Author Topic: Inductive Kickback (Read 159868 times)

citfta · « **Reply #75 on:** November 21, 2015, 02:54:33 AM »

Quote from: synchro1 on November 21, 2015, 02:47:28 AM

This should help clear everything up:

I do like that picture! Every time I see it it makes me feel like my brain is tilted a little. LOL

tinman · « **Reply #76 on:** November 21, 2015, 04:27:46 AM »

Here you go Synchro-test one complete.

What do you see?
1- a 30% on time from the FG(switch closed)
2-a 70% off time(switch open)

During the 30% on time(switch closed) we can see that the voltage(yellow trace) across the inductor(supply voltage) is on the top side of the zero volt line.The current is flowing in a forward direction,as the current trace(blue trace) is also above the zero volt line.

During the 70% off time(switch open) we can now see that the voltage is inverted across the inductor/resistor series circuit-!BUT! the current is still flowing in the same direction as it was during the 30% on time,as the current trace is still above the zero volt line.
You will also notice that the current continues to flow through the load(the LED) through the complete 70% off time,and never stop's flowing right up until the next on time period. This also shows you that the magnetic field dose not collapse completely before the next on time cycle starts.

So !Bub!
Please show us where exactly the current flow reverses direction?
And please tell us how a current continues to flow(in the same direction

) during the 70% off time if the magnetic field around the inductor collapses instantaneously ?
Show us where this !new! energy is?.

MileHigh · « **Reply #77 on:** November 21, 2015, 04:51:46 AM »

Brad,

That's an awesome posting and my sincere compliments.

I will just throw in some "flywheel" colour commentary:

You can see how there is always current flowing through the inductor -> the flywheel never stops turning.

The flywheel increases in speed when you apply positive torque (positive voltage) to it.

The flywheel decreases in speed when you apply braking to it. So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel. When the brakes are applied that is negative torque (negative voltage.)

Bonus round:

What happens when you disconnect the 10-ohm CVR from the circuit? Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage. All of the energy in the coil is burned off in the hot plasma. The current stops flowing through the coil.

What is the equivalent for the flywheel?

You clamp down on the brake calipers very fast and with maximum pressure. There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel. All of the energy in the flywheel is burned off in the hot brake pads. The flywheel stops spinning.

Bonus bonus round:

The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current. The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.

The big spike of negative torque happens because the flywheel has rotational inertia. The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.

tinman · « **Reply #78 on:** November 21, 2015, 05:15:55 AM »

Quote from: MileHigh on November 21, 2015, 04:51:46 AM

Brad,

That's an awesome posting and my sincere compliments.

I will just throw in some "flywheel" colour commentary:

You can see how there is always current flowing through the inductor -> the flywheel never stops turning.

The flywheel increases in speed when you apply positive torque (positive voltage) to it.

The flywheel decreases in speed when you apply braking to it. So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel. When the brakes are applied that is negative torque (negative voltage.)

Bonus round:

What happens when you disconnect the 10-ohm CVR from the circuit? Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage. All of the energy in the coil is burned off in the hot plasma. The current stops flowing through the coil.

What is the equivalent for the flywheel?

You clamp down on the brake calipers very fast and with maximum pressure. There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel. All of the energy in the flywheel is burned off in the hot brake pads. The flywheel stops spinning.

Bonus bonus round:

The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current. The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.

The big spike of negative torque happens because the flywheel has rotational inertia. The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.

I couldnt agree more MH with your pulsed flywheel analogy

. The relationship between the two are very similar.

One thing i need to say or would like to discus is this current flow from the inductive kickback in regards to the resistive load value-reference post 51. I am not seeing this on the scope. As i increase the value of the resistive load,the current flowing through the CVR decreases-along with the instantaneous current spike across the CVR. As i increase the load resistance, the current drop's,and the voltage rises. I am not seeing this instantaneous current flow remaining the same as the input current?.

Brad

MileHigh · « **Reply #79 on:** November 21, 2015, 05:25:44 AM »

Brad:

Sure, this is pretty easy to explain using the circuit that you just posted with modifications. I will modify the schematic.

Then just run the same test, but start increasing the value of the resistor. Say, 10 ohms, then 47 ohms, then 100 ohms, then 200 ohms. Something like that.

You can see it's very simple, when the input pulse energizes the inductor at the end of the ON time, let's assume for example that 100 milliamps are going through the coil.

So, when the pulse goes OFF, you have 100 milliamps flowing through the coil. We will ignore the current that also flows through the resistor.

If the resistor is 10 ohms, you should see an exponential decaying waveform that starts at -1 volt in amplitude.
If the resistor is 47 ohms, you should see an exponential decaying waveform that starts at -4.7 volts in amplitude.
If the resistor is 100 ohms, you should see an exponential decaying waveform that starts at -10 volts in amplitude.
Etc, etc.

The higher the value of the resistor, the faster the waveform will decay.

Ideally, your pulse train will have a long enough OFF time to let the current in the coil decay to zero.

So, you notice that this experiment will confirm that as long as you energize the coil for the same amount of time, and as long as there is no current flowing through the coil when you start the pulse, that the initial current flow through the coil is always the same when it discharges through different resistive loads.

MileHigh

MileHigh · « **Reply #80 on:** November 21, 2015, 05:47:01 AM »

And what about the flywheel? Well, if you are getting the flywheel analogy, then figuring out what will happen with your electrical circuit becomes a no-brainer when you do the same experiment in your head with a flywheel.

Recall, that current flow is equivalent to the rotational speed of the flywheel.

So, here is the equivalent experiment done on a flywheel:

You always start with the flywheel spinning at 100 RPM. Then you apply the brakes with different brake pressures. First you apply the brakes gently (10 ohms.) Then you apply the brakes with increased pressure. Then you apply the brakes one more time with even more pressure.

You observe what happens to the flywheel when you try different brake pressures. That's it.

P.S.: Obviously, the higher the brake pressure, the higher the initial negative torque (negative voltage) on the spinning flywheel.

MileHigh · « **Reply #81 on:** November 21, 2015, 05:56:15 AM »

Quote

I couldnt agree more MH with your pulsed flywheel analogy (http://overunity.com/Smileys/default/cool.gif). The relationship between the two are very similar.

The "secret" is that for all practical intents and purposes they are literally identical. They have identical sets of equations that describe their respective behaviours.

What it means is that Mother Nature decided to keep it simple. It turns out that electrical components act in the same way as physical things that we see and use in our daily lives.

tinman · « **Reply #82 on:** November 21, 2015, 06:12:54 AM »

Quote from: MileHigh on November 21, 2015, 05:25:44 AM

Brad:

Sure, this is pretty easy to explain using the circuit that you just posted with modifications. I will modify the schematic.

Then just run the same test, but start increasing the value of the resistor. Say, 10 ohms, then 47 ohms, then 100 ohms, then 200 ohms. Something like that.

You can see it's very simple, when the input pulse energizes the inductor at the end of the ON time, let's assume for example that 100 milliamps are going through the coil.

So, when the pulse goes OFF, you have 100 milliamps flowing through the coil. We will ignore the current that also flows through the resistor.

If the resistor is 10 ohms, you should see an exponential decaying waveform that starts at -1 volt in amplitude.
If the resistor is 47 ohms, you should see an exponential decaying waveform that starts at -4.7 volts in amplitude.
If the resistor is 100 ohms, you should see an exponential decaying waveform that starts at -10 volts in amplitude.
Etc, etc.

The higher the value of the resistor, the faster the waveform will decay.

Ideally, your pulse train will have a long enough OFF time to let the current in the coil decay to zero.

So, you notice that this experiment will confirm that as long as you energize the coil for the same amount of time, and as long as there is no current flowing through the coil when you start the pulse, that the initial current flow through the coil is always the same when it discharges through different resistive loads.

MileHigh

Should the circuit not be like the one below,where we can measure the current flowing through the inductor during the on time,and then measure the current flowing through the VR during the off time? This would show us what current flows through the inductor during the on time,and we can then see what current is flowing through the load during the off time,and how the different resistive loads change the flyback current delivered to those loads. This is what i was referring to,as we are looking into the flyback power delivered to various loads.

What you have depicted in your diagram is a current loop.

Brad

MileHigh · « **Reply #83 on:** November 21, 2015, 06:40:53 AM »

Brad:

You can experiment with that if you want but do you really need it to be that complicated? I think the use of the diode has merit if your signal source is your unamplified signal generator and you will be using low resistance values.

We know that the current going through the coil is going to be an increasing ramp that will eventually level out. Then, when the applied voltage to the coil goes OFF, the voltage across the variable resistor will tell you the current going through the coil provided that you measured the value of the variable resistor beforehand. Do you see what I mean? You can make an argument that the 10-ohm CVR is redundant. The variable resistor is the CVR with the caveat that you cannot see the current increasing through the coil during the energizing cycle. However, do you really need to see the increasing current through the coil while you are energizing it? If you always energize the coil with the same pulse timing, then the final current through the coil at the instant the energizing pulse goes OFF will be the same.

MileHigh

MileHigh · « **Reply #84 on:** November 21, 2015, 07:01:19 AM »

The attached graphic should make everything much clearer.

tinman · « **Reply #85 on:** November 21, 2015, 07:32:51 AM »

Quote from: MileHigh on November 21, 2015, 06:40:53 AM

Brad:

You can experiment with that if you want but do you really need it to be that complicated? I think the use of the diode has merit if your signal source is your unamplified signal generator and you will be using low resistance values.

We know that the current going through the coil is going to be an increasing ramp that will eventually level out. Then, when the applied voltage to the coil goes OFF, the voltage across the variable resistor will tell you the current going through the coil provided that you measured the value of the variable resistor beforehand. Do you see what I mean? You can make an argument that the 10-ohm CVR is redundant. The variable resistor is the CVR with the caveat that you cannot see the current increasing through the coil during the energizing cycle. However, do you really need to see the increasing current through the coil while you are energizing it? If you always energize the coil with the same pulse timing, then the final current through the coil at the instant the energizing pulse goes OFF will be the same.

MileHigh

MH
If we do it your way,then the flyback current is not a current source for a load-it is a current loop back onto itself. As we wish to measure the current as a supply source from the flyback,then that source should be dissipated across a load,in this case the VR.
The CVR will show us the current flowing into the storage unit(the inductor),and it will also show us the current of that stored energy being delivered to the load. If we are to look at the current flowing into the inductor,and then the current flowing from the inductor to a load,then the diagram you posted will not show this-it will only show the current flowing through the current loop,and not to a load. For a P/in and P/out measurement,your circuit will not show this,as there is no load,but only a loop. We want the engine to spin up the flywheel,and then for that stored energy in the flywheel to be delivered to a load-not back to the motor.

What point is there in saying the current in and out of the inductor will be the same value,when that current only loops back to where it came from?.

As i said,the current from the flyback is dependent on the load resistance. You then say it is not,and that i am wrong. But then you post a diagram showing a current loop,where the current from the flyback is sent straight back to where it came from. The current then is not a source,as it is a loop. What i was referring to is the current from the flyback being a source of current for a load,and in this case i am correct in saying that the current flowing from the inductor will be dependent on the load resistance.

Brad

woopy · « **Reply #86 on:** November 21, 2015, 10:52:41 AM »

Quote from: MileHigh on November 21, 2015, 12:06:42 AM

For an inductor, you can measure the current flowing through it, and the voltage across it.

Look at the picture of the laboratory flywheel. You can see there is a smaller disk so you can use your hands to spin the flywheel.

When you apply voltage across an inductor, the current does not flow instantly. Instead, it slowly rises.

If you have a stationary flywheel, and then you want to make it spin by applying torque to it with your hands, it does not spin at full speed right away. Instead, the rotational speed slowly rises.

This is the first thing you have to understand:

is like "pressure" on the coil.
Applying torque to the flywheel makes the flywheel spin, starting from zero. The torque is a pressure on the flywheel.

Therefore voltage is like torque.

If you can understand that, it's a good start.

Hi MH

Yes very nice analogy with the flywheel.

But i have a question. If i look at the scope shot i have posted earlier, at the end of the pulse , when the voltage stops or in your analogy when the torque stops spinning up the wheel, the current fall down instantly to zero.

So to me in your analogy it would be the same as if something bloqued instantly the wheel. What happen in this case, immediately after the stop, the really high inertia of the spinning wheel will be transformed in a huge and sudden toppling a the complete system which will jump down the table. This is the flybackspike energy.

So question is, is it possible that the "flywheel" stops instantly when the voltage stops ?

Just a supposition

Laurent

synchro1 · « **Reply #87 on:** November 21, 2015, 12:22:22 PM »

Quote from: tinman on November 21, 2015, 04:27:46 AM

Here you go Synchro-test one complete.

What do you see?
1- a 30% on time from the FG(switch closed)
2-a 70% off time(switch open)

During the 30% on time(switch closed) we can see that the voltage(yellow trace) across the inductor(supply voltage) is on the top side of the zero volt line.The current is flowing in a forward direction,as the current trace(blue trace) is also above the zero volt line.

During the 70% off time(switch open) we can now see that the voltage is inverted across the inductor/resistor series circuit-!BUT! the current is still flowing in the same direction as it was during the 30% on time,as the current trace is still above the zero volt line.
You will also notice that the current continues to flow through the load(the LED) through the complete 70% off time,and never stop's flowing right up until the next on time period. This also shows you that the magnetic field dose not collapse completely before the next on time cycle starts.

So !Bub!
Please show us where exactly the current flow reverses direction?
And please tell us how a current continues to flow(in the same direction ) during the 70% off time if the magnetic field around the inductor collapses instantaneously ?
Show us where this !new! energy is?.

@Tinman,

How do you get a 30% percent on time with a Reed switch? You are artificially starving the pulse of sufficient power to even achieve a field collapse! You need to load the coil to saturation to get a violent field collapse and current reversal, along with peak magnetic field strength for maximum rotor acceleration. Gotoluc is running a secondary coil on flyback. How is your pulse starvation approach supposed to help run his pulse motor and auxiliary Flyback power coil more efficiently? Milehigh has already exposed you as a hoaxer on the initial current fallacy.

Woopy's test was uncontrived and straight forward. You, on the other hand, have intentionally distorted the test conditions to confuse people and appear as some kind of savant. Your results are worthless. I see right through you! You're just a stinking charlatan and a retarded wanker.

synchro1 · « **Reply #88 on:** November 21, 2015, 01:25:49 PM »

@Woopyjump,

"A generator in a power plant produces electromotive force by moving magnets past coils of wire; the relay coil produces electromotive force as the collapsing magnetic field moves past the wires in the coil".

citfta · « **Reply #89 on:** November 21, 2015, 01:35:45 PM »

@synchro

KNOCK IT OFF WITH PERSONAL ATTACKS OR OUR DISCUSSION IS OVER!!

@Brad

Would you please rerun your test with a longer off time to allow the current to drop to zero?

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