Re: Chopper max voltage?

--- In CAD_CAM_EDM_DRO@egroups.com, "Mariss Freimanis"
<geckohall@h...> wrote:

> Hi,
>
> Here's my 2 cent's worth on the great chopper frequency debate. The
> choice of switching frequency is based on the best balance between
> two conflicting goals, motor losses vs. drive losses:
>
> 1) High frequency = low motor losses.
> 2) Low frequency = low drive losses.
>
> 1) Motor iron heating is caused by AC currents. Chopping

contributes

> to this by causing a "ripple current" AC component on top of the DC
> current passing thru the winding. The higher the amplitude of
> this "ripple current" the hotter the motor will get.
>
> The amplitude of this current is equal to V / 2fL. "V" is the power
> supply voltage, "L" is the motor inductance and "f" is the chopping
> frequency.
>
> The higher the chopping frequency "f", the lower the motor heating.
>
> 2) Transistors in a chopping drive get very hot when there is
> simultaneous high current and high voltage across them. This

happens

> breifly everytime the output changes from high to low or visa versa.
>
> This are called "switching losses", and the occur on every output
> change edge.
>
> Higher frequencies mean more edges per second, and thus more

heating

> in the drive transistors.
>
> The L297 uses Darlington output transistors. They have high gain

but

> are very slow in switching. They also have a high "on" voltage.

Both

> contribute to heating of the drive.
>
> Mariss
>
>
>
> P.S. Here's how an L298 compares to a MOSFET (IRF530) bridge at

24V,

> 2A and 20kHz:
>
> IRF530:
>
> Conductance losses (I squared R)= 4 X (.15 + .15) = 1.200W
> Switching losses 2f(IV)t = (2 X 20kHz X 24 X 2)100nS = .192W
> ------
> Total heating (spread out over 2 125W packages) 1.372W
>
> L298:
>
> Conductance losses (Von times Icc) = (2.7 + 2.3)X 2 = 10.000W
> Switching losses 2f(IV)t = (2 X 20kHz X 24 X 2)1uS = 1.920W
> ------
> Total heating (in one 15W package) 11.920W

Greetings....

Mariss makes a good point about the RDSon-resistance of the Mosfet,
dictating the conductance losses. Lets take it one-step further and
add that...by spending another $1 inside the G201 , the driver would
not even need an extra heatsink. This is based on the IRF530's RDSon
of .11-ohms and the IRF540's RDSon of a mere .052-ohms. The losses
during conduction would drop to about half and given that at 20khz.
the bulk of the losses are conduction....it would truely drop total
losses to half. But, since the IRF530's total gate charge is only
44nC and the IRF540's being 94nC....there would be a slight drive
supply penalty. On the order of 4ma added to the entire dual-H Bridge.
Based on Qc(44nC)x F(20khz.)= .9ma for the IRF530 and Qc(94nC)x F
(20khz.)= 1.9ma for the IRF540. I'm sure this benefit would only hold
true up to about 6 amps/phase. But, it does bring about an
interesting question,that is,whether to spend inside or outside the
package. Either way,Mariss has a series of drivers that are a Marvel !

Regards....
Steve