Schottky Diode Ratings (Re: duty cycle)

[TimNJ]

Hi everyone,

Just got back from school, so I thought I'd get started on a project. I'm building a DC electronic load. I'm following Dave's design and Chris Whittenburg's revision.

I thought I ought to give it  reverse polarity input protection. After weighing my options, I think a Schottky diode would be a good solution.

I'm between a $0.51 STPS3150 and a $1.38 MBRF10H100. The MBRF diode looks hugely overkill, since I've only speced the load for up to 2A continuous. However, the STPS was designed for switching applications so all its graphs are in terms of the duty cycle used. It says it's good for up to 3A forward current, but that's when delta = 0.5.

If I go down to Figure 1. (the first graph), you see forward power vs. forward current. My main question here is: Why is the forward power at a given current greater for a smaller % duty cycle and less for a greater % duty cycle? If the power is on for less of the time, how could it be a greater dissipated power?

And finally, could the STPS handle ~2A DC?

Thanks!!

STPS3150 datasheet: http://www.st.com/web/en/resource/technical/document/datasheet/CD00003323.pdf

MBRF10H100: http://www.vishay.com/docs/88667/mbr10hxx.pdf

[tggzzz]

You might like to look at the simple P-channel MOSFET circuits used to protect against reversed input polarity. They have a surprisingly low on resistance.

Googling for those obvious (once you know them!) search terms quickly brings results

[T3sl4co1l]

The power averages down over time.

[TimNJ]

You might like to look at the simple P-channel MOSFET circuits used to protect against reversed input polarity. They have a surprisingly low on resistance.

Googling for those obvious (once you know them!) search terms quickly brings results

Thanks! I actually read about them, but I decided that I wouldn't be able to keep the MOSFET conducting with a low voltage. (Don't  you usually need like ~4V volts to turn it on?) Perhaps I am wrong to think that. So I thought a Schottky would be good because of it's <1V forward voltage. I won't be using the load for anything less than 1V, I am sure.

[TimNJ]

The power averages down over time.

Thanks. But I'm not quite sure what that means.

[tggzzz]

The power averages down over time.

Thanks. But I'm not quite sure what that means.

And as you increase that time, the average power will asymptotically approach zero.

So no, it wasn't a particularly illuminating statement

[TimNJ]

The power averages down over time.

Thanks. But I'm not quite sure what that means.

And as you increase that time, the average power will asymptotically approach zero.

So no, it wasn't a particularly illuminating statement

Haha okay thanks!

[IanB]

If I go down to Figure 1. (the first graph), you see forward power vs. forward current. My main question here is: Why is the forward power at a given current greater for a smaller % duty cycle and less for a greater % duty cycle? If the power is on for less of the time, how could it be a greater dissipated power?

In short--because of math.

Ignore for a moment that it is a diode and just consider a resistor of say, 0.1 ohms with an average current of 10 A.

At DC, the average power in the resistor is the DC power, given by 0.1 x 10² = 10 W

At 50% duty cycle for the same average current the instantaneous current is 20 A (since 50% of 20 A is 10 A).
The average power is now 50% of 0.1 x 20² = 20 W.

Similarly, at 25% duty cycle the average power will be 25% of 0.1 x 40² = 40 W.

Diodes are non-linear, but the principle is the same. Power dissipated goes up as the duty cycle goes down if the average current remains unchanged.

[TimNJ]

If I go down to Figure 1. (the first graph), you see forward power vs. forward current. My main question here is: Why is the forward power at a given current greater for a smaller % duty cycle and less for a greater % duty cycle? If the power is on for less of the time, how could it be a greater dissipated power?

In short--because of math.

Ignore for a moment that it is a diode and just consider a resistor of say, 0.1 ohms with an average current of 10 A.

At DC, the average power in the resistor is the DC power, given by 0.1 · 10² = 10 W

At 50% duty cycle for the same average current the instantaneous current is 20 A (since 50% of 20 A is 10 A).
The average power is now 50% of 0.1 · 20² = 20 W.

Similarly, at 25% duty cycle the average power will be 25% of 0.1 · 40² = 40 W.

Diodes are non-linear, but the principle is the same. Power dissipated goes up as the duty cycle goes down if the average current remains the same.

Thanks Ian! Very interesting indeed. I guess I won't try to think about it intuitively...yet. Makes sense with some math. Now I can go to sleep tonight.