I am fascinated by your use of a pulse. I’m definitely going to try this. I’ve written a program in Fortran to find the needed unknowns and would be happy to share it.
well i have absolute respect for FORTRAN, i however have a humble spreadsheet in Excel that i'm happy to share with anyone........... Obviously after the Fortran option is exhausted and purely as a last resort :P
@@kennethstover4427 Indeed. i reduces a perhaps 15 minutes of calculation to finding the value of the inductor to perhaps (Considering you need to user cursors on the scope to get the frequency) Perhaps it reduces the time down to say 10 or 15 secods per inductor. Now when you have 5 inductor it doesn't really matter, when you however need to go through a bucket of 200 or so that you need to test, it matters quite a lot, it can be the difference between taking 1 week to go through it vs perhaps 1 to 2 hours. I say a person should do it manually a few times first so as to not get lazy , but once you do say 10, it's ok to do the other 190 with a spreadsheet because the spreadsheet gets the formulae in your head
This will be difficult, as the 56 pF in serie with the 50 ohms output of the pulgenerator will affect the Q seriously, I expect. Even the ringing frequency is affected by the dempening, although I do not expect that to be of great effect.
How do you maximize the time that the inductor takes to establish a field and the time that it takes to discharge in order to repeat immediately to eliminate the short circuit?
You feed a rectangle wave into it and modify the duty cycle. so where a square wave would be 50% Duty cycle, Try putting your function gen on Rectangle Wave, 1KHz and put it on like 5% Duty cycle, but then in certain inductors that rings will over lap, so you'll need to use between 8% - 12% or maybe 10% Duty Cycle, Have a play around with the Duty cycle, that's your answer Note : on a rigol this waveform is not called Rectangle, even though it should be, it's called PULSE
Thank you for this instructive video. Can you explain in a bit more depth why the 56 pF cap is needed? And does it not interact with the other components to shift the resonance frequency?
The 56 pF is just a way to insert the excitation into the tank circuit. I expect however that this method will seriously affect the ringing frequency. I would like to see the result when using a straight resistor, much larger than the impedance of the tank components. Say 10 Kohms. I addition, I wonder how accurate the 290 pF is. It is a major factor to the accuracy here.
waveform is "clean", but it's a mix of generator frequencies and tank circuit - the only "truly clean" form is a continuous sine with a constant amplitude - and amplitude of yours is not constant - and it's a result of mixing generator's and tank's frequencies and because of above you can't judge tank's frequency from max signal of spectrum - it has nothing to do with probes' setting - probe should be properly compensated with that variable capacitor on your oscilloscope schematics - resonant frequency shouldn't shift with properly compensated probe - only uncompensated probe will benefit in this regard
No mate, Probes are relevant , You try this say on 10x Probe setting but your probe is on say 1X , You're going to have a problem, or if both probes are on 10X for example. the probe coming from the Siggen is going to be 1X, the probe going into the scope needs to be 10X if you don't do that, you'll get a shit signal Also, A Constant AC is not going to do the job as what we are trying to do is ring out the inductor for it's resonant frequency, AC is not going to reveal that
I am fascinated by your use of a pulse. I’m definitely going to try this. I’ve written a program in Fortran to find the needed unknowns and would be happy to share it.
Fortran! I thought that had gone the way of Latin ;+}
I've been programming in Fortran since 1975 and it's been a long time since I've heard from someone who still used it who wrote, bravo👍
well i have absolute respect for FORTRAN, i however have a humble spreadsheet in Excel that i'm happy to share with anyone........... Obviously after the Fortran option is exhausted and purely as a last resort :P
@@martinkuliza I can see where Excel would work very well for this task.
@@kennethstover4427
Indeed. i reduces a perhaps 15 minutes of calculation to finding the value of the inductor to perhaps (Considering you need to user cursors on the scope to get the frequency) Perhaps it reduces the time down to say 10 or 15 secods per inductor.
Now when you have 5 inductor it doesn't really matter, when you however need to go through a bucket of 200 or so that you need to test, it matters quite a lot, it can be the difference between taking 1 week to go through it vs perhaps 1 to 2 hours.
I say a person should do it manually a few times first so as to not get lazy , but once you do say 10, it's ok to do the other 190 with a spreadsheet because the spreadsheet gets the formulae in your head
Thanks for the details!
You bet!
Thanks for your videos, interesting
Glad you like them!
I 've learned a lot
Amazing!🙌
Thank you 🙌
Thanks for the detail.
You can also calculate the Q of the inductor. By counting how many Cycles it Goes down to 0.707 of the Peak voltage Times 4.53
This will be difficult, as the 56 pF in serie with the 50 ohms output of the pulgenerator will affect the Q seriously, I expect. Even the ringing frequency is affected by the dempening, although I do not expect that to be of great effect.
I didn't understand it that well because its so hard but nice video mister
Thanks
Very nice clean demo N2CUA. I especially like the theory intro and conclusion. AK4GG
+J. Phillips Thanks
Thanks for the video.
Did you account for the presence of the 56 pF? I guess the L will be roughly 15 % lower than given here.
nice memo mister😍
So nice☺️
Thanks 😊
IF ANYONE NEEDS HELP PUTTING THIS CIRCUIT TOGETHER, it's actually pretty easy, and i'm happy to help you, Let me know
How do you maximize the time that the inductor takes to establish a field and the time that it takes to discharge in order to repeat immediately to eliminate the short circuit?
You feed a rectangle wave into it and modify the duty cycle.
so where a square wave would be 50% Duty cycle, Try putting your function gen on Rectangle Wave, 1KHz
and put it on like 5% Duty cycle, but then in certain inductors that rings will over lap, so you'll need to use between 8% - 12% or maybe 10% Duty Cycle, Have a play around with the Duty cycle, that's your answer
Note : on a rigol this waveform is not called Rectangle, even though it should be, it's called PULSE
anInductorWithIronCoreHasVariableInductance,DependentOnTheCurrentItCirculates?IfTheCurrentItsHigherTheInductanceLowersBecauseIronCoreSaturates?
Thank you for this instructive video. Can you explain in a bit more depth why the 56 pF cap is needed? And does it not interact with the other components to shift the resonance frequency?
The 56 pF is just a way to insert the excitation into the tank circuit. I expect however that this method will seriously affect the ringing frequency. I would like to see the result when using a straight resistor, much larger than the impedance of the tank components. Say 10 Kohms. I addition, I wonder how accurate the 290 pF is. It is a major factor to the accuracy here.
New subscriber, Jade T. Banguis
Welcome aboard!
Ma'am si Angel Mae D. Padayogdog ni ma'am
Hi maam ashley c. Fajardo cypress
waveform is "clean", but it's a mix of generator frequencies and tank circuit - the only "truly clean" form is a continuous sine with a constant amplitude - and amplitude of yours is not constant - and it's a result of mixing generator's and tank's frequencies
and because of above you can't judge tank's frequency from max signal of spectrum - it has nothing to do with probes' setting - probe should be properly compensated with that variable capacitor on your oscilloscope schematics - resonant frequency shouldn't shift with properly compensated probe - only uncompensated probe will benefit in this regard
I am afraid you are mixing up the transfer function of a well adjusted probe with the load thatvany probe will always have on the circuit under test.
... that any probe will...
You have to realize the amplitude is not constant because it is being pulse which is also not constant
No mate, Probes are relevant , You try this say on 10x Probe setting but your probe is on say 1X , You're going to have a problem, or if both probes are on 10X for example.
the probe coming from the Siggen is going to be 1X, the probe going into the scope needs to be 10X
if you don't do that, you'll get a shit signal
Also, A Constant AC is not going to do the job as what we are trying to do is ring out the inductor for it's resonant frequency, AC is not going to reveal that
@@calculatingcapibara2914
You didn't have to do that, You just go into your comment with the 3 dots on the side and select EDIT and change it
are you w2aew bother? :D
No, but thanks .. thats one of the nicest things anyone has ever commented
He does sound like w2aew.
Its very hard to do
Not really