A simple dI/dt confusion, Anyone like to try & solve it ?

[Sanjay kumar]

According to calculation : -
1uH coil is giving me 50A/uS for 100uS,
If I use 4uH coil, it will give me 12.5A/uS for 400uS

What I am trying to ask specifically is : -
Will total time get increased by 4x ?
Or It will only take extra 4uS for current to reach 50A, then for the rest of the time - 100uS, It will continue at 50A ????

"It will continue at 50A"
By this I meant, 50A constant current flow for the rest 100uS.
Not rising 50A/uS.

 

[Julie.]

According to calculation : -
1uH coil is giving me 50A/uS for 100uS,
If I use 4uH coil, it will give me 12.5A/uS for 400uS

What I am trying to ask specifically is : -
Will total time get increased by 4x ?
Or It will only take extra 4uS for current to reach 50A, then for the rest of the time - 100uS, It will continue at 50A ????

"It will continue at 50A"
By this I meant, 50A constant current flow for the rest 100uS.
Not rising 50A/uS.

I am not sure what you are asking, is this an inductance subject to a step change?

If so, think of it long term. In this case the current is not changing therefore the L di/dt would be zero - the only thing resisting any flow of current would be the resistance of the circuit not the inductance so to obtain a 50A flow of current you would need a 1 ohm resistance if the voltage is 50V, 2 ohm for 100V etc.

However at the step point, you are asking the current to change from zero amps to 50A in a sharp step, at time t=0 I = 0, and then time t+0 I = 50A ~ a change of 50A in 0 seconds this gives an infinite di/dt therefore L di/dt is also infinite and the current would be the voltage divided by (L di/dt) - giving zero.

If you integrate L di/dt over time you can see that the value of L di/dt will slowly reduce as the current slowly increases, this increase in current follows an inverse exponential curve, rising fairly rapidly at first, and slowing down as time goes on until you have no change in current, and you are in the long term conditions I described above.

Over the growth of the current at the start - from the step change through to the steady state (long term) conditions the current is not constant, but is changing all the time. You will never have any constant rate of change such as 2A per second for 1 second etc - it may be 1.84A/sec at 3.4 seconds and 2.16A/sec at 4.4 seconds (which averages out at 2A/sec over a second)

Any values you calculate over this growth period will be instantaneous only as the current, and rate of change of current will be changing all the time

 

[Lucien Nunes]

Yes as Julie points out you might be overlooking the fundamental relationship V=L.dI/dt, so the rate of change of current is only constant if the voltage across the inductor is constant. An ideal inductor connected across an ideal voltage source will pass a current increasing at a constant rate towards infinity. In reality the resistance will become significant and an exponential decay will occur, or the inductor will saturate and reduce in inductance, within the bounds of realistic currents.

 

[Vortigern]

I haven't a clue as to exactly what you are talking about, but it sounds very intelligent.

 

[Julie.]

Ok, bit of a further reply, I thought it would be worth posting here as it does illustrate why we need ALL the information rather than snippets because it can change the question completely.

Basically the idea is to discharge a capacitor with 945V stored via an inductor.

In this case – pictured here

It is not a simple constant voltage step change, the capacitor will have a store of energy, as the switch is closed it will discharge into the coil, this would produce a back emf – this interaction between the decay of voltage from the capacitor and growth of current through the inductor “fight against each other” – a bit like dropping a weight on a spring – the weight will fall then the spring will pull it back, then the weight will fall etc – basically discharging a capacitor into an inductor will oscillate!

Rather than just L di/dt the equation becomes:

Which is a second order differential equation – hence the oscillation.

There are a number of shortcuts to solving this, depending on the relative values of C, L & R – as here:

So forcing a low resistance will cause a decaying oscillation, but increasing the resistance beyond 2 Sqrt (L/C) will cause a non-oscillatory decay!





If I was solving this for the generic solution, I would use laplace transforms to solve the differential equation as a set of linear equations, I may do this later if the football is rubbish!

Probably the easiest solution these days is to download a program such as LTSpice – it’s free and allows time domain modelling of any circuit – not just L & C, but actually any SCR you may wish to use as the switching device.

This requires little mathematical knowledge.

 

[telectrix]

Ok, bit of a further reply, I thought it would be worth posting here as it does illustrate why we need ALL the information rather than snippets because it can change the question completely.

Basically the idea is to discharge a capacitor with 945V stored via an inductor.

In this case – pictured here

View attachment 61188

It is not a simple constant voltage step change, the capacitor will have a store of energy, as the switch is closed it will discharge into the coil, this would produce a back emf – this interaction between the decay of voltage from the capacitor and growth of current through the inductor “fight against each other” – a bit like dropping a weight on a spring – the weight will fall then the spring will pull it back, then the weight will fall etc – basically discharging a capacitor into an inductor will oscillate!

Rather than just L di/dt the equation becomes:

View attachment 61190


Which is a second order differential equation – hence the oscillation.

There are a number of shortcuts to solving this, depending on the relative values of C, L & R – as here:

View attachment 61189


So forcing a low resistance will cause a decaying oscillation, but increasing the resistance beyond 2 Sqrt (L/C) will cause a non-oscillatory decay!





If I was solving this for the generic solution, I would use laplace transforms to solve the differential equation as a set of linear equations, I may do this later if the football is rubbish!

Probably the easiest solution these days is to download a program such as LTSpice – it’s free and allows time domain modelling of any circuit – not just L & C, but actually any SCR you may wish to use as the switching device.

This requires little mathematical knowledge.

lost me half way down the first paragraph.

 

[Julie.]

Anyway, it was this or helping my husband trim the ivy - swift avoidance of Ivy to be had.. so:

This is the model in LTspice - simple Capacitor of 2000uF with the initial conditions of 945V discharging through an inductor of 200uH and series resistor - critical damping would be at 2x SQRT(2000/200) = 2 sqrt (10) = 6.32 ohm

In each case the switched is closed at 10ms

Case 1 Resistance = 6mOhm

As you can see the discharge oscillates as it decays down slowly

Case 2 60mOhm

This is the same as before, but the decay down is much quicker

Case 3 - in this case we have R = 6.4 Ohm - i.e. critical damping

As to be expected - straight forward decay with no oscilation.

If you want - here is the LTspice model, play to your heart's content.

It is a .txt file - you will need to rename it .asc before you load it into LTspice

 

[Sanjay kumar]

Ok I will try, I never used LTspice.

Can you please try with these specs
945v
400uf
1uH
Coil resistance 1milli Ohm
Total Discharge time 100uS

 

[Julie.]

Ok I will try, I never used LTspice.

Can you please try with these specs
945v
400uf
1uH
Coil resistance 1milli Ohm
Total Discharge time 100uS

download LTspice rename the file I posted as .asc and load it in to LTspice.

You would need to run using the simulate tab > Run, then once you have the graphical output screen (which will be empty) right click and select add Traces, you can then select any parameter to view - Vdepreswitched is the voltage on the capacitor, I(L1) is the current through the inductor etc.


to change component values hover over the component - then right click and you should get a screen like this:

Just change the values - resistance in this case so for 1 meg ohm enter 1M; for 10 mili ohm enter 10m or 0.01 or 10000u etc Then press OK, and re-run the simulation - the plots you selected will refresh

 

[Sanjay kumar]

If you do try it with above specs
Please send me that file also, since What I want is that or anything close to it.

Can I use this SCR - VS-ST330C16L

Can I use this circuit without adding any resistor?
[automerge]1601815994[/automerge]
Ok, I will try.

But please confirm that I can do what I am doing.

Simce resistance is very low in coil.
Can It create sparks in capacitors?

 

[Julie.]

400uF 1mH - runtime of 100ms

Zoomed in over the first 150us:

LTspice is easy to use once you have the model


OK, added the current to the last trace - check out the values - what do you think that will do to an SCR?

Yes that is ~18kA at 7.9kHz

 

[Sanjay kumar]

Ok, I will try.

But please confirm that I can do what I am doing.

Since resistance of coil is very low.
Can It create sparks in capacitors?


In order to get 1uH, Can I use 150mm wire ?

So what is dI/dt of above circuit??

do i need to worry about 1890A/uS rise time??

 

[Julie.]

Ok, I will try.

But please confirm that I can do what I am doing.

Since resistance of coil is very low.
Can It create sparks in capacitors?


In order to get 1uH, Can I use 150mm wire ?

So what is dI/dt of above circuit??

do i need to worry about 1890A/uS rise time??

I edited my previous post to include the current - at initial contact around 18kA at 8kHz or so


As for the suitability of electronic components - not my area!

Basic circuit theory - yes OK at that!

 

[Julie.]

spice model using the figures 400uF and 1mH

 

[Julie.]

Anyway, it was this or helping my husband trim the ivy - swift avoidance of Ivy to be had.. so:

This is the model in LTspice - simple Capacitor of 2000uF with the initial conditions of 945V discharging through an inductor of 200uH and series resistor - critical damping would be at 2x SQRT(2000/200) = 2 sqrt (10) = 6.32 ohm

In each case the switched is closed at 10ms

Case 1 Resistance = 6mOhm
View attachment 61196

As you can see the discharge oscillates as it decays down slowly

Case 2 60mOhm
View attachment 61198

This is the same as before, but the decay down is much quicker

Case 3 - in this case we have R = 6.4 Ohm - i.e. critical damping

View attachment 61199

As to be expected - straight forward decay with no oscilation.

If you want - here is the LTspice model, play to your heart's content.

It is a .txt file - you will need to rename it .asc before you load it into LTspice

It has just been pointed out to me - that I can't calculate!!

The critical damping should be 2 Sqrt (L/C) - I did C/R in error

So 2 Sqrt (0.1) = 0.632 ohm, not 6.32 ohm!

so 0/10 - must improve!

Here is the output for 0.64 ohm anyhow