How many mH? - Page 1
Forum section: electronics / beginners

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 08, 2015, 04:14:29 pm
content: Hi,I took that on a board. It has 25 turns 20mmm diameter, 10mm center diameter, 11 mm thick. I tested it with a resonnant frequency to know how many mH. No satisfying result:2.5 mH.I tried calculations....My multimeter gives 7.5 mH. Does that make sense?So, just like that, what would be your guess on it's number of mH, or uH?Thanks for your attentionJPD

author: bktemp
author_rank: Super Contributor
timestamp: June 08, 2015, 04:32:35 pm
content: If you get 7.5mH with only 25 turns it must be some type of ferrite or iron core. Because it has no air gap it can not handle much current. Your resonant circuit probably saturates the core and therefore the inductance decreases.Guessing is impossible without knowing what type of core you have used. It can be anything from nH for an air core upto some mH for a high permeability ferrite or iron core.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 08, 2015, 08:29:06 pm
content: The only clues I have are:a- it is greenb- two saw cut at 2mm gives 700 OhmsWell)ThanksJP

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 09, 2015, 03:28:51 pm
content: Quote from: AcHmed99 on June 08, 2015, 11:55:48 pmYes !ThanksJPD

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 09, 2015, 08:35:25 pm
content: I built your device with a 60V 12A N Mosfet. For the .001 Ohm, it is a small wire I suppose.My power supply was freaking, showing alternatively 3.5 A and 0. The voltage was freaking too. Anyway, I tried at 2H and 178 Hz. Replacing the 150 uH coil with my unknown one.Was that correct?How do I get the amps with that?I feel I am wrong ;-/JPD

author: bktemp
author_rank: Super Contributor
timestamp: June 09, 2015, 09:18:23 pm
content: The expected saturation current is in the 100mA to 10A range. Using a 1mOhm shunt, you expecting to see a signal in the 0.1-10mV range. 5V/div is a bit high to see anything useful.If you have a shunt with 10-100mOhm it could be better for such a small inductor.And the signal going to the mosfet must have a low duty cycle, <5%.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 09, 2015, 09:32:53 pm
content: I don't know what the hell your first two traces are...looks like several AM radio stations?Please read up on correct oscilloscope operation -- if the signal you're expecting is very slow, repetitive, and a strong signal, then you most certainly will not find it at the lowest of voltage and time settings!Poor probe technique allows radio stations to get introduced, especially if there are strong stations nearby.If I'm reading your trace correctly, this is the part you should zoom in on. I've drawn a green line suggesting what I expect is the function generator output. The function generator should be set for a very short duty cycle, a few percent, and probably a higher frequency, such that the on-time is not much longer than the time taken to reach saturation. Then you will not be shorting out the power supply.The important quantities to measure are: exactly what voltage is applied to the inductor during this period, for how long, and what current is achieved in that time. The first product is the flux (V*t). The quotient is the inductance, V*t / A = L.You should also have a local 4700uF electrolytic bypass capacitor as close to the transistor and clamp diodes as possible (which should also be as close together as possible).Tim

author: Andy Watson
author_rank: Super Contributor
timestamp: June 09, 2015, 09:39:58 pm
content: Quote from: J4e8a16n on June 09, 2015, 08:35:25 pmPut a resistor in series with the gate. Anything between 100 and 500 ohms should do. This will stop the high frequency parasitic oscillations.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 09, 2015, 10:57:51 pm
content: I thought it was only a clue.I found the diode was cooked.I have a 1n4148 150MHz and 4pf (would that create an lc ?).and a shotkey 1n5818Thanks for your answer.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 10, 2015, 06:30:15 pm
content: v = L di/dtIt is the formula you usev *dt/ L = di15*(0.000005)/(0.001) = .075AI will play with that)I am builted thatI dont know why I did not thought of v *dt/ L = diJPD

author: vk6zgo
author_rank: Super Contributor
timestamp: June 15, 2015, 03:59:59 am
content: Quote from: bktemp on June 08, 2015, 04:32:35 pmThree ways of testing:-(1)Borrow an LCR Bridge,& use that.(2)Place a known capacitor in parallel with the inductor, place the resulting assembly in series with one side of an ac signal from a Signal Generator,& monitor it with your Oscilloscope.Adjust the Sig Gen's frequency---at Resonance you will see a "null".Use the Resonance formula to find the Inductance.(3) Using the same circuit as above,connect it across the output of a square wave generator..Your Oscilloscope will show ripple after the transition---Determine the frequency of this ripple & use theResonance formula as before.At the low signal levels used,saturation will not occur.Do all three tests & take the average,if you are fussy!

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 15, 2015, 12:51:12 pm
content: Thanks to all. I am working at it.Questions and I should ask more.When are amp-turns important? For calculating v=L di/dt is di amp-turns? Etc.Are these v of the core, L, di of the core, dt of the core?When we say 'saturation' do we say saturation of the core, saturation of the transistor?Is the flux finding simple as thatQuoteordt * v /N = dFluxhttp://info.ee.surrey.ac.uk/Workshop/advice/coils/Faraday/index.html#fgivevIs it the voltage of the core * by the period of the core?Regards,JP

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 15, 2015, 03:20:12 pm
content: V, L, I and dI/dt are all in-circuit parameters. They do not contain units of turns.EMF, A_L, NI and Phi are core-and-winding level parameters. They do contain units of turns.Saturation of the core is the point where inductance (L, as measured at the coil terminals) or inductivity (A_L, as measured at the core) drops sharply. Analogously, the meaning of saturation for a transistor is where the V-I curve drops sharply.Flux (in circuit) is simply,integral V dtwhich for a square pulse, is simply the product of height Vpk and width t.Flux (in the core) isPhi = integral EMF dtNote that this carries a unit of "per turn", so that when multiplied by N, you get the in-circuit flux and voltage.V = EMF * N and all that.I should review this some day for accuracy and usability, it's kind of old. It should still be useful though. I wrote this in terms of using "turns" as a unit, so that as long as you follow dimensional analysis, you should get the correct results:http://seventransistorlabs.com/tmoranwms/Elec_Magnetics.htmlTim

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 16, 2015, 04:33:11 pm
content: I estimated the L with a multimeter and a resonnance setup for 8.5 mHAs you see , I have calculated the flux ... = .0003784 weber ... ?Are the A in your flux formula Amps? = .0445 ...Can I think that the 550 us space for the red line at 800mV is the charging time for the coil?The PS supplies .005 A * 25 turns this should gives .125 A-Turns ?Quoteathttp://www.prc68.com/I/JouleThief.shtml#Xtransistor is a 2n2504JP

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 17, 2015, 07:58:33 pm
content: Maybe an easier one to read.Power supply 1 volt less than 1 mA.Sorry, if I am buggy

author: XFDDesign
author_rank: Frequent Contributor
timestamp: June 17, 2015, 08:13:59 pm
content: Things are being made a bit too needlessly complicated.If you have a signal generator source, grab yourself a 47-51.1ohm resistor. Insert your unknown inductor in series from the generator to the resistor. Tie the resistor to ground. With your scope as your reference, set the frequency to around 10Hz and adjust the voltage until you get 1V amplitude (2Vpp) across the resistor. Increase frequency until the amplitude reads 0.7071V or there-about. At this frequency, your inductor is: L= Resistance/(6.283*frequency). If your inductor is 7.5mH, your 0.7071v (3dB) point will be around 1kHz.Trying to calculate flux, webers, coupling, etc, is going to send you spinning in circles.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 17, 2015, 09:30:53 pm
content: Thank you very much not letting me in confusion.Why did they created all these maths for?Although decibels are more maths I dont know ... ;o)I am on my way to test you method .JPD

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 17, 2015, 10:31:02 pm
content: Here I am,The following tester gave me 6.5 mH see imageI estimated the L with a multimeter and a resonnance setup for 8.5 mH1533 Hz5.2 mHI guess 5.2 is the oneJean Pierre

author: XFDDesign
author_rank: Frequent Contributor
timestamp: June 18, 2015, 03:21:07 am
content: Quote from: J4e8a16n on June 17, 2015, 09:30:53 pmThe physics of creating an inductor is a lot more elaborate than just using one. It's the difference between every-day engineering and physics.Typically in "napkin" engineering your inductor is going to be L = (num_turns^2 * premability_constant)/factor. So if you were to pick up a T50-6 core (12.7mm diameter or 1/2", made of #6 material) you have a constant of 40uH/100 turns^2 and a factor of 10^10. 10 turns? 0.4uH. 30 turns? 3.6uH.Then you have the "every-day" engineering level. The inductance calculation still holds, but then things like ESR or Q get factored in.Then "careful" level. Now add in power considerations, behavior over temperature and systemic effects. Mutual coupling to other devices, radiated EMI, etc.When you escalate to the highest degree, you need all of the Electromagnetic equations to get the desired outcome.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 18, 2015, 11:38:34 am
content: Well O0I will go with all of this.N = 10Another small thing: if I have .1 A*N does it mean that the resistor wich follow the coil will get 1 amp-turns ? or 1 amp/N ?And lets say I have 1 amp. The led takes 40mA. Where the hell is going the 960 mA left?Jean Pierre

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 20, 2015, 03:26:49 pm
content: UpdateIt comes from

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 24, 2015, 11:21:16 am
content: Question:Does v = L di/dt formula works for a square wave?If a Joule Thief wave is created by the on-off of a transistor we get a pulsed current no?Does the formula works for such a current , wave?JP

author: XFDDesign
author_rank: Frequent Contributor
timestamp: June 24, 2015, 10:21:12 pm
content: Either it works for the device, or it doesn't.In this case, what part about V(t) = L di/dt reads as subject to waveform?Suppose you had a physically impossible, perfect squarewave switching between -5 and 5V at 100kHz. The period is 10us which means 5us on, and 5us off. And we have a physically impossibly-perfect 10uH inductor.At time 0+, we apply 5V across the inductor. We can divide both sides by L to find the slew rate of the current:5V/10u = di/dt500,000A/s = di/dtNow, multiply both sides by the differential dt5*10^5 dt = diIntegrate both sides5*10^5*t + C= i(t)Let C=0 since we're just energizing it. After a perfect 5us, we have 5*10^5 * 5 * 10^-6 = 25 * 10^-1 or 0.25A stored in the inductor after 5us.Now, repeat the same thing but with -5V as applied voltage. Since you have an initial current of 0.25A, use that as your constant after integration. You will find that the current has returned to 0 after 10us.This is only for the perfect system as described, and is "pretty close" for a real world system.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 25, 2015, 01:57:03 pm
content: QuoteWhat is going to light the led?If the inductor discharge in the led .... ha... there is no alternating current it is dc.If I want to measure the emitter current with the current sensor (see attachment), how do I proceed?Is there a tutor or a message (I did not found one really on that subject) about that?Thanks for all.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 26, 2015, 05:50:01 pm
content: I tried to built one temporaryly. Is it the way to use it?

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 06, 2015, 03:04:28 pm
content: I found a good site :http://ludens.cl/Electron/Magnet.htmlandhttp://ludens.cl/Electron/lmeter/lmeter.html

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 08, 2015, 04:24:19 pm
content: Hi,Can we calculate the inductance L from these data has I did?Does the XL interfere with the current available from the power supply to the coil?Does it have to be added to the wire resistance wich would have almost no influencesqrt((130.624)^2+1^2) = 130 Ohms1.1 volt / 130 Ohm(xl) = 0.008 A in the coil?It fits with (.888/2) / 50 = 0.008As the current goes in the coils the flux climbs to 1.9 volts or more and discharge the current in the led when the transistor goes in saturation.If in the first time constant, let's say there is .5 A and the led uses only 50mA, does the flux 'wait'?ThanksJP