Inductive Kickback

[citfta]

MH,Verpies-anyone

When we calculate the dissipated power from the below diagram,during the 70% off time,the circuit should be split in half,and the dissipated power from each half should be equal-correct?.
So if we have say 13,68mW being dissipated across the inductor and resistor,we should also have 13.68mW being dissipated across the LED--?--see pic below.

So should it now be 13.68 + 13.68 X 70% for our power out?
Which would be 19.15mW ?

Hi Brad,

No disrespect to anyone else's opinion about how to analyze this circuit.  Almost any way to analyze could be correct depending on what you are looking for.  But I would approach it from a slightly different angle.  I would divide the circuit into the source and load.  During the off time until the coil's energy is depleted the coil is the source and the rest of the circuit is the load.  The source (coil) has to supply the same amount of energy as the load dissipates.

Carroll

[forest]

I think to resolve all issues  use a DPDT relay and external coil and measure coil current  then discharge coil separately into capacitor and scope current and voltage.
If you know any such video tutorial on youtube with scope measurement , let me know....

[tinman]

@Tinman,

"
Sometimes it helps to simplify to increase understanding as in the case of ideal componants like capacitors and inductors without involving all the messy issues that accompany them.

Your attitude has been repugnantly condescending, not only towards me and Erfinder,

I have shown you no disrespect synchro,despite your disrespect toward myself-E.G--> post 94 quote: Verpies and Milehigh are both full of shit just like you!

An "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy. However real inductors have side effects which cause their behavior to depart from this simple model".

And i use real inductors ,as i have no ideal inductors.

There's nothing I don't understand about the "Magnetic Field Inertia" issue you keep trying to teach me. Additionally, your "concealed coil shorting circuit" has caused me to grow suspicious of you.

Nothing new there synchro. I suspect there are many who think the same way. We humans tend to attack that of what we dont understand.

On top of that, and more importantly, you seem to fail to understand sufficiently how the rate at which spark gap contacts separate effects the level of Flyback Voltage.

You do understand synchro that any spark or arc is power being dissipated. There for an ideal switch would have no such arcing.

My opionion of you as a "Mylow" will remain unshaken until such time as you divulge your selfishly "Bogarted" secret!

I will keep that in mind, although i doubt i will loose much sleep over it.

[tinman]

Offhand I can't.  Especially since I believe you stated the math function doesn't really work on your scope.  If you explain why it might hit me, not sure.  Can you explain why you want to use an LED?

I dont think we need the math (although the math trace works quite fine on my scope)function to work out our P/in and dissipated power when the wave forms are very clean.

I am using an LED for a load for the very reason that the I and V curve is non linear.
It al;pws me to increase or decrease the inductive kickback current while maintaining very close to a set voltage during the off cycle. I can then measure the light output from the LED via way of a light box-->a small solar panel inside a box with the LED,and then we place a resistor across the solar panel output. I can now see if the LED's output rises with the increase of current,while the voltage remains much the same across that LED. For this particular LED,it seems that the optimum voltage is 3.2v,and any additional current makes no difference to the output of the LED. The other thing is i can now see what happens when i increase the voltage on the input--this is how i increase the current on the flyback. Now we have a situation where only the current on both the input and output rises,but both the input and output voltage remain the same,even though i am increasing the voltage on the input. On the flyback side this is expected with the LED in play,but why dose only the current rise on the input,and not the voltage?.

[tinman]

Hi Brad,

I can agree with splitting the circuit in half as you show but I disagree with the assumption that the dissipated power in each half should be equal.
IF your measurements indicate this half-half power split, I think it is only a coincidence in this particular circuit.  Just imagine you replace the LED with a normal Si diode, then the voltage and current conditions change in that half part of the circuit and this causes a change in the other half part too: why would the two changes have (again) an equalizing effect?

Under the conditions you show with the measured voltage and current (I suppose you measured them), the voltage drop across the 10 Ohm should be V= 4.8mA*10 Ohm=48 mV, right? (Of course I do not know the DC resistance of the coil.)
If you consider the LED DC resistance it represents in the moment when its current and voltage are 4.8mA and 2.85V, this gives R=593 Ohm.   (I neglect LEDs may have even some hundred pF self capacitance.)

It is okay that the total dissipation is approached by considering each components in the closed circuit and adding their dissipations up.   Perhaps a close approach would be to consider the coil's DC resistance too and adding up the losses for the 3 components: the LED, the 10 Ohm and the copper loss of the coil and then consider the duty cycle issue.

Gyula

Hi Gyula
Yes,the measurements are from the scope shot below.
If we look at the input side of thing's,it is saying our inductor has a resistance of around 1890 ohms at that frequency when we subtract the 10 ohm CVR. The reason i asked that question is that the LED dose not need that amount of current to get the 2.85 volts across it at that 70% duty cycle. So with this circuit,how do we measure the power being dissipated by the LED?.-->circuit below.