boost converter

Hello Michael,

Figure 5 shows just a graph. Duty cycle versus performance depends on the output load (voltage and current), it cannot be generalized. 80% typically happens when the output voltage needs to be 4-5 times higher than the input.

Nope, that would increase the losses because it would be like degrading the Rdson by 1ohm. When we design switchers there is always that hunt for the least in Rdson that a nickel or two can buy. No free lunches there.

There really isn't a short. The inductor current begins to ramp up. At some point the switch opens and the inductor dumps its current into the load side capacitor. "Some point" means before the inductor core saturates. Else there will be an impressive and loud pyrotechnical show followed by a plume of smoke and wailing fire alarms. Same if it goes into continous current mode (CCM) and the inductor current "ratchets up" for some reason.

There's no short circuit. When the switch is on, the inductor limits the current, which builds up slowly in the inductor. Well, relatively slowly, anyhow. See Fig. 3 in your article.

When the switch opens, the inductor "flies back" and dumps its energy into the output filter & load. Again, current ramps down, relatively slowly.

Introducing resistance in the high current paths will cause I^2*R losses, decreasing efficiency. Resistance is not your friend.

Best, James Arthur

Ok, thanks for the reply.

Loud pyrotechnical show, smoke, wailing fire alarms... just *not* what I'd want in a boost converter...

On that thought... put a fuse in the same path where I'd previously recommended a resistor?

Michael

This is a sci.electronics.basics question, so I am setting followups to there.

The inductor is not a short circuit in the sense you imagine. There is an impedance to it, if not a lot of resistance. And you need to imagine in a dynamic (time derivative) sense.

Let's say the inductor starts out with zero current when the switch is closed. The switch closure would attempt to have the current suddenly jump to some large value. But the inductor refuses to allow sudden changes. Instead: dI = V/L * dt, or dt = L/V * dI

So for each fraction of time, only a small fraction of current change is permitted. Since it starts at zero in this case, it cannot go instantly to a large value. Assuming the voltage across the inductor remains constant, and it does in this ideal case you cite, a fixed change in current will require a fixed change in time. In other words, it cannot happen right away. To achieve this, energy is gathered up and stored in a magnetic field around the inductor.

If you wait short enough time, the current will never manage to rise up too high. It will get to some point and then the switch opens. When that happens, the inductor will alter the voltage between its leads in any way required in order to maintain that current. Of course, the current will then have to flow through the diode and into the load. (Usually, there is a capacitor across the load at the other end of that diode in models like this to absorb the energy and supply current while the switch is closed.) This means that the energy stored in the magnetic field, as it now collapses, will be consumed.

Again the law remains, dI = V/L * dt, or dt = L/V * dI. So it will also take time to decline to zero current. However, in this case, the V will not be the V of the battery, but instead the V that the inductor creates as the field collapses, needed to ensure a gradual decline in the current instead of a sudden decline.

Does that make sense?

Jon

Hello Michael,

You could but by the time it blows the transistor is most likely already gone. A fuse is typically placed before the whole converter, in an attempt to prevent a major explosive event.

I've found inductors made from 5mm Copper tubing are ideal for use during much of the development process. (So trouble free, I've even pondered on specifying as the final component) john

Yes. Murphy says your $$ transistor will blow in microseconds, protecting your 10-cent fuse.

Cheers, James Arthur

What did you wrap them on? (Iron? Just a PVC pipe?)

Hello James,

A microsecond is actually a long time. A laser diode can go from fully functional to lalaland in one nanosecond. No kaboom, nothing, it's just dead.

Hello John,

An air core of that size can trigger the Federales to show up at your lab in a jiffy ;-)

the

show

up"

during

A bit of scrap plastic pipe. Then wind on turns to give the wanted inductance. A few dabs of hot glue to hold it together. Warm at 100Amps but no farting about with airgaps and saturation. john

Joerg a écrit :

Don't think so. John's in UK :-)

Hello Fred,

Well, then it would be her Majesty's guards :-)

The inductors are never run in a saturated condition, the voltage is applied for a relatively short time compared to what it takes to saturate the reactor.

If you just closed the switch for a long time with respect to the size of the inductor - yeah, you'd be wasting power. Below saturation or well below saturation most of the energy is give up to the load when the switch opens.

A resistor would decrease efficiency since all it can do is drop voltage and make heat. That energy is lost and not available to the load.

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Joerg a écrit :

Guard... as in shield?

Hello Fred,

I guess they don't use iron clad body armor anymore these days ;-)