In this circuit the input voltage is 11-12V, output voltage is 14V @ 3A. I'm trying to understand the inductor in this image. Based on this image the inductor is rated at 6.8A. Does that mean that for the circuit to output 3A as it is built, it has to be able to draw 6.8A from the source? Is that correct?

 

Definitely not 6.8A continuous current, but the inductor has to be rated to handle the maximum PEAK current and a bit extra so that rating is reasonable. In a boost converter, the peak current is a lot higher than the average input current.

 

Is there a common formula that can be used to estimate input current if you know the inductor peak current in a circuit? Basically I am trying to figure out if my source current is up to snuff for this circuit to output 3A.

 

The average input current is equal to the output current times the ratio of the output to input voltage divided by the circuit efficiency.

 

You are going to need a lot more than 100uF of input capacitance to handle the peak currents that thing will be pulling.

 

You are going to need a lot more than 100uF of input capacitance to handle the peak currents that thing will be pulling.

Perhaps. It greatly depends upon the switching frequency.

 

It's about 100 kHz, but I guarantee you need more capacitance for that kind of boost to feed the switch or the VIN line will be sagging like crazy every time it switches ON. And the input cap had better be very low ESR types.

That schematic looks like the SS software kicked it out, those are set for the cheapest design not the best performance.

 

If you approach the current limit of a coil, it will begin to saturate. This will cause efficiency to go down, and will cause heat developement inside the coil + the switcher IC, and maybe also the flyback diode.

At 3 Amps current, you can get some decent heat dev. with an underrated coil. Just inside the coil. A 6.8 Amps coil is appreciate for 3 Amps. Even if there might be the case where you can get away with a coil rated for lower current. It highly also depends on the frequency, and magnetic material.

100uF is not much for 3 Amps. Also depends mainly on the frequency.

Also in your circuit the voltage difference is not so much. This maybe can mean lower capacity can be used.

For your circuit maybe a 5 Amps coil is good enough.

 

I would not derate the coil if the design software picked it. I was at national semi when we developed that software, it was pretty accurate about that. It's designed to cheapskate the parts as much as possible down to the safe minimum for the application.

 

It is worth trying if you have different coils around for testing.

 

It's about 100 kHz, but I guarantee you need more capacitance for that kind of boost to feed the switch or the VIN line will be sagging like crazy every time it switches ON. And the input cap had better be very low ESR types.

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For an 11V input and 14V output @ 3A the average input current will be 4.77A with an assumed 80% converter efficiency. The maximum sag at 100kHz for this current and a 100μF input capacitor is 0.477V. Whether that's "sagging like crazy" or not I leave to your interpretation.

 

For an 11V input and 14V output @ 3A the average input current will be 4.77A with an assumed 80% converter efficiency. The maximum sag at 100kHz for this current and a 100μF input capacitor is 0.477V. Whether that's "sagging like crazy" or not I leave to your interpretation.

You are imagining an ideal cap, none such exist. each time the switch draws a big peak current, the input voltage sags by the amount of that current multiplied times the cap's ESR. There is also a shorter duration "spike dip" caused by the cap's ESL.

Believe me as I stated before: it needs more capacitance, and it needs very good (low ESR) caps. I would use a premium electrolytic like Oscon paralleled by a solid Tantalum cap and maybe a good ceramic.

Build it up and look at the garbage on the input line with a typical 100uF aluminum cap.

 

For an 11V input and 14V output @ 3A the average input current will be 4.77A with an assumed 80% converter efficiency.

But that's not how a boost works. The switch closes and input current flows into the coil, then it opens and dumps the energy. The input current may be an average of 4.77A, but it only flows maybe 60% of the time and it has a ripple ramp which probably means it peaks out at close to 9 or 10A. That 100uF cap is going to get flogged like a rented mule.

 

For an 11V input and 14V output @ 3A the average input current will be 4.77A with an assumed 80% converter efficiency. The maximum sag at 100kHz for this current and a 100μF input capacitor is 0.477V. Whether that's "sagging like crazy" or not I leave to your interpretation.

I discovered that the Webench Designer application has every conceivable value relevant to a circuit it you design. The average input current is stated to be 4.53A so you were pretty much right on the money.

 

I discovered that the Webench Designer application has every conceivable value relevant to a circuit it you design. The average input current is stated to be 4.53A so you were pretty much right on the money.

But the cap (and inductor) have to be strong enough to handle the peak currents.

The input current (shown below) is the lowest trace. The switch ON time is in red. Note that input current flows during that time. When the switch is off, the inductor is dumping the energy it stored during the ON time into the load.

So, calculate: if the average input current is 4.5A, but input current only flows about 40% of the time (see below) the peak input current value must be more than 10A.

Both the inductor and input cap on the design shown in the OP are undersized and not what I would ever use. But, the design software is made to cut it close to make the design cost lower.