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Light LED with 4 nanoamps

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Ash Small

Sat Jul 19 2014, 02:18PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

If you only have one pulse a minute or so, this equates to a frequency of ~1/60 Hz, so you don't need ferrite, do you?

Surely you can use a high permeability core, as used in 50-60Hz transformers, etc, as the frequency is so low?

(The reason these can't be used at high frequency is due to heating, but at these low frequencies this won't be an issue)

You don't need to use gapped ferrite here, any old laminated iron core will work.

Someone please correct me if I'm mistaken.

Newton Brawn

Sat Jul 19 2014, 03:43PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Tony

I have some old notes that says the density of energy in a air gap is about " 400000 joules x induction squared per cubic meter of air gap"

E/v = 398000 x B^2 [J/m^3]

E = Energia, joules [ J ]
v = air gap volume, cubic meter [m^3]
B = induction weber /m^2 = tesla [T]

has somebody a better figure? or a comment??

Tony Matt

Sat Jul 19 2014, 04:16PM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

The inductor is using a tv flyback core with air gap.

" If you are using a gapped high mu core, it will define the minimum volume of the air gap."

Yes, using a high mu core GAPPED, and assuming 0.3T as the Bmax in the core and air gap
and based on Dr Slack and Newton information,

the air gap volume for storage of the 156 microjoule could be:

J/m^3 = 398000 x B^2 >>> m^3 = J / (398000 x B^2)

m^3 = 0.000156 / (398000 x 0.3 x 0.3) = 4.4 x10^-9 ...m^3

but the air gap area = 0.0002m^2 = 2 x 10^-4

min. air gap lenght, d:

d= 4.4 X 10^-9 / (2 x 10^-4) = 2.2 x 10^-5 = 0.000022m = 0.022mm

air gap relutance:

R= l/(u x a) = 0.000022/ (4x pi x 0.0000001 X 0.0002) = 87500 amper turn per weber

or

Al = 11.4 uH/turn (600turns result in 4.11H)

EDIT: AL=11.4uH/turn^2

Newton Brawn

Sun Jul 20 2014, 02:33AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi Ash,

Well, the pulse lenght is much less tha 1 minute,

The repetition rate is about 167 second, ( The calculed time for 3 nanoamper charge a 0.001uF with 500V is about 167 seconds, say 2.8 minute ).

The available energy in a 0.001 uF capacitor charged with 500V could be ;

E= 0.5 x C x V^2 = 0.000125J

Applying this energy in a 2.5V 2mA LED. the LED time on will be:

0.000125/(2.5 x 0.002) = 0.025 seconds

The inductance could be :

V = Ldi/dt >>> L= Vdt/di = 2.5 X 0.025 / 0.002 = 31H

Is a too large inductance, the number of turns is above my limiite....

Dr. Slack

Sun Jul 20 2014, 09:38AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

I'm not sure you've divided the right thing by th eright thing there Newton.

If we take 125uJ in the cap, and work out what inductance is needed to store the same energy at 2mA in an inductor, then it's 62H. This is a lot. Unless you want to be up to your elbows in very fine wire, it sounds like rethinking the LED current, or the cap energy, would be in order.

2mA might be the rated contionuous current for the LED, but what's the pulse current? Can it take 20mA for a few mS once every 3 minutes? That would reduce the inductance to a more manageable 620mH. 50mA? Now you'd be talking 100mH.

Iron or ferrite? 1nF * 62H resonates at 640H, so you might just get away with iron, with a lot of losses, unless you can find aircraft grade 400Hz cores. Using 620mH puts you up at 6.4kHz, defintely ferrite territory.

Now were you to design a transformer to do it, then either ...

a) it's a flyback, and you'd need still to store the 125uJ in the core. Using a lower primary inductance would draw a higher current from the cap, which is all fine. But what about the secoindary? If you want to deliver 2mA max into the LED, you are back with a current step-down, or voltage step-up transformer, so an up to your elbows in fine wire secondary.

b) it's a flux-coupled conventional transformer, with a large step-down in voltage. Either you'd need signaificant leakage inductance to control the LED current, or a wasteful series resistor. With a uni-polar input, the core must still be volt.seconds rated to tolerate the rise in flux over the length of the pulse.

Ash Small

Sun Jul 20 2014, 09:50AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Inductors of several H are readily available from the power supply circuits of 'vacuum tube era' equipment. They are reasonably compact and don't require being 'up to your elbows' in fine wire.

I have one here rated for 463V. Looks like it's ~10H, but I've not measured it yet.

Dr. Slack

Sun Jul 20 2014, 05:32PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Ash Small wrote ...

Inductors of several H are readily available from the power supply circuits of 'vacuum tube era' equipment. They are reasonably compact and don't require being 'up to your elbows' in fine wire.

I have one here rated for 463V. Looks like it's ~10H, but I've not measured it yet.

50Hz iron is it?

Ash Small

Sun Jul 20 2014, 05:48PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

50Hz iron is it?

Yes, but isn't this circuit operating at ~1/60Hz, or so?

EDIT: And if you dump a capacitor into an inductor, does it matter what the core is made from? Won't the discharge rate be entirely dependant on the inductance? (assuming the ESR of the capacitor doesn't dominate)

Resistance losses in the copper obviously have to be considered, but these 'iron core' inductors will have less losses than a ferrite cored inductor, which will require more, presumably thinner, copper wire for the same inductance, won't it?

Dr. Slack

Mon Jul 21 2014, 07:03AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

If the 1nF cap and 60H inductor ring up one quarter cycle in 400uS, what frequency do you think the eddy current losses in the iron are interested in? the 640Hz fundamental frequency, or the once per 3 minutes that it happens? You're not worried about it getting hot, you're worried about the efficiency as energy out to LED divided by energy fromm the capactiro, or at least I am.

Ash Small

Mon Jul 21 2014, 10:19AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

If the 1nF cap and 60H inductor ring up one quarter cycle in 400uS, what frequency do you think the eddy current losses in the iron are interested in? the 640Hz fundamental frequency, or the once per 3 minutes that it happens? You're not worried about it getting hot, you're worried about the efficiency as energy out to LED divided by energy fromm the capactiro, or at least I am.

Ok, I think I get it, Neil. I'm not familiar with calculating eddy current losses, but I think I can see that at 450V there won't be a lot of resistance to the small amount of current we're considering here.

If I=V/R, then eddy current losses are entirely dependant on V and frequency, for any particular inductor?

I'm not sure I completely follow this, as the voltage is only applied for a relatively short period, every so often, and energy stored in a capacitor is a function of 1/2C and V^2, if I remember correctly, although I think I can see that, for any voltage, eddy current will be a constant, per unit of time. (EDIT: although V will be divided by the number of turns)

I can see that reducing eddy current losses will greatly improve efficiency, although using ferrite will presumably increase copper (ohmic) losses.

Would a powdered iron toriodal core be an inprovement over a gapped ferrite core? You presumably wouldn't need the same amount of copper, for example, due to the much higher permeability of the powdered iron.

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