Hi,

If I understand you right, you are going to charge up an inductor of 1.75mH with a DC source that takes the current up to 6 amps, then discharge it into a 10 Ohm resistor.
For this the overall circuit function changes from E=L*di/dt because there is also resistance in the circuit, although that is still how the inductor itself behaves.

The formula for the circuit would be:
iL=-i0*e^(-t*R/L)
where
iL is the decreasing current through R over time,
i0 is the initial inductor current (6 amps),
R is the resistor value (10),
L is the inductor value (1.75mH),
t is time in seconds.
You can ignore the minus sign and use:
iL=i0*e^(-t*R/L)

Using the values you provided this comes out to:
iL=6*e^(-5714.285714285715*t)
What this tells us is that after one time constant (175us) the current will be down to about 2.2 amps, and after 5 time constants (875us) the current will be less than 0.05 amps.

It's also interesting that the voltage will start out at 6*10=60 volts and gradually decrease over time in a similar manner to how the current decreases. Ohms Law: E=i*R, with 'i' decreasing over time.

The current (and also the voltage across the resistor) will be almost zero in less than 1 millisecond, is this something you want to happen?

 

No...... Sounds like the inductor L is too small, and input power/current should be higher for duration of load to about 10 milliseconds.

 

I visited post #1 again and now I am wondering if the purpose is to check that "meter"?
" I have neon bulbs type variable voltmeter 110 to 470v. Is there a simple diagram to test this as actually flashing? E = L X di /dt ". to verify such a meter usually requires a second meter of known accuracy and better resolution. If that is the case I suggest that the TS construct a voltage doubler circuit in accordance with high voltage safety standards, to check at voltages up to at least 350. Extreme caution with such a circuit is mandatory as those voltages can be fatal in an instant. To check the highest voltage will require a voltage tripler circuit, producing over 500 volts DC. These assembled circuits should then be taken apart after the neon voltmeter circuit is tested.
Caution must be taken when suing such high voltages, the "HAZARD OF DEATH" is real.

 

I visited post #1 again and now I am wondering if the purpose is to check that "meter"?
" I have neon bulbs type variable voltmeter 110 to 470v. Is there a simple diagram to test this as actually flashing? E = L X di /dt ". to verify such a meter usually requires a second meter of known accuracy and better resolution. If that is the case I suggest that the TS construct a voltage doubler circuit in accordance with high voltage safety standards, to check at voltages up to at least 350. Extreme caution with such a circuit is mandatory as those voltages can be fatal in an instant. To check the highest voltage will require a voltage tripler circuit, producing over 500 volts DC. These assembled circuits should then be taken apart after the neon voltmeter circuit is tested.
Caution must be taken when suing such high voltages, the "HAZARD OF DEATH" is real.

The purpose was to test an inductor for high voltage by test for Back EMF. A neon bulb would flash and go off. I never saw a circuit like that work, and I need 100-150 volts from a coil. I have a neons bulbs meter with several neons at differing levels of voltage. This newer on the market gauge meter is similar to the very higher KV meters that farmers use on electric fence tests.

 

Using 6 amps and 35 volts and a series resistor is not a very efficient way to produce a spark. Besides that the kickback voltage is across the same coil that is connected to that supply.
A hand-wound spark coil can do a better job with less input power, and the added advantage of being isolated from the power source. And probably it will have a better form factor as well. But now the question is about that 100 to 150 volts from a coil that you seek. There are even rather less expensive transformers around that can do the job with a lower current. Is the goal to flash a neon tube?? or ??? Probably somebody who reads these posts has alreadydone it and can share the information.

 

I am still not clear as to why the power source must be 35 volts at 6 amps and why the load is a 10 ohm resistor. If you need a 100 volts across a 10 ohm resistor then the current through it must be 10 amps. (That is 1000 watts) To do that the initial current through the inductor must be 10 amps.
This is because the inductor tries to maintain the current that was passing through it before the power source is removed. I think using an inverter to step up the 35 volts to 100 volts and use that to charge a capacitor. The capacitor could be conneted to the 10 ohm resistor via an SCR. When the SCR was triggered it would discharge through the 10 ohm resistor. The duration of the pulse would depend on the value of the capacitor. As you have not specified the required pulse duration SCRs on not very fast so if you require a pulse of less than a few micrseconds this suggestion will not work.

Les.

 

I see now that there is another thread addressing the same issue from a different point of view..

 

No...... Sounds like the inductor L is too small, and input power/current should be higher for duration of load to about 10 milliseconds.

Hello again,


Well, remember that quoted 1 millisecond is when the voltage is very close to zero, so that means it does not exactly last for 1 millisecond. In fact, using an inductor of 2mH, after around 138 microseconds the current will be down to around 1/2 of the initial current. That means it will be only around 3 amps after 138 microseconds.

Now going to an inductor of 10 times larger (20mH) means the times will increase by 10 times. That means the current will be down to about 3 amps after about 1.38 milliseconds.

I am not sure this is what you want either. Remember also that with just 6 amps and a 10 Ohm load, the maximum voltage will be just 60 volts. That does not sound high enough for your testing needs.
If you need twice that, you would need to increase the current to 12 amps, and if 3 times that, then 18 amps. That's the initial current though, and that means after a short time the current will be much less.

What you will need to do is specify the current or voltage you need and also the duration of that pulse. It may mean you would have to do this in a different way.

 

We could also choose to call the spike generated when the current is halted an "Inductive spike", and choose to not argue about easy stuff any more.

AND, I amazed looking at a catalog of clothing used to protect individuals from "EMF", which in that publication is used to describe those deadly "Electro-Magnetic-Fields" that are poisoning us constantly. That whole fraudulent business is a testimony as to why we all should have a lot more science education in the very early grades of school.

Hi,

Language is a continuously evolving way of communicating. The words or phrases that are accepted or rejected over time always amaze me, and then we have political correctness on top of all that.
Remember how long it took for the word "ain't" to be 'officially' accepted into the English language (and is it still yet fully accepted). Usage precedes acceptance sometimes by a long time. I guess it is the "lexicographers" who introduce new words into the dictionary. I wonder how long they argued for and against "ain't"

One thing I can say though is that when we talk about motors, we 'almost' exclusively use "back emf". That's also not a hard and fast rule though as I've read otherwise.

 

OK, MrAL.
Now in this post, and the other related one that was just created recently, there is a whole lot that is far from clear as to just exactly what the ultimate goal is. And my mind-reading abilities have become even poorer recently, so even guessing at the ultimate desired results does not yield anything that makes sense.
Any comparison to an ignition system high voltage pulse especially.

 

The basic concept of the TS query is flawed. A 35v supply with a 1.75mH inductor and a 10ohm series resistor is 3.5A. While the inductor open-circuit may produce 1000's of volts, connected to a 10ohm resistor the resulting back EMF is 35v and current is 3.5A.

However, the TS then said:

The purpose was to test an inductor for high voltage by test for Back EMF. A neon bulb would flash and go off. I never saw a circuit like that work, and I need 100-150 volts from a coil. I have a neons bulbs meter with several neons at differing levels of voltage. This newer on the market gauge meter is similar to the very higher KV meters that farmers use on electric fence tests.

A circuit configuration that might suit is shown below. The max inductor current is set by the width of the on pulse, here 310uS gives 6A (R1 is the inductor internal resistance and helps define the rate at which the current builds). When the switch opens, the inductor discharges through the diode and load resistor R2, with an equivalent current. R2 therefore defines the back emf voltage across the inductor. 10ohms gives 60v, 100ohms shown here gives 600v. But a 10mS wide pulse needs a completely different approach.

 

"The purpose was to test an inductor for high voltage by test for Back EMF" That does not make sense to me. What sort of high voltage test? Insulation breakdown?? internal arcing" continuity? Under some driving conditions some inductance s will deliver a kickback spike that will flash a neon bulb. But here is an interesting fact: The breakdown voltage in a neon bulb will drop a fair amount if you shine a light on it. Just a bright flashlight can drop the trigger voltage several volts. So a neon bulb can be a photo sensor. So the test could be a pass or a fail depending on the ambient light level. And the sensitivity seems to increase as the neon bulbs age.

 

The basic concept of the TS query is flawed. A 35v supply with a 1.75mH inductor and a 10ohm series resistor is 3.5A. While the inductor open-circuit may produce 1000's of volts, connected to a 10ohm resistor the resulting back EMF is 35v and current is 3.5A.

However, the TS then said:


A circuit configuration that might suit is shown below. The max inductor current is set by the width of the on pulse, here 310uS gives 6A (R1 is the inductor internal resistance and helps define the rate at which the current builds). When the switch opens, the inductor discharges through the diode and load resistor R2, with an equivalent current. R2 therefore defines the back emf voltage across the inductor. 10ohms gives 60v, 100ohms shown here gives 600v. But a 10mS wide pulse needs a completely different approach.

View attachment 321610

Hi,

Not sure what you are getting at with the first part of your reply, but the second part makes sense except I would just assume that the power source has a 6 amp current limit. That means, in theory, you could leave it connected for 10 minutes and the max current would be 6 amps. Of course power dissipation becomes an issue too, but that's the simplest way to think about it I think.
Once the circuit is opened, the inductor voltage of course jumps up because the inductor is still trying to force 6 amps to pass through it. For that microsecond or so, 6 amps will flow, then gradually decrease exponentially. Unfortunately it decreases so fast with a small value inductor that I don't see any use for that, and the TS replied that they probably need a larger inductor. 0.002H is too small, 0.020H is probably also too small, 0.200H might be getting close.

 

"The purpose was to test an inductor for high voltage by test for Back EMF" That does not make sense to me. What sort of high voltage test? Insulation breakdown?? internal arcing" continuity? Under some driving conditions some inductance s will deliver a kickback spike that will flash a neon bulb. But here is an interesting fact: The breakdown voltage in a neon bulb will drop a fair amount if you shine a light on it. Just a bright flashlight can drop the trigger voltage several volts. So a neon bulb can be a photo sensor. So the test could be a pass or a fail depending on the ambient light level. And the sensitivity seems to increase as the neon bulbs age.

Hi,

I just took it in the simplest possible way and then explained what would happen to see if the TS liked the way it was going to behave. So I did not have to know exactly what they were doing, just what the circuit that was described was going to do. After presenting some evidence as to what would happen, the TS decided that they probably need a larger value inductor.

The only other thing I would worry about is if the inductor is not connected correctly it could produce a VERY high voltage and at even 1 amp that could be very dangerous. There has to be some assurance that the load will never be disconnected or blow open for any reason like overheating.

 

The Inductor should never be allowed to create a discharge with no Load.
The Voltage could go so high that it may blow a hole right through the Insulation of the Windings.

If this happens, the Inductor is now an expensive Paper-Weight with no purpose in Life.

Automotive-Ignition-Coils are commonly ruined in the same manner.
.
.
.

 

The Inductor should never be allowed to create a discharge with no Load.
The Voltage could go so high that it may blow a hole right through the Insulation of the Windings.

If this happens, the Inductor is now an expensive Paper-Weight with no purpose in Life.

Automotive-Ignition-Coils are commonly ruined in the same manner.
.
.
.

What brand of automotive coils are ruined by having the secondary open circuited??? I am aware that some brands do use inferior parts to reduce costs.
I have had that happen several times over the years, with various cars, that a plug wire came off, either accidentally or as the result of not having a locking hood. And never suffered a damaged ignition coil.

 

Sometimes its hard to determine whether or not this damage has actually occurred,
and it's always been on a poorly maintained Ignition-System that had signs of Carbon-Tracking
on the exterior of the Coil-Tower, or a damaged Distributor-Rotor, or
a disconnected Spark-Plug-Wire that's just flopping around loose.

The symptoms are high-rpm, high-load, mis-fires, or occasional back-firing during cranking,
even after replacing all other Ignition-Components, Spark-Plug-Wires, and Spark-Plugs.

The only "proof" is that replacing the Ignition-Coil eliminated the un-explainable problems,
and that the engine would run perfectly fine under light-load conditions where the
peak-Ignition-Voltages were much lower.

The required Ignition-Coil-Voltage increases with increased
Cylinder-Pressure, or increased Spark-Plug-Gaps,
or any other situation where the distance between the Coil-Output,
and the closest nearby Ground, is greatly increased,
causing the peak-Coil-Voltage to go sky-high.
.
.
.

 

How much back EMF volts can I get from positive 35v DC on+ off, at 6 amps DC , to inductor 1.75 millihenry to 10 ohms resistance to ground? I need = > 100volt over the resistance. Fast type diodes inverse parallel. The Formula depends on speed of a switch type as solid state transistor, (from it's data sheet part number), and amount of inductance. I have neon bulbs type variable voltmeter 110 to 470v. Is there a simple diagram to test this as actually flashing? E = L X di /dt
Or maybe the main line has resistor only, and charges coil L on the side, same time. Then coil L reverses back and sends high volts through resistor to ground in same original direction as positive power supply.

I think you are describing a boost converter. When interrupting 6 amps through an inductor the inductor continues that 6 amps. In the case of 6 amps and a 10 ohm load the voltage developed across the resistor is 6A x 10V = 60 volts.

 

One more thought is that a spark-ignition coil is a two winding transformer with the design goal being to produce an adequate spark intensity to ignite the fuel/air mix even when cranking with a feeble battery. So not really a fair target for spike reduction.