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High energy discharge but not as you know it.

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Andy

Sat Mar 07 2015, 05:08PM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Trying to understand that page got two foumlas
Vl = Vs*e(Rt/L)
(1-(1/(e*t)))*Vs

With these values
0.036 ohm
0.001 Sec
0.012 H
12 volt

I get 0.097 and 0.903(negtive), as a percent 1.289 volt.
Am I even on the right field?

mister_rf

Sat Mar 07 2015, 05:51PM

Registered Member #4465 Joined: Wed Apr 18 2012, 08:37AM
Location: Bucharest, Romania
Posts: 145

Inductor transient response online calculator:

Andy

Sat Mar 07 2015, 08:03PM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Im trying to make a oneshot were the energy stored in the inductor is set to a certain level, and then discharged, ive made the below circuit, but im not shore its working as its meant to.

Andy

Sun Mar 08 2015, 12:06AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

This is the thing im trying to make work, ill would like 50-100amp at high voltages, but im not sure it will make, it makes higher and high volts, but would like amps. Could this help anyone?

Newton Brawn

Sun Mar 08 2015, 01:58AM

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

"Got the Xl/2*pie*f from practical electronics book, but they are showing different values?"

>> The formula IS CORRECT, BUT refrase as
L=Xl/(2*pi*F)
were
L= inductance,
Xl = Inductive reactance of the winding in ohms,
pi as 3.14,
F as frequency in herts, Hz

If you appply 10.9V 50Hz across a coil and get a 2.36A as a current through the coil, the coil reactance will be:

Xl = E/I = 10.9/2.36 = 4.619 ohms

them the inductance = L= Reactance/(2*pi*F)
L=4.619/(2*pi*50)
L=0.0147 H
Of course we have desconsidered the DC resistence of coil, that is very small compared with the coil reactance.

Typical value for a 120A 50Hz wellding buzzbox rated 10% service factor or duty is primary coil inductance of 0.17 H , secondary coil inductance as 0.01H.
The primary inductance is known as magnetizing inductance, and produces a primary current of 4-5 amper when connected to 220V, 50Hz, with no load at secondary.

The second calculation that you wrote

"Or
314.2*4.61 = 1448.46
1/1448.46 = 0.00069H "
I could not understand, more precise, I get lost.

Please define what you want measure:
1- the primary coil inductance,
2- the secondary coil inductance,
3- the primary-secondary leaking inductance,
4- or what ?

regards

mister_rf

Sun Mar 08 2015, 08:31AM

Registered Member #4465 Joined: Wed Apr 18 2012, 08:37AM
Location: Bucharest, Romania
Posts: 145

Andy wrote ...

Im trying to make a oneshot were the energy stored in the inductor is set to a certain level, and then discharged, ive made the below circuit, but im not shore its working as its meant to.

I think the basic components of the switching circuit you have proposed, can be rearranged to form a step-up (boost) converter. Thereâ€™s a need for feedback and control circuitry nested around these circuits to regulate the energy transfer and maintain a constant output within normal operating conditions. The energy stored by the inductor in a switching regulator topology can be transformed to output voltages that can be greater than the input (boost). A basic boost configuration is described here:

Andy wrote ...

This is the thing im trying to make work, ill would like 50-100amp at high voltages, but im not sure it will make, it makes higher and high volts, but would like amps.

Thatâ€™s a cascade concept of converter, and can help to implement the output voltage increasing, but I think you are aiming too high. For sure this project is growing too big for your budget.

Andy

Sun Mar 08 2015, 05:44PM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Im thinking off a boost converter circuit.

Ive got some supercapactors and I might beable to charge them with some AA batterys, they should beable to supply 50 amps for oneshot into the converter, the backvoltage would proable destroy them thought.

Conundrum

Mon Mar 23 2015, 07:41AM

Registered Member #96 Joined: Thu Feb 09 2006, 05:37PM
Location: CI, Earth
Posts: 4062

Schottky diodes?
I have found that a lot of nice ones can be pulled from defunct SMPS units and often they are rated to 30 odd amps *per diode* !

hen918

Mon Mar 23 2015, 06:21PM

Registered Member #11591 Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556

normal diodes should do. you just have to be careful they don't explode! look at the pulse current (usually 10ms) in the datasheet.

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