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Calculating current in pulsed/square wave DC circuits

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Gregary Boyles

Tue Mar 21 2017, 11:07AM

Registered Member #9039 Joined: Wed Dec 26 2012, 03:31PM
Location: Epping, Victoria, Australia
Posts: 117

My question concerns the following schematic by Steve Ward.

For the convenience and lower cost of having both inverting and non inverting outputs on the one chip, I want to use UCC27425 instead of the two separate chips that Steve has specified.

I am trying to figure out what sort of current will be flowing through C6 and the UCC2745 chip to see if its lower max current of 4.5A will not be exceeded.

But I have had great difficulty in finding a web page with a simple explanation, with a worked example, of how to do this with pulsed /square wave DC, as opposed to pure AC.

Can anyone help me out?

MiniSSTCfnlsch

Sulaiman

Tue Mar 21 2017, 12:00PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

The impedance of C6 can be considered to be zero at the switching frequency,
the transformer is 1:1:1, each secondary winding has a load of (Rgate + Cgate)

the driver output voltages are Voh and Vol, effectively each consisting of a voltage drop and a series resistance from Vs and 0V
for this type of driver, the output resistance dominates the performance,

so at switching transitions, the peak current will be;

2x Vs/(Roh + Rol + 0.5xRgate)

e.g. 2x 18/(1.4 + 1.1 + 2.5) = 7.2 A pk ... neglecting other effects

Gregary Boyles

Tue Mar 21 2017, 01:10PM

Registered Member #9039 Joined: Wed Dec 26 2012, 03:31PM
Location: Epping, Victoria, Australia
Posts: 117

Sulaiman wrote ...

The impedance of C6 can be considered to be zero at the switching frequency,
the transformer is 1:1:1, each secondary winding has a load of (Rgate + Cgate)

the driver output voltages are Voh and Vol, effectively each consisting of a voltage drop and a series resistance from Vs and 0V
for this type of driver, the output resistance dominates the performance,

so at switching transitions, the peak current will be;

2x Vs/(Roh + Rol + 0.5xRgate)

e.g. 2x 18/(1.4 + 1.1 + 2.5) = 7.2 A pk ... neglecting other effects

Thanks for this it is a terrific start.
OK Voh and Vol are straight from the datasheet and the voltage that appears across the primary of the GDT will be 11.5V or there abouts....assuming I have read the datasheet correctly.

Rol and Roh are also straight from the datasheet.

The bits I don't yet understand is how you get Rgate and why x 0.5? Because their are two secondary windings I would guess for starters? I have the inductance of the GDT primary - 120mH but how do you convert that to an impedance with a square wave. With pure AC it is just X = 2PifL.

I can see that part of the equation is simply Ohm's law but why is the x2 in front of it?

And one other thing - can I use C6 to introduce some impedance (if necessary) at my secondaries resonant freq of 432KHz in order to keep the GDT primary current within 4.5A max for UCC27425. Or should I instead just introduce a resistor of appropriate value?
If I need to do any of this then I assume I can just play with the values of R3 and R4 to compensate.

Mads Barnkob

Tue Mar 21 2017, 01:51PM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
Location: Denmark, Odense C
Posts: 1968

I wrote a IGBT gate drive calculator:

It should be fine to use for MOSFETs aswell

Sulaiman

Tue Mar 21 2017, 02:01PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Gregary Boyles wrote ...

Rgate is the external resistor in series with the gate, (R3, R4) 5 Ohms each
the transformer effectively puts the two loads in parallel, seenfrom the primary side,
hence the 0.5

Just before switching the gate capacitances will be charged,
one to about +Vs, the other to -Vs (the gate current will be near zero by the end of a switching cycle)
so at the moment of switching, driver polarities reverse,
the starting voltages on the gate capacitances will be of opposite polarity to the new driving voltage,
hence the 2x (close enough)

P.S. in my example above, I used Vs=18V instead of 12V, so 7.2A becomes (12/18)x7.2 = 4.8 Apk

P.P.S. if you try to model the driver and output stages there are myriad parasitic interactions ;)

P.P.P.S. I use calculator apps. for many things, but for me
simulation is better,
researching, doodling and pondering increase understanding
experimentimg is even better,
(in theory, practice and theory are the same, in practice, ...)

Gregary Boyles

Tue Mar 21 2017, 03:05PM

Registered Member #9039 Joined: Wed Dec 26 2012, 03:31PM
Location: Epping, Victoria, Australia
Posts: 117

Using UCC37322/21 I get...

2 * 12 * (15 + 1.1 + (0.5 * 5)) = 1.2A

Using UCC27425 I get...

2 * 12 /(30 + 2.2 + (0.5 * 5)) = 680mA

So that makes sense to me because UCC27... has half the max current of UCC37...

If I was to remove the 5 Ohm resistors then it would be 745mA peak

I could raise my Vcc to 16V (I am using a CMOS 555 that has a max Vcc of 18V) if necessary to get that peak current closer to 1.2A - I calculate it at just under 1 A with wire links in place of R4 and R5.

But in your view is it necessary?

I mean how much current do you practically need to get IXTK62N25 gates to switch at 432kHz of there abouts.

According to the datasheet these have Ciss = 6.6nF, Coss = 1.125nF, Crss = 0.27nF

By the way with Rol and Roh and max Vcc of 16 how is it actually possible to hit the 4.5A max with UCC27425?

And I really appreciate the trouble you have gone to to explain this to me. I have been trying to figure it out for quite a while from piecemeal info of the web and gaps in my own knowledge due to the fact I am nor formally trained in electrical engineering.

I have bookmarked this thread so I can refer to it in future.

Gregary Boyles

Tue Mar 21 2017, 04:16PM

Registered Member #9039 Joined: Wed Dec 26 2012, 03:31PM
Location: Epping, Victoria, Australia
Posts: 117

I found this http://www.edaboard.com/thread41454.html

Tsw=Qgd/Ig

Tsw: switching speed in seconds
Qgd: from the datasheet in coulombs (85 nC for IXTK 62N25)
Ig: required current in amps

1/432000 = 0.000000085 / Ig
0.00000231 = 0.000000085 / Ig
Ig * 0.00000231 = 0.000000085
Ig = 0.0368A or 36mA

IRFP260 has a Qgd of 110
So the authors 1.2A into the gates of his FETs seems like overkill anyway.

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