Author Topic: Joule Thief 101 (Read 944418 times)

MileHigh · « **Reply #1035 on:** March 21, 2016, 03:19:21 PM »

Brad:

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Are you unaware that the base current is additive to the collector current?,and if we increase the current to the base,then we increase the current flow through the inductor in the case of the common JT circuit.

A typical standard Joule Thief has a 1K-ohm base resistance, correct? A standard Joule Thief is designed to have the transistor act as an ON-OFF switch.

That means that with a 1K base resistor that the transistor will be fully ON, and the collector-emitter voltage will be at a minimum. If you reduce the base resistance to 700 ohms, then you will increase the base current, but the transistor as a switching device will still be fully ON, with the same minimum collector-emitter voltage.

When the transistor is acting like an ON-OFF switch, when it is ON and the battery voltage is constant, in the case of a 1K-ohm base resistor, or in the case of a 700-ohm base resistor, there will be no increase in current flow through the inductor, it will be the same. The factor that is limiting the current flow is the resistance of the inductor, it has nothing to do with the amount of base current flowing into the transistor.

In addition, if the battery voltage is 1.2 volts, then the maximum current flowing through the coil will be proportional to (1.2/coil_resistance). If the battery voltage decreases to 1.0 volts, then the maximum current flowing through the coil will be proportional to (1.0/coil_resistance). Therefore, there will be a decrease in the maximum current flow through the coil when the battery voltage is lower, resulting in a decrease in the initial current flow through the LED and therefore a dimmer LED for a lower battery voltage.

MileHigh

MileHigh · « **Reply #1036 on:** March 21, 2016, 03:33:40 PM »

Brad:

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1-we can increase the current flowing to the base of the transistor, as we decrease the base resistance along with the supply voltage drop.
2-which switches the transistor on harder and longer, resulting in a pulse width increase,
3-thus maintaining a magnetic field of the same value as the battery voltage drops

1. You can increase the current flowing into the base of the transistor, but it will just be "wasted current" that does nothing because the transistor is already fully ON.

2. The transistor will not switch on "harder" because it is already fully ON. Why should it switch on longer? Why? What is the mechanism that determines the length of time that the transistor stays ON? It is certainly not "switching harder" that makes the transistor supposedly stay on longer.

3. The current flow though the inductor (a.k.a. the "magnetic field") will principally be controlled by (battery_voltage/inductor_resistance) because when the transistor is switched fully ON, there is a very low and constant voltage drop across the collector-emitter junction and that collector-emitter voltage drop will be the same if you switch the base resistance from 1K-ohm to 700 ohms.

MileHigh

tinman · « **Reply #1037 on:** March 21, 2016, 03:42:12 PM »

Quote from: MileHigh on March 21, 2016, 03:19:21 PM

Brad:

A typical standard Joule Thief has a 1K-ohm base resistance, correct? A standard Joule Thief is designed to have the transistor act as an ON-OFF switch.

That means that with a 1K base resistor that the transistor will be fully ON, and the collector-emitter voltage will be at a minimum. If you reduce the base resistance to 700 ohms, then you will increase the base current, but the transistor as a switching device will still be fully ON, with the same minimum collector-emitter voltage.

When the transistor is acting like an ON-OFF switch, when it is ON and the battery voltage is constant, in the case of a 1K-ohm base resistor, or in the case of a 700-ohm base resistor, there will be no increase in current flow through the inductor, it will be the same. The factor that is limiting the current flow is the resistance of the inductor, it has nothing to do with the amount of base current flowing into the transistor.

In addition, if the battery voltage is 1.2 volts, then the maximum current flowing through the coil will be proportional to (1.2/coil_resistance). If the battery voltage decreases to 1.0 volts, then the maximum current flowing through the coil will be proportional to (1.0/coil_resistance). Therefore, there will be a decrease in the maximum current flow through the coil when the battery voltage is lower, resulting in a decrease in the initial current flow through the LED and therefore a dimmer LED for a lower battery voltage.

MileHigh

MH
You need to have a good look at the basic JT circuit--the one you like,or call the JT circuit.
You have two coil's that provide current to build the magnetic field--not one.
As you decrease the resistance to the base,you increase the current flowing to the base,and this current is additive to the collector current due to the way the two coils are linked together,and thus the magnetic field strength can be maintained by reducing the base resistance value, so as to provide the same amount of power flowing through the LED from the kickback as the battery voltage drop's.

I will draw up a quick schematic of my test setup,and post it here. When you see the test setup,you will know that the measurements taken,and the statements i provided are correct.

Brad

tinman · « **Reply #1038 on:** March 21, 2016, 04:13:32 PM »

@MH

Below is my test setup schematic.
I have added a diode and cap to the JT circuit,so as the LED see's only a stable DC current at all time's.
The LED is a 2.6 volt LED,and i have used a 2.5k VR instead of the 1k resistor. This is needed so as i can get the voltage across the cap down to the rated 2.6 volts of the LED,when the circuit is supplied with 1.5 volt's. If i use a 1k pot,the cap will charge up to 3.1 volt's,and the LED passes out. At 1.2 volt's,we are very close to the 1k ohm resistance value,and so very close to your standard JT circuit.

I have a 10 ohm CVR,as a 1 ohm is far to noisy at these low power levels,and the 10 ohm CVR allows for very clear scope traces.

I have the LED across a 4700uf cap,and this gives us a very clear/clean voltage value across the LED.

With this setup,i can clearly show a maintained voltage across the LED as the supply voltage is reduced. This can only mean that the magnetic field strength is being maintained as the input voltage is dropped.

With the 10ohm CVR,i can also clearly show the pulse width increase-and there for the current increase, as the base resistance is reduced along with the input voltage--all while maintaining 2.6 volts across the LED, with a smooth DC current flow through the LED.

Brad

tinman · « **Reply #1039 on:** March 21, 2016, 04:28:46 PM »

Quote from: MileHigh on March 21, 2016, 03:33:40 PM

Brad:

MileHigh

Quote

1. You can increase the current flowing into the base of the transistor, but it will just be "wasted current" that does nothing because the transistor is already fully ON.

MH
1.2v/1kohms is 1.2mA. This is not enough current to fully switch on the transistor,as the transistor is a current device,unlike the FET which requires very little current,but a higher voltage.

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2. The transistor will not switch on "harder" because it is already fully ON.

It will not be fully on at a low base current of 1.2mA.

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Why should it switch on longer? Why? What is the mechanism that determines the length of time that the transistor stays ON? It is certainly not "switching harder" that makes the transistor supposedly stay on longer.

The reason it stays on longer, is because the transistor is switching on harder,in that the required switching current to the base has now increased.

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3. The current flow though the inductor (a.k.a. the "magnetic field") will principally be controlled by (battery_voltage/inductor_resistance) because when the transistor is switched fully ON, there is a very low and constant voltage drop across the collector-emitter junction and that collector-emitter voltage drop will be the same if you switch the base resistance from 1K-ohm to 700 ohms.

As i stated,the lower the supply voltage get's,the less base current there is available to switch the transistor fully on. There is also the fact that the base current is additive to the collector current in this situation. So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

Brad

seychelles · « **Reply #1040 on:** March 21, 2016, 04:53:51 PM »

this is just one transistor circuit what the fuck..you guys should deal in uhf circuitry running at 450 mhz.. you guys make me fucking sick of your petty childish fucking arguments..

TinselKoala · « **Reply #1041 on:** March 21, 2016, 08:08:22 PM »

Quote from: tinman on March 21, 2016, 04:13:32 PM

@MH

Below is my test setup schematic.
I have added a diode and cap to the JT circuit,so as the LED see's only a stable DC current at all time's.
The LED is a 2.6 volt LED,and i have used a 2.5k VR instead of the 1k resistor. This is needed so as i can get the voltage across the cap down to the rated 2.6 volts of the LED,when the circuit is supplied with 1.5 volt's. If i use a 1k pot,the cap will charge up to 3.1 volt's,and the LED passes out. At 1.2 volt's,we are very close to the 1k ohm resistance value,and so very close to your standard JT circuit.

I have a 10 ohm CVR,as a 1 ohm is far to noisy at these low power levels,and the 10 ohm CVR allows for very clear scope traces.

I have the LED across a 4700uf cap,and this gives us a very clear/clean voltage value across the LED.

With this setup,i can clearly show a maintained voltage across the LED as the supply voltage is reduced. This can only mean that the magnetic field strength is being maintained as the input voltage is dropped.

With the 10ohm CVR,i can also clearly show the pulse width increase-and there for the current increase, as the base resistance is reduced along with the input voltage--all while maintaining 2.6 volts across the LED, with a smooth DC current flow through the LED.

Brad

Have you thought about using a suitable PNP transistor, with the proper feedback to its base, in place of the variable resistor?

MileHigh · « **Reply #1042 on:** March 21, 2016, 08:43:17 PM »

Brad:

Oops, I notice a much more subdued Brad and much less trash talk after my two postings.

Pre my two postings:

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MH
It is painfully obvious that you really do not know how a transistor operates

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I am starting to think some brainless twat has hacked your account

Now, your words come back to haunt you. You don't really know how a transistor works. You also don't know how to analyze how a transistor works when doing a basic switching function like you see in a Joule Thief. Think about that. You have been doing this stuff for six years, and all this time you have never reviewed basic transistor switching circuits to understand them and use them properly. You don't know how to determine the proper base resistor for a given switching function. And you clearly have some serious gaps in your understanding of how a transistor works.

You have heard me complain about the "continuous affirmation" environment that you guys set up for yourselves, and the "Straitjacket of Agreement" where you are all paralyzed and can only agree with each other like a bunch of bobbing rubber ducks in a pond.

And look at the results. Combine the "continuous affirmation" and the "Straitjacket of Agreement" and the bobbing rubber duckies and your "I am Brad and I am never wrong" and "I am Brad and I take the lazy route when I can" attitudes and here you are six years later and you can't properly analyze a basic switching function in a five-component circuit like a Joule Thief, nor do you truly understand how a Joule Thief works..

Instead, you play this ridiculous trash talk game and you are as fake-ass as a three-dollar bill when you assume the role of a bad actor and play your ridiculous "MileHigh you are making mistakes everywhere" game.

Now that that has been said, I will comment on the technical in my next posting. However, I am not going to spoon feed you anything. You can go online or order a few books about electronics from Amazon and undertake to educate yourself.

MileHigh

EMJunkie · « **Reply #1043 on:** March 21, 2016, 09:10:23 PM »

Quote from: seychelles on March 21, 2016, 04:53:51 PM

this is just one transistor circuit what the fuck..you guys should deal in uhf circuitry running at 450 mhz.. you guys make me fucking sick of your petty childish fucking arguments..

Nicely said Seychells!!!

MileHigh is wrong again - No one takes him seriously anymore anyway!!!

Chris Sykes
hyiq.org

MileHigh · « **Reply #1044 on:** March 21, 2016, 09:20:51 PM »

Brad:

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You have two coil's that provide current to build the magnetic field--not one.

Take a look at the attached schematic for a standard Joule Thief. L1 is the power coil that gets energized and then illuminates the LED. L2 is the feedback coil and is essentially an EMF source to switch the transistor ON and OFF via the base resistor. Your statement above is nonsensical. L2 does not provide any energy towards the lighting of the LED. Only one coil, L1, "builds up a magnetic field" with energy that gets discharged into the LED.

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As you decrease the resistance to the base,you increase the current flowing to the base,and this current is additive to the collector current due to the way the two coils are linked together,and thus the magnetic field strength can be maintained by reducing the base resistance value, so as to provide the same amount of power flowing through the LED from the kickback as the battery voltage drop's.

As you increase the current flowing into the base input of the transistor because of a decreased value of base resistor, that represents expending more energy to switch on the transistor than you have to. If you put more current than you need to though the L2 coil, than that means that the battery can supply less current to the L1 coil (when factoring in a higher battery output impedance for a nearly dead battery), and that translates into less energy available to light the LED. An excessively low value of base resistor just makes the Joule Thief less efficient. If you ignore the issue of the battery output impedance then you are just siphoning extra energy away from the battery to switch on the transistor. That extra energy could be put to better use by saving it to light the LED via L1.

I have already stated that when the battery voltage has dropped, you can't escape the simple V/R limiting factor for the amount of current that can be induced to flow through L1, and that means less current to light the LED. Even if somehow the transistor stays on longer, the V/R current limiting factor is what really counts and the LED will be dimmer.

With respect to your setup, as far as I am concerned it deviates too far away from a standard Joule Thief to extract any useful information about the behaviour of a standard Joule Thief. Your 10-ohm CVR may be introducing a voltage bounce to the whole circuit that throws off the feedback. The filtering capacitor is not needed and also is probably throwing off the feedback. If you know what you are doing, just probing a standard Joule Thief and trying varying the base resistor should be enough.

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1.2v/1kohms is 1.2mA. This is not enough current to fully switch on the transistor,as the transistor is a current device,unlike the FET which requires very little current,but a higher voltage.

One kohm is in the standard Joule Thief schematic and I will take their word that it is a correct value based on the indicated recipe for winding the Joule Thief transformer and the associated resistance of L1. You completely forgot the base-emitter voltage drop in your comment. Saying "XX current is not enough current to fully switch on the transistor" is just you revealing that you don't understand the issues around how a transistor switching circuit works like I already stated. Perhaps go to Amazon and do some online shopping.

MileHigh

MileHigh · « **Reply #1045 on:** March 21, 2016, 09:44:23 PM »

Brad:

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The reason it stays on longer, is because the transistor is switching on harder,in that the required switching current to the base has now increased.

This was a covered in detail earlier on in this thread and apparently none of it registered with you. The switching timing is due to the positive feedback. Your comment above is wrong, and it shows that you don't truly understand how a Joule Thief works. I hate to say it and it probably infuriates you but it is the truth. What you need to do is take those lemons, educate yourself, and turn them into lemonaid.

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As i stated,the lower the supply voltage get's,the less base current there is available to switch the transistor fully on. There is also the fact that the base current is additive to the collector current in this situation. So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

The Joule Thief is designed such that as the battery voltage drops, the value of the base resistor is chosen such that the transistor is still switched fully on - within certain limits of supply voltage. As we know, the whole switching mechanism falls apart blow a certain voltage. The base current is added to the emitter current, not the collector current. Assuming that the transistor is fully switched on then the resistance of the L1 coil is what determines the limiting factor for how much current passes through L1.

I will single this one out:

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So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

No, no, no, no, and no. This is just you blindly believing that "more base current equals more inductor current." You are completely ignoring the V/R current limiting factor assuming that the transistor is fully switched ON. This is basic basic stuff and I have covered this point a few times in these postings. You need to take a step back and really think about this stuff.

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Do you need an answer to understand the need for a variable base resistor MH?-or will your batteries simply remain at the rated voltage of 1.5 volt's?.
That was a bit of a silly statement by your self MH.

I think that enough has been said to make the point that your statement quoted above is wrong. It's just a bunch of fake swagger and you not truly understanding all of the switching issues around a Joule Thief circuit. A standard Joule Thief circuit is designed such that a conscious decision is made for the value of the base resistor. There is no point in lowering the value of the base resistor beyond a certain point. There is a relationship between the large-signal gain of the transistor, the resistance of the L1 coil, and the EMF that L2 presents to the base resistor that allows the Joule Thief designer to make a conscious decision for the value of the base resistor and clearly you are not aware of these issues. Hence I strongly advise you to get a mastery of basic transistor switching circuits and understand what "fully ON" really means.

I think it is time to move on, and if you are a keener, then you have some homework to do.

MileHigh

tinman · « **Reply #1046 on:** March 22, 2016, 12:34:04 AM »

author=MileHigh link=topic=8341.msg477969#msg477969 date=1458593063]

MileHigh

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I hate to say it and it probably infuriates you but it is the truth. What you need to do is take those lemons, educate yourself, and turn them into lemonaid.

No MH,the below is what is troubling.

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The Joule Thief is designed such that as the battery voltage drops, the value of the base resistor is chosen such that the transistor is still switched fully on

It is hard to believe,but you are having arguments with your self.
One minute you say that lowering the base resistance will not change a thing as the battery voltage drop's,and in the next breath you are saying that the base resistor has to be chosen in accordance with battery voltage value

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No, no, no, no, and no. This is just you blindly believing that "more base current equals more inductor current." You are completely ignoring the V/R current limiting factor assuming that the transistor is fully switched ON.

More base current dose equal more inductor current,as there are two conductors/coil's wrapped around the core-not one. You are also forgetting resistive losses,and those losses are reduced when the base resistance is reduced.
Your assumption that the transistor is fully switched on,is your downfall on your V/R limit argument.

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With respect to your setup, as far as I am concerned it deviates too far away from a standard Joule Thief to extract any useful information about the behaviour of a standard Joule Thief. Your 10-ohm CVR may be introducing a voltage bounce to the whole circuit that throws off the feedback. The filtering capacitor is not needed and also is probably throwing off the feedback. If you know what you are doing, just probing a standard Joule Thief and trying varying the base resistor should be enough.

The setup was designed to show exacting result's in the clearest way,and the same results are had using the circuit without the cap and CVR.

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This was a covered in detail earlier on in this thread and apparently none of it registered with you. The switching timing is due to the positive feedback. Your comment above is wrong, and it shows that you don't truly understand how a Joule Thief works.

I think you are the one that dose not understand the effects of the positive feedback MH. The lower the resistance in the feedback coil,the higher the current flow through the feedback coil,and the higher the current flow through the feedback coil,the stronger the magnetic field built by that feedback coil-->and we know what than means for the current flowing through the drive coil.

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within certain limits of supply voltage. As we know, the whole switching mechanism falls apart blow a certain voltage.

Yes we do. With the 1kohm resistor in place,the circuit shutdown voltage would be around!lets say! 600mV. If we replace that 1kohm resistor with say a 100ohm resistor,then the circuit shutdown voltage would be lower-say 450mV. The reason for this is we now have less voltage drop across the base resistor,and thus the transistor can still switch on at lower voltages.

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The base current is added to the emitter current, not the collector current.

When the transistor is switched fully on,the collector and emitter are one-the switch is closed,so either is correct.

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Assuming that the transistor is fully switched on then the resistance of the L1 coil is what determines the limiting factor for how much current passes through L1.

And that is where you reach the limit of your knowledge--in the assumption that the transistor is fully switched on--which is incorrect when the battery voltage reaches a certain limit-->and hence the need for a drop in resistance of the base resistor--so as the transistor can switch fully on--switch on hard.

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This is basic basic stuff and I have covered this point a few times in these postings. You need to take a step back and really think about this stuff.

MH
You really need to get back onto the bench,and teach yourself the difference between facts and fiction.

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I think that enough has been said to make the point that your statement quoted above is wrong. It's just a bunch of fake swagger and you not truly understanding all of the switching issues around a Joule Thief circuit. A standard Joule Thief circuit is designed such that a conscious decision is made for the value of the base resistor.

This comment is idiotic.
That resistor value changes as the supply voltage drop's,and so the reasoning behind a VR on the base. I find it quite comical that you dont understand voltage drops across resistor's,and how the relates to the switching of the transistor in reference to supply voltage.
If you have a set 1k base resistance,as the supply voltage drop's,so too will the available current and voltage required to switch on the transistor fully. The voltage is not really a problem due to the positive feedback,but enough must be there to start to switch on the transistor to start with before the positive feedback can switch the transistor on hard.

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There is no point in lowering the value of the base resistor beyond a certain point. There is a relationship between the large-signal gain of the transistor, the resistance of the L1 coil, and the EMF that L2 presents to the base resistor that allows the Joule Thief designer to make a conscious decision for the value of the base resistor and clearly you are not aware of these issues. Hence I strongly advise you to get a mastery of basic transistor switching circuits and understand what "fully ON" really means.

The base resistor value cannot be a common value throughout the supply voltage range.
You also fail to take into account that the L1's resistance will increase with frequency,and that frequency increases as the supply voltage drop's. This is where your V/R limit also falls apart-->this is another factor you have failed to take into account,and one that makes every thing i claimed to be correct.

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I think it is time to move on

No--i think it is time that you were taught the truth,so as you discontinue to peddle rubbish,due to your lack of understanding of a circuit that you think you know all about.

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, and if you are a keener, then you have some homework to do.

It is you that needs to do some homework MH.

You are truly lost MH.

Brad

Magluvin · « **Reply #1047 on:** March 22, 2016, 02:00:28 AM »

While looking through some JT circuits I found the one below to be interesting. It uses resistors from the pos and neg of the battery.
Going to try it tonight. Having the coil between the base and the voltage dividing resistors seems like it will give greater control while not having to increase the resistance too high. Will see.

Mags

Pirate88179 · « **Reply #1048 on:** March 22, 2016, 02:18:10 AM »

Back in the old JT topic area, we learned early on that using a base vr was beneficial in several ways. Yes, it could help maintain the brightness of the led as the battery voltage dropped but, as the battery "died" down to around .4 volts, this would cause the frequency of the circuit, which had previously been high enough that the human eye could not see the on/off switching of the led, to dip low enough that the led would appear to flash on/off. A little tweak of the base vr and...Bob's your Uncle...the led would now once again appear to be on constantly.

So, I do know from experience that it is useful to use a vr on the base rather than choosing a fixed resistor that is a poor compromise over the entire range of the battery voltage. There is no single fixed resistance that can give you the longevity of non-flashing, bright light from the led across this range.

Just my 2 cents from having built many of these circuits over the years. Once you get to where the output from your AA battery JT is over 300 volts, other things become more important to consider as well. (Like not getting zapped!) As I mentioned early on here, it all depends upon your goal...brightest light possible or longevity of the light from your "dead" battery.

Bill

Magluvin · « **Reply #1049 on:** March 22, 2016, 03:29:55 AM »

Also on resonance. Will make a thread on this in the resonance board.

I had made this base a while back. Work started consuming my time and it sat on top of my tv.

Anyway, the base is 3/4in particle board with 1/8in black plexy, the aluminum base pieces and a stainless strip. The 2 mags on top are 1/2 by 3/8 N52. The diametric mag in the drill is also N52. I will get into showing things in a day or so. But here is a pic below.

One thing I can say is, it seems the best time to take output from the device while its in resonance it when the wave is near peak. And just take enough not to severely disturb it oscillation. So what I did was set the mag on the drill about 3in away like where it sits in the pic and adjust the trigger speed for close to max resonance, then I held the screwdriver with the butt end toward the mag on that side of the oscillation. If I could hold the cam, drill and screw driver, Id be doin a vid. need to make a setup.

But with the screwdriver close enough for the bag to hit near peak, it gets 'hammered' pretty good. And the closer I move the screwdriver in toward tdc, the oscillation practically ceases far before getting to tdc. So. Something learned on the bench,

and this should translate to the electronic version.

So we can take hard hits of output near peak, and not destroy the oscillation of the resonance.

Now. If I lower or raise the speed on the drill, there are some movements of the wiggler, but nothing near what we have at resonance. So it seems, if we are not working with resonance, we are stuck right here where the books tell us we are suppose to be.

Mags

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