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Artikbot

Thu Sept 30 2010, 08:49PM

Registered Member #3247 Joined: Mon Sept 27 2010, 09:42AM
Location: Spain
Posts: 137

Okay. So hands down to that parasitic resistance, my MOSFET no longer blows up!

Now it is my capacitor (obvious) because there isn't any current limiting part, so ti overcharges and BAM.

I'm trying to connect the OpAmp there, but
1: Yenka doesn't have potentiometers so I'm a bit screwed. I've calculated 3MOhm resistance for the OpAmp to receive 3V from the capacitor and 10k from the voltage source to receive 3V to use as reference, but somehow it doesn't work and it lets voltage grow up on the capacitor, blowing it up.

Apart from the opamp issue, everything is working fine now ^^

Side note: Is it normal that my IC works with negative current? It works anyway O.o

Hoi, I feel this is starting to light up :)

Thanks a lot!

DerAlbi

Thu Sept 30 2010, 11:08PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

ScotchTapeLord, i see the misunderstanding now

Of course a capacitor in seieres to the inductor would not work for the boost topology..
The cap-idea was for my 4 transitor-circuit. There its is real AC, and then they would limit DC current if the circuit fails. In boost ciruict current limiting would be stupid, since the inductur needs the current to build up the enrgy that is released within the other 3% of time.
Maybe my English is worse than i thought.

Artikbot

Thu Sept 30 2010, 11:37PM

Registered Member #3247 Joined: Mon Sept 27 2010, 09:42AM
Location: Spain
Posts: 137

Oh wait I've deduced something.

If the inductor builds up (!) current for the 97% of the time and then releases it in a 3%... the current output is going to be INSANE! Maybe reducing the duty cycle also reduces the current output? That would be good...

Increasing the input voltage will do the trick I guess. But without exceeding the 555 voltage treshold... Zeners there we go :P

But for now, I'm focusing on fixing the OpAmp issue.

ScotchTapeLord

Fri Oct 01 2010, 12:14AM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
Location:
Posts: 635

DerAlbi:

It had nothing to do with English. I just made an inaccurate assumption!

Artikbot:

You're starting to understand the topology a little better, but you need to think of the energy stored as Joules of energy, not Amps of current. The energy stored is a function of the voltage, current, and time. It is also released as a function of voltage, current, and time. With the Energy in Joules fixed, lowering the time increases the power, as in both voltage and current, not just current. This is why you get 300V out with only 9V in!

Also, the output does not flow through the MOSFET, since it is discharged when the MOSFET turns off, but flows only through the diode and the capacitors you are charging.

If you lower the duty cycle you will decrease the charging current but also the output voltage, so it's not a good thing. For charging capacitors quickly, high currents are good. Remember, however, they are still limited by many factors that aren't accounted for in your simulator.

Please look this up: V = L*di/dt
It means the voltage output is based on the product of the inductance and the rate of change of current. We achieve a high rate of change by raising the current as high as we can, and then shutting it off as quickly as we can!

GhostNull

Fri Oct 01 2010, 01:22AM

Registered Member #2648 Joined: Sun Jan 24 2010, 12:45PM
Location: Australia
Posts: 291

Really great work Artikbot! It's wonderful to see you improving and learning.

So your current problems are:
Op Amp
High Current Output

I'm not really sure about this stuff but I'll tell you what I think (so don't take this as rock solid information =S)

Could you show us how you wired up the op amp?
Op amps are supposed to output negative current when inverting input is high, I'm not so sure though. Yenka does have a Potentiometer, try poking around a bit more.

I don't really understand what you mean by a high current pulse
In a boost converter the energy in the inductor is being transformed from high current, low voltage to higher voltage, lower current. So you won't get high voltage high current if that is what you are thinking. You can't get Low voltage, High current and turn it into higher voltage, high current. The energy has to come from some were.
P = V * I
Where:
P = Power in watts
V = electro motive force in volts
I = current in Amps

So for example if you put in 12V at 1 Amp you get:
P = 12 * 1
P = 12watts

You can't get 300V 1 Amp output
P = 300 * 1
P = 300watts

That would be impossible since you can't put in 12 watts and get 300watts out.
Probably not the best explanation but my point is that you can't get more energy out than what you put in. If you put in low voltage and want high voltage, you are going to sacrifice some current to transform it into the higher voltage.

But if there is high current, it will be a short pulse so consider the pulse rating of you components, not the average or RMS. For example, i'm sure you found with your SCR, a 300amp SCR may have a surge rating of 5000 amps for 10mS but that is not to say that all SCRs or components have a great surge rating.

Also consider the inductive reactantance of the inductor -> limiting current:

I don't have the time right now to explain it but if you need me to just ask but it's best you do your own learning =D

ScotchTapeLord

Fri Oct 01 2010, 03:47AM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
Location:
Posts: 635

I may have been misleading when I said there would be rise in both current and voltage...

There isn't much of a rise in voltage until the capacitor is charged to a decently high voltage. The charging current begins high and gets lower as the capacitor fills. The voltage is the inverse of that, low to high. The average current will, of course, not exceed the input current. I was referring to the individual pulse, where the values can reach high peaks.

Artikbot

Fri Oct 01 2010, 07:42AM

Registered Member #3247 Joined: Mon Sept 27 2010, 09:42AM
Location: Spain
Posts: 137

@ScotchTapeLord: Just like happens with P-channel MOSFET? So essentially the stupid Yenka is exaggerating with the current flow. And this time I got it, I need to use an N-channel because electrons transfer energy WAY faster than holes :)

Woah I failed at thinking stuff about joules. I was only thinking about Amps... but if supply can deliver 9V with no amp limiting, happens the problem I have I guess :P

@GhostNull:
Okay, I'll try limiting the current input to 1A and see what happens. Otherwise yenka is (oh surprise) screwing me.

Ye, I knew that input/output stuff, ty anyways :)

The OpAmp is wred this way:

RST pin----< (+) pin goes to +VDC with a 10kOhm resistor and (-) pin goes to capacitor terminals, wired this way:

(+)3MOhm resistor(to OpAmp)3MOhm resistor(-)

Resistors are calculated (at least I did it that way) for the OpAmp to receive both 3V signal from supply and from fully charged caps.

I can't post a pic because Yenka says I'm not connected to Internet (LOL to that).

Anyways, I'llkeep trying during the day and see if it wants to connect.

Oh wait! Negative voltage thru the 555 timer... OpAmp is wired not only to pin 4, but also to pin 8 (because they are both interconnected). That essentially means that pin 8 is both fed with positive voltage from the supply AND from the OpAmp signal? It sounds to me like it can't be any good, huh.

Thanks a lot!

GhostNull

Fri Oct 01 2010, 10:29AM

Registered Member #2648 Joined: Sun Jan 24 2010, 12:45PM
Location: Australia
Posts: 291

Your wiring explanation is a bit hard to understand. Try taking a screenshot instead.

When I said that the inductor limits the current that's not what I meant =S

Inductive reactance:
When the current through an inductor increases an electromagnetic field builds up. The creation of an electro magnetic field impedes the flow of current, taking up energy; similar but NOT the same way a resistor resists current flow. When current through an inductor decreases the electromagnetic field collapses, the collapsing field induces current in the inductor. This phenomenon is known as inductance. In a DC circuit this does not happen because there is not change in current. The greater the electromagnetic field produced by the inductor the greater this effect (inductance) is. The faster the current through an inductor changes (frequency) the greater the impedance. A boost converter relies on the phenomenon of inductance to create a higher voltage.
In the boost converter, current through the inductor is being varied at high frequency. This means that the current and therefore electromagnetic field constantly has to quickly build up and due to this the impendence of the inductor is greater. Again not the best explanation, but just to let you know what inductive reactance is.

The point is that an inductor switching at a high frequency will impede current flow and I doubt Yenka includes this in it's simulation so take that into account

Artikbot

Fri Oct 01 2010, 11:21AM

Registered Member #3247 Joined: Mon Sept 27 2010, 09:42AM
Location: Spain
Posts: 137

12:40 So, when the core collapses it acts as a resistor? Aaaaha now I get why do we have to discharge it and charge it again! Hey, that's great, I've finally understood why the single inductor boosts voltage!

Thanks alot for that :)

Concerning the OpAmp, Yenka works again :)

Wires are going from one edge to another because Yenka will make nodes if I cross two wires with a very little distance between them.

So far the 4.7Ohm resistor works flawlessly! 2Amp limited current, everything happens slowly and therefore more controlably :)

I think I'll search for a beefy 4.7Ohm resistor to put in my supply so no excessive power goes across the circuit.

If only the OpAmp would work, my circuit would be working flawlessly on Yenka, meaning even better on the board (Yenka doesn't count in peak voltages/currents, and most of my components relay on peak, not sustained).

I've searched in the nets, they say that on the (-) part I should fit the inverting signal, I guess it means the one to aplify, therefore the one coming from the cap bank; and the (+) part the non-inverting current, I guess directly from the power supply with its adequate resistor.

Something to note about the Op-Amp: It RECEIVES current from the timer via the output pin. And then I say: What on earth!? But then it comes to my mind that both 4 and 8 are connected, so maybe that's the problem. But then, if I separate pin 4 and only connect it to the OpAmp output, the circuit doesn't start to produce a pulsed signal.

12:55 Wait... I've been observing my circuit and... Is it normal that every single pin but Vcc and GND outputs a pulsed current? I mean, no solid current flow thorugh any pin, just pulsed. I'll upload the circuit so you can see what I mean.

13:01 I'll update myself (LOL I write/read/write/read that's why my posts show an evolutive line :P).

So, essentially pin 4 is connected to Vcc if and ONLY if no other source provides +0.7V to it, right? So, if my comparator was working properly I wouldn't have any need to put pin 4 directly connected to Vcc right?

O'sir, now we gotta fix why the OpAmp (or shall I say the resistors) don't receive the correct voltage and everything solved!!
Dis sir, is pure excitement. I now know why did I chose to study electronics engineering ^^

13:13 Oh sir I now got the reference voltage for my OpAmp!! I wasn't applying the voltage dividers properly ^^

Circuit upgraded. I need halp with the resistor values, somehow my calculations go bonkers and I get really weird voltage values :S

Thanks an epic amount of times!!!

GhostNull

Fri Oct 01 2010, 01:35PM

Registered Member #2648 Joined: Sun Jan 24 2010, 12:45PM
Location: Australia
Posts: 291

Artikbot wrote ...
Dis sir, is pure excitement. I now know why did I chose to study electronics engineering ^^

I don't know if you are being sarcastic or not =S
No one ever said this was going to be easy but the elation when you finally get it to work will far out weigh the stresses and troubles you put in. I hope =S

Artikbot wrote ...

12:40 So, when the core collapses it acts as a resistor? Aaaaha now I get why do we have to discharge it and charge it again! Hey, that's great, I've finally understood why the single inductor boosts voltage!

NO!!! >.<
I was trying to explain inductive reactance in a simple sense but I think I messed up and made you think something else

So I've done some of my own research to give you a better explanation >.<

The point I was trying to make was that the inductor and the AC nature of the circuit would cause something called impedance and this would in a sense "resist" electrical flow.

When determining current flowing in AC circuits there is more to consider than just pure DC resistance . There are also other factors that effect the current flow, the collective factors that oppose current flow in an AC circuit (including pure DC resistance) are expressed as impedance.

To determine current in AC circuits something similar to Ohm's law applies; in a slightly, different, equivalent form :
I = Z / V
Where:
I = Current in Amps RMS
Z = Impedance in Ohms
V = Electromotive force in Volts RMS
note: RMS is like an average, it is a way of stating a varying quantity but more complex. This might help you understand RMS:

Inductive reactance is part of impedance. Inductive reactance is the opposition to AC current flow caused by an inductor/inductance. This happens due the reasons I tried and failed to describe before

The inductive reactance of an inductor can be calculated by
XL = 2*pi*f*L
Where:
XL = Inductive reactance in Ohms
f = frequency in Hertz
L = Inductance in Henrys

Using the two previous formulas we can find the approximate current in your circuit considering only the inductor, frequency and voltage.
LX = 2 * pi * 10000 * 5*10^-4
LX ~ 31.4 ohms

I = Z/V We assume the voltage from the power source is constant
I = 31.4 / 9
I = 3.48 Amps RMS

So overall my point is that due to something called impedance, you won't get a huge amount of current in your circuit.

Need to read:
Inductance:

If you want to learn more about impedance:
Wiki:

AllAboutCircuits:

Helpful in understanding impedance:

So to find the values you need for the voltage divders you need to this should help:

As for getting your components, you are in Spain right? Here is farnell's Spain site:

Farnell is a big electronics distributor with pracitally all the components you would ever need. Just don't always assume all the prices are cheap. Some of things are REALLY over priced others are okay. I suggest you find the components you plan to use and input their values into your sim.

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