional to relative permiability then the inductance reduces also, very close to zero. if you are using a core in the inductor/transformer design it is because you are after large energy storage, usually large inductance values too. thus, when i say inductance zero, i mean in comparison to the original value it is orders of magnitude smaller.

why does it do this? think of the B-H curve. as you increase in magnetization strength the flux levels off and doesn't increase anymore. since the permiability is the slope of the B-H curve, and the slope is zero (or close to zero) at the saturation point then it follows that the permiability is zero. thus inductance falls away.

V = L*di/Dt

here, V stays the same, dT stays the same.. but L is falling. to compensate for this then the only variable left is dI - which must rise.

this is the begining of the end because as current increases it forces the saturation even more until even the voltage begins to breakdown.

it all comes down to seeing how V= L*di/dT is formed

L = u * n^2*A / l_m

when flux density can no longer increase with increasing current then the rate of change of flux density falls away to zero.

this is Faraday's law you mention --> v = n*A*(dB_sat/dT) = 0
since v = 0 you have a short circuit... the current is now limited only by series resistance.

now you have very high current rate of change.. it was once being slowed down (di/dt) by the inductance but now that inductance is gone... so it suddenly increases its ramp rate dramatically. you will see this on the oscilliscope because the ramp slope increases. a lot of current limited by small resistance equals a lot of heat. lots of heat equals damaged inductor. although from exeperience i can tell you that inductors/transformers are very robust.. you have to leave the high current running for a LONG time before any permanent damage is realized. ("Long time" is a relative termn.. use it wisely)

Mr.Cool
 

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how to find di/dt in matlab when you know i

Mr.Cool said:
when the core saturates the permiability is reduced to very small number. since inductance is proportional to relative permiability then the inductance reduces also, very close to zero. if you are using a core in the inductor/transformer design it is because you are after large energy storage, usually large inductance values too. thus, when i say inductance zero, i mean in comparison to the original value it is orders of magnitude smaller.
Agree. When core is saturated, permeability of the core reduces (see attached figure), and hence the inductance of the inductor/transformer reduces as inductance is proportional to permeability.
  • Inductance, L = (u x n^2 x A) / length
    u = permeability
    n = number of turn
    A = cross section area
Mr.Cool said:
think of the B-H curve. as you increase in magnetization strength the flux levels off and doesn't increase anymore. since the permiability is the slope of the B-H curve, and the slope is zero (or close to zero) at the saturation point then it follows that the permiability is zero. thus inductance falls away.
If I'm not mistaken, at any point on the curve, permeability is given by the ratio (B/H) and not by dB/dH (the slope of the curve), right? ( Pls refer https://www.ee.surrey.ac.uk/Workshop/advice/coils/mu/#mu )


Mr.Cool said:
V = L*di/Dt

here, V stays the same, dT stays the same.. but L is falling. to compensate for this then the only variable left is dI - which must rise.

this is the begining of the end because as current increases it forces the saturation even more until even the voltage begins to breakdown.

it all comes down to seeing how V= L*di/dT is formed

L = u * n^2*A / l_m

when flux density can no longer increase with increasing current then the rate of change of flux density falls away to zero.

this is Faraday's law you mention --> v = n*A*(dB_sat/dT) = 0
since v = 0 you have a short circuit... the current is now limited only by series resistance.
I see the picture now. I get your points now. Great !
 

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