Measuring Inductance of a Transformer - Page 1
Forum section: electronics / beginners

author: fatalelieberi
author_rank: Contributor
timestamp: August 10, 2020, 02:45:13 pm
content: Hi guys,I found some high frequency transformers in a broken ballast. Out of curiosity, I want to know the inductance of each coils in the transformer. My idea is that by knowing the inductance, I can find the turn ratio between the windings. However, since I don't have any LCR meter, I use my signal generator and oscilloscope. I wired the transformer in series with a known resistor, in my case I use 1K resistor with an actual value of 975 Ohms, forming an high pass RL filter.My theory is that I can measure the output voltage at a frequency to find the inductive reactance of the coil. From that point, I can calculate the winding inductance. You guys can see below my test result. Green is the measured value and yellow is the calculated value.As you can see, the result is rather funny. I don't understand why the inductance has an imaginary part. Also, the second winding gave higher voltage than the input voltage. Can you guys enlighten me or point me which theory I can read? Any pointer is really appreciated.

author: TimFox
author_rank: Super Contributor
timestamp: August 10, 2020, 04:19:38 pm
content: L1 and L2 are not pure inductances, but each has a series resistance. If you calculate the complex impedance, the imaginary term represents the inductive reactance. An inductance calculated from that will be real and the resistance will be imaginary. The reactance of an inductor should be positive, while a capacitive reactance is negative. I can't audit your spreadsheet remotely, but there is a sign error somewhere.With transformers, you need to measure the "magnetizing" inductance of both windings, with the other winding open-circuited, and the "leakage" inductance of one winding with the other short-circuited to characterize the transformer. From the two magnetizing inductances, the turns ratio is the square root of the inductance ratio.There are many references: my favorite is Kenneth R Clarke and Donald Y Hess,Communication Circuits: Analysis and DesignReading, Mass. Addison-Wesley, 1971, chapter 2 (reprinted 2002 by Krieger Publishing).

author: fatalelieberi
author_rank: Contributor
timestamp: August 11, 2020, 01:09:08 am
content: So, I looked again the calculation after you said I may had sign error, and I did put wrong sign at the phase difference. I mistook that leading phase as negative where the correct one should be positive. Also, for the second winding I forgot to remove millivolt sign from the V_out. Finally, I found something very stupid. The oscilloscope probe for V_out was accidentally set at 1X instead of 10X!Therefore, I re-measured it once more and I got more sensible result. You can see it below.So, based on this result, I think the second winding is faulty or at worst, there's no second winding at all. It is actually an inductor but using a transformer body. Is my guess correct?

author: fatalelieberi
author_rank: Contributor
timestamp: August 11, 2020, 06:39:59 am
content: Okay, I noticed I had done something stupid, again. I forgot that it was an RL filter which meant at 1 kHz, it may had passed beyond its 3db point when measuring the second winding reactance. So, I re-measured it again, this time at 5 Hz. And i think it is correct now, but still has a weird negative real impedance.

author: TimFox
author_rank: Super Contributor
timestamp: August 11, 2020, 01:20:20 pm
content: Your second-to-last line, labeled XL, should really be ZL = R + jX, the complex impedance where X is the imaginary part and R is the real part.If your measurements are correct: Since your measurement of L2 does not change when shorting L1, and vice-versa, there is zero coupling between them. A negative real value for a passive component is impossible.Important suggestion: displaying 15 significant figures calculated from data that only have 3 significant figures is counter-productive. It makes it difficult to follow, and there is absolutely no information contained in the extra dozen numbers.

author: Mecanix
author_rank: Frequent Contributor
timestamp: August 11, 2020, 02:16:00 pm
content: https://youtu.be/ajycdvKyCEwL = R/(2*PI*F)

author: TimFox
author_rank: Super Contributor
timestamp: August 11, 2020, 02:52:17 pm
content: Quote from: Mecanix on August 11, 2020, 02:16:00 pmThat should be L = X / (2*pi*f)

author: fatalelieberi
author_rank: Contributor
timestamp: August 15, 2020, 05:09:23 am
content: Because of the strange result of my measurements, I re-did it again. This time I found out that the coupling of the channels of my oscilloscope were different. Channel 1 (measuring V_in) was in DC, while Channel 2 (measuring V_out) was in AC. I don't know why, but if the couplings were different, there was a phase shift when measuring the second winding. Also, the second channel still had some DC offset, about 5-10mV when measuring V_out, even though it was AC coupled. However, if it measured V_in, the offset was completely gone. So, I set them both to AC coupling and had this result.My interpretation is that the transformer does not have a secondary winding since at any frequency, it does not have any attenuation at all. Also, that explains the 100% leakage inductance. Am I correct?

author: TimFox
author_rank: Super Contributor
timestamp: August 15, 2020, 01:21:13 pm
content: I don't understand your column headings: did you get the identical results with the other winding open and shorted? In that case there is no coupling and the total inductance is "leakage", since none of the flux went to the other winding.In general, when measuring magnetizing inductance, it is better to use low frequencies, to avoid possible resonance with winding capacitance, especially on the un-shorted secondary.

author: fatalelieberi
author_rank: Contributor
timestamp: August 16, 2020, 10:57:57 am
content: Yes, the column heading means no difference when the other winding is shorted or open. How low the frequency should be? I'm sorry if it sounds really dumb, I'm just restarting my electronic hobby (that was also short-lived before) after a few years and I can only do it in my spare time due to my long hour work.

author: TimFox
author_rank: Super Contributor
timestamp: August 16, 2020, 01:26:02 pm
content: The short answer is to go low enough that the measured inductance doesn't change when you go lower. Your data for L1 seem to be stabilizing below 8 kHz.In your case, however, it's not clear what you have connected as "L2", since you could not measure an inductance there and it does not affect "L1".

author: fatalelieberi
author_rank: Contributor
timestamp: August 20, 2020, 11:13:53 am
content: Hi TimFox,Regarding your question, here is how I setup my measurement. This is the transformer that I found and I want to measure. I've numbered the pins and checked with winding resistance. For pin 1-4, I can get 0.8 Ohm of resistance, but for pin 2-3, and 1-2, I cannot get any resistance measured. However, assuming that the usual winding configuration for this transformer is 2 windings, I assumed that pin 2-3 is the second winding L2, and pin 1-4 is L1.Transformer(?) pinoutMeasuring resistance of pin 1-2Measuring resistance of pin 1-4Measuring resistance of pin 2-3Here is I how I measured it with my oscilloscope and function generator, with open and closed second winding.Open L2Closed L2Then, I measured it with the following parameters and the resulting waveforms.01 L1 (Open L2) 1 KHz02 L1 (Open L2) 500 Hz03 L1 (Open L2) 100 Hz04 L1 (Open L2) 50 Hz05 L1 (Open L2) 10 Hz06 L1 (Closed L2) 1 KHz07 L1 (Closed L2) 500 Hz08 L1 (Closed L2) 100 Hz09 L1 (Closed L2) 50 Hz10 L1 (Closed L2) 10 Hz11 L2 (Open L1) 1 KHz12 L2 (Open L1) 500 Hz13 L2 (Open L1) 100 Hz14 L2 (Open L1) 50 Hz15 L2 (Open L1) 10 Hz16 L2 (Closed L1) 1 KHz17 L2 (Closed L1) 500 Hz18 L2 (Closed L1) 100 Hz19 L2 (Closed L1) 50 Hz20 L2 (Closed L1) 10 HzMy interpretation is the second winding L2 actually does not exist. However, there's some error when measuring L1 when the second winding is closed.

author: TimFox
author_rank: Super Contributor
timestamp: August 20, 2020, 01:19:49 pm
content: "L2" might just be two extra pins for mechanical mounting purposes. There may be some stray capacitance to the pins that affect your measurements slightly.

author: fatalelieberi
author_rank: Contributor
timestamp: August 20, 2020, 02:02:51 pm
content: Okay, then can I just take the average of inductance of the lowest frequency, which in this case the average of inductance at 10Hz?

author: Jan54
author_rank: Newbie
timestamp: February 04, 2023, 04:58:29 pm
content: I measured a IKEA lamp transformer (20W, 12V, 230V).The internal DC R=0,4 Ohm.When i put 15,4 V (Fleischmann transformer) with serial 18 Ohm 5W resistor on it, I measure 10,4 V on the secundairy windings and 6,5V on the resistor.Calculating with complex math I come to 18 Ohm and 70 mH. The math is from an EE book, so that is right; examples of examinations work out well.The 18 Ohm must be wrong. The 70 mH I cannot judge.When I repeat the measurements with a 6V transformer, the same big internal DC R pops out.When I repeat measuring a TL lamp ballast inductor, with a 1k8 serial resistor, I calculate a 1k8 internal resistor and 28H. Again, the 1k8 Ohm result must be wrong (I measure 38 Ohm).Have you any explanation of this weird results? Is this leak capacity? I did not expect that so huge with 50 Hz.